UNIT I PART A - Vel Tech HighTech

VELTECH VEL MULTITECH VEL HIGHTECH


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UNIT – I

PART – A

1. Define the term Thermal Engineering. Thermal Engineering is the science that deals with the energy transfer to practical applications such as energy transfer power generation. Refrigeration, gas compression and its effect on the properties of working substance. 2. What is meant by thermodynamics system? How do you classify it? Thermodynamic system is defined as the any space or matter or group of matter where the energy transfer or energy conversions are studies. It may be classified into three types. a) Open system b) Closed system c) Isolated system 3. What is meant by closed system? Give an example. When a s*9*2 1ystem has only heat and work transfer, but there is no mass transfer, it is called as closed system. Example: Piston and cylinder arrangement. 4. Define an open system. Give an example. When a system has heat, work and mass transfer, it is called as open system. Example : Air compressor 5. Distinguish between Open and Closed systems. (Ann Univ. Oct‟ 02)

S.No. Closed System Open System

1, There is no mass transfer. Only heat and work will transfer.

Mass transfer will take place, in addition to the heat and work transfer.

2. System boundary is fixed one. System boundary may or may not change.

3. Ex: Piston & cylinder arrangement, thermal power plant.

Air compressor, boiler.

6. Define an Isolated System. Isolated system is not affected by surroundings. There is no heat, work and mass transfer takes place. In this system total energy remains constant.



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7. Define specific heat capacity at constant pressure. It is defined as the amount of heat energy required raising or lowering the temperature of unit mass of the substance through one degree when the pressure kept constant. It is denoted by Cp. 8. Define specific heat capacity at constant volume. It is defined as the amount of heat energy required to raising or lowering the temperature of unit mass of the substance through one degree when volume kept constant. 9. What is meant by surroundings? Any other matter out side of the system boundary is called as surroundings; 10. What is boundary? System and Surroundings are separated by an imaging line is called boundary. 11. What is meant by thermodynamic property? (MU- Apr. 2001) Thermodynamic property is any characteristic of a substance which use used to identify the state of the system and can be measured, when the system remains in an equilibrium state. 12. How do you classify the property? Thermodynamic property can be classified in to two types.

1. Intensive or Intrinsic and 2. Extensive and Extrinsic property

13. Define Intensive and Extensive properties? [MU-Oct. 96, Oct. 98] The properties which are independent on the mass of the system is called intensive properties. E.g: Pressure, Temperature, Specific Volume etc. The properties which are dependent on the mass of the system is called extensive properties. Eg: Total energy Total volume, weight etc. 14. Differentiate Intensive and Extensive properties. [MU-Apr. 99, Apr.2001] Sl. No.

Intensive Properties Extensive Properties

1. Independent on the mass of the system. Dependent on the mass system.

2. If we consider part of the these properties remain same e.g. Pressure, temperature, specific volume, etc.

If the consider part of the system it will have a lesser value e.g. Total energy, Total Volume, weight, etc.



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15. What do you understand by equilibrium of a system? When a system remains in equilibrium state, it should not under go any changes to its own accord. 16. When a system is said to be in “Thermodynamic Equilibrium”? (Anna Univ. Apr.04 (EEE), MU-Apr. 98; MSU-Apr. 96) When a system is in thermodynamic equilibrium, it should satisfy the following three conditions a. Mechanical : Pressure remains constant b. Thermal Equilibrium : Temperature remain constant c. Chemical equilibrium : There is no chemical reaction 17. Define Zeroth law and first law of thermodynamics. (Anna Univ. Nov. 03 Apr.05 (EEE), MU-Nov. 94, Apr. 2001; BRU – Apr. 96) Zeroth law of thermodynamics states that when two systems are separately in thermal equilibrium with a third system, then they themselves are in thermal equilibrium with each other. First Law of thermodynamics states that when system undergoes a cyclic process net heat transfer is equal to work transfer

Q = W 18. (a)(i) State First Law of thermodynamics and any two of its corollaries. [Anna Univ. Apr.04 (EEE)] Zeroth law of thermodynamics states that when two systems are separately in thermal equilibrium with a third system, then they themselves are in thermal equilibrium with each other. First Law of thermodynamics states that when system undergoes a cyclic process net heat transfer is equal to work transfer.

Q = W Corollaries of first law of thermodynamics Corollary I There exists a property of a closed system such than a change in its value is equal to the difference between the heat supplied and the work done any change of state. Corollary II The internal energy of a closed system remains unchanged if the system is isolated from its surroundings. Corollary III A perpetual motion machine of first kind (PPM-1) is impossible.



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19. What is meant by “Perpetual Motion Machine of First Kind”? [Anna Univ. Apr. 03 (EEE)] PMM of first kind delivers work continuously without any input it violates first law of thermodynamics. It is impossible to construct an engine working with this principle. 20. Prove that for an isolated system, there is no change in internal energy. [Anna Univ. Apr. 03 (Mech)] For any isolated system. There is no heat, work and mass transfer. Q=W+=0 According to the first law of thermodynamics,

Q=W+U

U=0

21. Determine the molecular volume of any perfect gas at 600 N/m2 and 30C. Universal gas constant may be taken as 8314 j/kg mole – K. [Anna Univ. Apr. 03 (Mech)] Given data: P = 600N/m2

T = 30C=30+273=303K R=8314 J/kg mole – K To find: Molecular volume, V=? Solution: Ideal gas equation, PV=mRT

600 V = 1 8314 303 V=4198.57 m3/kg mole Result: Molecular volume, V=4198.57 m3/kg mole 22. An insulated rigid vessel is divided into two parts by a membrane. One part of the vessel contains air at 10 Mpa and other part is fully evacuated. The membrane ruptures and the air fills the entire vessel. Is there any heat and / or work transfer during this process? Justify your answer. [Anna University Nov‟03 (March)] For rigid vessel and unrestrained expansion Change in volume dV = 0

Work transfer. W = pdV = 0 For insulated vessel, heat transfer, Q = 0



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According to the first law of thermodynamics the sum of work transfer is equal to the sum of heat transfer.

W = Q = 0 23. Define the term process: [MKU – NOV‟96] It is define as the change of state undergone by a gas due to energy flow. 24. Define the term cycle: [MKU – NOV‟96] When a system under goes a series of processes and return to its initial condition, it is known as cycle. 25. What is meant by open and closed cycle? In a closed cycle, the same working substance will recirculate again and again. In a open cycle, the same working substance will exhausted to the surrounding after expansion. 26. What is meant by reversible and irreversible process? [MU – Apr‟2001; BNU - Nov‟94] A process is said to the reversible. It should traces the same path in the reverse direction when the process is reversed. It is possible only when the system passes through a continuous series of equilibrium state. If a system does not passes through continuous equilibrium state, then the process is said to be irreversible. 27. What is meant by point and path function? [MU-Oct. 2000] The quantity which is independent on the process or path followed by the system is known as path functions. Example: Heat transfer, work transfer 28. What is Quasi- static Process? [MU-Oct. 98, Apr. 2000, Apr. 2001] The process is said to be quasi – static, it should proceeds infinitesimally slow and follows continuous series of equilibrium states. Therefore the quasi-static process may be a reversible process. 29. Define the term enthalpy. [MU-Oct. 99] The combination of internal energy and flow energy is known as enthalpy of the system. Mathematically, enthalpy (h) = U + pVkJ Where U –Internal energy P – Pressure V – Volume

In terms of Cp&T H = mCp (T2 – T1)kJ



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30. Define the term internal energy. [MKU – Apr. 96] Internal energy of a gas is the energy stored in a gas due to its molecular interactions. It is also defined as the energy possessed by a gas at a given temperature. 31. What is meant by thermodynamic work? It is the work done by the system when the energy transferred across the boundary of the system. It is mainly due to intensive property difference between the system and surrounding. 32. Sketch isothermal expansion on p-V diagram and state the properties that remain constant. [Anna Univ. Apr. 2004 (EEE)] The following properties remain constant. 1. Temperature 2. Internal energy 3. Enthalpy 33. Prove that the difference in specific heat capacities equal to Cp – Cv = R. [Anna Univ. Apr. 03 (Mech)] Consider a gas heated at constant pressure So, heat supplied, Q = mCp (T2 – T1) Work done, W = p(V2 – V1)= m R (T2 – T1)

Change in internal energy, U = mCv (T2 – T1) According to the first law of thermodynamics,

Q = W + U So, mCp (T2 – T1) = mR (T2 – T1)+ mCv(T2 – T1)

Cp = r + Cv Cp – cv = R



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PART – B

1. A mass of gas is compressed in a quasi-static process from 100 kPa 0.3m3 to 500 kPa, 0.05m3. Assuming that the pressure and volume are related by PV” = constant, find the workdone by the gas system.

System : Closed system Process : PV” = C Known : Initial pressure = 100 kPa Initial Volume = 0.3m3

Final Pressure = 500 kPa Final Volume = 0.05m3 To find : Work done Diagrams : 0.3 m3 0.05 m3 100 kPa 500 kPa Initial State Final State

P (kPa) 2

500 pVn = C 100 1

2

1

pdV

0.05 0.3 V(m3

P-V Diagram

In this equation all the values are known except the polytrophic index n.

Consider the process relation P1V1

n = P2V2n

1 2

2 1

n

V p

V p

Taking natural log on both sides



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n In 1 2

2 1

V pIn

V p

1 2

2 1

( / )

( / )

In V Vn

In p P

Substituting the numerical values

500

100

0.3

0.05

IN

n

In

n = 0.898 Hence the work done

1W2 = 2

1

PdV

2 2 1 1

1

500 0.05 100 0.3

0.898 1

49.01

p V pV

n

KJ

Comment Negative sign indicates that the work is done on the system. 2. A certain fluid expands from 10 bar, 0.05m3 to 2 bar, and 0.2m3 according to linear law. Find the work done in the process. System : Closed system Process : Linear Law Known : Initial pressure = 10 bar = 1000 kPa Initial volume = 0.05m3 Final pressure = 2 bar = 200 kPa Final volume = 0.2m3 To find : Work done Diagrams: 10 bar 2 bar 0.05 m3 0.2 m3 Initial State Final State



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P (kPa) 1 1000 A1 200 2 A2 0.05 0.2 V (m3)

P-V Diagram

Analysis : 1W2 = 2

1

pdV

= Area A1 + Area A2

= 1

(0.2 0.05)(1000 200) (0.2 0.05)2002

= 90KJ 3. Following diagram shows a cycle executed by a closed system on a p-V diagram. Calculate the work done by the system for each of three processes and also the network done. P(kPa) 1 1000 500 3 2 a b

0.1 0.2 V(m3)

p-V Diagram System : Closed System Known : Initial and final pressure, Initial and final Volume of all the processes. Process 1-2 Linear Law 2-3 Isobaric (p=C) 3-1 Isochoric (V=C)

Work done 2

1

pdV in KJ

Area 12ba 1

1(0.2 0.1) (1000 500) (0.2 0.1) 500

2

75

= 500(0.1– 0.2) -= -50

=0 = p2 (V3 –V2)



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Network: Wnet = 1W2 + 2W3 + 3W1 = 75 – 50 + 0 = 25KJ 4. A spherical balloon of 0.5 m diameter contains air at a pressure of 500 kPa. The diameter increases to 0.55m in a reversible process during which pressure is proportional to diameter. Determine the work done by the air in the balloon during this process. Also calculate the final pressure. System : Closed system Process : Expansion of air in a spherical balloon such that pressure is proportional to the diameter. Known : 1. Initial diameter = 0.50m 2. Initial Pressure = 500 kPa 3. Final diameter = 0.55m To find : 1. Work done 2. Final weight Diagram : P (kPa) 2 1 500

0.5 0.55 dia (m)

Air Dia = 0.5m P = 500 kPa

Air

Dia = 0.55m



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Analysis:

1. Work done

2W1 = 2

1

pdV

From the condition of the process P = CD Where p is the pressure (kPa) C is a constant D is the diameter (m) From the initial condition

500 = C 0.5 C = 500 / 0.5 = 1000 Therefore P = 1000 D Substituting p we get,

1W2 = 2

1

1000DdV

To substitute dV in terms of dD Consider the volume of the sphere

V = 3

6D

Upon different we get

dv = 236

D dD

Substituting dV in terms of dD we get

1W2 = 2

2

1

1000 36

D D dD

0.55 33

0.50

0.554

0.50

4 4

1000

6

5004

500[0.55 0.50 ]

4

11.39

D dD

D

KJ

2. Final pressure



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P2 = 100 D2

= 1000 0.55 = 550 kPa 5. A rubber balloon flexible is to be filled with hydrogen from a storage cylinder which contains hydrogen at 5 bar pressure until it has a volume of 1 m3. The atmospheric pressure may be taken as 101.32 kPa. Determine the work done by the system comprising the hydrogen initially in the bottle. System : Closed Process : Hydrogen expanding from a cylinder against the atmospheric pressure. Known : 1. Volume of expansion dv = 1 m3

2. Atmospheric pressure Patm = 101.32 kPa 3. Storage cylinder pressure = 500 kPa.

Diagram: Balloon after filling Balloon before filling Hydrogen Cylinder

To find : Work done by the hydrogen gas

Analysis : Work done = 2

1

pdV

2

1

0.10

22 1

1 0.15

52.0

5 2.0( ) 100

V

VIn V V

V

As the pressure is in bar, it has to be multiplied by 100, to get it in kpa Substituting the limits we get

1W2 = 5 In 0.10

2.0(0.1 0.15)0.15



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= 2.027 + 0.1

= 1.027 KJ 6. A compressor receives 10kW of power at 600 rpm from a motor. What is the torque acting on the shaft connecting the motor and the compressor? System : Compressor receiving power from a motor through a shaft Known : 1. Power = 10kW 2. Speed = 600 rpm To find : 1. Torque Diagrams:

Analysis : 1. w = 2nt Rate of work transferred

= 10 103 KJ/s Speed = 600 rpm = 600 / 60 rps = 10 rps

3

/ 2

10 10 =

2 10

= 159.15 N m.

T W n

7. An elastic linear spring of spring constant 144 N/cm is compressed from an initial unconstrained length to a final of 6cm. If the work required on the spring is 648J, determine the initial length of the spring in centimeters: System : Elastic spring Process : Stretching the elastic spring Known : 1. Spring constant = 144 N/cm 2. Final length = 6 cm 3. Work supplied = 648 KJ To find : Initial length in cm

Compressor

Motor



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Fin

al Le

ngth

Diagrams: -

Analysis :

1

2

144648

2

W k

cm

Initial length = final length + deflection () = 6 + 3 = 9 cm 8. Calculate the work required to lift 25 kg from an evaluation of 208m above mean sea level to an elevation 80m higher in (a) min (b) 10 min. System : A body of mass 25 kg Process : Lifting the body against gravity Known : 1. Mass = 25 kg 2. Initial height = 208m

3. Change in height = 80m 4. Time a) min b) 10 min

To find : Work done Final state Initial state 80 m 208 m Mean sea level

Fre

e L

en

gth



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Analysis : Work done = -mgh : The negative sign indicates that work is done on the system. Work is independent on time and hence for both cases work done is – 19620J. 9. Determine the work required to accelerate a body of mass 100 kg from rest to a velocity of 100 m/s. System : A body of mass 100 kg Process : Acceleration of a body from rest Known : 1. mass = 100 kg 2. Initial velocity 3. Final velocity = 100 m/s. To find : Work Analysis

2 2

2 1

2 2

1( )

2

1 100 (100 0 )

2

= 500

W m C C

KJ

comment : The negative sign indicates that work done on the body to accelerate it. 10. A perfect gas for which the ratio of specific heats is 1.4, occupies a volume of 0.3m3 at

100 kpa and 27C. The gas undergoes compression to 0.06m3. Find the heat transfer by the gas in each of the following process.

a) Isobaric b) pV1.1 = Constant Assume R = 0.287 KJ/ kgK.

Case(a) System : Closed system Working fluid : A perfect gas Process : Isobaric Known : 1. Initial volume (V1) = 0.3m3 2. Initial pressure (p1) = 100 pa

3. Initial Temperature (T1) = 27C = 300 K

4. Ratio of sp. Heat () = 1.4 5. Final Volume (V2) = 0.06m3

6. Characteristic gas constant = 0.287KJ/kg K To find : Heat transfer Q12



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Diagram:- 0.3 m3 0.06 m3 100 kPa 100 kPa 300 K Initial State Final State P (kPa) 2 1

2

1

pdV

0.06 0.3 V(m3)

p-V diagram

Analysis : Q12 – 1W2 = U

Where 1W2 = 2

1

pdV

= p(V2 – V1) for an isobaric process = 100 (0.06 – 0.3) = - 24KJ

U = 2

1

dU

= 2

1

. 3.23mC dT from eqn

= mC(T2 – T1) To find T2 Since the working substance is an ideal gas P1V1 = mRT1 P2V2 = mRT2 By rearranging the above equation

1 1 2 2

1 2

pV PV

T T

Applying the given process condition p1 = p2



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1 2

1 2

22 1

1

2

0.06300

0.3

60

V V

T T

VT T

V

T k

To find C, Cp – Cv = R

1

1

1

0.287

0.4

0.717

p

v

v

v

p

v

CC R

C

RC

C

C

R

v

KJ

kgK

To find m

m = 1 1

1

pV

RT [ The given substance is a perfect gas]

= 100 0.3

0.287 300

= 0.348 kg Substituting the values of m and T2

U = 0.348 0.717 (60 – 300) = 59.93KJ

Therefore Q12 = 1W2 + U = - 24 – 59.93 = - 83.93KJ Comment: The negative sign in work indicates work is Done on the system.

The negative sign in U shows that internal energy decrease. The negative sign in heat transfer shows that heat is rejected by the system.



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Case (b) System : Closed system Process : Polytropic of index 1.1 (pV1.1 = C) Working fluid : A perfect gas Known : 1. Initial volume (V1) = 0.3m3 2. Initial pressure (p1) = 100 kPa 3. Initial Temperature (T1)

4. Ratio of sp. heat () = 1.4 5. Final Volume (V2) = 0.06m3

6. Characteristic gas constant = 0.287 KJ/kg K. Diagram:- 0.3 m3 100 kPa 0.06m3 300 K Initial State Final State P (kPa) 2 pV1.1 = C 1

2

1

pdV

0.06 0.3 V(m3)

p-V diagram

Analysis : Q12 – W2 = U

1W2 = 2

1

pdV

= 2 2 1 1

1

p V pV

n

for a polytropic process

since the working fluid is a perfect gas

1W2 = 2 1

1

mRT mRT

n



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= 2 1( )

1

mR T T

n

To find m

m = 1 1

1

[ ]pV

RTThe given substance is a perfect gas

100 0.3

0.287 300

0.348 kgb

To find T2 Since the working substance is an ideal gas p1V1 = mRT1 p2V2 = mRT2 By rearranging we get

1 1 2 2

1 2

pV p V

T T

or

2 2 2

1 1 1

T p V

T pV

The process equation gives p1V1

n = p2V2n

2 1

1 2

n

p V

p V

Combining the above equations



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2 2 2

1 1 1

1

1 1

2 2

1

1

2

1

12 1

2

0.40.3

3000.06

571.1

n

n

n

T P V

T P V

V V

V V

V

V

VT T

V

K

Substituting the values of m and

1W2 = 0.348 0.287(571.1 300)

1.1 1

T2 = 270.76KJ

U = mCv (T2 – T1)

= 0.348 0.717 (571.1-300) = 67.64 KJ

Q12 = 1W2 + U

= 270.76 + 67.64

= 203.12 KJ Comment : The negative sign in work indicates that work is done on the gas. The negative sign in heat transfer shows that heat is coming out of the system. 11. Calculate the change in internal energy, heat transferred and change in enthalpy for 0.5

kg of air expanding according to law pv1.2 = C from 1 Mpa and 300C to 100 kPa. What will be the workdone by air during the expansion? System : Stationary system of fixed mass Process : Polytropic (pV1.2 = C) Work fluid : Air (ideal gas) Known : 1. Mass (m) = 0.5 kg

2. Initial pressure (p1) = 1Mpa = 1000kPa

3. Initial Temperature (T1) = 300C = 573 K

4. Ratio of sp. heat () = 1.4

To find : 1. Change in internal energy U 2. Heat transferred Q12

3. Change in enthalpy H 4. Work done 1W2



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Diagrams: 1 Mpa 100 kPa

300C Final State Initial State Analysis :

U = 2

1

dU

= mCv (T2 – T1) To find T2 Since the working substance is an ideal gas P1V1 = mRT1 P2V2 = mRT2 By rearranging we get

2 2 2

1 1 1

T PV n

T pV

From the process equation p1V1

n = P2v2n

P

(kPa)

1000

100

1

pV1.2

= C

2 2

1pdv

V(m3)

p-V Diagram



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1

2 1

1 2

1

2

1

n

n

V P

V P

P

P

combining the above equation

2 2 2

1 1 1

1

2 2

1 1

11

2

1

1

22 1

1

0.2

1.2

P =

100 573

1000

390.4

n

n

n

n

T P V

T P V

P

P P

P

P

PT T

P

K

12 1 2

12 1 2

0.5 0.7178 (390.4 573)

65.47

New U

KJ

Q W U

Q can be found only if W is known

1W2 = 2

1

pdV

2 2 1 1

2 1

1

( )

1

0.5 0.287(390.4 573)

1.2 1

PV PV

n

mR T T

n


since the working fluid is an ideal gas

= + 131.02 KJ

1Q12 = 1W2 + U = 131.02 + (-65.47) = 65.53 KJ



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Comment : The positive sign in work indicates that work is done by the system. The positive sign in heat indicates that heat is given into the system. 12. A space vehicle has a mass of 500kg and is moving towards the moon. Calculate the kinetic and potential energies relative to the earth when it is 40 km from the launching place and travelling at 2400 km / hr and when the acceleration of the earths gravitational field is 792.5cm/sec2. System : A moving system of fixed mass Known : 1. Height h = 40 km

= 40 103 m 2. Velocity C = 2400 km/hr.

32400 10

3600

666.67 / .

m

s

m s

.

3. Acceleration due to gravity = 792.5 cm /sec2 = 7.925 m/Sec2

To find : a) Kinetic energy b) Potential energy Analysis : a) Kinetic energy

2

2

6

1

2

1500 666.67

2

111.11 10

111.11

mC

J

MJ

b) Potential energy = mgh

= 500 7.925 40 103

= 158.5 106 J = 158.5MJ

13. During a short period of time the following observation were made for a system of mass 50 kg. a. Heat received = 107 J

b. Work produced = 4 106 J



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c. Initial Velocity = 10 km/s. d. Final velocity = 25 m/s e) Initial elevation = 20 m f) Final elevation = 12 m Find the change in internal energy of the system. Solution: System : A moving system of fixed mass Known : a. Heat received = 107 J

b. Work produced = 4 106 J c. Initial velocity = 10 km / s d. Final velocity = 25 m /s e. Initial elevation = 20 m f. Final elevation = 12 m

To find : U change in internal energy Diagrams :

Analysis : 1Q12 – 1W2 = E

Note: Since a moving system of fixed mass is analyzed here E U

Q12 = 1W2 + U + KE + PE

Initial State

Q = 1 107 J

10 103 m/s

50 kg

w = 4 106 J

20m

25 50 kgs.

during the process 12m

Datum



25

or U = Q12 – 1W2 - KE - PE

2 2

12 1 2 2 1 2 1

2 27 6 3 3

7 6 10

10

1( ) ( )

2

11 10 4 10 50 25 10 10 10 50 9.81(12 20)

2

1 10 4 10 1.3125 10 3924

1.3 10

Q W m C C mg Z Z

U J

14. Air enters a compressor with a velocity of 60 m/s, pressure 100 kPa, temperature 40C

and leaves the compressor with a velocity of 90 m/s, 500 kPa and 120C. Consider the system as adiabatic. Find the power of the motor for a mass flow rate of 40 kg per minute. Write the assumption made. System : Open System Working Fluid : Air Process : Steady flow, adiabatic Known :

Property At inlet At exit

Velocity 60 m / s 90 m / s

Pressure 100 kPa 500 kPa

Temperature 40C 120C

and the mass flow rate = 40 kg / min. Diagrams :-

Analysis : [ ]Q W m h ke pe

Q = 0 sincde the flow is adiabatic.

Unless the height from the datum of the inlet and out are mentioned pe = 0.

In

Q = 0

Control

Out

W

Volume



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Therefore W = 2 2

1 21 2

2

C Ch h

2 2

1 21 2

2 2

2

60 9040 1005 40 120

2

C Cm Cp T T

6

6

3.306 10min

3.306 10

60

55.1 /

55.1

J

J

s

KJ s

kW

Comment : Since the working fluid is an ideal gas h1 - h2 is replaced by Cp (T1 – T2). 15. In a steam power station, steam flows steadily through a 0.32 m diameter pipeline from the boiler to the turbine. At the boiler end, the steam conditions are found to be P = 4 Mpa, t

= 400C, h = 3214 kJ / kg and v = 0.073 m3 / kg. At the turbine end p = 3.5 Mpa, t = 392C, h = 3202.6 kJ / kg and v = 0.084 m3 / kg. There is a heat loss of 8.5 KJ/kg from the pipeline. Calculate the steam flow rate. System : Open system Working fluid : Steam Process : Steady flow process Known :

Property At inlet At exit

Pressure 4 Mpa 3.5 Mpa


Enthalpy 3214

KJ

kg 3020.6

KJ

kg

Sp. Volume 0.073 m3 / kg 0.84 m3 / kg

and heat loss of Q = 8.5 KJ/ kg. To find : The flow are in



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Diagram:-

Analysis : Q – W = m [h + ke + pe] W = 0 Since it involves only the flow through pipes. As the flow is steady mass flow rate is same all the inlet and exit.

in outm m

. .

. .

in exit

in exit

Volume flow rate Volume flow rate

Sp Volume Sp Volume

Area velocity Area velocity

Sp Volume Sp Volume

1 21 2

1 2

22 1

1

1

2 1

2 2

2 1

2 2

1 1

2

1

2

1

0.084

0.073

1.15

1

2

11.15

2

10.32

2

0.16

C CSince A A

V V

VC C

V

C

C C

ke C C

C C

C

C

Control volume

Exit Inlet

Boiler Turbine



28

where

11

1

2

0.073

0.2

2.32

mVC

A

m

m

ke = 0.16 (2.324 m)2 = 0.86 m2

pe = 0

Now Q – W = m [h + ke + pe] becomes

2

2 1

2

2 3

0 0.86 0

8.5 103 = (3202.6 - 3214) 103 + 0.86m

0.86 2.9 10

58.07 / .

Qh h m

m

m

m kg s

Comment: Though the flow is steady through the pipe the increase in velocity is due to in increase in specific volume. 16. The stream of air and gasoline vapour in the ratio of 14:1 by mass enters a gasoline

engine at a temperature of 30C and leaves as combustion products at a temperature of

790. The engine has a specific fuel consumption of 0.3 kg/k Whr. The net heat transfer rate from the fuel – air stream to the jacket cooling water and to the surroundings in 35 KW. The shaft power delivered by the engine is 26KW. Compute the increase in the specific enthalpy 0 fuel air stream. System : Open system Process : Steady flow, process Known : 1. Air fuel ratio = 14:1

2. Inlet temperature t1 = 30C

3. Exit temperature t2 = 790C 4. Specific Fuel Consumption = 0.3kg/kwhr.

5. Heat rejection rate 35Q KW

6. Shaft work output W = 26 KW

To find : Increase in specific enthalpy h



29

Diagram :-

Analysis : [ ]Q W m h ke PE

ke and pe terms can be neglected unless the relevant data are given.

To find m

Specific fuel consumption = 0.3 kg/kwhr One whr is equivalent to 3600KJ. Hence, fuel consumption is 0.3 kg for an output of 3600 KJ. Since the output is 26kW, the rate of fuel consumption.

3

0.326

3600

2.17 10 /

r

kg kJm

kJ s

kg s

The rate of air consumption = 14 rate of fuel consumption

3

3

14 2.17 10

0.03033 /

0.03033 2.17 10

0.0325 /

r

a f

m

kg s

m m m

kg s

Substituting the numerical values in the SFEE we get

- 35 – 26 = 0.0325 h

h = -1877 KJ / kg

Comment: Each kg of the mixture contain 1

15kg of fuel and

14

15kg of air. This mixture

undergoes combustion in the engine. Out of the energy liberated during combustion only 1877 KJ is extracted within the engine and the remaining is taken away by the exhaust gases. This is the reason for higher exhaust temperature.

Fuel air Stream

26kW

ENGINE Shaft work W =

Exhaust

35Q kW



30

17. A steam turbine operates under steady flow conditions receiving steam at the following state: Pressure 15 bar. Specific volume 0.17m3/kg internal energy 2700 KJ/kg velocity 100 m/s. The exhaust of steam from the turbine is at 0.1bar with internal energy 2175 KJ/kg. specific 15 m3/kg and velocity 300 m/s. The intake is 3 m above the exhaust. The turbine develops 35 kW and heat loss over the surface of turbine is 20 KJ /kg. Determine the steam flow rate through the turbine. System : Open System Process : Steady flow process Known :

Properties At inlet At exit

Pressure 15 bar = 15 103 Pa bar = 0.1 103 Pa

Internal energy 2700 103J/Kg 2175 103 J/Kg

Sp. Volume 0.17 m3 / kg 15 m3 / kg

Height from the datum 3m 0m

Velocity 100 m/s 300 m/s

Work output W = 35 103 W

Heat rejected work 320 10 /Q

q J kgm

Diagram:

Analysis : [ ]Q W m h ke pe

(or)

2 2 2 2

2 2 2 1 1 1 2 1 2 1

3 5

3 5 3

2 2

1( )

2

2175 10 0.10 10 15

(2700 10 15 10 0.17) 20 10

1300 100 9.81(3)

2

wq u p v u p v C C g Z Z

m

w

m

In

q = 20 103 J/Kg

Out

Control Volume

W = 35 103W



31

3

3

3

3

2325000 2955000 4000 29.43 20 10

569.970 10

35 10

569.970 10

0.0614 / .

wm

m

w

m

m

m kg s

18. Steam enters a nozzle operating at steady state pressure of 2.5 Mpa and a temperature

of 300C (h1 = 3008.8 KJ/kg) and leaves at a pressure of 1.7 Mpa with a velocity of 470 m/s. The rate of flow of steam through the nozzle is 1360 kg/hr. Neglecting the inlet velocity of the steam and considering the flow in the nozzle to be adiabatic, find: a) the enthalpy hexit b) the nozzle exit area, if Vexit = 0.132 m3/kg. System : Open system Process : Steady flow, adiabatic Known :

Properties At inlet At exit

Pressure 2.5 mPa -

Temperature 300C -

Enthalpy 3008 KJ/kg -

Velocity negligible 470 m/s

Sp. Volume - 0.132 m3/kg

The mass flow rate in = 1360 Kg/hr

= 1360

/3600

kg s

= 0.378 kg/s To find : Exit enthalpy – h2 Exit Area – A2

Diagram :

In Out



32

Analysis : Q W m h ke kpe

Flow is given as adiabatic hence 0Q

Nozzle will not involve W

Change in potential energy can be neglected in nozzles and diffusers

h = ke

2 2

1 2

2 2

1 2

2

2

C C

C Ch

As the inlet velocity is negligible

h2 – h1 = 2

2

2

C

= 2470

2

= - 110450 J/kg

h2 = h1 2

2

2

C

= 3008.8 103 – 110450

= 2898.4 103 J/kg = 2898.4 KJ/kg

b)

2 2

2

22

2

4 2 2

0.378 0.132

470

1.062 10 1.062

A Cm

V

mVA

C

m or cm

19. Air enters the diffuser of a jet at 10C, and 80 kPa, with a velocity of 200 m/s. The inlet area of the diffuser is 0.4m2. Neglecting exit velocity of air determine. a. the mass flow rate of air b. the exit temperature of air System : Open system Process : steady flow Known : At the inlet



33

Temperature T1 = 283 K Pressure P1 = 80 kPa Velocity C1 = 200 m/s Area A1 0.4 m2 To find : 1. Mass flow rate m

2. Exit temperature T2 Diagram :

Analysis : 1. .

Area Velocitym

Sp Volume

1 1 1 1

1 1 1

/

AC ACor

V RT P

= 80 0.4 200

0.287 283

= 78.8 kg/s.

2. Q W m h ke pe

Heat transfer across the control surface and change in potential energy can be neglected in a diffuser.

A diffuser will not involve w

h = -ke

(h2 – h1) = - 2 2

2 1

2

C C

Since air is an ideal gas

2 2

2 12 1

2

12 1 1

( )2

( ) 2

p

p

C CC T T

CC T T Since C is negligible



34

2

12 1

2

2

200 283

2 1005

303

p

CT T

C

K

20. Hellium at 300 kPa, 60C, enters a nozzle with negligible velocity and expands steadily without heat transfer in a quasi equilibrium manner to 120 kPa. The process is such that Pv1.67 = constant. Calculate the exit velocity. System : Open system Process : Steady flow, adiabatic, pV1.67 = C Known : 1. Inlet pressure (p1) = 300 kPa

2. Inlet Temperature (t1) = 60C 3. Exit pressure (p2) = 120 kPa Diagram:

Analysis : Q – W = m [h + ke + kpe]

We can‟t proceed with this equation since the value of enthalpy drop is not known. It can be substituted by Cp (T2 – T1), but still Cp is an unknown. In problems of this kind, Equation can be used.

2 2 2

2 12 1

1

( )2

flow

C CW vdp Z Z g

Since these is no work interaction in a nozzle and (Z2 – Z1) term can be dropped.

22 2

2 1

12

C Cvdp

Where C1 is negligible

22

2

12

Cvdp

In Out

Control Volume



35

From the process equation pVn = constant 1

1

2 1 12

2

1

21

11

2

1

2

11

n

n n

nn

Cv

p

CC p dp

pC

n

1 11

2 1

1 1 1

1 1 2 1

1 1 11

1 1 1 2 1

1 1

2 11 1 1 1

1 1

1

1

1( )

1since

n n

n nn

n n

n n n

n n

n n n

n n

n n

n n

n n

p pC

n

n

np v p p

n

np p v p p

n

p pnp v

np p

1

21 1

1

1

22 1 1 1 1

1

0.67

1.67

11

21 [ )

1

2 1.67 8314 120 333 1

1.67 1 4 300

1030 / .

for an ideal gas

n

n

n

n

pnp v

n p

pnC RT p v RT

n p

n

m s

21. Imagine two curves intersecting at some point on a p – v diagram. The curves are representing reversible processes undergone by a perfect gas; one is an adiabatic process and the other an isothermal process. Show that the ratio of slope of the adiabatic curve to that of the isothermal curve at the point of intersection is equal to:-



36

System : Closed / Open Process : case (i) Isothermal case (ii) Adiabatic Diagram :

To prove : Adiabatic

Isothermal

dp

dv

dp

dv

Proof : For case (i) Isothermal Process pv = RT = constant By differentiating we get, Pdv + vdp = 0 vdp = dv

(1)Isothermal

dp p

dv v

For case (ii) Adiabatic Process

pv = constant Taking natural log we get

In (p) + In (v) = In (C) By differentiating we get

(2)Adiabatic

dp dv

p v

dp p

dv v

case (i)

case (ii)

P

V



37

From (1) & (2) we get Adiabatic

Isothermal

dp

dv

dp

dv

Hence proved. 22. A piston cylinder arrangement as shown in the diagram contains 5g of air at 250 kPa,

300C. The piston is of mass 75 kg and diameter 10 cm. Initially the piston is kept pushed

against the stops. The atmospheric pressure is 100 kPa and temperature is 20C. The

cylinder now code to 20C. Calculate the heat transfer. Process : Constant volume / constant pressure

When heat is removed from the system temperature decreases and volume remains constant until the upward pressure exerted on the bottom side of the piston just balances the atmospheric pressure and the weight of the piston. Further cooling results in decreases of volume at constant pressure.

Known : T1 = 300C P1 = 250 kPa DP = 0.1 m Patm = 100 kPa Tatm = 293 K T2 = 293 K Diagram : To find : Heat transfer Q12 Analysis : To find the pressure which just balance the atmospheric pressure and the weight of the piston.

Pair APiston = mPiston g + Patm APiston

Stop

Piston

Air Cylinder

P

2

1

21

V



38

3

2

3

75 9.81 100 10

0.1

193.68 10

193.68

Piston g

air atm

Piston

mP P

A

Pa

kPa

Therefore cooling from 250 kPa to 193.68 kPa will be taking place at constant volume. Temperature at the stage can be computed as follows:-

2 2 1 1

2 2

2 1

22 2

1

193.68 573

250

443.9

PV PV

T T

V V

PT T

P

K

since

Therefore further cooling from 443.9 K to 293K will be a constant pressure process. Thus, Q12 = Q12‟ + Q2‟-2 Q12 = 1W2 + 2W2 + (U2 – U2‟) + (U2 – U1) Q12 = 1W2 + 2W2 + (U2 – U1) Since the process 1-2‟ is a constant volume process 1W2 = 0 and as the process 2‟-2 is isobaric.

2‟W2 = P2‟ (V2‟ – V2) where V2 = Substituting we get

2‟W2 = 193.68 (2.171 10-3 – 3.2 10-3) = - 0.199 KJ. U2 – U1 = mCv (T2 – T1)

= 5 10-3 0.717 (293-573) = 1.0038 KJ. Therefore Q12 = -0.199 – 1.0038 = - 1.2028 KJ.



39

23. A closed system undergoes a cycle consisting of two processes. During the first process, 40 KJ of heat transferred to the system while the system does 60 KJ of work in the second process, 45 KJ of work is done on the system. a) Determine the heat transfer during the second process. b) Calculate the network done and net heat transfer for the cycle. Solution:- System : Closed System Process : Cycle consisting of two processes Known :

To find : Q21, Q, W Analysis : 1

Q21 = 2W1 + U2-1 Where 2W1 = -45 KJ.

To find (U2-1) For cyclic process the algebraic sum of any property of the system must be zero.

(U12) + (U21) = 0

(U12) = (U)12 = - [Q12 – 1W2] = -[40 – 60] = +20 KJ

U21 = -45 +20 Q21 = -25KJ 2 Network done = (1W2 + 2W1) = 40 – 25 = 15 KJ Net heat transfer = (Q12 + Q21) = 60 – 45 = 15 KJ. Comment : The negative sign indicates Q21 that 25 KJ of heat is supplied by the system to the surroundings during the process 2 to 1.

Process Q(KJ) W(KJ)

1-2 2-1

40 unknown

60 -45



40

24. A system is capable of executing a cyclic process as indicated in the pV sketch; it may be executed either clockwise abca or counter clockwise adca. a. When going clockwise to state C, 80 KJ of heat flow to the system and 35 KJ of work are done by it. When returning to state a for c, 60 KJ of heat flow from the system. Find work along the path ca. b. When going counter clockwise to state C, 70KJ of heat flow to the system. Find the work during the process adc. Solution : System : Closed (as the total volume changes) Process : Cyclic with two possible cycles. Known :

Cycle abca Cycle adca

Process Q(KJ) W(KJ) Process Q(KJ) W(KJ) a – c c – a

80 -60

35 Unknown

a – c c – a

70 Unknown

Unknown Unknown

To find : 1. Work done in the process c – a in path abca 2. Work done in the process c – a in path adca

Analysis : 1. dQ dw for the cycle abca

80 – 60 = 35 +cWa cWa = 20 – 35 = 15 KJ 2. Qac = Wac + Uc – Ua Wac – Qac – (Uc - Ua) To find : Uc – Ua.

Irrespective of the path followed U between also are some.

(U)ac in cycle abc = (U) ac in cycle adc (Uc – Ua) in adc = (Qac – aWc) in path abc

P

b

a d

c

V



41

= 80 – 35 = 45 KJ

cWa = Qac – (Uc – Ua) = 70 – 45 = 25 KJ. 25. For a certain system executing a cyclic process 250J of heat is absorbed by the system and 100J of heat are rejected. The system also receives 30W-s of electrical power while it moves a 3 kg mass vertically by means of a pulley arrangement. How far does the mass move? Take local g=9.65 m/s2. System : Closed / Open Process : Cyclic Known : Qin = 250 J Qout = 100 J EEin = 30 w-s To find : Height through which is a mass to be lifted. Analysis : Energy in = Energy out 250 + 30 = 100 + W W = 180J This must be equivalent to the increase in potential energy of the mass that is mgh = 180.

180

3 9.64

6.22 .

h

h m

26. A three process cycle operating with nitrogen as the working substance has constant temperature compression 1-2 (T=40C, p1 = 110 kPa); constant volume heating 2-3; and polytropic expansion 3-1 (n=1.35), the isothermal compression requires – 67 KJ/kg of work. Determine. a. P, T and v around the cycle: b. The heats in and out; c. The net work. Take Cv = 0.7431 KJ/kg. System : Closed System Process : Cycle consisting of there processes Known : P1 = 110 kPa T1 = 313K

Process Type QKJ/Kg WKJ/Kg

1 – 2 2 – 3 3 – 1

Isothermal Isochoric PV1.35 = C

- - -

-67 - -



42

To find : 1. p.T.v. at state points 1, 2 and 3 2. Q, w in all the processes 3. net work State 1: Analysis : a) Finding p, T, v of the state points

Consider state 1 and process 1-2 (Isothermal) P1 = 110 kPa T1 = 40 + 273 = 313 K

11

1

3

1 2

8.314313

28

110

0.845 /

67 /

RTv

p

m kg

W KJ kg

2since N , is an ideal gas

That is

21 1

1

2

1 1 1

67

110 0.845

2 1

3

67

67

0.411 /

vp v In

v

vIn

v p v

v v e

m kg

Also p1v1 = p2v2

1 11

2

12 1 2 12

226.17

[

67 /

p vp

v

kPa

and Q W U

KJ Kg

12

since the process is isothermal]

Q

Consider state 3 and the process 2-3 (Isochoric)

2W3 = 0



43

V3 = V2 0.411 m3 /Kg T2 = 313 K From process 3-1 )pV1.35=C) p1v1

1.35 = p3v3 1.35

1.35

13 1

3

1.350.845

1100.411

291.0

vp p

v

kPa

Knowing p3, T3 can be found

T3 = T2 3

2

p

p

23 3 2

31 1 2

291.0313

226.17

402.72

( )

0.7431(402.72 313)

66.67 /

( )

0.7431(313 402.72)

66.67 /

v

v

K

U C T T

KJ kg

U C T T

KJ kg

1 1 3 33 1

1

p v p vW

n

31 3 1 31

110 0.845 291 0.411

1.35 1

76.2

76.2 66.67

9.5 / .

KJ

Kg

Q W U

KJ Kg



44

Tabulating the property values:

State P(kPa) T(K) V(M3/kg)

1. 110 313 0.845

2. 226.17 313 0.41

3. 291.0 402.72 0.411

b) Heats in and out.

Process QKJ/Kg W(KJ/kg) U(KJ/Kg)

1 – 2 - 67 - 67 0

2 – 3 66.67 0 66.67

3 – 1 9.5 76.2 - 66.67

c) Wnet = - 67 + 76.2 = 9.2 KJ/Kg

Check : Q = - 67 + 66.7 – 9.5 = 9.8 KJ/Kg Qnet = Wnet from first law

U = 0 + 66.7 – 66.67 Also

U = 0 27. In a steady flow process air is passing through a series of components as shown in the following diagram. Assuming the heat addition process is reversible isobaric and all others are reversible adiabatic processes obtain the exit velocity. The power developed by the turbine is just enough to run the compressor. System : Open Process: Steady flow through a compressor, a heat exchanger, a turbine and a nozzle in sequence. Known : P1 = 100 kPa T1 = 300K M = 0.2 Kg/s P3 = 600KPa P5 = 100 Kpa

o o

r cW W

To find : Exit velocity from the nozzle



45

Diagram :- Analysis : Process : 12

2 1

1 1

322 1 1

1 1

0.4

1.4

( )

600300

100

500.6

o o

c pW mC T T

PPWhere T T T

P P

K

(ii) Process 23

23

3 2

3

3

100

( ) 100

100500.6

0.2 1.005

993

o

o

p

Q kw

mC T T

T

T K

(iii) Process 34

4 3 2 1( )

998 (500.6 300)

7975

T T T T

K

100 kPa

Heat exchanger

2 3

1

Com

pre

ssor

Turb

ine

4

5

Nozzle

Q = 100 kW

Common Shaft

600 kPa

600 kPa 100 K 0.2 kg/s



46

o o

T cW W

3 4 2 1( ) ( )o o

p pmC T T mC T T

3 4 2 1

4 3 2 1( )

998 (500.6 300)

7975

T T T T

T T T T

K

Also 1

4 4

3 3

P T

P T

P4 =

1.4

0.47975600

998

= 273.6 Kpa (iv) Process 4-5

1

5 5

4 4

0.4

1.4

5

1007975

273.6

598.2

T P

T P

T

K

For the flow through the nozzle

2 2

5 44 5

2

C Ch h

Neglecting C4 compared C5 we get

5 4 52 ( )pC C T T

2 1005(7975 598.2

632.9 / .m s

The exit velocity of air from the nozzle is 632.9 m/s. 28. An air compressor is used to supply air to a rigid tank that h as a volume of 4m3.

Originally, the air pressure and temperature in the tank are 101 kPa and 35C. The supply pipe is 5 cm diameter and the velocity of air in the inlet pipe remains at 15 m/s. The

pressure and temperature of the air in the inlet pipe are constant at 750 kPa and 35C. Calculate the following:



47

a) the time rate of change of mass inside the tank.

b) the mass of air added to the tank if the pressure of the tank reaches 400 kPa and 50C.

c) time required to reach the tank pressure of 400 Kpa at 50C. System : Open Process : Unsteady flow Known : 1. Volume of the tank V1 = 4m3 2. Initial pressure P1 = 101 Kpa 3. Initial temperature T1 = 273 + 35 = 308 K 4. Supply pipe diameter D1 = 0.05 m 5. Velocity of inlet air Ci = 15 m/s. 6. Pressure at the inlet Pi = 750 Kpa 7. Temperature at the inlet Ti = 273 + 35 = 308 K Diagram :- To find : 1. Rate of increase in mass of the control volume.

2. Mass of the air in the tank when the pressure and temperature reach 400 kPa and 50C respectively.

3. Time required for the pressure of the tank to reach 400 kPa at 50C. Analysis :

1. Rate of increase in mass of the control volume = Rate of mass inflow into the control volume.

That is 1 1

dmAC

dt

Where i

i

p

RT

= 750

0.287 308

= 8.48 kg/m3

Air from the compressor

Tank of volume 4m3 Insulation



48

2

2

3 2

0.05

1.963 10

i iA D

m

Therefore 38.48 1.963 10 15cv

dm

dt

= 0.25 kg/s. 2. Mass of air added = Final mass – Initial mass = (m2 – m1)cv

Where m1 = 1 1

1

pV

RT

2 22

2

101 4

0.287 308

457Kg

p Vm

RT

Final volume V2, is the same as V1 since the vessel is rigid.

400 4

0.287 323

17.26 kg

Substituting m1, and m2 we get (m2 – m1) = 17.26 – 4.57 = 12.69 kg

3. Time required to reach 400 kPa and 50C

2 1

( )

12.69

0.25

50.76

cv

cv

m m

dm

dt

Seconds

Increase in mass

Rate of air supply



49

29. A rigid vessel of volume 250 litre containing at air 1 bar and 300 k is being filled adiabatically by connecting it to a pipe supplying air 5 bar and at unknown temperature T1. Once the vessel completely filled, it is disconnected from the main temperature and pressure of the air inside the vessel when it is full is T2 and 5 bar, respectively. Determine the temperature T2. Take T1 = T2. System : Open Process : Unsteady flow Known : 1. Volume of the vessel V1 = 250 lit = 0.25 m3 2. Initial pressure p1 = 1 bar = 100 Kpa. 3. Initial temperature T1 = 300 K 4. Final pressure p2 = 5 bar = 500 KPa 5. Pressure of the supply air p1 = 5 bar = 500 KPa 6. Final Temperature T1 = Supply air temperature. Diagram :- To find : Final temperature Analysis : Balancing mass over the filling process (m2 – m1)cv = mihi Balancing energy over the filling process

(E)cv = mihi

m2u2 – m1u1 = mihi m2CvT2 – m1CvTi = m1CpTi

m2T2 – m1T1 = miTi Since T2 – Ti and mi = m2 – m1

m2 T2 ( - 1) = m1T2 - m1T1

m2T2 ( - 1) = m1T1 2

1

T

T

Consider this equation

m2T2 = 2 2p V

Rand m1T1 = 1 1pV

R

substituting in the above equation we get

Vessel or Volume 250 litre

Supply air



50

2 2 1 1 2

1

( 1) ( 1)p V pV T

R R T

As we vessel in rigid V2 = V therefore

22 1

1

12 2 1

1

22 1 1

1

2 12 1

1

1 2

1

( 1) ( 1)

( 1) ( )

( 1)

1( 1)

1 ( 1)

Tp p

T

pp T T

T

pT T T

p

PTT T

P

T P

P

substituting the numerical values we get

2

300 5001 (1.4 1)

1.4 100

642.9

T

K

30. What are the Limitations of First Law thermodynamics?

(1) First Law of thermodynamics does not specify the direction of flow of heat and work. i.e., whether the heat flows from hot body to cold or from cold body to hot body.

(2) The heat and work are mutually convertible the work can be converted fully into heat energy but heat cannot be converted fully into mechanical work. This violates the foresaid statements. A Machine which violates the First law of thermodynamics is known as perpetual motion machine (PMM - !) of the first kind which is impossible.

PMM – 1 is a machine which delivers work continuously without any input. Thus, the machine violates first law of the thermodynamics. 31. During a flow process a 5 kW paddle wheel work is supplied while the internet energy of the system increases in one minute is 200 kJ. Find the heat transfer when there is no other form of energy transfer. Given data: Work done, W=5kW(since work is supplied to the system)

Internal energy, U=



51

2003.33 /

60kj see ]

From first law of Thermodynamics.

Q=W+U Q=5+3.33 Q=-kW Heat transfer, Q=-1.67kw (Note:-ve sign indicates that the heat is transferred from the system.) 32. A liquid of mass 18 kg is heated from 250C to 850C. How much heat transfer is required? Assume Cp for water is 4.2kJ/kgK. M=18kg T1=250C=25+273=298K T2=850C=85+273=358k Heat transfer, Q=mCp(T2-T1) Q=18x4.2x(358-298) Q=4536kJ 33. A closed system receives un input of 450kJ and increases the internal energy of the system for 325kJ. Determine the work done by the system. Given Data Heat received, Q=450kJ

U=325kJ. By fist law of thermodynamics

Q=W+U 450=W+325 W=450-325=125kJ 34. During the compression stroke of reciprocated compressor, the work done to the air in the cylinder is 95kJ/kg and 43lJ/kg of heat is rejected to the surroundings. Determine the change in internal energy. Given data: W=-95kg(-ve sign indicates that the work is done on the system) Q=-43kJ(since the heat is rejected form the system)



52

Solution: By first law of Thermodynamics

Q=W+U

-43=-95+U

U=-43-(-95)

U=52kJ/kg The internal energy increases by 52kj/kg. 35. Calculate the distance moved by a locomotive from consuming 2 tone of coal if 10% of the heat generated by the coal is converted into to coal gas then into mechanical work. The tractive effort required is 30N/tonne of dead mass of the locomotive where the dead mass of the locomotive is 2500 tones. Assume 1 kg of coal liberates 35000kJ of heat on burning. Given data: Mass of coal, Q=2 tonne =2000kg Available mechanical work=10%Q Tractive effort =30N/tonne Mass of the locomotive =25000 tonnes. For 1kg of coal, 35000kJ of heat is generated.

Calorific value of coal. CV=35000kJ/kg Heat generated, Q=mcxCV=2000x35000=70x6kJ

6

5 8

1070 10

100

70 10 70 10

W

kJ J

we know that, work done = Force x distance Where work done is nothing but available mechanical work 70x105x103=75000xx x=93333.3m =93.3km. Distance moved by a locomotive, x=93.3km 36. The following data refer to a closed system, which under goes a thermodynamics cycle consisting of four processes.

Process Heat transfer kJ/min Work transfer kJ/min

a-b b-c c-d d-a

50,000 -5,000

-16,000

- 34,200 -2,200 -3,000



53

Show that the data is consistent with the first law thermodynamics and calculate:

a) Net rate of work outputs is MW b) Efficiency of the cycle

Given data: Qa-b =50,000kJ/min Qb-c =50,000kJ/min Qc-d =-16,000kJ/min Qd-a=0 Wd-a=0 Wb-c=34,200kJ/min Wc-d=-2,200kJ/min Wd-a=3,000kJ/min To Find

1. ?W and

2. =? Solution: Cyclic heat transfer of the cycle,

Q=50000-5000-16000+0 =29000kJ/min similarly, cyclic work transfer of the cycle,

W=0+34200-2200-3000 =29000kJ/min

From first law of the thermodynamics Q=W

The gives data is consistent with the first law of thermodynamics.

29000Net work output, W= / sec

60

=483.3kW

=0.48MW

kJ

Heat supplied, Qs = 50,000kJ/min(Taking the positive heat only)

Efficiency of the cycle,=

supplied

29000 =

50,000

=58%

s

WWork done

Heat Q



54

Result:

1. Total work output of the cycle, W=0.48MW

2.Efficiency of the cycle, =58%

37. A paddle wheel fixed to the shaft of an engine revolves in a closed hollow vessel containing water. This closed vessel is connected freely on the shaft and restraint to its turning moment is proved by mass attached to its side. Find the temperature rise for the following observations. Engine rpm=650 Load applied = 60kg at a leverage of 1.2m Quantity of water =200kg. Duration of test = 20minutes Given data: Speed, N=650kg

Load, W=60kg

Leverage,1-1.2m

Mass of water, mw=200kg.

Time, t=20 minutes. Torque=WxI

=60x10x12 [1kg=10N] =720N-m

Break power or work done = 2x650x720x20 =58810614.48J

Heat gained by the water in 20min Q=mwCpw tw

58810614.48=200x4.18x103xtw [Cpw=4.18kJ/kgK=4.18x103J/kgK]

tw=70.350C. Result:

Rise in temperature, tw=70.350C



55

38. During a non flow process, the temperature of the system changes form 150C. the work done by the system and heat transfer/0C rise in temperature, at each temperature is given by

0 06 0.065 / 1.005 / .w Q

T kJ C and kJ CdT dT

Determine change in internal energy of the system during the process. Given data : T1=150C=15+273=288k T2=500C=50+273=323k

0

0

6 0.065 /

1.005 / .

wT kJ C

dT

QkJ C

dT

Solution:

.............(1.2)

1.005 ..............(1.3)

6 0.065 ...........(1.4)

Q W

U Q dT

W T dT

Substituting (1.3) & (1.4) in (1.2)

2 2

1 1

2 2

1 1

2 2

2 1 2 1 2 1.

2 2

1.005 6 0.065

0.0651.005[ ] 6

2

0.0651.005[323 288] 6 323 288 323 288

2

520.188

T T

T T

T T

T T

U Q W

dT T dT

T T T T T T

kJ

Result :

The change in internal energy, U=520.188kJ.



56

39. A 1kW string motor is applied to tank or water. The tank contains 15kg of water and the string action is applied for ½ hr. if the tank is perfectly insulated calculate the change in internal energy of water assuming that the process occurs at constant pressure and that Cv for water may be taken as 4 kJ/kg K. Also calculate rise in temperature. Given data: Power P=1kW Time, t=1/2 hr. Mass of water, mw =15kg. For perfect insulation, Q=0. Cv=4. 18kJ/kg K Solution: By First law of Thermodynamics,

Q=W+U Q=0 for perfect insulation

W=-U Work done, W=1 x ½ x3600[1 hr= 3600 sec] =1800kJ

U=1800kJ -ve sign indicates that the internal energy is decreased. For calculates, we can consider only magnitude.

U=mwCwwt

t=28.7080C Result:

1. The change in internal energy U=-1800kJ

2. The rise in temperature, t=28.7080C 40. During summer season a room measuring 10x13x6m3 is cooled electrically from initial temperature 280C to 20C. the air pressure inside the room is same as that of surroundings and is equal to 72cm of Hg. The pressure remains constant during the cooling process. The cooling capacity of furniture and wall is 35kJ/k. The specific heat of air is 1.005 kJ/kg k. calculate the amount of electric energy needed cooling the room. How much air comes out through gaps and windows during cooling period?



57

Given data: Volume of air =10x13x6m3 Temperature range =280C to 20C at 72cm of Hg Cooling capacity of a/c and wall = 35kJ/K Cpof air, Cpa = 1.005kJ/kg k. Solution: We know that, 76cm of Hg = 1.013 bar

For 72cm of Hg = 1.013

7276

=0.9596 bar Volume of air, Va = 10x13x6 =780m3

1

1

1

PVm

RT

Mass of air before cooling, 1

0.9596 100 780

0.28 301

=888.097kg

m

1

1

2

PVm

RT

Mass of air after cooling, 1

0.9596 100 780

0.28 273 2

=972.06kg.kg

m

The amount of air coming into the room during cooling. M=m2-m1 M=972.06-888.097 =83.963kg Cooling effect required from 280C to 20C Q=mCp(T1-T2)

1 2

1

m 888.097 972.06 mass, m=

2 2

=930.08kg

Q 930.08 1.005(28 2)

=24302.99kJ

mMean

Cooling effect required to cool the walls and furniture Q2=35x(28-2) =910kJ



58

The total cooling effect Q=Q1+Q2 Q=24302.99+910 =25212.99kJ

Q needed, E=

3600

25212.99 =

3600

=7kW-hr

Electricity

Result:

1. Amount of electricity needed, E=7kW-hr. 2. Amount of air comes through gaps & windows 83.963kg.

41. 25 people attended a farewell party in a small room of size 10x8m and have a 5m ceiling. Each person gives up about 350 kJ of heat per hour. Assuming that the room is completely sealed off and insulated, calculate the air temperature rise occurring within 10 minutes. Assume Cv of air 0.718kJ / kg k and R=0.287kJ/kg k and each person occupies a volume of 0.05m3. Given data: No. of persons =25 Room size = 10x8m Ceiling height =5m Heat /Person=350kJ/hr. Time, t==10min Cv0.718kJ/kg K. R=0.287kJ/kg K. Each person volume, Vp=0.05m3 Solution: Volume of the room, Vr=10 x 8 x 5 =400m3

The volume of air, Va=Vr -Vp x n =400-0.05 x 25 =398.75M3

0

1.013 100 398.75 of air,=

0.287 293

=472.29kg

[ Assuming, p=1.013 bar & T=20 ]

apVm

RT

Mass

C

By first law of thermodynamics,

Q=W+U Assume heat addition at constant volume process



59

W=0

Q=U=Heat/ person x no. of persons =350 x 25 =8750 kJ/hr

Heat loss for 10 minutes, 8750

10 1458.3360

Q kJ

Heat gained by air, Q=mCv=T

1458.33=472.29x0.718xT

T=4.220C. Result:

The rise in temperature, T=4.220C. 42. Mass of 15kg of air in a piston cylinder device is heated from 250C to 900C by passing current through a resistance heater inside the cylinder. The pressure inside the cylinder is held constant at 300 kPa during the process and heat loss of 60kJ occurs. Determine the electrical energy supplied in kW – hr and changed in internal energy. [Madras univ. Apr‟99] Given data: M=15kg T1=250C90+273=298K T2=900C=90+273-363K P1=P2=300Pa=300kN/m2 Q=-60kJ To find:

(i) Electrical energy supplied (ii) Change in internal energy

Solution: Work done, W=mR(T2-T1) =15x0.287(363-298) =279.825kJ Work done in kW-hr= Work done x 3600 =279.825x3600 =1007.37x103kW-hr Change in internal energy

U=Q-W =-60..279.825=-339.825kJ.



60

Result:

(i) Electrical energy =1007.37x103kW-hr

(ii) Change in internal energy,U=-339.825kJ 43. In a vessel 10kg of O2 is heated in a reversible non-flow constant volume process, so that the pressure of O2 is increased two times that of initial value. The initial temperature is 200C. calculate the final temperature, change in internal energy, change in enthalpy, and heat transfer. Take R=0.259kJ/kg K and Cv=0.625 kJ/kg K for oxygen. [ Madras Univ.-Oct‟98]. Given data: M=10kg Process: constant volume P2=2p1 T1=200C=20+273=293K R=Cp=Cv=0.259kJ/kgK So1 Cp=R+Cv=0.259+0.652=0.911kJ/kg K To find:

T2U,H and Q Solution:

(i) temperature at the end of the process (T2): for constant volume process

1 1

2 2

1

1 2

2

293

2

586

P T

P T

P

P T

T K

(b) Change in internal energy

U=mxCvx(T2-T1) =10x0.625x(586-293) =2669.23kJ By first law of thermodynamics

Q=W+U For constant volume process, w=0

Q=U Q=1910.36kJ



61

Result: Final temperature T2 =586K

Change in internal energy, U=1910.36kJ

Change in enthalpy, H=26669.23kJ Work done, w=0 Heat transfer, Q=1910.36kJ. 44. A certain gas of volume 0.4m3, pressure of 4.5 bar and temperature of 1300C is heated in a cylinder to a 9 bar when the volume remains constant. Calculate (i) Temperature at the end of the process, (ii) the heat transfer, (iii) change in internal energy (iv) work done by the gas, (sv) change in enthalpy and (vi) mass of the gas, assume Cp = 1.005kJ/kg K. and Cv = 0.7kJ/kg K. Given data: V1=0.4m3 P1=4.5bar=450kN/m2 T1=1300C=130+273=403K P2=9bar=900kN/m2 Cp=1.005kJ/kg K Cv=0.71 kJ/kg K and It is given that the process ion constant volume process To find:

T2,Q,U, W,H and m. Solution:

(i) Temperature at the end of the process (T2): Relation between p.v.T for constant volume process is

1 1

2 2

2

2 1

1

900403 806

450

P T

P T

PT T k

P

Heat transfer (Q): Q=mCv(T2-T1)kJ First find out the mass of the gas by using characteristic gas equation. P1V1=mRT1 Where R=Cp-Cv=1.005-0.71=0.295kJ/kg K



62

1 1

1

450 0.41.51

0.295 403

PVm kg

RT

Then, Q=1.51 x 0.71 x(806-403) =433.2kJ

(c) Change in internal Energy, (U):\

From First law, Q=W+U In this process, W=0

Therefore, Q=U

U=433.2kJ. (d) Work done by gas (w): For the constant volume process work done by the gas zero. W=0 (e) Change in Enthalpy(H)

H=mCv(T2-T1) =1.51 x 0.71 (806-403)=432.06kJ

H=432.06kJ (f) Mass of the gas (m): Already, we find that mass, m=1.51kg. Result:

(i) Temperature at the end of the process (T2)=806K (ii) The heat transfer, (Q)=433.2kJ

(iii) Change in Internal energy, (U)=433.2kJ (iv) Work done by the gas, (W)=0

(v) Change in Enthalpy, (H)=432.06kJ (vi) Mass of the gas, (m)=1.51kg.

45. 2 5kg of air at 400 C and 1 bar is heated in reversible non-flow constant pressure until the volume is doubled. Find (a) change in volume (b) work done (c) change in internal energy and (d) change in enthalpy. Given data: M=5kg T1=400C P1=1bar =100kN/m2 V2=2V1 P=constant To find:

1. V2-V1



63

2. W=?

3. U=?

4. H=?

Solution: From ideal gas equation

1 1 1

1

3

5 0.287 313

100

=4.49m

p V mRT

V

The final volume V2=2v1 =2x4.49 =8.98m3

(i) Change in volume: V2-V1=8.98-4.49 =4.49m3

(ii) Work transfer:

2

1

2 1

=100(8.98-4.49)

=449kJ

v

v

W pdv p v v

(iii) change in internal energy:

U=mCv(T2-T1) For constant Pressure process

2 2

1 1

2

2 1

1

1

1

2V =313

V

=626K

V T

V T

VT T

V

U=5x0.714(626-313) =1117.41kJ

(iv) Change in enthalpy

H=mCp(T2-T1) -5x1.005(626-313) =1572.825kj.



64

Result: 1. Change in volume, V2-V1=4.49m3 2. Work done, W=449kJ

3. Change in internal energy, U=1117.41kJ

4. Change in enthalpy, H=1572.83kJ. 46. One kg of gas is expands at constant pressure from 0.085 m3 to 0.13 m3. If the initial temperature of the gas is 2250C, find the final temperature, net heat transfer, change in internal energy and the pressure of gas. Given data: M=kg V1=0.085m3 V2=0.13m3 T1=2250C=25+273=498K Assume Cp=1.005kJ/kg k; Cv=0.71kJ/kg K. To find:

T2,Q,U&P Solution:

(i) Final temperature (T2): By using p, V&T relation

1 1

2 2

2

2 1

1

2

0.13 498

0.085

761.6

V T

V T

VT T

V

T K

(ii) Heat transfer(Q): Q=mCp(T2-T1) =1x0.7(761.6-498) Q=264.9kJ

(iii) Change in Internal energy (U):

U=mCv(T2-T1) =1x0.7(761.6-498)

U=187.16kJ.



65

(iv) Pressure(p):

p1V1=mRT1

11

1

2

1 2

1 0.295 498 =1.005-0.71

0.085 R=0.295kJ/kg K

1728.3 /

p v

p

p p

R C CmRT

V

kN m

Result: (i) Final Temperature (T2=761.6K (ii) Heat transfer (Q) =264.9kJ

(iii) Change in internal energy (U)=187.16kJ (iv) Pressure, p1=p2=1728.3kN/m2

47. 0.25kg of air at a pressure of 1 bar occupies as volume of 0.3m3. if this air expands isothermally to a volume of 0.9m3. Find (i) the initial temperature, (ii) the final temperature (iii) external work done, (iv) Heat absorbed by the air, (v) change in internal energy. Assume R=0.29kJ/kg K. Given data: M=0.25kg P1=1 bar = 100kN/m2 V1=0.3m3 V2=0.9m3 To find:

1. T1=? 2. T2=? 3. W=? 4. Q=?

5. U=? Solution: From ideal gas equation. P1V1 = mRT1

1

100 0.3T

0.25 0.287

For isothermal process T1 = T2 = 418.12K



66

Work done, W, mp1V, in 1

2

Vp1 orp1v1in

p2 V

0.9100 0.3 in

0.3

=32.96kJ Heat absorbed, Q = W = 32.96kJ

Change in internal energy, U = 0. Result:

1. Work done, W = 32.96kJ 2. Heat absorbed, Q = 32.96kj

3. Change in internal energy, U = 0. 48. A mass of 1.5kg of air is compresses in a quasi static process from 0.1MPa to 0.7 MPa for which pV = constant. The initial density of air is 1.16kg/m3. Find the work done by the piston to compress the air. Given data: M = 1.5kg Process = Quasi – static, pV = C (isothermal) P1 = 0.1MPa = 100kN/m2 P2 = 0.7MPa = 700kN/m2

Initial density (1) = 1.16kg/m3 To find: Work done, w Solution:

1

3

mass mDensity =

Volume V

mass m 1.5Volume 1.293m

Density 1 1.16

Volume (V1) = 1.293m3

21

1

VForisothermalprocess,W p1Vin

V

0.1847100 1.293 in 251.6kJ

1.293



67

(-ve sign indicates that the work is done on the system) Result: Work done, W = -251.6kJ 49. 2 kg of gas at a pressure of 1.5 bar occupies volume of 2.5m3. If this gas compress isothermally to 1/3 times the initial volume. Find (i) initial temperature (ii) final temperature (iii) work done and (iv) heat transfer. Assume R = 0.287 kJ/kg K. Given data: M = 2kg P1=1.5 bar = 150k/m2 V1=2.5m3

3

2 1

1 1V V 2.5 0.83m R 0.287kJ / kgK

3 3

To find: T1, T2, W and Q Solution: (i) Initial and final temperature (T1 & T2) P1V1 = mRT1

11

p1V 150 2.5T

mR 2 0.287

T1=653.3K

This is a isothermal T1 = T2 = 653.3K (ii) Work done (W):

21

1

VW mRT in

V

0.832 0.287 653.3 in

2.5

W = -413.5kJ Note: Here, negative sign indicates that the work is done on the system. (iii) Heat transfer (Q):

From First law, Q = W + U

For isothermal process, U = 0



68

Q = W = -413.5kJ Note: Here negative sign indicates that the heat is rejected by the system. Result:

i. Initial and Final temperature, T1 = T2 = 653.3K ii. Work done (W) = -413.5kJ iii. Heat transfer (Q) = -413.5kJ

50. 10kg of gas at 10 bar and 4000C expands reversibly and adiabatically to 1 bar. Find the work done and change in internal energy. (Manonmaniam sundaranar univ. – Apr‟96). Given data: M = 10kg P1 = 10bar = 1000kN/m2 T1 = 4000C = 400 + 273 = 673K P2 = 1 bar = 100 kN/m2 To find

W and U Solution By adiabatic relation,

1

2 2

1 1

14 1

14

2

T p

T p

100T 673 348.58K

1000

Work done,

1 2mR T TW

1

10 0.287 673 348.582327.74kJ

1.4 1

Change in internal energy

U = m x Cv x (T2 – T1) = 10 x 0.718 x (348.58 – 673) = -2327.74kJ



69

Alternately, from first law, Q = W + U In adiabatic process, Q = 0

W = -U = -2327.74kJ Result:

1. Work done, W = 2327.74 kJ

2. Change in internal energy, U = -23 27.74kJ 51. 1.5kg of certain gas at a pressure of 8 bar and 200C occupies the volume of 0.5 m3. It expands adiabatically to a pressure of 0.9 bar and volume 0.73 m3. Determine the work done during the process, gas constant, ratio of the specific heats, values of two specific heats, change in internal energy and change in enthalpy. Given data: M = 1.5kg P1 = 8 bar = 800kN/m3 V=1 = 0.15m3 T1=200C = 20+273 = 293K P2=0.9bar = 90kN/m2 V2 = 0.73m3

To find: W,R, Cp / Cv, Cp, Cv, U and H Solution: From characteristics gas equation P1V1 = mRT1

1

1

p1VR

mT

800 0.15R

1.5 293

R = 0.273kJ / kgK Ratio or two specific heat.

p

v

10

110

2

C

C

p2log

p1

Vlog

V



70

10

10

90log

800

0.15log

0.73

p

v

p v

p v

v v

v

v

p

1.38.

C1.38

C

C 1.38C

R C C

1.38C C

0.273 C 1.38 1

0.273C 0.718kJkgK

0.38

C 1.38 0.78 0.99kJ / kgK

By using Relation between T & V for isentropic process

1 1

2 1 12 1

1 2 2

1.38 1

2

T V VT T

T V V

0.15T 293 160.59K

0.73

Change in internal energy (U)

U=mCv(T2 – T1) =1.5 x 0.718 x (160.59 – 293)

H = -196.6kJ Work done (W):

1 2p1V p2V 800 0.15 90 0.73W

1 1.38 1

W .89kJ

Alternately, from first law of thermodynamics, Q = W + U For isentropic process, Q = 0

W = -U = 142.5kJ Results:

i. Work done (W) = 142.89kJ

ii. Gas constant = 0.273kJ/kg K.

iii. Ratio of two specific heat p

v

C1.38

C



71

iv. Values of two specific heat, Cp=0.99kJ/kg Cv = 0.718kJ/kg K

v. Change in internal energy (U) = -142.5kJ

vi. Change in enthalpy (H) = -196.6kJ 52. A gas of mass 0.35g pressure 1535 kN/m3 as temperature of 3350C is expanded adiabatically to pressure of 126 kN/m2. The gas is than heated constant volume until it reaches 3350C, when its pressure is found to be 275kN/m2. Finally the gas is compressed isothermally until the original pressure 1535 kN/m2 obtained. Draw the p-V diagram and find out the following (i) the value of adabatic index (ii) change in internal energy during adabatic process and (iii) heat transfer during constant volume process Take Cp = 1.005kJ/kg k. Given data: M = 0.35kg P1 = 1535 kN/m2 Process 1-2: Adiabatic T1 = 3350C = 608K P2 = 126kN/m2 Process 2 – 3: Constant volume T2 = T1 = 608K P3=275kN/m2 Process 3-1: Isothermal To find:

0U and Q Solution: Consider process 2-3: Constant volume process

pV=C

pV=C

p1

2

V1 V2

Volume (V)

Pre

ss

ure

(p

)



72

2 3

2 3

2

2 33

p p

T T

p 126T T 608

p 275

T2 = 278.57K Consider process 1-2: Adiabatic process

13

1

2

2

T p

T p

Taking log on both sides

110 10

2

10 10

T 1 p1log log

T p2

608 1 1535log log

278.57 126

10.339 1.0857

0.339 10.3122

1.0857

0.3122 1

0.6877 1

11.454

0.6877

1.454

Change in internal energy (U) during adiabatic process

U=mCv(T2 – T1)

p

v

v

C

C

C kJ/ kgK

U = 0.35 x 0.691 x (278.57 – 608)

U = -79.69kJ Heat transfer during Constant volume process

U=mCv(T3 – T2) = 0.35 x 0.691 x (608 – 278.57)

U=79.69KJ



73

From first law of thermodynamics, Q = W + U

Q = U [W=0] =79.69kJ Result:

i. Value of adiabatic index, = 1.454

ii. Change in internal energy during adiabatic process (U) =-79.69kJ iii. Heat transfer during constant volume process (Q) = 79.69kJ

53. A cylinder contains 1 m3 of gas at 100 kPa and 1000C, the gas is polytropically compressed to a volume of 0.25 m3, the final pressure is 600kPa. Determine (a) mass of the gas (b) the value of index „n‟ for compression (c) change in internal energy of the gas (d)

heat transferred by the gas during compression. Assume R=0.287 kJ/kg and 1.4. [Bharathiyar univ, - Nov‟95] Given data: V1 = 1m3 P1 = 100kPa = 100KN/m2 T1 = 1000C = 100+273=373K V2 = 0.25m3 P2 = 600kPa = 600kN/m2 R = 0.287kJ/kgK

= 1.4 To find:

M,n U and Q Solution: By general gas equation

P1V1 = mRT1 ====>

1

1

p1Vm

RT

100 1m 0.93kg

0.287 373

The value of index „n‟ is given by

10 10

11010

2

p1 600log log

p2 1001.29

1Vloglog

0.25V

From polytrophic equation,



74

n 1

n2

1

1

1.29

2

T p2

T p2

600T 373 558K

100

Change in internal energy.

U = m x Cv x (T2 – T1) = 0.93 x 0.718 x (558 – 373) =123.53kJ

1 2mR T TW

n 1

0.93 0.287 373-558Work done, =

1.29 1

170.19 1

nQ W

1

1.4 - 1.9Heat transfer, = -170.1 46.802kJ

1.4 - 1

Result: Mass of gas, m = 0.93kg Value of index, n = 1.29

Change in internal energy, U = 123.53kJ Heat transfer, Q = -46.802kJ 54. An ideal gas of molecular weight 30 and specific heat ratio 1.4 is compressed according to the law pV1.25 = C from 1 bar absolute and 270C to a pressure of 16 bar (abs). Calculate the temperature at the end of compression, the heat received or rejected, work done on the gas during the process and change in enthalpy. Assume mass of the gas as 1 kg. Given data: Molecular weight, M = 30

p

v

C1.4

C

m 1kg

p1 1bar 100kN/ m2

p2 16bar 1600kN/ m2



75

0

1

1.25

T 27 C 27 273 300K

PV C

To find: T2, Q, and W Solution: For polytropic process, the p,V and T relation

n 1 n 1

n n2

2 1

1

1.25 1

1.25

2

T p2 p2T T

T p1 p1

600T 300 522.33K

100

Work done:

1 2

u

mR T TW

n 1

R Universal as constantGas constant, R=

M molecular weight

0.277kJ / kgK

1 0.277 300 522.33W

1.25 1

246.34kJ

(-ve sign indicates that the work is done on the system) Heat transfer,

nQ W

1

1.4 1.25246.34 92.3782kJ

1.4 1

(-ve sign indicates that the heat is rejected from the system) Change in enthalpy:

p 2 1H m C T T

1 1.005 522.38 300

233.49kJ

Result: Temperature at the end of the compression, T2=522.38K Work done on the system, W = -246.34kJ Heat transferred from the system, Q = -92.378kJ

Change in enthalpy, H = 223.49kJ



76

55. A perfect gas for which ratio of specific heats is 1.4 occupies a volume of 0,3 m3 at 100 kPa and 270C. The gas undergoes compression of 0.06m3. Find the heat transfer during the compression for the following methods (a) pV = constant (b) Isentropic and (c) pV1.1 = C. Given data: P1=1000kPa=100kN/m2 V1=0.3m3 T1=270C+273=300K

= 1.4 V2=0.06m3 Process (i) Constant pressure

(i) pV = C (ii) pV1=C and (iii) pV1.1 = C

To find: Heat transfer (Q) Solution:

a. pV = C Heat transfer,

1 2pIV p2V nQ

n 1 1

100 0.3 587.3 0.06 1.4 1.1

1.1 1 1.4 1

39.285kJ

Result: Heat transfer for a. pV = C process, Q=-48.28kJ b. pV1 = C process, Q=0 c. pV1.1 = C process, Q=-39.285kJ

56. In steady flow process, 125 kJ of work is done by each kg of working fluid. The specific volume, velocity and pressure of the working fluid at inlet are 1.41 m3/kg,15.5 m/s and 6 bar respectively. The inlet is 31m above the ground, and the exhaust pipe is at ground level. The discharge conditions of the working fluid are 0.64 m3/kg, 1 bar and 263 m/s. The total heat loss between inlet and discharge is 8.7 kJ/kg of fluid. In flowing through this apparatus, does the specific internal energy increases or decreases and by how much? Given data:



77

W = 125kJ/kg V1=0.41m3/kg C1=15.5m/s P1=6bar=600kN/m2 Z1=31m Z2 = 0 V2 = 0.64m3/kg C2 = 263 m/s P2 = 1 bar = 1000kN/m2 Q = 8.7 kJ/kg To find: Whether the internal energy increases or decrease and how much? Solution: The steady flow energy equation is

2 2

1 2

1 1 1 1 2 2 2 2

2 2

1 2

1 2 2 1 2 2 1 1

2 2

3 3

1 2

1 2

1 2

2 1

2 2

2

263 15.59.81 0 31 10 10 100 0.64 600 0.4

2

125 8.7

0.304 34.46 182 125 8.7

14.14 /

14.14 /

C Cgz u Pv Q gz u Pv W

C Cu u g z z Pv Pv W Q

u u x

U U

U U kJ kg

U U kJ kg

Result: Internal energy increases by 14.14kJ/kg

57. 50 kg/min of air enters the control volume in a steady flow statement at 2 bar and 100 and at an elevation of 100m above the datum. The same mass leaves the control volume at

150m Elevation with a pressure of 10 bar temperature of 300C. the entrance velocity as 2400 m/min. during the process 50,000 kJ/hr. calculate the power developed. (Madras University Apr 96). Given data:

m = 50 kg/ min = 50

60=0.83kg/sec

p1 = 2 bar = 200kN/m2



78

T1=100C = 373K Z1= 100m Z2 = 150m P2 = 10 bar = 1000 kN/m2

r2 = 300C = 573k

C1m = 2400m / min = 2400

60= 40m/sec

C2m = 1200m / min = 1200

60s=20m/sec

Q = 50,000kJ / hr = 50,000

3600= 13.89kJ / sec

To Find: Power developed, P = ? Solution: SFEE is

2 2

1 21 1 1 1 2 2 2 2

2 2

1 22 1 1 2

2 23 3

2 2

2

40 209.81 100 150 8 10 13.89 10

2

C Cgz u p v Q gz u p v W

C CW g z z h h Q

W

W=5999.5J/kg = 6kJ/kg K Power developed, P = W x mass = 6.0 x 0.83 P = 4.98kJ / sc-4.98kW Result: Power developed P = 4.98kW

58. A boiler produces steam from water at 35C. enthalpy of steam is 2675kJ/kg. calculate the head transferred per kg. neglect the potential and kinetic energies. Give data:

Tw=35C = 35+273 = 308k H2 = 2675 kJ/kg Cpw = 4.19kJ/kg K. To find:



79

Heat transferred, Q Solution: Enthalpy of feed water

1

1

4.19 308

1290.52 /

pu wh C T

h kJ kg

SFEE for boiler heat transferred (Q) Q = h2-h1=2675 – 1290.52 Q = 1384.48 kJ/kg Result: Heat transferred per kg, Q = 1384.48 kJ/kg 59. In a steady flow of air through a nozzle, the enthalpy decreases by 50 kJ between two sections. Assuming that there are no other energy changes than the kinetic energy determine the increases in velocity at sectiuon2, if the initial velocity is 90m/s. Given data: Enthalpy decrease, (h2-h2) = 50kJ = 50x103J Velocity at section (1), C1 = 90m/s To find: Increase in velocity, (C2 – C1) = ? Solution: Steady flow energy equation for a nozzle is

2 2

2 11 2

2

C Ch h

2

2 1 2 1

3 2

2

at exit, C 2

= 2 50 10 90

C =328.78m/s

velocity h h C

Increase in velocity C2 – C1 = 238.78m/s Result: Heat transferred per kg. Q=1384.48 kJ/kg. 60. At the inlet of the nozzle, the enthalpy and velocity of the fluid are 3000 kJ/kg and 50m/s respectively. There is negligible heat loss from the nozzle. At the outlet of the nozzle enthalpy is 2450 kJ/kg. if the nozzle is horizontal, find the velocity of the fluid at exit. (Bharathidasan University – Nov 94).



80

Given data: h1=3000 kJ/kg C1= 50m/s Q = 0 z1= z2 h2 = 2450kJ/kg C2-C1 = 328.78-90 = 238.78m/s To find Velocity of the fluid at exit C2 = ? Solution: The SFEE for nozzle

2 2

1 21 1

2

2 1 2 1

2 2

2

2 2

C 2

= 2 3000 2450 10 50

C 1050 /

C Ch h

h h C

m s

Result: Velocity of fluid at exit, C2=1050 m/s. 61. In a thermal power station, the steam flows, steadily through a 0.3 m diameter pipeline from boiler to turbine. A boiler exhausts the steam at a pressure of 4.2 bar the temperature

of 420C and the enthalpy of 3216.3 kJ/kg. The specific volume of the steam boiler outlet is 0.076m3 /kg. After flowing through the turbine, the steam conditions are measured as the

pressure of 3.1 bar, temperature of 379C, the enthalpy of 3201.7 kJ/kg and the specific volume of 0.086 m3/kg. There is a heat loss of 8.3 kJ/kg from the pipeline between boiler and turbine. Calculate the rate of flow of steam. Give data: d1= 0.3m p1= 4.2 bar = 420kN/m2

T1=420C h1= 3216.3kJ/kg v1 = 0.076m3/kg p2 = 3.1

T2 = 379C h2 = 3201



81

v2 = 0.086m3/kg Q =-8.3kJ/kg (heat loss) To find: Rate of flow of steam, m=? Solution: By mass flow equations

1 1 2 2

1 2

AC A C

v v

Since, pipe diameter 0.3m is constant (A1=A2)

1 2

1 2

1 12 2

1

2 1

0.0860.076

C 1.13

C C

v v

C CC v

v

C

The steady flow energy equation for the above system is

2 2

1 22 2 1 2

2 2

2 11 2

2 2

2 1

2 2 3

1 1

2 3

1

1

: 02 2

2

= 3216.3-3201.7 8.3

=6.3kJ/kg

C 2 6.3 12.6 /

1.13 12.6 10 /

45.5 10

213 /

C Ch Q h z z w

C Ch h Q

C kJ kg

C C j kg

C

C m s

Mass rate of flow statement

2

1 1

1

0.3 2134 198

0.076

ACm kg

v



82

Result: Mass rate of flow of steam, m=198kg/sec 62. A turbine operates under steady flow conditions receiving steam at the following state:

pressure = 1.2 mpa, temperature = 188C. The steam leaves the turbine at the following state. Pressure = 20 kpa, Enthalpy = 2512 kJ/s. if the rate of steam flow through the turbine is 0.42 kg/s, what is the power out put of the turbine in kW? Given data: p1=1.2MPa

T1= 188C h1= 2785kJ/kg. C1= 33.3m/s z1=3 p2= 20 kpa h2 = 2512kJ/kg C2=100m/s z2 = 0m Q=-0.29kJ/s. m=0.42kg/s. To find: Power output W=? Solution: SFEE

2 2

1 21 1 2 2

2 2

2 2

33.3 9.81 3 1000.42 2785 0.29 0.42 2512 0

2000 1000 2000

C Cm h z g Q m h z g W

W

Result: The power out put of the turbine, W = 112.52W.

63. A steam turbine operates under steady flow conditions. It receives steam 7200 kg/hr from the boiler. The steam enters the turbine at enthalpy of 2800 kJ/kg, a velocity of 400 m/min and an elevation of 4m. The steam leaves the turbine at enthalpy of 2000 kJ, a velocity of 8000 m/min and an elevation of 1m. Due to radiation the amount of heat losses from the power output of the turbine. (madras university Apr 95)

Given data:



83

2

1

1 2

2

2

7200/ / sec

3600

2800 /

400400 / min 6.67 /

60

4 & 1

2000 /

8000 / min 133.3 /

1580 / 0.438 / sec

m kg hr kg

h kJ kg

C m m s

z m z m

h kJ kg

C m m s

Q kJ hr kJ

To find: Power output of the turbine, p=? Solution: The SFEE for the above system is

2 2

1 21 1 1 1 2 2 2 2

2 2

1 21 2 1 2

2 2

3 3

2 2

2

6.67 133.3# 9.81 4 1 2800 2000 10 0.438 10

2

C Cg u p v Q gz u p v W

C CW g z z h h Q

W = 791605.23 J/kg = 791.6kJ/kg Power output from the turbine, P=W x mass = 791.6 x 2 P=1583.2 kW Result: Power output from the turbine, P=1583.2 kW 64. Air flows steadily at the rate of 0.5 kg/s through an air compressor. Entering at 7 m/s velocity, 100 kPa pressure and 0.95 m3/kg, volume and leaving at 5m/s, 700kPa and 0.19 m3/kg. The internal energy of the air leaving is 90kJ/kg greater than that of the air entering. Cooling water in the compressor jackets absorbs heat from the air at the rate of 58kW. (a) Compute the rate of shaft work input is the air in kW. (b) Find the ration of the inlet pipe diameter to the outlet pipe diameter.



84

Given data:

1

1

3

1

2

2

3

2

2 1

m 0.5kg/ s.

C 7m/ s.

P = 100kpa

v = 0.95 m /kg

C = 5m /s

P = 700kPa

v 0.19m /kg

U U 90kJ/kg

Q 58kN.

To find:

1. Work input, W =?

2. 1

1

D?

D

Solution: SFEE,

2 2

1 21 1 1 1 2 2 2 2

2 2

1 21 1 1 1 2 2 2 2

1 2

2 2

C Cu p v z g Q m u p v z g w

2 2

C Cp v z g Q m (u u )p v z g w

2 2

Assumez z

7 50.5 (100 0.95) 58 0.5 90+(700 0.19)+ W

2000 2000

work input, W=-5.995K W

( -ve sign indicates that the work is done on the system) From continuity equation,



85

1 1 2 2

1 2

1 2 1

2 1 2

2

4 1

2

4 2

2

1

2

1

2

A C A C

v v

A C v 53.57

A C v 7 0.19

D3.57

D

D3.57

D

D1.89

D

Result: 1. Work input, W = -5.994k W

2. The ratio of the inlet to outlet pipe diameter, 1

2

D1.89

D

65. Air is compressed from 100 kpa and 220C to a pressure of 1 Mpa while being cooled at the rate of 16 KJ/kg by circulating water through the compressor casing. The volume flow rate of air inlet condition is 150m3/min and power input to compressor is 500 kW Determine (a) mass flow rate (b) temperature of air exit. Neglect datum head . [Madras univ. Apr‟97] Given data:

1

o

1

2

3

1

P 100kPa

T 22 C 22 273 295K

p IMPa

Q 16kJ/kg.

C 5m /min

W =500kw.

To find:

1. Mass flow rate, m=? 2. Temperature of air exit, T2=?

Solution:

2 2

1 21 1 2 2

C Cm h z g Q m h z g W.............(1)

2 2

From ideal gas equation, P1V1= mRT1



86

1 2

2 2

1 2

1 2

100 150m

0.287 295

177.17kg/min

2.953kg/ sec

Neglect datum head, Z Z 0

C CAssume change in velocity head, 0

2

The equation (1) reduces to

m(h ) Q m(h ) W

1 2

2 2

1 2

1 2

p

p 1 p 2

100 150

0.287 295

177.17kg/min

2.953kg/ sec

Neglect datum head, z z 0

C CAssume change in velocity head, 0

2

Theequation (1) reduces to

m(h ) Q m(h ) W

AssumeC 1.005kJ/kg K for air.

M(c T Q) mC T W

2

2

2

.953 (1.005 295-16)=(2.953 1.005T ) 500

T 110.6K

Result:

1. Mass flow rate, m=2.953kg/sec. 2. Temperature of air exit, T2=110.6K.

66. During the working stroke of an engine the heat transferred out of the system was 150kj/kg of the working substance, determine the work done, when the internal energy is decreased by 400 kj/kg. Also state whether the work done on or by the engine. [Nov-01] Given data: Q=-150 kJ/kg (Heat is transferred out0

U=-400kj/kg. To find: Work done, W=?



87

Solution: By first law of thermodynamics

Q=W+U

W=Q-U =-150-(-400) =250kJ/kg Result: Work output, W=250kJ/kg (Positive work indicates that the work done is by the engine) 67. A fluid is confined in a cylinder by a spring-loaded frictionless piston so that the pressure in the fluid is a linear function of the volume (p=a+bv). The internal energy of the fluid is given by the following equation u = 34+3. 15pV, where us is kJ, p is in kPa and V in m3

. If the fluid change from an initial state of 170kPa. 0.03m3 to a final state of 400 kPa, 0.06m3, with no other work than that done on the piston, find the direction and magnitude of the work and heat transfer. Given data: P. a+bV U = 34+3.15pV

P1=170kPa V1=0.03m3 P2=400kPa V2=0.06m3

To find: Work transfer, W =? Heat transfer =? Solution: Change in internal energy,

2 1 2 2 1 1

2 2 1 1

U U (34 3.15p V ) (34 3.15p V )

= 3.15(p V p V )

= 3.15(400 0.06-170 0.03)

= 59.535kJ

From pressure equation. P=a+bV P1= a+b V1.........(1.24) P2=a+bV2……….(1.25)



88

At initial state, 170=a+b(0.03) =a+0.03b………….(1.26) At final state, 400 =a+0.06b………….(1.27) Equation (1.27)-(1.26) results, s30=0.03b B=7666.67 Substituting the value of b in (1.26) 170=a+0.03(7666.67) a=60.0 The pressure equation, p=-60+7666.67V

2 2

1 1

v v

v V

0.062

0.03

2 2

W p.dV ( 60 7666.67V)dV

V = -60+7666.67

2

7666.67W 60(0.06 0.03) (0.06 0.03 )

2

= 8.55kJ

By first law of Thermodynamics.

Q = W +U Q=8.55+59.535 =68.085kJ Result: 1. Work transfer, W=8.55kJ (+ve sign indicates that the work is done by the system.) 2. Heat transfer, Q = 68.085 J (+ve heat transfer indicates that the heat is transferred into the system.) 68. A Piston and cylinder machine contains a fluid system which passes through a complete cycle of four processes. During the cycle, the sum of all heat transfers is – 170kJ. The system completes 100cycles per minute. Complete the following table showing the method for each item, and compute the net rare of work out put in kW.

Process Q(kj/min) W(kj/min) E(kJ/min) a-b 0 2,170 - b-c 21,000 0 - c-d -2,100 - - d-a - - -



89

Solution : Process a-b: By first law of thermodynamics

Q=W+UE

0=2170+E

E=-2170kJ/min Process b-c

Q=W+E

21,000=0+E

E=21,000kJ/min Process C-d -2,100=W-36,600 W=34, 500kJ/min Process d-a:

Q = -170k(given) No. of cycles/min = 100 =-17,000kJ/min

cycle Q=Qa-b+Qb-c+Qc-d+Qd-a

-17,000=0+21000-2100+Qd-a

Qd-a=-35900kJ/min. We know that, in a cyclic process

Q=W -17,000=Wa-b+Wb-c+Wc-d+Wd-a

=2170+0+34500+Wd-a

Wd-a = -53670kJ/min. Change in internal energy,

E=Q-W =-35900 –(-53670) =17770kJ/min.

Rate of work output,

W=-1700kJ/min 17000

60

=-283.3kW.



90

Result:

Process Q(kJ/min) W(kJ/min) E(kJ/min)

a-b 0 2,170 -2170

b-c 21,000 0 21,000

c-d -2,100 34,500 -36,600

d-a -35,900 -53670 17770

69. Five kg of air is compressed poly tropically (n-1.3) from 1 bar and 270C to 3 bar. Find (i) work transfer (ii) heat transfer (iii) change in internal energy. Given data: P1=1bar=1100kN/m2 T1=270C=27+273=300K P2 = 3 bar = 300kN/m2 N=1.3 To find:

1. Work transfer, W=? 2. Heat transfer, Q=? 3. Change in internal energy, U=?

Solution: During poly tropic process,

n 1

n2

2 1

1

1.3 1

1.3

PT T

P

300 =300

100

=386.57K

1 2mR T TWork transfer, W=

n 1

1 .287 300-386.57 =

1.3 1

=-82.82kJ/kg

r-nheat transfer, Q= W

r-1

1.4-1.3 = 82.82

1.4-1

Q=20.705kJ/kg.



91

By first law of thermo dynamics, U=Q-W =-20.705-(82.82) =62.1155KJ/kg. Result:

1. Work transfer, W=-82.82KJ/Kg 2. Heat transfer, Q=20.705kJ/Kg 3. Change internal energy, U=62.115kJ/kg

70. 5kg of air at 40C and 1 bar is heated in a reversible non-flow constant pressure until the volume is doubled. Find (a) change in volume (b) work done (c) change in internal energy and (d) change in enthalpy. Given data: M=5kg

T1=40C P1=1bar=100kN/m2 V2=2V1 P=Constant To find:

1. V2-V1=? 2. W=?

3. U=?

4. H=? Solution: From ideal gas equation. P1V1=mRT1

1

5 0.287 313

100V

=4.49m

The final volume V2=2V1 =2 x4.49 =8.98m3 Change in volume V2-V1=8.98-4.49 =4.49m3



92

2

1

r

2 1

r

transfer, W= ( )

100(8.98 4.49)

449 .

Work pdV p V V

kJ

Change in internal energy

U=mCv(T2-T1) For constant Pressure process

2 2

1 1

22 1

1

1

1

2313

626

5 0.714(626 313)

1117.41

V T

V T

VT T

V

V

V

K

U

kJ

Change in enthalpy

H=mCp (T2-T1) =5 x 1.005(626-313) =1572.825kJ. Result:

1. Change in Volume, V2-V1=4.49m3 2. Work done, W=449kJ

3. Change in internal energy,U=1117.41kJ

4. Change in enthalpy,H=1572.83KJ

71. A gas whose original pressure volume the temperature were 140kNm2,0.13 and 25 C respective. It is compressed such that new pressure is 700kN/m2 and its new temperature is

60C. Determine the new volume of the gas.[Nov-01] Given data: P1=140kN/m2 V1=0.1m3

T1=25C=25+273=298K P2=700kN/m2

T2=60C =60+273=333K To find: V2=?



93

Solution: From ideal gas equation,

1 1 2 2

1 2

2

3

2

700140 0.1

298 333

0.0223

pV p V

RT RT

V

V m

Result: New volume, 3

2 0.0223V m

72. 0.25kg of air at a pressure of 1 bar occupies a volume o 0.3m3. If this air expands isothermally to a volume o 0.9m3. Find (i) the initial temperature , (ii) The final temperature (iii) External work done, (iv) Heat absorbed by the air, (v) change in internal energy Assume R=0.29kJ/K. Given data: M=0.25kg P1=1 bar =100kN/m2 V1=0.3m3 V2=0.9m3 To find:

1. T1=? 2. T2=? 3. W=? 4. Q=?

5. U=? Solution: From ideal gas equation, P1V1=mRT1

1

100 0.3

0.25 0.287

=418.12K.

T

For isothermal process T1=T2=418.12.12K



94

11 1

2

V1, 1 1 In

2 V

0.9100 0.3

0.3

32.96 .

pWorkdon W mp V In orp v

p

In

k J

Heat absorbed, Q=W =32.96kJ

change in internal energy ,U=0. Result:

1. Work done, w=32.96kJ 2. Heat absorbed, Q=32.96kJ

3. Change in internal energy ,U=0. 73. In a steady flow system a working substance at a rate of 4kg/s enter a pressure of 620 kN/m2 at a velocity of 300m/s. The internal energy is 2100 kJ/kg and specific volume 0.37m3/kg. It leaves the system at pressure of 130kN/m2, a velocity of 150m/s, Inter energy of 1500 kJ/kg and specific volume of 1.2m3/kg. During its passage in the system, the substance has a heat transfer of loss of 30kJ/kg to its surroundings. Determine the power of the system. State that it is from (or) to the system. Given data: M=4kg/s P1=620kN/m2 C1=300m/s U1=2100kJ/kg V1=0.37m3/kg P2=130kn/m2 C2=150m/s U2=1500kJ/Kg V2=1.2m3/kg Q=-30kJ/kg To find: Power of the system, W=? Solution; SFEE



95

2 2

1 21 1 1 1 2 2 2 2

1 2

2 2

1 21 1 1 1 1 2 2 2

2 2

2 2

2 2

300 1504 2100 (620 0.37) 30 4 1500 (130 1.2)

2000 2000

2708.6

C Cm u p v Z g Q m u p v Z g W

AssumeZ Z

C Cm u p v Z g Q m u p v Z g W

W

W KW

(+Ve sign indicates that the work is one by the system.) Result: Work output from the system, W=2708.6kW

74. A mass of air is initially at 260C and 700kPa and occupies 0.028m3. the air is expanded at constant pressure to 0.084m3. A polytrophic process with n=1 is then carried out, followed by a constant temperature process. All the process are reversible.

1. Sketch the cycle in the p-V and T-s planes 2. Find the heat received and heat rejected in the cycle. 3. Find the efficiency of the cycle[Apr-03]

Given data;

T1=260C=273+260=533K P1=700kPa=p2 V1=0.028m2 V1=0.084m2 Process 2-3 is polytrophic Process 3-4 is constant temperature. To find:

1. Sketch p-V and t-s diagram 2. Heat received and heat rejected, Q=?

3. Efficiency of the cycle, =? Solution; Process 1-2: Constant pressure process.



96

1 1

1 2

22 1

1

0.084533

0.028

1599

v T

T T

VT T

V

Mass of air,

1-2 2 1

1-2 2 1

700 0.028 =

0.287 533

=0.128kg

Work done, W ( )

=700(0.084-0.028)

=39.2kJ

Heat transfer, Q ( )

p

pVm

RT

p V V

mC T T

p

=0.128 1.005 (1599-533)

C air =1.005kJ/kgK

=137.13kJ

of

Process 2-3 Polytropic process



97

1

3 3

2 2

1.5 1

1.53

3

33

3

3

1 1 2 22 3

533

1599 700

25.93

0.128 0.287 533

25.93

0.755

,1

700 0.084-25.93 0.755 =

1.5 1

n

nT p

T p

P

P kPa

FrompV mRT

mRTV

P

m

pV p VPolytropicwork W

n

23 2 3

=78.446kJ

Heat transfer. Q1

=19.612kJ

nW

Process 3-1

Constant temperature process, T1=T3=260C=533K

Work transfer, W3-1=P3V3 In 1

3 1

P

P

25.9325.93 0.755

700

64.52 ( )

64.52 [ U=0]

In

kJ Workinput

Q kJ

Heat received in this cycle, Qs=137.13kJ (consider only +ve heat) Heat reject in this cycle QR=19.612+64.52(consider only –ve heat)=84.132kJ



98

1-2 2 3 3 1

work done of the cycle, =

sup

W =

39.2+78.446-64.52 =

137.13

s

EfficiencyHeat plied

W W

Q

R

s

=38.74%

Or

QEfficiency of the cycle, =

Q

84.132 =1.

137.13

=38.74%

Result: 1.p-V and T-s planes and drawn 2. Heat received by the cycle, Qs=137.13kJ 3.Heat rejected by the cycle, QR=84.132kJ

4. Efficiency of the cycle,=38.74%

75. Air at a temperature of 15C passes through a heat exchanger at a velocity of 30m/s and

expands until the temperature falls to 650C. On leaving the turbine, the air is taken at a

velocity of 60m/s to a nozzle where it expands until the temperature has fallen to 500C. If the air flow rate is 2kg/s, calculate: (i) The rate of heat transfer to the air in the heat exchanger, (ii) The power output from the turbine assuming no heat loss, and (iii) The velocity at exit from the nozzle, assuming no heat loss. Take the enthalpy of air as h=Cpt, where Cp is the specific heat equals to 1.005kJ/kg K and „t‟ the temperature. Given data:

T1=15C=273+15=288K C1=30m/s

T2=800C=273+800=1073K C2=30m/s

T3=650C=273+650=923K C1=30m/s

T4=500C=273+500=773K m=2kg/s



99

To find:

(i) The rate of heat transfer in the heat exchanger, Q1-2=? (ii) The velocity at exit from the nozzle C4=? (iii) Power out put from the turbine, W2-3=?

Solution: SFEE between 1-2,

2 2

1 21 1 2 2 1 2.............(1 28)

2 2


Assuming that, Work transfer ,W1-2=0 and Z1=Z2 So, the above equation reduces to

2 2

1 21 1 2 2

1 2

1 2 2 1

2 1

22

322 2 2 3 2 3 2 3

2 2

,

( )

( )

2 1.005 (1073 288)

1577.85

.....(1.29)2 2

p

C Cm h Q m h

Here C C

Q m h h

mC T T

W

CCm h z g Q m h z g W



100

Assuming that Q-3=0 and Z2=z3 Equation (1.29) reduces to

22

322 3 2 3

2 2

2 3

2-3

2 2

30 602 1.005 1073 2 1.005 923

2 1000 2 1000

W 298.8

CCm h m h W

x W

kW

Similarly, applying SFEE between 3-4

2 2

3 43 3 3 4 4 4 3 4.....(1.30)

2 2


Assuming that Z3=Z4, Q3-4=0 (adiabatic nozzle) and W3-4=0

The Equation (1.30) reduces to

2 2

3 42 3

22

4

4

2 2

602 1.005 923 2 1.005 773

2 1000 2 1000

C 552.36 /

C Cm h m h

C

m s

Result:

1. The rate o heat transfer in the heat exchanger, Q1-2=1577.85kW 2. The power output from the turbine , W2-3=298.8kW 3. The velocity of nozzle exit, C4=552.36 m/s

76. A gas of mass 1.5 kg undergoes a quasi-static expansion which follows a relationship p=a+bv, when a and b are constants. The initial and final pressure are 1000kPa and 200kPa respectively and the corresponding volumes are 0.2m3 and 1.2m3. the specific internal energy of gas is given by the relation u=1.5pV-85kJ/kg. where p is in kPa and V is in m. Calculate the net heat transfer and the maximum internal energy of the gas attained during expansion [Nov-03].

Given data: M=1.5 kg P=+bV P1=10000kPa P2=200kPa V1=0.2m3 V2=1.2m3



101

U=1.5 pV-85 To find:

1. Heat transfer, Q=?

2. Maximum internal energy,U=? Solution: P1=a+bV1 and p2=a+bV2 1000=a+0.2b………(1.31) 200=a+1.2b………(1.32) (1.32)-(1.31),b=800 From (1.31),a-1160

the relationship becomes P=1160-800V

2 1.2

1 0.2

1.22

0.2

2 2

transfer, W= (1160 800 )

800V = 1160V

2

=1160(1-2-0.2)-(1.2 0.2 )

=600kJ

Work pdV V dv

change in internal energy, U=U2-U1

=(1.5p2V2-85)-(1.5p1V1=85) =(1.5p2V2-1.5p1V1) =1.5(200 x 1.2-1000 x 0.2) =60KJ According to the first law of thermodynamics

Q=W+U =600+60 =660kJ Maximum internal energy, U2=mU2

1.5 200 1.21.5 85

1.5

232.5KJ



102

Result: Heat transfer, Q=660KJ

Maximum internal energy, U=232.5kJ 77. A room for four person has two fans, consuming 0.18kW power and three 100W

lamps Ventilation air at the rate of 80kg/hr enters with the enthalpy of 84kJ/kg and leaves with an enthalpy a 59KJ/Kg. If each person puts out heat at the rate 630kJ/hr. Determine the rate at which eat is to be removed by a room cooler so that steady state maintained in the room[ Nov-03]

Given data: Np=(persons)s, nf=2 Wf=0.81kW (each) W1=100 W(each)

Mass of air, m=80kg/hr=80

0.22 / sec3600

kg

Enthalpy of air entering, h1=84kJ/kg Enthalpy of air leaving, h2=59kJ/kg Qp=630kJ/hr (each person) To find: Rate of heat is to be removed =? Solution; Rate of energy increase = Rate of energy inflow-Rate of energy outflow

2 2

1 21 1 2 2 .....(1.33)

2 2

C CE m h z g Q m h z g W

Assuming that,

2 2

1 2

1 2

1 2

1 2

02

( ) 0

, equation (1.33) reduces to Q=Em (h )

6304

3600

0.7 / 0.7

80( ) (84 59)

3600

p p

C C

z z g

Now the h W

n Q

kJ s kW

m h h



103

0.55kJ/s=0.55kW W= electrical energy input =nfWf+n1W

1002 0.18 3

1000

0.66

0.70.556 0.66

1916

kW

Q

kW

Result; Rate of heat to be removed , Q=-1.916kW 78. Air flows steadily at the rate of 0.5kg/s through air compressor entering at 7m/s

velocity, 100kpa Pressure, and 0.953/kg specific volume, and leaving at 5m/s. 700kPa, and 0.19m3/kg. the internal energy of air leaving is 90kJ/Kg greater than that of the air entering. Cooling water in the compressor jackets absorb heat at the rate of 58kW. Calculate the rate of shaft work input to the compressor [Nov-04]

Given data: M=0.5kg/s. C1=7m/s P1=100kPa V1=0.95m3/kg C2=5m/s P2=700kPa V2=0.19m3/kg U2-U1=90KJ/kg Q=58kN. To find:

1. Work input, W=?

2. 1

2

?D

D

Solution: SFEE,



104

2 2

1 21 1 1 1 2 2 2 2

2 2

1 21 1 1 1 2 2 2 2

1 2

2 2

2 2

2 2

Z

7 5100 0.95 58 0.5 90+ 700 0.19)+

2000 2000

input W =-5.995kW

C Cu p v z g Q m u p v z g W

C Cp v z g Q m u u p v z g W

Assume Z

S W

work

(-ve sign indicates that the work is done on the system from continuity equation)

1 1 2 2

1 2

1 2 1

2 1 2

53.57

7 0.19

AC A C

V V

A C v

A CV

2

4 1

2

4 2

2

1

2

1

2

3.57

3.57

1.89

D

D

D

D

D

D

Result; Work input , w=-5.994 kw. 3. Work input, W =5.994kW

4. The ratio of the inict to outlet pipe diameter 1

2

D16

D



105

UNIT – II

PART – A 1. State the Kelvin – Planck statement of second law of thermodynamics? [ MU – Nov‟94,

Oct ‟97, Oct‟ 2000 MKU – Nov‟95, Apr ‟96 ]

Kelvin – Plank states that it is impossible to construct a heat engine working on cyclic process, whose only purpose is to convert all the heat energy given to it an equal amount of work.

2. State Clausius statement of second law of thermodynamics? [MU-Oct‟97, Oct‟99,

Apr‟2000;MK – Nov‟95] It states that heat can flow from hot body to cold body without any external aid but heat cannot flow from cold body to hot body without any external aid.

3. Write the two statements of the Second law of thermodynamics? [ Anna Univ.Apr‟03] Kelvin planck statement: It is impossible to construct an engine working on an cyclic process which converts all the heat energy supplied to it into equivalent amount of useful work. Clausis statement Heat cannot flow from cold reservoir to hot reservoir without any external aid. But can flow from reservoir to cold reservoir without any external aid. 4. State / Carnot‟s theorem?

No heat engine operating in a cyclic process between two – fixed temperatures can be more efficient that a reversible engine operating between the same temperature limits.

5. What are the Corollaries of Carnot theorems?

i. All the reversible engines operating between the two given thermal reservoirs with fixed

temperature have the same efficiency ii. The efficiency of any reversible heat engine operating between two reservoirs is

independent of the nature of the working fluid and depends only on the temperature of the reservoirs.

6. Define – PMM of second kind

Perpetual motion of second kind draws heat continuously from single reservoir and converts it into equivalent amount of work. Thus it gives 100% efficiency.



106

7. What is difference between a heat pump and refrigerator.

Heat pump is a device which operating in a cycle process, maintains the temperature of a hot body at temperature higher that the temperature of surrounding.

A refrigerator is a device which operating in a cycle process maintains the temperature of a cold body temperature lower than the temperature of the surrounding.

8. What is mean by heat engine?

A heat engine is a device which is used to convert the thermal energy into mechanical energy.

9. Define the term COP

Coefficient of performance is defined as the ratio of heat extracted or rejected to work input

Heat extracted rejected

COPWork input

10. Write the expression for COP of a heat pump and a refrigerator? COP for heat pump

2HP

2 1

THeat rejectedCOMP

Work input T T

COP for refrigerator

1ref

2 1

THeat extractedCOP

Heat input T T

11. Why Carnot cycle cannot be realized in practice?

i) In a Carnot cycle all the four processes are reversible but in actual practice there is no process is reversible.

ii) There are two processes to be carried out during compression and expansion. For

isothermal process the piston moves as fast as possible. This speed variation during the same stroke of the piston is not possible.

iii) It is not possible to avoid friction between moving parts completely

12. Name two alternative methods by which the efficiency of a Carnot cycle can be increased.

i) Efficiency can be increases as the higher temperature T2 increases ii) Efficiency can be increases as the lower temperature T1 decreases



107

13. Why a heat engine cannot have 100% efficiency

For all the heat engines there will be a heat loss between system and surroundings. Therefore we can‟t convert all the heat input into useful work.

14. When the Carnot efficiency will be maximum

Carnot cycle efficiency is maximum when the temperature is OK

15. What are the processes involved in Carnot cycle? Carnot cycle consist of

i. Reversible adiabatic compression ii. Reversible isothermal heat addition iii. Reversible adiabatic expansion iv. Reversible isothermal heat rejection

16. Sketch the p – V and T – s diagram for Carnot cycle. PV Diagram T-S Diagram Process 1 -2: Isentropic Compression Process 2 – 3: Isothermal heat addition Process 3 – 4: Isentropic expansion and Process 4 – 5: Isothermal heat rejection 17. Write the expression for efficiency of the Carnot cycle.

Carnot = 2 1

2

T T

T

18. Is the second law independent of first law? Explain. Yes. The second law is independent of first law. The second law speaks about the quality of energy.



108

19. Define entropy? Entropy is an index of unavailability or degradation of energy. 20. Define change of entropy. How entropy is compared with heat transfer and absolute

temperature?

The measure of irreversibility when the energy transfer takes place within the system or between system and surrounding is called a change of entropy. It is simple known as unaccounted heat loss.

21. Define the terms source, sink and heat reservoir?

Source:

The part where the heat to be rejected to absorbing or work developing device is called source

Sink:

The part which receives heat from work absorbing or working developing device is called sink.

Reservoir:

The part which supplies or receives heat

Continuously without change in its temperature is called as reservoir.

22. Why the performances of refrigerator and heat pump are given in terms of C.O.P. and

not in terms of efficiency? The performance of any device is expressed in terms of efficiency for work developing machines. But heat pump and refrigerator are work absorbing machines. So, the performance of those devices based on C.O.P. only. 23. Comment on the statement “The entrophy of universe tends to be maximum?

If the entropy of universe tends to be maximum the irreversibility will be more due to friction between moving parts. 24. Write down the equation for Carnot C.O.P of a heat pump which works between two heat

reservoirs of temperature T1 and T2 if T1>T2

Carnot C.O.P. of heat pump = 1 2

1

T T

T



109

25. What is meant by principle of increase of entropy? For any infinitesimal process undergone by a system, change in entropy, dS>dQ/T

For reversible dQ = 0, hence, dS=0 For irreversible ds>0

So the entropy of an insolated system would never decrease. It will always increase and remains constant if the pressure is reversible is called as principle of increase of entropy.

26. What do you mean by “Calusius inequality”? It is impossible for a self acting machine working in a cyclic process unaided by any external agency to convey heat from a body at a low temperature to a body at a higher temperature. 27. Explain briefly clausius inequality.

dQ0

T is known as inequality of clausius

If 1. dQ

0T

, the cycle is reversible.

2. dQ

0T

, the cycle is irreversible and possible

3. dQ

0T

, the cycle is impossible (Violation of second law).

28. For compression process between same and states, which work will be more, reversible or irreversible. Irreversible work will be more in the compressor. Generally for compression, the actual work given will be higher than the calculated work (Wrev). 29. A heat pump pumps 10MJ/KW whr to the high temperature reservoir. What is the C.O.P.?

C.O.P. = Heat Supplied

Work input

310 10

2.783600



110

30. Find the entropy of universe when 1000 KJ of heat is transferred from 800K to 500K.

Entropy of universe, univ

1 2

Q QS

T T

1000 1000

800 500

0.75KJ/K

31. Give the expressions to find change in entropy during constant pressure and polytropic process. Show on T-S diagram. For constant pressure process,

22 1 p

1

TS S S mC In

T

For polytropic process,

2 22 1 PIn

1 1

2 22 1 vIn

1 1

T PS S S m C RIn

T P

or

T VS S S m C RIn

T V

32. Explain the term “Reversibility”. If the process traces the same path during the process is reversed is called as reversibility. 33. Can entropy of universe ever decrease? Why? Entropy of universe can not ever decrease. It will be remains constant or will increase due to irreversibility. 34. What is the essence of the second law of thermodynamics?

1. To know the feasibility of process 2. To know about the quality of energy



111

35. If Carnot engine efficiency is 50%. Find C.O.P. of Carnot refrigerator working between same temperatures.

1 2

1

2

1

2 1

1 2

T TH.E. 50%

T

T1 0.5

T

T T0.5 2

T T

COP of refrigerator 2 1

11 2

2

T T

TT T1

T

1

2 1

1

36. Define the term absolute entropy. The change entropy of the system with respect to ambient conditions or any other standard reference condition is known as absolute entropy.



112

PART – B

1. Two heat engines operating on Carnot cycle are arranged in series. The first engine A

receives heat at 927C and rejects heat at a constant temperature T2. The second engine B

receives the heat rejected by the first engine, and in turn rejects heat to a reservoir at 27C. Calculate the temperature T2, in degree Celsius, for the situation where (a) the work output of the two engines are equal and (b) the efficiency of two engines are equal.

System : Two Carnot engines A & B operating in series between 927C and 27C

Known : T1 = 927C = 1200 K

T3 = 27C = 300 K To find : Intermediate temperature T2 when

(a) WA = WB

(b) A = B Diagram :-

Analysis : case (a) WA = WB Since it has been already proved that total work output from any number Carnot engines operating in series is equal to that of a single Carnot engine operating between the same reservoirs.

Reservoir at 1200 K

Engine A

Engine B

Reservoir at 300 K 3 (a)

WA

WB

Q2 T2

Q3



113

1

3001

1200A BQ W W

Since WA + WB

1

21 1

2

2

2

3001 2

1200

2 11200

1200 29002

1200 1200

1200 9001200

2 1200

750

A

H

H L

Q W

T TQ Q

T T

T

T

T K

Case (b) When

2

2

12

2

2

3001 1

1200

(1200 300)

600

A B

T

T

T

T K

2. Two Carnot refrigerators are arranged in series. One receives 300 KJ/cycle from a heat source at 300K. The heat rejection from this refrigerator serves as the heat input to a second refrigerator, which delivers its output heat to reservoir maintained at 1000 K. If the two refrigerators have the same COP, determine a) Heat rejection to the 1000 K reservoir b) The intermediate temperature between the two refrigerators in Kelvin and c) COP of the two refrigerators. System : Two Carnot refrigerators operating in series Known : Q3 = 300 KJ/Cycle T3 = 300 K T1 = 1000 K To find : (a) Q1 – heat rejection to the reservoir at 1000 K

(b) T2 – Intermediate temperature in K (c) COP



114

Diagrams:

(b) (COP)Ref1 = (COP)Ref2

3 2

2 1 1 2

2

2 2

2

2 2 2

2

2

300

300 1000

300 1000 300 300

300 1000

547.7

T T

T T T T

T

T T

T T T

T

T K

(a) 2 2

2 1 2

547.7

1000 547.7

Q T

W T T

2

2

22

1 2

1 2 2 2

1.211

1.211

0.826

1.826

Q

W

QW

W Q

Also Q Q W Q

Where Q2 = Q3 + W1

Reservoir 1 @ 1000 K

Q1

W2

W1

Q2 T2

Ref2

Ref1

Q3

Reservoir 3 @ 300



115

33

2

2

1.211

1.211

QQ

Q

W

Ref2 Ref1since COP = COP

1

1

1

11

1.211

300 1.826

547.7 /

Q 1.826 547.7

1000.14 /

Q

KJ Cycle

Q KJ Cycle

(c) COPref1 = COPRed2 = 1.211

3. A Carnot engine receives 90KJ from a reservoir at 627C. It rejects heat to the

environment at 27C. One-fifth of its work output is used to derive a Carnot refrigerator. The

refrigerator rejects 60 KJ to the environment at 27C Find:

a) The work output of the engine. b) The efficiency of the heat engine c) The temperature of the low temperature reservoir for the refrigerator in degree

Celsius. d) The COP of the refrigerator.

System : A cyclic heat engine operating a cyclic refrigerator both working on Carnot cycle. Known :

Heat engine Refrigerator

TH = 627C = 900 K

TL = 27C = 300 K Qin = 90 KJ

WRef = 1

5 Wheat engine

Qout = 60 KJ TH = 300K

To find : (a) Wheat engine

(b) heat engine

(c) TL refrigerator (d) COP Refrigerator



116

Diagrams :-

1

5heat engineW

Analysis : (a) Wheat engine = 1 Lin

H

TQ

T

3001 90

900

60 KJ

(b) Heat engine = 1 L

H

T

T

3001

900

0.667 (66.7%)

(c) Wref =

1

5Heat engineW

R Re

ReRe

160

5

12

60 12

48

out ef f

inL

fH L f

KJ

Q Q W

KJ

QT

T T W

in Ref

Source at 900K Environment at 300 K

90 KJ 60 KJ

HE Ref

Sink at 600 K Refrigerator space at TL Ref

4

5 Wheat engine



117

484

300 12

4(300 )

240

L

L

L L

L

T

T

T T

T K

d) COPRef = Re

in

f

Q

W

48

412

4. An irreversible heat pump is designed to remove heat from the atmosphere at 7C and to

supply 43,200 KJ/hr of heat to a constant temperature reservoir kept at 52C. The heat pump is of COP 80% of the maximum possible between the two reservoirs. Power required

running the heat source kept at 1000 K and the reservoir at 52C which is receiving heat

from the heat pump. Taking the efficiency of the heat engine at 70% of supplied to the 52C reservoir and also the heat extracted by the heat engine from the reservoir at 1000 K. System : An irreversible heat pump driven by an irreversible engine. Known : Qout HP = 43,200 KJ/hr

43,200

3,600

12 /

52 273 325

7 273 280

1000

52 273 325

H HP

L HP

H HE

L HE

KJ

S

KJ S

T K

T K

T K

T K

Source at 1000 K

HE

HP

Qin HE

WHE

WHP

Qout HP

QOut HE

ab

Reservoir at 52C

Reservoir at

7C



118

To find : (a) Qsupplied to 52% reservoir (b) Qin for the heat engine

Analysis : (a) COPHP = 0.8 HP

HP HP

H

H L

T

T T

3250.8

325 280

5.78

5.78

5.78

HP

HP

out

HP

HP

Out

HP

QAlso COP

W

QW

12

5.78

2.08 /KJ s

3250.7 1

1000

0.4725

2.08

0.4728

HE

HE

HE

HEHE HE HP

in

HPin

HE

WAlso and W W

Q

WQ

h

= 4.40-2.08 = 2.322 KJ/s

HE HEin outQ Q

Hence the total heat supplied to the

Reservoir kept at 52°C

= 15 + 2.322 = 17.322 KJ/s Or

4.40 /

15.840 /

HEinQ

KJ S

Or

KJ hr

Hence extracted by the heat engine

from the reservoir at 1000K



119

5. A heat engine operates between the maximum and minimum temperature of 671C and

60C respectively with an efficiency of 50% of the appropriate Carnot efficiency. It drives a

heat pump which uses river water at 4C to heat a block of flats in which the temperature

difference of 10C exists between the working and the river water on the one hand, and the required room temperature on the other hand, and assuming the COP of heat pump to be 50% of the ideal COP that can be obtained under the same working conditions, find the heat input the engine per unit heat output from the heat pump. Why is direct heating thermodynamically more wasteful? System : A cyclic heat engine operating a cyclic heat pump with their efficiency / COP have been defined in terms of ideal efficiency/COP. Known : For heat engine TH = 671 + 273 = 944K TL = 60 + 273 = 333 K

HE = 0.5Carnot For the heat pump

Tsource = 4C

Tsink = 20C

A temperature difference of 10C required on either side. Analysis :

671C 20C

Q1

HE HP

T = 10C

W

Q2 T = 10C

Q3

60C

Q4



120

. .

1

333 1

944

0.647

0.5 0.647

0.5 0.647

0.323

LCarnot

H

H E

T

T

COPHP = H

H L

T

T T

Where TH = Tsink + 10C = 20 + 10

= 30C = 303 K

TLTSource-10C = 4-10

= -6C COPHP = 0.5COPIdeal

4

4

1

3030.5

30 ( 6)

3030.5 4.2

36

1

1 0.238

4.2

0.238 0.736

0.323

HP

HE

When Q KJ

QW

COP

KJ

WQ

Result : Per unit head output from the heat pump, 0.736 KJ of heat is to be given to the heat engine.



121

Comment: It can be understood from the result that with the help of 0.736 KJ. 1KJ is supplied to the conditioned space whereas in direct heating whatever the quantity required it has to be directly supplied. Moreover direct heating results in degradation of energy. 6. A reversible engine works between three thermal reservoirs. A, B and C. The engine absorbs an equal amount of heat from the thermal reservoirs A and B kept at temperature TA and TB respectively and rejects heat to the thermal reservoir C kept at temperature TC. The efficiency of the engine, which works between the two reservoirs A and C. prove that

(2 1) 2(1 )A A

n C

T T

T T

System : A reversible heat engine operating between three reservoirs A, B and C Known : 1. Heat supplied to the engine from reservoirs A and B are equal.

2. Efficiency of the given engine is equal to times the efficiency a reversible engine operating between reservoirs A and C.

To prove : (2 1) 2(1 )A A

B C

T T

T T

Diagram : Analysis : As the given heat engine is reversible

0

1 1

CA B

A B C

A B

CA

A B C

C C C

A A B

QQ Q

T T T

Since Q Q

QQ

T T T

Q T Tor

Q T T

TA TB

TC

HE QA QB

QC

Wnet



122

Also it is given that

1 1

1 12

C C

A B A

C C

A A

Q T

Q Q T

Q T

Q T

C

A

QSubstituting for we get

Q

11

2

2 1

C C C

A B A

C C C

A B A

T T T

T T T

T T T

T T T

Multiplying A

C

T

T by we get,

(2 1) 2(1 )A A

B C

T T

T T

7. A solar powered refrigerator receives heat from a solar collector at Th and rejects heat to the atmosphere at Ta, and removes heat from a space at Tc. The three heat transfers are Qh,

Qa and Qc respectively. Derive an expression for the minimum ratio of C

A

Q

Q, in terms of the

three temperatures. If Th = 400K, Ta = 300K, Tc = 270K and Qc = 10KW, what is the minimum Qa? If the collector captures 0.2KW/m2, what is the minimum collector area required? System : A refrigerator driven by solar heat Known : Tn = 400K Ta = 300K Tc = 270K Qc = 10 KW

20.2 /CQKW m

A

To find : (i) An expression for h

c

Q

Q

(ii) To find the minimum are of the collector



123

Diagram

Analysis : (1) The ratio represents the performance index of the refrigerator which will be minimum if the processes are reversible. Therefore,

0h C A

n C A

Q Q Q

T T T

Dividing by QC and rearranging, we get

1 1 1h a

C h C C a

Q Q

Q T T Q T

From energy balance Qh + QC = Qa Therefore

1 1 1

1 1 1 1

h h C

C h C C C a

h h

C h C C a a

Q Q Q

Q T T Q Q T

Q Q

Q T T Q T T

1 1 1 1

/

( /

h

C a h C a

a c a ch

C h a a h

h h a C

C C h a

Q

Q T T T T

T T T TQ

Q T T T T

Q T T T

Q T T T

Solar Collector Tn

Atmosphere Ta

Qh

Qa

Ref

Qc

Cold Space Tc



124

The expression represents the ratio of h

C

Q

Q

in terms of the three temperatures.

(ii) Substituting the numerical values for Th, Ta, Tc and Qc, we get 400 300 270

10 270 400 300

hQ

2

4.4

0.2

h

h

Q kW

Q kwGiven

A m

24.4 22

/ 0.2

h

h

QTherefore A m

Q A

Minimum collector are required is 22m2.

8. It is required to maintain a house at a temperature of 17C where the ambient

temperature falls to 0C in winter. A device working on reversed Carnot cycle is proposed to maintain the temperature of the room. It heat transfer across the walls and roof is estimated as 2000 KJ per degree temperature difference between inside and outside, estimate the power required?

If the same device is used to cool the room during summer by supplying the same

amount of power, what is the maximum outside temperature upto which we can maintain

the room temperature at 17C? Case (i) System : A heat pump working on reversed Carnot cycle Known : TL = 273 K TH = 300 K

2000hQ KJ

T h

Diagram

(Room)T1

HP

(Atmosphere) T2

1Q

2Q

inW



125

Analysis :

1

1

2000( )

2000 17

2000 17

3600

9.44 /

HP

in

Q T

KJ

h

KJ

s

KJ S

QCOP

W

where

HHP

H L

TCOP

T T

300

17.65300 273

9.44 0.535

17.65

Lin

HP

QW

COP

KW

Case (ii) System : A refrigerator working on reversed Carnot cycle. Known : TL = 300K

0.535inW KW

Q2 = 2000 (TH – TL) To find : TH Diagram :

inW

(Atmosphere) TH

Ref

(Room) TL

Q1

Q2



126

Analysis : It is given that

2

2

2000 ( )

2000 ( 300)( )

3600

H L

H

Q T T

TQ a

As reversed Carnot cycle is being followed

2

2

300( )

300

L

in H L

H

Q TCOP

W T T

Q bT

From equations (a) and (b) we get

2

2000 ( 300) 300 0.535

3600 ( 300)

300( 300)

2000 0.535 3600

317 (44 )

H

H

H

H

T

T

T

T K C

9. An insulated rigid vessel is divided into two chambers of equal volumes. One chamber contains air at 500 K and 2 Mpa. The other chamber is evacuated. If the two chambers are connected d, what would be the entropy change? System : Closed system Process : Unresisted expansion Known : T1 = 500K

P1 = 2 103 kPa To find : Entropy change Diagrams: Analysis :

Air Vacuum Air

Initial State Final State



127

2 22 1

1 1

2 22 1

1 1

( )

P

V

T PS S C In R In or

T P

T VS S C In R In

T V

After expansion air will occupy the entire volume of the container.

V2 = 2V1 Also 1W2 = 0 since it is an unresisted expansion Q12 = 0 since the vessel is insulated Applying the first law of thermodynamics

1 2Q U W

Therefore U = 0 For air

2 1

2 1

( ) 0

. .

VCM T T

i e T T

Hence S2 – S1 = CV in 2 2

1 1

T V

R InT V

= 0.287 In 1

2

2V

V

= 0.199 KJ/kgK Comment: Though the process is adiabatic entropy increases as the process involving unresisted expansion is an irreversible process. It also proves the fact that.

( ) 0dQ

Ds or dsT

10. An adiabatic chamber is partitioned into two equal compartments. On one side there is

oxygen at 860kPa and 14C. On the other side also, there is oxygen, but at 100 kPa and

14C. The chamber is insulated and has a volume of 7500 cc. The partition is abruptly removed. Determine the final pressure and the change in entropy of the universe. System : Closed Process : Adiabatic Mixing Known :



128

Subsystem I Subsystem II

Fluid Oxygen Oxygen

Initial pressure 850 kPa 100 kPa

Initial temperature 14C 14C

Initial Volume 7500

2

cc

7500

2cc

Diagrams: Analysis : Here the energy interaction is taking place only between the two fluids and therefore the energy lost by one fluid should be equal to the energy gained by the other fluid. Taking tF as the final temperature we get M1C1 (t1 – tF) = m2C2 (tF – t2) Since the same fluid is stored in both the systems at the same temperature C1 = C2 and

t1 = t2 = 14C

Therefore the final temperature will also be 14C

After removing partition total mass of oxygen is occupying the entire 7500cc at 14C. Hence the final pressure can be computed as given below :

1

1 1

1m

PV

RT

Mass of oxygen

in the subsystem1

6850 3750 10

8.314287

32

0.0427 kg

O2

850 kPa

14C

O2

100 kPa

14C

O2

Initial State Final State



129

2

2 2

2

6

100 3750 10

8.314287

32

0.00503

m

PV

RT

kg

Mass of oxygen

in the subsystem 2

To find the final pressure

1 2

1 2

1 2

( )

8.314(42.7 5.03) 287

32

7.5

475

F F

F

rF

r

F

System

P Vm m

RT

m m RTP

V

P

kPa

S S S

= m1

2

1 1 2 2

f ftF

V V

V VTTC In R In m C In R In

T V T V

3

8.314 8.3140.0427 2 0.00503 2

32 3

8.596 10

0

8.596

Surroundings

Universe

In In

KJ

K

S

JS

K

11. Two vessels, A and B each of volume 3m4 may be connected by a tube of negligible

volume. Vessel A contains air at 0.7 Mpa, 95C while vessel B contains air at 0.35 Mpa,

205C. Find the change of entropy when A is connected to B by working from the first principles and assuming the mixing to be complete and adiabatic. System : Closed Process : Adiabatic mixing Known :



130

Properties Subsystem A Subsystem B

Fluid Air Air

Pressure 0.7 MPa 0.35 MPa

Volume 3 m3 3 m3


Diagrams : After Mixing Analysis : Since the energy interaction is taking place only between the two fluids energy lost by one fluid is equal to the energy gained by the other fluid.

QA = QB Taking t2 as the final temperature after mixing ma Ca (t2 – t1a) = mb Cb (t1b – t2). Since in both A and B the same fluid is stored, Ca = Cb Also ma

1

700 3

0.287 368

19.9

A A

A A

p V

R T

kg

mb = 1

B B

B B

p V

R T



131

2 2

2 2

2 2

2

350 3

0.287 478

7.65

19.9 ( 95) 7.65 (205 )

2.6 ( 95) 205

2.6 205 2.6 95

125.6

kg

t t

t t

t t

t C

Entropy Change = SA + SB

2 2

1 1

125.6 273 619.9 0.717 0.287

95 273 3

5.08

A A V

A A

T VS m C In R In

T V

In In

KJ

k

2 2

1 1

Bm V

B B

T VS C In R In

T V

125.6 273 67.65 0.717 0.287

205 273 3

0.525

In In

KJ

K

5.08 0.525

5.605

0

5.605

sys

surr

universe

S

KJ

K

S

KJS

K

Final pressure P2 = 2

2

mRT

V

(19.9 7.65) 0.287 (125.6 273)

6

525 kPa



132

12. Air enters a turbine at 400C, 30 bar and velocity 160 m/s. It leaves the turbine at 2 bar,

120C and velocity 100 m/s. At steady state it develops 200 KJ of work per kg of air. Heat transfer occurs between the surroundings and the turbine at an average temperature of 350K. Determine the rate of entropy generation. System : Open Process : Steady flow Known :

Properties Inlet Outlet

Pressure 30 bar 2 bar

Velocity 160 m/s 100 m/s


Ambient temperature = 350 K Work output = 200 KJ/kg Diagram: To find : Rate of entropy generation

Analysis :

( ) ( )surr CV

Rate of entropyS S

generation

2 1( ) ( )CVS m S S

For unit mass 2 2

1 1

( ) CV P

T PS C In R In

T P

393 21.005 0.287

673 30

0.236 /

( ) sursur

sur

In In

KJ kgK

QS

T

160 m/s 30 bar

400C

Ambient @ 350 K

400C

w = 200 KJ/kg of air

2 bar 120C 100 m/s



133

Where 2 2

2 12 1 )

2000CV

V Vsur Q W m h h

Q

2 2100 160200 1 1.005(393 673)

2000

89.2 /

89.2( )

350

0.255 /

sur

sur

Q KJ kg

S

KJ kgK

Rate of entropy generation = 0.255 – 0.236 = 0.019 KJ/kgK. 13. A turbine operating at steady state receives air at a pressure of p1 = 3.0 bar and temperature of 390K. Air exist the turbine at a pressure of p2 = 1.0 bar. The work developed is measured as 74 KJ/kg of air flowing through the turbine. The turbine operates adiabatically, and changes in kinetic and potential energy between inlet and exit can be neglected. Using ideal gas model for air, determine the turbine the turbine efficiency. System : Open Process : Steady flow Known : P1 = 3.0 bar P2 = 1.0 bar T1 = 390 K Wa =74 J/kg Diagrams:

Analysis : 1 2

1 2

t

s

h h

h h

1 2

12

T T

T T s

for an ideal gas

where

1 1

2 2 2

1 1 1

r r

r rs sT T P

T P P

Air out (2)

1 bar

Air in (1)

W

~

T

s

11

3 bar

2s

2



134

0.4

1.4

2

2

1390

3

284.9

s

s

T

T K

1 2

1 2

( ) 74

74

1.005

73.63

aP

WC T T

m

T T

K

Hence 1 2

1 2

t

s

T T

T T

73.63

390 284.9

0.7 ( 70%)or

14. A closed system is taken through a cycle consisting of four reversible processes. Details of the processors are listed below. Determine the power developed if the system is executing 100 cycles per minutes.

Temperature (K)

Process Q(KJ) Initial Final

1 – 2 2 – 3 3 – 4 4 – 1

0 + 1000

0 -

300 1000 1000 300

1000 1000 300 300

System : Closed Process : The system is executing cyclic process Known : Heat transfer in process 12, 23 and 34 and Temperature change in all the process.

No. of cycles per minute. To find : Power developed. Diagrams :

T(K)

1000

300

3

4

2

1

S



135

Analysis : To find power developed Wnet per cycle must be known. From I Law Wnet = Qnet which can be computed from the following table.

Process Q(KJ) Temperature (K)

S Initial Final

1 – 2 2 – 3

3 – 4 4 – 1

0 1000

0 -

300 1000

1000 300

1000 1000

300 300

0 1000

11000

KJ KJ

k k

0

S41

For a cyclic process = 0

where is any property

s = 0

(i.e) S12 + S34 + S41 = 0

0 + 1 + 0 + S41 = 0

S41 = 1KJ

K

Since the process 4-1 is isothermal

41

41

1300

300

Q

Q KJ

Therefore

12 23 34 41

0 1000 0 300

700

netQ Q Q Q Q

KJ per cycle

700 net netW Q KJ

And power developed

sec

100700

60

1166.7

netW Cycle

Cycle

KW



136

15. Two kilogram of air is heated from 200C at constant pressure. Determine the change in entropy. System : Open / closed Working : Air fluid Process : Constant pressure heating

Known : 1) t1 = 200C

2) t2 = 500C Diagram :

To find : Change in entropy s Analysis :

2 2

1 1

2

1

500 273 2 1.005

200 273

0.987 / .

p

P

T PS m C In R In

T P

TmC In

T

In

KJ K

16. A Carnot engine operated between 4C and 280C. If the engine produces 300 KJ of work, determine the entropy change during heat addition and heat rejection. System : Open / Closed Process : The working fluid is executing Carnot cycle

Known : 1) t1 = 280C

2) t2 = 4C 3) W = 300 KJ

t(C)

500

200

1

2 p = C

S



137

Diagram :

To find : 1. s during heat addition

2. s during heat rejection Analysis : In Carnot engine heat is added at constant temperature Therefore

1

2

1

T = 1 -

4 273 1 0.499

280 273

1.087 /

in

in

QS

T

WWhere Q

T

KJ K

Therefore Qin = 300

0.499 = 601.1 KJ

1

601.1

(280 273)

inQS

T

2. In Carnot engine heat rejection is also taking place at constant temperature. Therefore

Source @ 280C

Heat Engine

Sink @ 4C

300 KJ

Qin

Qout



138

p = 2 bar

t = 200C

Section A Section B

p = 1.5 bar

t = 150C

2

601.1 300

out

out in

QS

T

where Q Q W

301.1

301.1

(4 273)

1.087 /

KJ

S

KJ K

Comment:

In a Carnot change two isothermal process and two isentropic process. Therefore s during

heat addition must be equal to S during heating rejection so that.

0ds

which obeys Clausius inequality. 17. Air flows through a perfectly insulated duct. At one section A the pressure and

temperature are respectively 2 bar 200C and at another section B further along the duct

the corresponding values are 1.5 bar and 150C. Which way the air flowing? System : Open Process : Steady flow process Known : 1. P1 = 2 bar

2. t1 = 200C 3. P2 = 1.5 bar

4. t2 = 150C To find : To know flow direction Diagram :



139

Analysis : This problem cannot be solved by simple application of first law of thermodynamics. Because there is nothing to tell us whether the fluid is expanding from A to B or being compressed from B to A.

However, since the duct is insulted the inference is that there is no heat transfer to or from the environment and therefore there is no change of entropy in the environment. But in any real process change of entropy of the system plus the surroundings must be positive. In other words

SAB > 0.

B BB A p

A A

T PS S C In R In

T P

273 150 1.5 1.005 0.287

273 200 2

0.02966

In In

KJ

kg K

Thus SA > SB and the flow is from B to A. Even though entropy cannot be measured directly it can still be used to find the sense of flow in a well insulated duct given two salient states as above. 18. A certain fluid undergoes expansion in a nozzle reversibly and adiabatically from 500 kPa, 500 K to 100 kPa. What is the exit velocity?

Take 1.4 0.287KJ

and RkgK

System : Open Process : Reversible adiabatic expansion Known : 1. Inlet pressure = 500 kPa 2. Inlet temperature = 500 K 3. Exit Pressure = 100 kPa 4. The ratio of specific heats = 1.4

5. Characteristic gas constant = 0.287 KJ

kgK

To find : Exit velocity Diagram



140

Analysis: applying steady flow energy Equation

Q W m h ke pe

therefore 2 2

2 11 2

C Ch h

2

2 p 1 2C 2C T T

Where Cp and T2 unknowns

To find Cp P

1C 1 R

p

RC

1

Substituting and R we get cp = kJ

kgK

It is stated in the problem that the process of expansion reversible. Therefore,

dQ

dsT

Also the process is given as adiabatic. That is

(1) (2)

Flow diagram



141

2 1

3p

1 1

22 1

P 1

e

dQ0

T

(or) ds 0

S S 0

PTC In Rin 0

T P

pRT T e In

C p

0.287 100 500 In

1.005 500

315.8 K

Substituting numerical values for T2 and Cp, we get

2C 2x1005 (500 315.08)

608.5 m / s

19. Show from the first principle that, for a perfect gas with constant specific heat capacity expanding polytropically (Pvn = constant) in a non-flow process, the change of entropy can be expressed by

22 1

1

PnS S xIn

1 P

Gaseous methane is compressed polytropically by a piston from 25 and 0.8 bars to a pressure of 5.0 bar. Assuming an index of compression of 1.2, Calcutta the change of entropy and work done, per unit mass of gas. The relative molecular weight of methane is

16 and = 1.3. System : Closed Process : Polytrapic (pVn = C) Known : 1. T1 = 298 K 2. P1 = 80 kPa 3. P2 = 500 kPa 4. n = 1.2 5. M = 1.6

6. = 1.3 To find:



142

1. 1W2 – Work done

2. S – change inentropy Analysis : a) To prove

22 1

1

Pn RS S x xIn

1 n P

From first law of Thermodynamics

Q12 = 1w2 + U

2 2 1 1v 2 1

2 1

v 2 1

v 2 1v 2 1

v 2 1

v 2 1

v 2 1

p v p vC T T

n 1

R T TC T T

n 1

C ( 1)T TC (T T )

n 1

11 C (T T )

n 1

1 n 1C (T T )

n 1

nC (T T )

n 1

In differential for

v

ndQ C dT

n 1


Therefore ds = dQ

T

v

n dTC

n 1 T

Upon integration we get

22 1 v

1

TnS S C In

1 T

From the process relation



143

n 1

n2 2

1 1

T P

T P

Substituting for 2

1

T

T we get

n 1

n2

2 1 v

1

22 1 v

1

PnS S C In

1 P

Pn n 1S S C x xIn

n 1 n P

We know that R = Cp - Cv

R = Cv ( - 1) 22 1

1

Pn RS S x xIn

n 1 n P

Cv = R

1

Substituting for Cv we get (2) Work done

2 11 2

p vW

n 1

2 1R(T T )

n 1

Where T2 = T1

n 1

n2

1

p

p

=

0.2

1.25298

0.8

= 404.45 K Substituting numerical values

1 2

8.3142(404.45 298)

16W

1.2 1

= -276.6 kJ/kg.



144

(3) Change in entropy

22 1

1

pn RS S x xIn

1 n p

1.2 1.3 8.314 /14 5 In

1.3 1 1.2 0.8

kJ 0.2645

kg K

Comment: The negative sing in work indicates that work is given into the system. The negative sign in entropy change indicates that there is heat rejection by the system to the ambient during compression. 20. A closed system undergoes the internally reversible process as shown below:- Computer the heat transfer. Process : Defined by a straight line on a T-S diagram. Known T1 = 200K T2 = 600K S1 = 1 KJ/K S2 = 3 KJ/K To find : Heat transfer Analysis : Q = Area Under the curve representing the process in a T-S diagram =

200 600

(3 1) 800 kJ2



145

21. In a refrigerant condenser superheated vapour of ammonia enters steadily at 1.4 Mpa,

70C. It leaves the condenser at 20C. At 1.4 Mpa condensation begins and ends at 36.28C.

Cooling water enters condenser at 10C and leave 15C. Determine.

a) The amount of heat rejected per kg of ammonia vapour condensed for the given inlet and exit conditions.

b) Mass of water to be supplied for each kg of ammonia vapour condensed. c) The change in specific entropy of ammonia d) The entropy generation per kg of ammonia

Take Cpvapour = 2.9 kJ/Kg K, Cpliquid = 4.4 KJ/KgK and latent heat of evaporation of ammonia at 1.4 Mpa is 1118 KJ/Kg. Also represent the process in a T-S. Diagram: System : Open Process : Steady flow process

Know : T1 = 70C P1 = 1.4 Mpa

T2 = 20C

Tw1 = 10C

Tw2 = 15C To find : a) the mount of heat rejected per kg of ammonia vapour condensed for the given inlet and exit conditions. (b) mass of water to be supplied for each kg of ammonia vapour condensed. (c) the change in specific entropy of ammonia (d) the entropy generation pr kg of ammonia. Diagrams :

Schematic diagram



146

Analysis : (a) Heat rejected per kg of ammonia Q12 = Q12a + Q2a – 2b + Q20.2

p vapour 1 2 p liquid 2b 2

Latent heatC t t C t t

of evaporation

2.9 (70 36.28) 1118 4.4 (36.28 20)

1287.42 kJ/kg.

(b) Water flow rate required per kg of ammonia.

p water(mC T) 1287.42

1287.42 m

4.186 (15 10)

kg of water 61.51

kg of ammonia

(c) Change in Specific entropy of ammonia

1 2a 2a 2b 2b 2

2a 2pvapour p liquid

1 2a 2b

S S S

T TLatent heatC xIn C x In

T T T

36.28 273 ( 1118) 2.9 In

70 273 (36.28 273)

(20 273)4.4 x In

36.28 273

kJ 4.153

kg



147

(d) water ammoniaS S

Where o

water p

m

TS mC In

T

28861.51 4.186 In

283

4.509

Substituting the values we get

universeS 4.509 ( 4.153)

kJ 0.356

kg of ammonia

Comment: As heat is removed from ammonia its entropy decreases where entropy of water increases as it receives heat. But total entropy change will be positive as heat is transferred through finite temperature difference. 22. The specific heats of gas are given by Cp = a + kT and Cv = b+kt, where a b and k are constants and T is in K. Show that for an isentropic of this gas. Tb Va-b ekT = constant System : Closed Process : Isentropic Known : 1. Cp = a + kT 2. Cv = b + kT To prove : Tb Va-b e kT = constant for an isentropic process. Proof : for a gas Cp – Cv = (a + kT) – b (b+kT) (or) R = a-b For an isentropic process Ds = 0

(or) v

dT dvC R 0

T v

Substituting for Cv and R

dT dv(b KT) (a b) 0

T v

Upon integration Bin T + KT + (a – b) Inv = constant Taking antilog TbeKT

Va-b = constant.



148

23. Calculate the entropy change of the universe as a result of the following process:

(a) A metal block of 0.8 kg mass and specific heat capacity 250J/kgK is placed in lake at 8C

(b) The same block, at 8C, is dropped from a height b of 100 m into the lake.

(c) Two such blocks, at 100C and 0C, are joined together. Case (a) System : A metal block Process : Cooling the metal block by dipping it in a lake. Known : 1. Initial temperature of the block (T1) = 100 + 273 = 373 2. Final temperature of the Block (T2) = 8 + 273 = 281 K 3. Mass of the metal block (m) = 0.8 Kg. 4. Specific heat capacity of the metal block.

J(C) 250

kgK

To find : Entropy change of the universe. Diagram :



149

Analysis :

(d) universe surroundingsS S

Where Swater = mCIn 2

1

T

T

2810.8 250 In

373

J56.6

K

surroundings

surroundings

surroundings

sur sys

2 1

surroundings

QS

T

WhereQ Q

m C (T T )

0.8 250 (281 373)

18400 J

18.400S

281

65.4

1

8K

Substituting the values we get

Ssurroundings = -56.6 + 65.48 = 8.84 J/K Comment: As discussed earlier the entropy change of the universe is positive. The reason is the irreversible is positive. The reason is the irreversible heat transfer through finite temperature difference from the metal block to the lake. Case (b) System : A metal block Process : Falling of the metal block into the ake and reaching equilibrium. Known : 1. Initial temperature = 281 K 2. Final Temperature = n281 K 3. Initial Height = 100m 4. Mass of the metal block (m) = 0.8 kg 5. Specific heat capacity of the metal block (C) = 250 j/kgK.



150

Where Ssystem = 0, as the system is at the same temperature at both the initial and final state.

surroundings

surroundings

surroundings

surroundings system

surroundings

surroundings

QS

T

S E

=mgh

=0.8 9.81 100=784.8J

784.8S 2.79J /K

281

S 2.79J /K

Comment: increase in entropy of the universe indicates that there is a irreversibility or degradation of energy. That is the high grade potential energy is converted low grade energy during this process. Case (c) Systems : Two metal blocks Process : Two metal blocks are brought in thermal

contact and allowed to reach equilibrium Known : Initial temperature of the blocks T1a =373K T1b = 273 K To find : Entropy change of the universe Diagrams: (a) (b) (a) (b) Initial State Final State

1000C 00C T2 T2



151

universe a b

2a n

1

2a

1

S S S

TAnalysis : where S mCI

Ta

T S mCIn

Tb

To find T2 Qa = -Qb

2 ta 2 1

1a 1b2

m T T m T Tb

T TT

2

373 273323K

2

2 2universe

2

T TS mc In In

Tfa T1b

323 = 0.8 250 In

373 273

J =4.85

K

Comment: In this process also the heat transfer through finite temperature difference makes the process irreversible which in turn results in increase in entropy of the universe. 24. Find the maximum work developed when air expands in piston-cylinder assembly from an initial state of 600 kpa and 150 kPa and 500C. Also find the availability at the initial and final states, max useful work and change in availability. Assume T0 = 300K, P0=100kPa. Worked Fluid : Air System :Closed Process : a polytropic process Known : P1 = 600kPa P2 = 150kPa T1 = 423 K T2 = 323 K To find: Max work, availability at the initial and final states, change in availability and max useful work

600 Kpa 423 K Air

Analysis: a) Max work = Wrev = (U1-U2) – T0(S1-S2)

150 Kpa 323 K Air



152

1 1v 1 2 0 p

2 2

T PC T T T C In RIn

T P

423 6000.717 (423-323) - 300 1.005In 0.287In

323 150

109.74kJ / kg.

b. Availability at the initial state

1 1 0 0 1 0 0 1 0

01 1 1v 1 0 0 p

1 0 0 0

U U P V V T S S

RTRT T PC T T P C In R In

P P T P

0.287 423 0.287 3000.717 423 300 100

600 100

423 600300 1.00In R 73kJ / kg

300 100

b. Availability at the final state

1 2 0 o 2 0 0 2 0

02 2 2v 2 0 0 P

2 0 0 0

U U P V V T S S

RTRT T PC T T P C In R In

P P T P

0.287 323 0.287 3000.717(323 300) 100

150 100

323 1501.003 In 0.287 In 4.83kJ / kg

300 100

d. Change in availability

1 2

73 4.83

68.17kJ / kg

e. Maximum useful work = change in availability = 68.17 kJ/kg 25. Determine the available energy of 80kg of water at 1000C. Temperature of the surroundings is 150C. System: Closed Process: (To determine the available energy) water is assumed to be cooled at constant pressure to a dead state. Known: 1. Mass of Water m = 80kg 2. Temperature of water = 1000C 3. Atmospheric Temperature = 150C



153

To find : Available Energy Diagram: Analysis : From the T-S diagram it is evident that AE = Area under AB – Area BCDE = Q – TB(SD – SC)

AA B B

B

AA B B

B

TmC T T T mCxIn

T

Tmc T T T In

T

373

80 4.186 100 15 288In288

3522.3kJ

Comment : to determine the available portion of total energy content, the system is assumed to undergo an imaginary process in which it brought to dead state that is state at which comes in equilibrium with the surroundings. 26. Consider the transfer of 1000kJ of heat from a reservoir at 1200K to 5 kg of a gas initially at 100kPa and 500 K in a closed tank. For the gas, Cv=0.8 kJ/kg K throughout the temperature range involved. The lowest temperature in the surroundings is 300K. Determine how much of heat removed from the reservoir is available and unavailable and how much of that absorbed by the gas is available and unavailable.

Case (i) Working Fluid: Air Case (i) Working Fluid: Air System A reversible heat engine is assumed to receive the heat from the reservoir at 1200K and rejecting heat to 300 k sink.

1000C

B

150C

UAE

AE E

C S D



154

Known : Quality of heat extracted = 100kJ To find : Available and unavailable part of this 1000kJ Diagrams:

AE – Available Energy, UAE – Un Available Energy Analysis a. Available Energy = Area ABCD in the Ts diagram.

= (1200 – 300) S

Qwhere s=

1200

10000.833kJ /K

1200

AE (1200 300) 0.833

750kJ

b. Unavailable Energy = 300 x s =300x 0.833 = 250 kJ Alternatively : UAE = Q – AE = 1000 – 750 = 250kJ Case (ii) System : Closed

Reservoir @ 1200K

Rev

HE

Reservoir @ 1200K

AE

JAE

S

T(K)

1200

Available Energy

D Unavailable

Energy

E F

B A

300



155

Process : 1000kJ of heat received by the system at constant volume

Known : P1 = 100kPa T = 500K Cv = 0.8kJ/kg T0 = 300K To find : Available and unavailable parts of the

1000 kJ given to the gas Diagrams:

Initial state Final state Analysis : a. AE = Area ABCE b. UAE = Area CDEF

2 2v 2 1

1 1

2

1

2

v 2 1

T VWhere s=m C In R In Since V V

T V

T5 0.8 In

T

To find T : Consider First Law

Q-W = U since dv = 0

U=mC T T 1000

5 Kg of gas 100 kPa 500@

5 Kg of gas @ V2 = V1

UAE D

E

AE

A

B

C

F

T(K)

500K



156

2 1

v

1000T T

mC

1000500

5 0.8

750

750s 5 0.8 In 1.62kJ /K

500

Hence UAE=300 1.62

=486.6kJ

AE=1000-486.6=513.4kJ

27. Helium enters an actual turbine at 300 kpa, 3000C, and to 100kpa, 1500C, heat transfer to the atmosphere at 101.3 kpa, 250C amounts to 7.0kj/kg. Calculate the entering stream availability, the leaving steam availability, and the maximum work. For helium Cp =5.2kJ/kgk and. Molecular weight = 4.003 kg/kg.mol System: open Process: Steady flow process Worked Fluid: helium Known: P1 = 300 kpa P2 = 100 kpa T1 = 473 k T2 = 423 k Qcv = -7.0KJ/kg P0 = 101.32 kpa T0 = 298k

To find: Availability at the inlet and outlet, maximum work Diagrams: Analysis: Availability at the inlet

Qcv=7.0kJ/kg

100kpa 1500C

in

300 kpa 2000C



157

1 0 0 1 2

1 1p 1 0 0 p

2 0

V H H T S S

T PC T T T C in R in on unit mass basis

T P

423 100=5.2 423-298 298 5.2 in 2.077 in

298 101.32

650 298 1.848

99.1kJ / kg

max imum 1 2W

866 99.1

766.9kJ / kg

28. An ideal gas having a constant pressure specific heat of 1.6 KJ/kgk and a molar mass of 30 KJ/kg. Mol, initially at a pressure of 6.7 Mpa and a temperature of 4250C, under goes a steady Flow process until it reaches a pressure of 1.3 MPa and a temperature of 1500C. If the environment is at 100KPa and 250C, find the reversible work that would be obtained in this process: System : Open Process : Steady flow process Working Fluid: An ideal gas Known : M =30kg/kg.mol Cp =1.6kj/kgK P1=6.7mPa P2 = 1.3 < Pa T1 = 4250C T2 = 1500C T0 = 250C P0 = 100 kPa To find: Reversible Work Diagrams:

P0 = 100KPa

T0 = 250C

P2 =100KPa T2 = 1500C

In

6.7 MPa 4250



158

Analysis: Wrev=(h1-h2) – T0 (S1 – S2)

1 1p 1 2 0 p

2 2

T PC T T T C In R In

T P

8.314 8.314where R= 0.277kJ / kgK

M 30

425 273 6.71.6 425 150) 298 1.6 In 0.277 In

150 273 1.

336.6kJ / kg

29. In an adiabatic mixing chamber, 80 kg of water at 1000C are mixed with 50 kg of water at 600C. Determine the decrease in available energy due to mixing. The ambient temperature = 1500C. System : There are two streams of water mixing adiabatically in a mixing chamber Known : 1. Temperature of stream 1 = 1000C 2. Temperature of steam 2 = 600C 3. Mass of stream 1 = 80 kg 4. Mass of stream 2 = 50 kg 5. Ambient temperature = 150C Diagram: Analysis: Decrease in available energy is the difference in the total available energy before mixing and after mixing Total available energy Available Energy of Stream 1

before mixing Available Energy of stream 2

MIXING

CHAMBER

80 Kg 1000C

50kg 600C

130kg T3 = ?



159

1 11 1 a a 2 2 a a

a a

T Tm C T T T in m C T T T in

T T

373 83380 4.186 100 15 288 in 50 4.186 60 15 288 in

288 288

4189.4kJ

Available energy after mixing

31 2 3 a a

a

Tm m C T T T in

T

Where T3 is the temperature after mixing which can be obtained through energy balance. Q1=Q2 m1C(T1 – T2)=m2C(T3-T2) m1T1+m2T2 = (m1+m2)T3

3

80 373 50 333T

80 50

357.6K

Therefore AE after mixing =

357.6

80 50 4.186 357.6 288 288 in288

=39951.3kJ Decrease in available energy due to mixing =4189.4 – 3951.8 = 237.6kJ Comment: Mixing is an irreversible process and hence available energy decreases due to mixing. 30. Oxygen gas is throttled from 5 bars and 270C to 1 bar through a well insulated valve. Determine the reversible work and irreversibility Take T0 = 288K System : open Process : Steady flow process Known : 1. Inlet pressure P1 = 500 kPa 2. Inlet temperature T1 = 300 K 3. Exit pressure P2 = 100 kPa To find : 1. Reversible work (Wrev) 2. Irreversibility (1)



160

Diagram: Analysis:

rev 1 2 0 1 2W h h T S S

For a throttling process h1 = h2 and T1 = T2

1 10 p

2 2

10

2

T PT C in R in

T P

PT R in

P

rev

8.314 500288 in

32 100

120KJ/ kg

Irreversibility (1) = W 120KJ/ kg

31. Air enters an adiabatic nozzle, operating at steady state with negligible velocity. At the inlet, the pressure is 180 kpa, and the temperature is 650C. The mass flow rate is 0.15 kg/s, and the exit pressure is kpa. If the exit velocity is 300 m/s, pressure is 100 kpa. If the exit velocity 300 m/s, determine:

a. The isentropic efficiency of the nozzle b. The exit temperature c. The irreversibility of the process d. The exit area of the nozzle

Assume T0 = 250C and P0 = 100 kPa. System : Open Process : Steady flow process Working Fluid : Air (ideal gas) Known: P0 = 100 kPa P1 = 180 kPa



161

P2 = 100 kPa T1 = 650C C2 = 300 m/s T1 = 338 K T0 = 298 K Diagrams:

To find:

1. istentropic 2. T2 3. I 4. A2

Analysis: As shown in the h – s diagram 1 – 22 Isentropic 1 – 2 Actual Since P2 = P2s

180kPa 650C in

100kPa

S

h

1

2

2s



162

1

2s 2s

1 1

1

21 1

1

0.4

1.4

2

T P

T P

PS T

P

100S 338

180

285.75K

Also from I law thermodynamics

2 2

2 11 2

2 2

2 1p 1 2

2 2

2 12 1

p

Q W m h ke pe

h ke

C Ch h

2

C CC T T

2

C CT T

2C

300

3382 1.005

T2 = 293.2K or T2 = 20.20C

1 2isentropic

1 2s

p 1 2

p 1 2s

h ha.

h h

C T Tsince air is an ideal gas

C T T

338 - 293.2=

338 285.75

= 0.86 T2 = 293.2k or b) Exit temperature

T2 = 20.2C

c) Irreversibility = T0 [(S2 – S1)sys+Ssurr] since the flow is adiabatic



163

2 20

1 1

293 1000.15 298 1.005 in 0.287

338 180

1.153 /

p

T PmT C in R in

T P

in

kJ S

d) Exit area

2

2 2

22

2

-4 2 2

m =

P

RT

0.15 =

100300

0.287 293.2

=4.21 10 4.21

mA

P C

C

m or cm

32. Carbon dioxide gas is contained in a 1.0 m3 tank initially at 1.2 bar 300k. The temperature is increased to 400 k by supplying heat from a reservoir at 500k. The ambient conditions are 1.0 bar and 300k. find the irreversibility. Take CP = 1.043kJ/kgk. And Cv =0.854kJ/kg System : Closed Worked : CO2 – ideal gas Process : Constant volume heating Known : P1 = 1.2 bar T2 = 400k T1 = 300k Tres – 500k T0 = 300k P0 = 100kpa Cp = 1.043 kJ/kgk Cv = 0.854 kJ/kgk To find : Irreversibility Diagram:

Reservoir

500K

Tank V = 1.0m3

Carbon dioxide



164

Analysis: I advance volume is remaining constant

0

2 0 1 21 2 0 11

R

TS S QR

TU U P V V T

02 21 2 0

1 1

1 1

1

p

R

V 2 1

T in 1

T

8.314 C

8.314 =

44

=0.1889kJ/kgk

120 1m=

0.1889 300

= 2.12kg

Q

=W

=mC

1=2

V P R

R

v

system

act

TVmC T T mT C Rm Q

V T

Vm

RT

R also Cmoleculareweight

Q

U

T T

hence

.12 0.854 300-400 2.12 0.854

300 3002.12 0.854 400 300 1

400 400

181 156 72.4

47.4

in

kJ

33. The inlet and exit conditions of a working fluid expanding in a turbine are listed below.

Heat is rejected by the fluid at an average temperature of 200C whereas the work developed is 650kJ/kg. Neglecting the changes in kinetic and potential energies and

assuming T0 = 25C and P0 = 100kpa find: a) The heat transfer b) The reversible work c) The irreversibility



165

Properties P mpa T C Hkj/kgk Skj/kgk

Inlet 2.5 400C 3240 7.0148

Outlet 50kpa 81.33 2530 7.2688

System : Open Process : Steady flow Working : Not known Known : Exit and inlet conditions W = 650kJ T0 = 298K P0 = 100kpa TRejection = 473K Diagrams: S = 7.0148 KJ/kgk S1 = 7.0147kJ/kgk

Analysis : Q – W = [ ]m h KE PE

on unit mass basis

Q = W+h = 650 + (2530 – 3240) = -60 kJ/kg Heat is rejected by the system to the surroundings b) Reversible work

0

0

1 0 0 1 0 Re

Re

1

1

2983240 2530 298 7.0158 7.688 60 1

473

763.2 /

in out

j

g

Tq

T

Th h T S S q

T

kJ kg



166

c) Irreversibility

763.2650

113.2 /

rev actW W

kJ kg

34. A lead storage battery of the type used in an automobile is able to deliver 5.2 mj of electrical energy. This energy is available for starting the car. Suppose we wish to use compressed air for doing an equivalent amount of work in starting

the car. The compressed air is to be stored at 7 mpa 25c. What volume of tank would be required to have the compressed air have an availability of 5.2mj? System : Closed existing in a thermodynamic

state Working fluid : Air (Ideal gas) Known : Pressure and Temperature P = 7Mpa T = 25+273 = 298K And the required total energy To find : Volume of the air required Analysis : Following steps can be adopted to

solve the problem 1. Finding the availability of 1 kg of air @ 7MPA and 298K 2. Computing Mass air required for the required 5.2MJ 3. Calculating the volume required from the mass.

Step 1: Availability @ 7MPa and 298K

0 0 0 0 0 0

0

0 0 0

0 0 0

0

in

T=T

0.287 298 0.287 298 7000100 298 0287

7000 100 100

279.2 /

v P

u u p V V T S S

RTRT T PC T T P T C Rin

P P T P

Since

In

kJ kg

Step 2: Mass required to supply 5.2MJ

35.2 10

279.2

18.625kg



167

Step 3: Required volume

3

18.625 0.28 298

7000

0.222

mRTV

P

m

35. At a certain location the temperature of the water supply is 15C. Ice is to be made form this water supply by the process shown in the following figures. The final temperature of

ice is -10C, and the final temperature that is used as cooling water in the condenser is

30C? What is the minimum work required to produce 1000kg of ice?

System: A refrigerator that converts water at 15C into ice at - 10C by pumping the heat from it

and reject the heat to the remaining water increasing its temperature from 15 to 30C Known:

1. Inlet temperature of water Tla = 288K and Tib = 288k

2. Exit temperature of ice T2a = 263k 3. Exit temperature of condenser T2b=303k

To find: Wmin Analysis: From energy balance across the refrigeration we get Q1 = Q2 +W where

2

1

heat1000 288 273 273 263

of fusion

= 1000 4186 288-273 335 2.09 273 263

= 418690kJ

Q 4.186 30 15

Pwater pice

cw

LatentQ C C

m

The term mcw is the mass of condenser cooling water which is an unknown its value corresponding to the minimum work can be obtained as follows. For the work to be minimum heat rejection to the condenser cooling water and formation of ice must be reversible processes. Thus the entropy change of the universe is zero.



168

0

0

universe

a b

S

S S

Where

22731000

273 273

273 335 2631000 4.186 2.09

288 273 273

a

pw piceala

TheatS C in C in

T

in in

1529.0

3034.186

288

0.212

cwb

cw

kJ

k

S m in

m

Substituting the results we get -1529.0+0.212mcw = 0

(or) mcw =1529

0.212

=451722kJ Wmin = Q1-Q2 =4517222.4 – 4186990 =33032kJ =33.03kJ 36. The exhaust from a gas turbine at 1.12 bar, 800k flows steadily into a heat exchanger which cools the gas to 700k without significant pressure drop. The heat transfer from the gas heat an air flow at constant pressure, which enter the heat exchanger at 470k. The

mass flow rate of air is twice that of gas and the surroundings are at 1.03 bar, 20. Determine a) The decrease in availability of the exhaust gases, and b) The total entropy production per kg of gas what arrangement would be necessary to make the process reversible and how much would this increases the power output of plant per kg of turbine gas? Take Cp for exhaust gas as 1.08 and for air 1.05 kJ/kgk. Neglect heat transfer to the surroundings and the changes in kinetic and potential energy System : open Process : steady flow Known

1) Inlet temperature of the gas Tg1 = 800k 2) Exit temperature of the gas Tg2 = 700k



169

3) Inlet temperature of the air Ta1 = 470k 4) mair = 2mg 5) Temperature of the surroundings T0 = 293k 6) Cpg = 1.08kj/kgk 7) Cpg = 1.05ij/kgk

To find:

a) Decrease in availability of the exhaust gas b) Entropy production c) Means of making the process reversible and the corresponding increase in power output.

Diagram: Analysis:

a) Decrease in availability of exhaust gas

g1 g2

1 2 0 1 2

h h T S S

g1

pg 1 2 0

g2

TC T T T In

T

= 1.08 ( 800-700 ) – 293 x In 800

700

= 66 KJ

kg

b) Entropy production

gas air

s S

Where

g2

psgasg1

TS C xIn

T



170

= 1.08 x In 700

800

= -0.1442 KJ

kgk

a2a p2air

a1

TS m C xIn

T

For each kg of the exhaust gas, heat transferred to air

= 1 x pg g1 g2C x T T

= 1x1.08 x In 800 - 700

= 108 KJ

kg

This heat is fully transferred to air therefore,

a pa g1 g2 m C T T 108

a2

108T 470

2x1.05

521k

Substituting Ta2, we get

air

521S 2x1.05xin

470

= 0.216 KJ

kg of gas k

Entropy production

= 0.2164 – 0.1442

= 0.216 KJ

kg of gas k

Entropy production = 0.2164 – 0.1442



171

=0.216 KJ

kg of gas k

c) In order to make the process of heat transfer reversible a heat engine must be introduced

as shown below: As all the heat are not given to air the exit temperature of air can no longer be 521 k the actual exit temperature must be such that the entropy change of universe is zero mathematically

gas air

S universe = 0

That is s s 0

g2 g1

pg pa

g1 g2

T TC in C in 0

T T

1.05 x In a2T 8001.08xin

470 700

Ta2=503k Work delivered form the heat engine is the increase output which can be obtained from the energy balance. That is W = Q1 – Q2 = mg Cpg ( Tg1-Tg2) –mgCpg ( Tg1-Tg2) = 1 x 1.08 ( 800 – 700 ) – 2 x 1.05 ( 503 – 470 )

= 38.7 KJ

kg of gas

37. An inventor claims to have developed an efficient hot engine which would have a heat source at 10000 C and reject heat to a sink at 500 C and gives a efficiency of 90% Justify his claim is possible or not. Given data:

0

H

0

L

T 1000 c = 1273k

T 50 C 323k

n 90%

Solution: According to Carnot theorem reversible engine gives maximum efficiency than other heat engine.



172

Maximum efficiency,

h Lmax

H

T T 1273 323n 0.476 74.6%

T 1273

Maximum efficiency ( 74.6% ) is less than proposed engine efficiency ( 90% ). Therefore his claim is impossible. Ans. Result: Inventor‟s claim is impossible. 38. An inventor claims that his propose engine has the following specification: Power developed = 50 kW Fuel burnt = 3 kg/hr Calorific value of the fuel = 75000 KJ/kg Temperature limits = 270 C and 6270C Find out whether it is possible or not. Given data: P = 50kW Fuel burnt = 3 kg/hr C.V. of fuel = 75,000 kj/kg

Temperature limits, 0

L

0

H

T 27 C 300k

T 627 C 900k

Solution: Heat supplied to the engine = Fuel burnt x C.V = 3 x 75000 = 225000KJ/hr QS=62.5kJ/sec=62.5kW Work done, p = 50kW From Carnot theorem cannot engine gives maximum efficiently than any other engine.

H Lcarnot max

H

T TCarnot efficiency, n n

T

900 300

0.66 66.6%900

Efficiency of inventor‟s claim engine,



173

actual

s

workdone W 50n 0.8 80%

Heat supplied Q 62.5

here the inventor‟s claim engine has the higher efficiency than maximum engine efficiency which is possible. Result:

Inventor‟s claim is impossible

39. Determine whether the following cases represent the reversible irreversible or impossible heat engines:

i) 900 kW heat are rejected ii) 560 kW of heat rejected iii) 108 kW of heat rejected

In each case the engine is supplied with 1120 kJ/sec of heat. The source and sink temperature are maintained at 560 k and 280 k. Given data:

L R

H R

s R

T 280 k i)Q 900kw

T 560 ke ii)Q 560kw

Q 1120KJ/ sec iii)Q =108kw

Solution: From Carnot theorem, cannot engine gives maximum efficiency than any other engine. Maximum efficiency or Carnot efficiency,

H Lmax

H

T T 560 280n 0.5 50%

T 560

Case(i)

s R1

s

1 max

Q Q 1120 900n 0.196 19.6%

Q 1120

n n The engine is possible heat engine.

Case (ii)

s RH

s

Q Q 1120 560n 0.5 50%

Q 1120



174

n1=nmax

Therefore, it is reversible heat engine because by II law all the reversible engines have same efficiency. Case (ii)

s Rm

s

m max

Q Q 1120 108n 0.90 90%

Q 1120

n n it is impossible

Result: Case (i) is possible heat engine 1 max n n

Case (ii) is reversible heat engine 1 max n n

Case(iii) is impossible heat engine m maxn n

40. A heat engine of 30% efficiently drives a heat pump of COP = 5. The heat is transferred both form engine and the heat pump to circulating water for heating building in winter. Find the ratio of heat transfer to the circulating water from the heat pump to the heat transfer to the circulating water from the heat engine. Given data: Nhe = 30% COP OF H.P = 5 To find:

s

s

Q?

Q

QR

'

RQ

Qs

'

sQ

H.E.

H.P.



175

Solution: Where,

'

sQ heat transfer to the circulating water from heat pump.

sQ - heat transfer to the circulating water from heat engine.

For Heat Engine For Heat pump

s RH.E

s

Q Qn 0.3

Q

s

s R

QCOP 5

Q Q

s R sQ Q 0.3Q s

s R

QQ Q

5

R R RQ Q 0.3Q 0.7Q

R

s

Q0.7

Q

ss R

QQ Q

5

s R s sW Q Q Q 0.7Q s R

1Q 1 Q

5

sW 0.3Q s R0.8Q Q

s

wQ

0.3 R

s

Q0.8

Q

R s sW Q Q Q 0.8QS'

sw 0.2Q

s

wQ

0.2

s

s

Q w 0.3 1.5

0.2Q 0.2w

0.3

Result:

Ratio s

s

Q1.5

Q

41. A reversible heat engine operating between reservoirs at 900k and 300k drives a reversible refrigerator operating between reservoirs at 300k and 250k. The heat engine receives 1800kJ heat from 900k reservoir. The net output from the combined engine refrigerator is 360kJ. Find the heat transferred to the refrigerator and the net heat rejected to the reservoir at 300 k. Given data :



176

T2 = 900 k T1=300 k T3= 250 k Qs1 = 1800 KJ To find Qs2 =? QR1+QR2=? W1 W2

H.E H QR.1 QR2 Solution : Maximum efficiency of heat engine

H L 2 1max

H 2

T T T Tn

T T

1max

s1

900 3000.66 66.6%

900

wn

q

1 s1 maxW Q xn 1800x0.66 1200KJ

1 s1 R1 R1 s1 1W Q Q Q Q W 1800 1200

R1Q 600kJ

T2 = 900 T3 = 250k

H.E H.P

.

T1 = 300 k



177

3Lref

H L 1 3

TTCOP

T T T T

250

5300 250

R2ref

s2 R2

QAlso, COP

Q Q

R2

2

Q5 ----------- (2.11)

W

It is given that W1-W2=360kJ W2=W1-360=1200-360 W2=840kJ From equation (2.11)

s2

R2

Q5

40

Q 5X840 4200kJ

Heat transfer to refrigerator

R2Q 4200kJ

we know that, W2 = Qs1 – QR2 840 = Qs2 – 4200 = 5040kJ Ans. Net heat transferred the reservoir at 300k = QR1+ Qs1 = 600+5040kJ Ans. Result:

i) Heat transfer to the refrigerator QR2=4200kJ ii) Heat rejection to the reservoir at 3000C=5640kJ

42. A carnot heat engine cycle works at maximum and minimum temperature of 10000C and 400C respectively. Calculate thermal efficiency and work one if Qs

= 1010kJ. Given data:



178

TH = 10000C = 1273k Tt = 400C=313 k Qs = 1010kJ To find: n, W Solution: Thermal efficiency,

H L

H

T Tn

T

1273-313 =

1273

= 75.4% Ans.

s

s

wn

Q

W nxQ 0.754x1010

W = 761.54kJ Result: n = 75.4% W = 761.54kJ 43. A Carnot heat engine receives heat from 600o source. The efficiency of the engine is 59%. Find the amount of heat supplied and heat rejected per kW of work output. Given data: n = 59% TH = 6000C W = 1kW To find: Qs = QR



179

Solution:

s

s

w W 1n Q 1.695 kw

Q n 0.59

s R

R s R

R s

R

w = Q Q

Q Q Q

Q = Q w 1.695 1

Q 0.695kW

Results:

1. Heat supplied, Qs = 1.695 kW 2. Heat rejected, QR = 0.695 kW

44. The temperature in a domestic refrigerator is to be maintained at 100C. The ambient air temperature is 300C, If the heat leaving through the refrigerator is 3 kW, determine the least power necessary to pump out this heat continuously. TL = 100c = 263 k TH=300C = 303 k Qs = kw To find: Power (w) QR W R Qs

Solution: A refrigerator removes heat from refrigerator at the same rate at which the heat leaks from it.

TH = 300C

Ref

TL = -100C



180

+-For reversible engine, condition of minimum power requirement.

sR

L H

LR s

H

R

s R

QQ

Q T

TQ xQ

T

263Q x3 2.6kW

303

w = Q Q 3 2.6 0.396kW

Result: Least power necessary to pump out this heat continuously, W = 0.396kW 45. Two-Carnot engines A and B are operated in series. The first one (a) receives heat at 870k and rejects to a reservoir at temperature T. The second engine (B) receives to a heat reservoir at 300k. Calculate the intermediate temperature T in 0C between two heat engines for the following cases. (iv) The work output of the two engines are equal and (v) The efficiencies of the two engines are equal. Give Data TH = 870k TL = 300k Case (i) WA = WB

Case (ii) A = B To find: T in C for (i) & (ii) case



181

Solution:

Let QSA and QRA – Heat supplied and rejected to H.E.-A QSB and QRB --- Heat supplied and rejected to H.E.-B.

Case (i) WA = WB

2

2 870 300

1170585

2

312

A A B B

A A B B

A A A B A B

A A B

A S R B S R

S R S R A B

S R R R R S

R S R

H L

W Q Q and W Q Q

Q Q Q Q W W

Q Q Q Q Q Q

Q Q Q

T T T Q is proportional to T

T k

t C

Where, T is the intermediate temperature between H.E.-A and h.e.-b. Case (ii)

870(2.12)

870

A A

A A

A B

S RAA

S S

Q QW T

Q Q



182

300(2.13)

870 300

870

B B

B B

S RBB

S S

Q QW T

Q Q T

T T

T

2

,

(870 ) 870 ( 300)

870 870 261000

A B equating above two equation

T T T

T T T

2 261000

510.88

237.88

T

T K

T C

Result :

(i) If WA = WB, the temperature T = 312C

(ii) If A = B, the temperature T = 237.88C 46. An office room which was heated by electric resistance heater consumer 1200 kW-hr of electrical energy in a winter month. Instead of this heater if the same office room is heated by a heat pump which is having 20% of COP of the ideal Carnot pump. The room

temperature is 24C while surrounding is at 0C. If heat supplied from the surrounding by the heat pump is 0.65 kJ, determine COP and money saved per month. Assume Cost of Electricity is Rs.1.75 kW/hr. Given data: Power consumed by electric heater = 1200kW/hr COP of H.P. = 20% of COP of ideal Carnot pump

TL = 0C =273k

TH = 24C = 297 Heat supplied by the Hp, Qs = 0.65 kW Cost of Electricity = Rs. 1.75/kW-hr To find :

(i) COP of heat pump (ii) Money saved per month

Solution: Electricity charge by using heater = 1200 × 1.75 = Rs.2100 per month. COP of Carnot pump



183

29712.375

297 273

H

H L

T

T T

COP of Heat pump = 20% of COP of Carnot pump

20

12.375 2.475100

COP of heat pump = sup

Heat plied

Work done

Work done = sup 0.65

. . 2.475

Heat plied

COP of H P

W = 0.26kW Power required by heat pump in kW-hr = work done × 3600 = 0.26 × 3600 = 945.45 kW-hr Electricity charge for running heat pump = 945 × 1.75 = Rs. 1654.50 Money saved per month = 2100 - 1654.50 = Rs. 445.50 Result : COP of heat pump = 2.475 Money saved per month = Rs.445.50 47. Briefly Explain CLAUSIUS INEQUALITY? Clausius inequality states that “when a system undergoes a cyclic process, the summation of dQ

Taround a closed cycle is less than or equal to zero.

Consider an engine operating between two fixed temperature reservoirs TH and TL. Let dQs. units of heats be sullied at temperature TH and dQR units of heat be rejected at temperature TL during a cycle.

Thermal efficiency, s R

s

dQ dQ

dQ

Thermal efficiency of any reversible engine working on the same temperature limit is given by

Thermal efficiency for reversible engine = H L

H

T T

T

The efficiency of an actual engine cycle must be less than that of a reversible cycle. Since no



184

engine can be more efficient than that of a reversible engine. Hence,

s R H L

s H

dQ dQ T T

dQ T

0

H

R L

s H

sR

L

sR

L H

dQ T

dQ T

dQdQ

T T

dQdQ

T T

for the entire cycle, dQ

0T

This equation is known as Clausius inequality. It provides the criterion of the reversibility of a cycle. If

dQ

0T

, the cycle is reversible

dQ

0T

, the cycle is irreversible and possible

dQ

0T

, the cycle is impossible.

Since the cyclic integral dQ

T is less than zero in a cycle, the cycle violates the second law

of thermodynamics. So, it impossible. We can apply the equality to the Carnot cycle since it is reversible cycle. Then the equation

becomes, dQ

0T

.

48. An inventor claims that his new engine will develop 3 kW for a heat addition of 240

kJ/min. The highest and the lowest temperature of the cycle are 1527C and 32C respectively. Would you agree his claim? Use Clausius inequality method. Given data:

W – 3kW = 30 60 = 1800 kJ/min Q1 = 240 kJ/min

T1 = 1527C



185

T2 = 327C To find : Agreement his claim = ? Solution: By clausius inequality,

1 2

1 2

Q QdQ

T T T

But, Work out, W = Q1 – Q2 1800 = 240 –Q2

Q2 = -1560kJ / min = 0

1 2

1 2

Q Q 240 15602.47 kJ/min.

T T 1800 600

Here dQ

0T

.

So, we agree his claim. Result: Inventor claim is accepted. 49. Entropy is an index of unavailability of degradation of energy. Heat always flows from hot body to cold body and thus becomes lesser value available. This unavailability of energy is measured by entropy. It is an important thermodynamics property of the working

substance = fp

L

TC In

T

By entropy principle

Entropy of universe, Suniv



186

f fP P

H H

2

fP

H L

f

2

fP

1 2

T TC In C In 0

T T

TC In 0

T T

T to be minimum

TC In should be equal to zero

T T

2

fP

1 2

2

fP P

1 2

TC In 0

T T

TC In 0 & C 0

T T

2

f

1 2

2

f

1 2

f 1 2

TIn Inl

T T

T1

T T

T T T

Maximum work, Wmax = P 1 2 1 2C (T T 2 TT

= 2

P 1 1 2C T 2 T T

Wmax = 2

1 2T T

50. 1.6 Kg of air compressed according to the law pV1.3C form pressure of 1.2 bar and

temperature of 20C to a pressure of 17.5 bar. Calculate (a) the final volume and temperature (b) work done (c) heat transferred and (d) change in entropy. Given data: M = 1.6 kg

PV1.3 = C n = 1.3

P1 = 1.2 bar = 120 kN/m2

T1 = 20C = 293 K P2 = 17.5 bar = 1750 kN/m2 To find :

V2, T2, W, Q, and S



187

Solution: From general gas equation.

1 1 1

311

1

1.3 1.3

1 1 2 2

1

1.32 1

1 2

1

1.3

2

3

2

p V mRT

mRT 1.6 0.287 293V 1.121m

P 123

p V p V

V p

V p

120V 1.121

1750

V 0.1427m

From polytropic process relation.

n 1

n2

2 1

1

1.3 1

1.3

PT T

P

1750293

120

543.82 K

Work done, W = 1 1 2 2p V p V

n 1

120 1.121 1750 0.1427W

1.3 1

W 384.02 kJ

Heat transfer.

nQ Wx

1

1.4 1.3Q 384.02

1.4 1

Q 96kJ

Change in entropy, S = mR in



188

1.121 543.83

1.6 0.718 In0.1427 293

S = 1.657 kJ/K. Result:

(i) Final Volume, V2 = 0.1427m3 (ii) Final Temperature, T2 = 543.826 K (iii) Work done, W = - 384.02 kJ (iv) Heat transfer, Q = -96 kJ

(v) Change in entropy S = 1.657 kJ/K.

51. Air expands from 11 bar at 550C to a pressure of 3 bar adiabatically. Determine temperature at the end of expansion and work done. Find also the change in entropy. Given data : P1 = 11 bar = 10 kN/M2

T1 = 550C = 550 = 273= 823 K P2 = 3 bar = 300 kN/m2

Process : adiabatic To find :

T2, W and S Solution:- The p, V & T relation for adiabatic process

1

2 22

1 1

1 1.4 1

1.42

1

1

2

T pT

T p

p 300 T 823

p 1100

T 567.78k

Work done, 1 1 2 2 1 2p V p V mR(T T )W

1 1

1 0.287 (823 567.78)

1.4 1

183.12kJ/kg.

Change in entropy S = 0 for adiabatic process.



189

Result :

(i) Final temperature T2 =567.78 K

(ii) Change in entropy S = 0 (iii) Work done, W = 183.12 kJ/kg.

52. 1 Kg of air is compressed according to the law p. V1.25 = c from 1 bar and 15C to 17 bar. Calculate the change in entropy. Cp = 1.005 kJ/Kg K and Cv = 0.72 kJ/Kg K. Given data : M = 1 Kg pV1.25 = C; n = 1.25 P1 = 1 bar = 100 kN/m2 P2 = 17 bar = 1700 kN/m2

T1 = 15C = 288 K Cp = 1.005 kJ/kg K Cv = 0.072 kJ/kg K R = Cp – Cv = 1.005 – 0.72 = 0.285 kJ/kg K. Solution : From polytropic relation

n 1

n2

2 1

1

1.25 1

1.25

2 2

pT T

p

1700T 288 T 507.55 K

100

Change in entropy (S)

1 2p

2 1

p TS mRIn mC In

p T

1 507.55S 1 0.285 e In 1 1.005 In

17 288

S 0.238 kJ/k

Result :

Change in entropy S=-0.238kJ/K.



190

53. One kg of air in a closed system initially at 50C occupying a volume of 0.3 m3 undergoes a constant pressure heating process to 1000C. There is no work other than pdV work. Find the work transfer heat transfer and the entropy change of the gas. [Madras Univ. Apr‟95] Given data: M=1kg T1=50C V1=0.3m3 T2=1000C Solution: Work transfer, W=P(V2-V1)=mR(T2-T1)=1x0.287(100-5) [Assume for air, R=0.287kJ/kg K]=27.26kJ Heat transfer, Q=mCp(T2-T1)=1x1.005(100-5)=95.48kJ

2

p

1

373 in entropy, S=mC 1 1.005 In

278

=0.295 KJ/K

TChange In

T

54. Ten grams of water at 200 C is converted into ice at -10C at constant atmospheric pressure. Assuming specific heat of liquid water to remain constant at 4.2 j/g K and that of ice ot be half of this value and taking the latent heat of fusion of ice at 00C to be 335 J/g. Calculate the total entropy change of the system Cp of ice = 2.093 J/gK.

[Madras Univ. Oct-95] m=10g Tw=200C Tice=-100C Cp1=4.2J/gk Hfg=335J/g Solution: Heat absorbed from the water to form 10g of ice. Q=Heat absorbed from the liquid + latent heat + heat absorbed from the solid phase.

Q=mwCpw(T1-0)+hfg+miceCpice(0-T2) =10x4.2(20-0)+335+10x2.093(0-(-10)) =4399.3J

system

4399.3Entropy change of atmosphere S

273

=16.12J/K.

Q

T

Entropy change of the system from 200 to 00C



191

0

1 w pw

1

S =m C

27310 4.2 In

293

=2.97 /K

TIn

T

Entropy change of the system from 0 to 100C

0

1 w pice

1

S =m C

26310 2.093 In

293

=0.781 /K

TIn

T

Total entropy change S=S1+S2+Ssystem

=-2.97-0.781+16.12 =12.369 J/K 55. A constant volume chamber of 0.3m3 capacity contains 1 kg of air at 560C. Heat is transferred to the air until the temperature is 1000C. Find the work transfer, heat transfer and the change in internal energy enthalpy and entropy. [Madras Univ.Oct‟95] Given data: V1=0.3 m3 M=1kg T1=560C T2=1000C Solution: Work transfer, W=0 for constant volume process Heat transfer = change in internal energy.

Q=U=mCv(T2-T1) =1 x 0.718(100-56) =31.59kJ

Change in enthalpy H=U=31.59KJ

2

1 v

2

change in entropy, S =mC

3731 0.718 In

329

=0.09kJ/K /K

TIn

T



192

56. 10 kg water 900C mixes with 2.5kg of water at 200C under adiabatic condition. Find the final temperature and entropy generation. [Madras Univ.Apr‟97] Given data: M1=10kg T1=900C M2=2.5kg T2=200C To find:

1. Tf=?

2. S=? Solution:

1 1 1 2 2 2

1 1 2 2

f

0

f

water=4.187kJ/kg K.

10 4.187 363 2.5 4.187 293T

1.4.187 2.5 4.187

=349K

T 76

f

p

m C T m C TT

m C m C

C of

C

1 1 2 2

1 2

Entropy generation, S=m C m C

349 34910 4.187 In 2.5 4.187 In

363 293

=0.184kJ /K

f fT TIn In

T T

Result:

1. The final temperature, Tf=760C

2. Entropy generation, S=0.184kJ/K 57. In a certain heat exchange, 45 kg/min of water is to be heated from 600C to 1150C by hot gases, which enter the heat exchanges at 225 and flow at the rate of 90kg/min. compute the net change of entropy. Assume specific heat for water and gases as 4.18 and 1.045kJ/kg K. [Madras Univ. Oct-98] Given data:



193

Mw=45kg/min Tw1=600C Tw2=1150C Tg1=220C Mg=90 kg/min. To find:

Net change of entropy, S=? Solution: By energy balance Heat gained by the water = Heat given out by the gas.

2 1 1 2

2

2

2

1

g

45 4.18(115 60) 90 1.045 225

115 .

Now change in entropy of gas, S

90 115 = 1.045

60 225

s

s

u p w w g p g g

g

g

g

g P

g

m C T T m C T T

T

T K

Tm C In

T

In

=1.052kJ/K.

2

1

wNow change in entropy of water, S

45 388 = 4.18

60 333

=0.479 kJ/K.

w

w

w P

w

Tm C In

T

In

Net change in entropy, S=Sg+Sw =-1.052+0.479 =0.573 kJ/K. Result:

Net change in entropy, S=-0.573kJ/K. 58. A perfect gas is contained in a cylinder and undergoes a controlled expansion according to the law p-A=A+BV, where A and B constants. P be the pressure in kgf/cm2 and v the volume in m3. The initial and final pressure are 8.4 kgf/cm2 and 2.8 kgf/cm2 and the corresponding volumes are 0.056 m3 and 0.168 m3.

Assume =1.39, R=0.28 kJ/kg K. The initial temperature is 150C. Determine (a) work done by the gas (b) heat transferred during the process in magnitude and direction (c) change in



194

entropy per kg of gas during expansion (d) maximum value of internal energy per keg reckoned from 00C. Given data: P= A+BV P1 =8.4kgf/cm2 P2=2.8 kgf/cm2 V1=0.056 m3 V2 = 0.168 m3. T1 = 150C.

=1.39 & R=0.28 kJ/kg.K. To find:

(i) Work done, W=? (ii) Heat transfer, Q=?

(iii) Change in entropy, S=?

(iv) Change in internal energy, U=? Solution: P=A+BV --------------------------(2.31) At p1-8.4 kgf/cm2 7 V1=0.056m3 8.4=A+0.056 B ------------------(2.32) At p2 = 2.8 kgf/cm2 & V2 =0.168m3 2. 8=A+0.168 B------------------(2.33) Subtracting eqn. (2.32) from (2.33) -5.6=0.112B B=-50

A=112

The equation (2.32) becomes p=11.2-50V



195

2

1

2

1

2

1

v

v

v 0 168

v 0 056

v 0 168

2

v 0 056

22

Work done, W= 100

= 11.2 50

50 =100 11.2V-

2

=100 11.2 .168-0.056 25[(0.168) 0.055

pdv

V dv

V

=62.72kJ.

1 1

1

p of the gas, m=

8.4 100 0.056 =

0.28 288

=0.583kg

VMass

RT

2

mRT know that, V= (2.34)[ ]

p

(2.34) (2.31)

0 (2.35)

We pV mRT

substituting in

BmRTp A

p

p Ap BmRT

For getting maximum temperature, differentiate the eqn. (2.35) and equate to zero. 2pdp – Adp + BmRT =0

20

2 0

/ 2

11.25.6 .

2

dT p A

dp BmR

p A

p A

p bar

From eqn(2.31) 5.6=112-50 V V=0.112m3 At p=5.6 bar & V=0.112m3

max

5.6 100 0.112

0.583 288

=384.2K

pVT

mR

Maximum Internal energy,



196

Umax=Cv(Tmax-T0)

max 01

0.28(384.2 273)

1.39 1

79.84 / .

[ transfer, Q=work transfer [ for pV=C]

RT T

kJ kg

Heat

=62.72kJ

2 2v

1 1

1 2

2

1

[ is supplied into the system]

Entropy, S=C my gas.

For T

In

0.1680.28 In

0.056

0.308 kJ/kg K.

Heat

T VIn Rin for

T V

T

VS R

V

Result:

(i) Work done, W= 62.72kJ (ii) Heat transfer, Q=62.72kJ

(iii) Entropy, S=0.308kJ/kg K

(iv) Max internal energy, Umax=79.84kJ/kg. 59. As single stage air turbine is to operate with air inlet pressure and temperature of 1 bar and 60k. During the expansion the turbine loses are 20kJ/kg to the surroundings which is at 1 bar and 300k. for 1 kg.of mass flow rate determine (i) decrease in availability (ii) maximum work (iii) the irreversibility. [Madras univ. Oct‟96] Given data: P1=1bar T1=600K Q=-20kJ/kg P2=1bar T2=300K M=1kg. To find:

(i) Decrease in availability 1-2=? (ii) Maximum work, Wmax=? (iii) Irreversibility, l=?



197

Solution: For any process, change in entropy

2 2

2 1 p

1 1

S=S -S =m C I

300 11 1.005 In 0.287 In

600 1

0.697 kJ/ K.

T PIn R n

T P

Decrease in availability,

1 2 1 2 1 2

1 2 1 2 =

=1 1.005 600-300 300 0.697

=510.6kJ/kg

o

p o

m h h T S S

m C T T T S S

Maximum work – decrease in availability Wmax = 510.6kJ/kg From SFEE, H1+Q=h2+W W=m[(h1-h2)+Q] =m[Cp(T1-T2)+Q] =1x[1.005 9600-300)-20] W=281.5kJ/kg Irreversibility, I=wmax-W =510.6-281.5 =229.1kJ/kg

Decrease in availability, 1-2=510.6kJ/kg Maximum work, Wmax =510.6kJ/kg Irreversibility, I=229.1kJ/kg 60. In a steam generator, the steam generating tubes receive heat from hot gases passing over the oxide surface evaporating water inside the tubes. Flue gas flow rate is 20kg/s with an average specific heat f 1.04kJ/kg K. The gas T decreases from 6500 C to 4000C while generating steam at 3000C water enters the tube as a saturated liquid and leaves with a quality of 90%. Assume environment temperature as T0=270C. Determine the water flow rate. Availability of hot fluid and cold fluid, irreversibility and second law efficiency. Given data: M1 =20kg/s C1=1.04kJ/kg K



198

T1=6500C T2=4000C T3 = 3000C (saturated liquid) X4=0.9 T0=270C To find:

1. m2=?

2. 1=?

3. 2=? 4. L=?

5. II =? Solution: For heat exchanger energy balance, Heat given out by the flue gas=Heat gained by the steam Corresponding to 3000C, read hf 7 hfg, Sf and Sfg H3=hf=1345kJ/kg,hfg=1406kJ/kg S3 =sf = 3.255kJ/kg k,sfg =2.453kJ/kg K

h4=hf+x4hfg =1345+0.9 x 1406 2610.4KJ/kg s4=sf +x4sfg =3.255+0.9x2.453 =5.46kJ/kg K m1C1(T1-T2)=m2(h4-h3) 20x1.04(650-400)=m2(2610.4-1345) m2=4.11kg/s Availability of hot fluid,

1 1 1 2 0 1 2

1

1 2 p

2

923 =20[1.40(650-40)-300x 1.40In

6T3

T3228.88 s -s =C In

T

pm C T T T s s

kW

Availability of cold fluid,



199

2 2 1 2 0 1 2

1 2

II

=4.11[ 2610.4-1345 -300 5.467 3.255

2473.39

,

=3228.88-2478.39

=755.49KW

AvailSecond law efficiency,

pm C T T T s s

kW

Irreversibility I

ability of cold fluid

Availability of Hot fluid

2473.39 = 100 76.6%

3228.88

Result:

1. Mass of stem, m2=4.11 kg/s

2. Availability of cold fluid 2=2473.39kW

3. Availability of Hot fluid 1=3228.88kW 4. Irreversibility Hot fluid, I=755.49kW

5. Second law efficiency,II =76.7% 61. Steam flows in a pipeline at 1.5 Mpa. After expanding to 0.1 Mpa in a throttling calorimeter, the temperature is found to be 1200C. Find the quality of steam in the pipeline and also calculate availabilities, irreversibility and second law efficiency. Assume T0=250C Given data: P1=1.5Mpa, and p2 = 0.1 Mpa T2 =1200C To find:

1. x1=?

2. 1=?

3. 2=? 4. I=?

5. II=? Solution: Corresponding to p2=1 bar & T2=1200C from the super heated Table, read H2=2716.2kJ/kg. S2=7.4606kJ/kg K For throttling process



200

H1=h2 H1=2716.2kJ/kg But at p1 = 15bar, read hf hfg, sf &Sfg H1=844.89kJ/kg Hfg=1947.3kJ/kg Sf=2.3315kJ/kg Sfg=6.448kJ/kg H1=hf+x1hg 2716.2 =844.89+x1(1947.3) x1=0.963 s1=sf+x1sfg s1=2.3315+0.963(6.4448) =8.538kJ/kg K

Availability at inlet, 1=h1=T0s1 =2716.2-298 x 8.538 =171.876 kJ/K

Availability at outlet, 2 =h2-T0s2 =2716.2-298 x 7.4606 = 492.94 kJ/kg

Irreversibility, I=2-1 =492.94-171.876 =321.064kJ/kg or =T0 (s1-s2) =298(8.538-7.4606) =321.065kJ/kg

1II

2

law efficiency,

171.876 = 100

492.94

=34.867%

out

in

ASecond or

A

Result:

1. Quality steam, X1=0.963

2. Availability of inlet, 1=492.94kJ/kg

3. Availability of Outlet 2=171.876kJ/kg 4. Irreversibility, I=321.065kJ/kg

5. Second law efficiency, II=34.87% 62. An inventor claims to have developed a refrigerating unit which maintains the refrigerated space at -60 while operating in a room where temperature is 270C and has COP 8.5. find out whether his claims is correct or not.



201

Given data: Inventor‟s refrigerating unit. TH=270C TL=-60C COP=8.5 To find: Decision = ? Solution:

1

H

inventor

T of carnot =

T

-6+273 =

27-(-6)

=8.09

here, COP

His claim is not correct.

L

carnot

COPT

COP

Result: His claim is not correct because COPcarnot is less than the inventor‟s claimed COP. 63. A heat engine is used to drive a heat pump. The heat transfer from the heat engine and from the heat pump is used to heat the water circulating through the radiators of building. The efficiency of the heat engine is 27% and COP of the heat pump is 4. (i) Draw the neat diagram of the arrangement and (ii) evaluate the ratio of heat transfer to the circulating water to the heat transfer to the heat engine.[Oct-95] Given data:

H.E =27% COPH.P=4 To find:

1 2

1

?R R

S

Q Q

Q

Solution:



202

1 1

.

1

1

1

1

1

1 1

.

R2

S1 1

R2

S1 1

done

sup plied

0.27 1

0.73

0.73 (2.38)

supplied

done

Q =

Q

Q 4= (2.39)

Q

(2.38) (2.39)

s R

H E

S

R

s

R

S

R s

H P

R

R

Q QWork

Heat Q

Q

Q

Q

Q

Q Q

HeatCOP

Work

Q

Q

Substituting in

R2

S1 S1

R2

S1

R2 S1

Q 4=

Q 0.73 4=Q

Q =

0.27Q

Q 108Q (2.40)

Total heat supplied to the water, Q=

1RQ +2RQ

1 1

1

1

1 1

0.73 1.08

1.81

1.81

=1.81

s s

s

s

s s

Q Q

Q

QQ

Q Q



203

Result: The ratio of heat transfer to the circulating to the heat transfer to the heat engine,

1 2

1

1.81R R

S

Q Q

Q

64. A carnot heat engine takes heat from an infinite reservoir at 5500C. Half of the work delivered by the engines used to run generator and the other half if used to run heat pump which takes heat at 2750C and rejects it at 4400C. Express the heat rejected at 4400C by the heat pump as % of heat supplied to the engine at 5500C. If the operation of the generator is 500kW, find the heat rejected per hour by the heat pump at 4400C. [R-92] Given data: T1=5500C T2=2750C T4=4400C Wg=500kw To find:

2

1

R2

1. ?

2. rejected, Q ?

R

S

Q

Q

heat

Solution: For cannot heat engine

1 1

1 2

1

1 1

2

1

R1

823 =

548

=1.502Q (2.41)

S R

S R

R

Q Q

T T

TQ Q

T

Q

Work done by the heat engine, W=Qs1-QR1 =1.502 QR1-QR1 =0.502 QR1

1g

R1

0.502 input, W

2

=0.251 Q

RQGenerator

Work input to the heat pump.



204

WHP=0.251 QR1 Heat rejected by the heat pump, QR2 = Qs2 + WHP = Qs2+0.251 QR1---------(2.42)

2 2

2 4

22 2

4

2

2 2

s2

For reverse heat pump,

543 =

713

0.77

Substitutinge Q (2.42)

S R

S R

R

S R

Q Q

T T

TQ Q

T

Q

Q Q

in

2 2 1

2 1

2 1

2

1

0.77 0.251

1 =0.77 0.251

1.502

0.727

72.7%

R R R

R s

R s

R

s

Q Q Q

Q Q

Q Q

Q

Q

The generator power input, Wg=500kw 0.251

1RQ =500

1RQ =1992.03kW

Qs1=1.502 1RQ

=1.502 1RQ

=2992.03 kW

2 0.727 2992.03

=2175.21kW

2175.21 =

1000

=7830.74MJ/hr

RQ

Result:

1. Heat rejected by the heat pump 2

1

72.7%R

s

Q

Q

2. Heat rejected by the heat pump if the generator has 500kW, 1RQ =7830.74 MJ/hr.



205

65. A heat engine operates between a source at 6000C and a sink at 600C. Determine the least rate of heat rejection per kW net output of the engine. [Nov-01] Given data: TH=6000C=273+600=873K TL=600C=273+60=333K To find: QR/net output=? Solution: Take W=1kW. At the least rate of heat rejection, the efficiency should be maximum. Carnot efficiency of heat engine,

.

H.E

873 333

873

=61.86%

1

0.616

1.616

=1.616-1

=0.616-1

=0.616/kW of the output

H LH E

H

carnot

s

s

s

R s

T T

T

W

Q

Q

Q

Q Q W

Result: QR/ net output = 0.616/kW

66. 5 kg of air at 2 bar and 30C is compressed to 24 bar pressure according to the law

p.V1.2= Constant. After compression air is cooled at constant volume to 30C. Determine, (i) Volume and temperature at the end of compression, (iii) Change in entropy during compression, (iii) Change in entropy during constant volume cooling. Take Cp=1.005 kJ/kg K, Cv= 0.718 kj/kg K, Cv=0.718 kJ/kg K. [Dec – 01] Given data:



206

M=5 kg P1=2 bar

T1=30C P2=24 bar pV1.2= Constant

T3=30C Cp=1.005kJ/Kg K Cv=0.718 kJ/kg K To find:

1. V2=? 2. T2=? 3. S2-S1=? 4. S3-S2=?

Solution: According to the law, pV1.2= C

p1V1=p2V2 From characteristic gas equation , pV=mRT

11

1

3

1.2 1.2

1 2 2

5 0.287 (303) =

200

=2.17 m

From p

mRTV

p

V p V

1.2

12 1

2

1

12

3

2 = 2.17

24

=0.274 m

pV V

p

Similarly,

1.2 1

1.22 2

1 1

0.2

1.2

2

24 T 303

2

=458.47K

=185.47 C

T p

T p



207

Change in entropy during compression,

2 2

1 1

2 22 1

1 1

2 1

458.47 0.2745 0.718 5 0.287I

303 .217

=1.483kJ/K

Process 2-3 is a constant volume process.

p

v

T pmC In mRIn

T p

T VS S mC In mRIn

T V

S S In n

33 1

2

in entropy, S

303 =5 0.287In

458.47

=1.487kJ/K.

v

TChange S mC In

T

Result;

1. Final volume at the end o compression, V2=0.274m3

2. The corresponding temp, T2=185.47C. 3. Change of entropy during compression S2-S1-1.483 kJ/K 4. Change of entropy during constant volume, S3-S2=1.487kJ/K

67. 0.2kg of air at 1.5bar and 27C is compressed to a pressure o 15 bar according to the law PV1.25= Constant. Determine work done on or by air, eat flow to or from the air, increase or decrease in entropy.[Oct-20002] Given data: M=0.2kg P1=1.5bar

T1=27C=300K P2=15bar pV1.25=C To find:

W, Qs,S=? Solution: By general gas equation P1V1=mRT1



208

11

1

3

1 1.25 1

1.252

2

1

1 1

1.251

2

2

3

0.2 0.28 300

150

=0.148m

15300

1.5

=475.5k

1.5V 0.148

15

=0.018195m

n

n

n

mRTV

p

PT

P

p

p

Work done,

2 2 1 1

5 5

1

-1.5 10 0.1148 15 10 0.018195 =

1.25 1

=40.29kJ

p V pVW

n

Heat transfer

1.4 1.2540.29

1 1.4 1

=15.1088kJ

nQ W

change in entropy

2 2

1 1

0.01148 475.50.2 0.287 0.2 0.718

0.018195 300

v

V TS mRIn mC In

V T

In in

=0.17187kJ/K Result:

(i) Work done=40.29kJ (ii) Heat transfer=15.1088kJ (iii) Change of entropy=0.17187kJ/kgK

68. A domestic food freezer maintains a temperature of -15C. the ambient air is at 30C. If heat leads into the freezer at a continuous rate of 1.75kJ/s, what is the least power necessary to pump the heat out continuously? [Apr -03] Given data:



209

TL=15C=273-15=258K

TH=30C=273+30=303K Qs=1.75kW To find: Least power. W=? Solution:

Carnot COP=258

303 258

=5.733

L

H L

T

T T

Actual COP of refrigerator = sQ

W

For minimum power required to pump the heat, Carnot COP = Actual COP

5.733=1.75

W

W=0.305 kW Result: Least power necessary to pump heat, W=0.305kW

69. Find the change in entropy of 1 kg of ice which I heated from ink - 5C to 0C. It melts

into water at 0C. CPice = 2.093kJ/kgK. The pressure during heating is maintained at 1 atm constant. Latent heat of fusion of rice =334.96kJ/kg. [Apr-03] Given data: M=1kg

Tice=5C=-5+273=268 K

T0=0C=0+273=273K Cpice=2.093 kJ/kg K L=334.96 kJ/kg To find

Change in entropy, S=/ Solution: Heat released by 1kg of ice, Q=mCpice (T0-Tice)+mL



210

=1x 2.093 x (273-268)+1x 334.96 =345.425kJ Entropy change,

273

0268

( )

273 1 334.961 2.093

268 273

1.266 /

fusionice

pice

S S S

dT mLmC

T T

In

kJ K

Result:

Change in entropy ,S=1.266kJ/K 70. Three identical bodies of A, B and C constant heat capacity are at temperatures of 300,300 and 100K. A heat engine is operated between A and B and a heat pump working as refrigerator is operated between B and C. The heat pump is operated by the output of heat engine. If no work or heat supplied from outside , find the highest temperature to which any one of the body can be raised by the operation of heat engine or refrigerator. [Apr -03] Solution; A heat engine and a refrigerator are operated between temperature limits 300,300 and 100 k.

HP

W=QS1-QR1

QS1

Tf A

300k

C

300k

HE

B 300K

Tf



211

Let Tf be the final temperature of bodies A and B. Tf be the final temperature of body C and C be the final heat capacity of these three identical bodies.

(S)univ > 0 -------(2.43)

(S)univ = (S)A +(S)B +(S)c +(S)H.E +(S)H.P

But (S)H.E and (S)Ref=0

(S)univ=(S)A+(S)B+(S)C >0

From (2.43). (S)A+(S)B +(S)c > 0

' 2

1

2 '

0 300 100 300

0300 300 100

f f fp

f f

T T T TCIn CIn CIn S C In

T

T TCIn

For minimum value of Tf (S)univ=0

2 '

6

2 '

6

.0

3 10

.1

3 10

f f

f f

T TCIn

T TCIn In

Since In 1 =0, the above equation becomes

2 , 6

s1 1 2

2 1

'

f

9 10 (2.44)

also Q

(300 ) ( 100) ( ' 100)

[ ( )]

300 100 ' 300

T 700 2 (2.45)

f f

R R

f f f

f f f

f

T T

and Q Q

C T C T C T

Q mC T T

T T T

T

From (2.44) and (2.45), Tf

2 (200-2 x Tf)=9 x 106 Tf=300K Tf=700 -2 x Tf =700-2 x 300 =100 K Result: The maximum temperature can be raised for 100K body and 300 K of B.



212

71. A reversible heat engine operates between a source at 800C and a sink at 30C. What I the least rate of heat rejection per kW network output of the engine? Given data:

TH=800C =273+800=1073K

TL=30C =273 +30 = 303 K To find: Least rate of heat rejection / KW of network output, Q=? Solution;

, 1

303 =1-

1073

Lcarnot

H

TCarnot efficiency

T

=0.7176 Work Output, W =1kW W=Qs-QR 1-Qs-QR Qs=1+QR ----------(2.46) But efficiency of the engine

,

1

0.7176 1

0.7176 11

0.3935

R

s

R

s

R

R

R

Q

Q

Q

Q

Q

Q

Q kW

Result; Least rate of heat rejection , QR=0.3935 kW

72. One kg o ice at -5C is exposed to the atmosphere which is at 20C. the ice melts and comes into thermal equilibrium with the atmosphere (i) Determine the entropy increase of the turbine. (ii) what is the minimum amount of work necessary to convert the water back to

ice at - 5C? assume Cp for ice as 2.093 kJ/kg K and the latent heat o fusion of ice as 333.3 kJ/kg. [Nov -03] Given data:



213

T1=5C=273-5=268K

Ta=20C=273+20=293K Cpi=2.093 kJ/kg K L=333.3 kJ/kg To find:

1. Entropy increase of universe, (S)univ =? 2. Minimum amount of work, Wmin=? Solution:

Heat absorbed by air from atmosphere, (Q)= Heat absorbed in solid phase+ Latent heat+ eat absorbed in liquid phase

0 0

pw

( ) ( ) ( )

min 1 kg and C 4.187 / K

Q=1 2.093(0-(-5))+1 333.3+1 4.187(20-0)

=427.535 kJ

pi i pW aQ mC T T mL mC T T

Assu gm kJ kg

Entropy change of atm.

( )

425.535 = 146 /

293

atm

QS

T

KJ K

Entropy change of sysem

(S)system=(S)ice +(S)fusion +(S)liquid

273 293

268 273

273 1 333.3 2931 2.093 1 4.187

268 273 273

1.556 /

pi pw

dT mL dTmC mC

T T T

In In

KJ K

Entropy of universe,

(S)univ =(S)sys +(S)atm =1.556-1.46 =0.096 kJ/K If water is to be converted back to ice using reversible refrigerator, heat to be removed from water.

Q=427.535 kJ

Now, (S)sys =1.556kJ/K



214

But, (S)atm =atm

Q W

T

(S)ref = 0 as the refrigerant operates in a cycle.

(S)univ =(S)sys +(S)ref +(S)atm 0

-1.536+0+

min

445.908

445.908 -427.535

So, W 28.373 KJ

atm

Q WkJ

T

W

Result:

1. Entropy increase of universe, (S)univ =0.096 kJ/K 2. Minimum amount of work, Wmin =28.373 kJ

73. Air is closed vessel of fixed volume 0.15m3, exerts pressure of 12bar at 250C. If the vessel is cooled so that the pressure falls to 3.5 bar, determine the final temperature, heat transfer and change of entropy. [Apr -05] Given data: V1 = 0.15m3 P1 = 12bar = 1200 kN/m2 T1 = 2500C = 273 + 250 = 523 K p2 = 3.5bar = 350 kN/m2 To find:

1. Final temperature, T2 = ? 2. Heat transfer, Q = ?

3. Entropy change, S = ? Solution: From ideal gas equation

1 1

1

p Vm

RT

1200 0.151.2kg

0.287 523

For constant volume process,



215

22 1

1

pT T

p

3.5523 152.54K

12

Heat transfer, Q = mC(T2 – T1) = 1.2 x 1.005 (152.54 – 523) (Cp of air = 1.005 kJ/kgK) = -446.78 kJ (-ve sign indicates that the heat is rejected from the system)

2 2p

1 1

2

1

T PEntropy change, S=mC In mR In

T P

T mCuIn forV C

T

152.54 3.51.2 1.005 In 1.2 0.287In

523 12

1.06kJ /K

Result:

1. Final temperature, T2 = 152.54 K 2. Heat transfer, Q = -446.78kJ

3. Entropy change, S = -1.06 kJ/K 74. Two reversible heat engines A and B are arranged in series. A rejecting heat directly to B. Engine receives 200kJ at a temperature of 4210C from hot source, while engine B is in communication with a cold sink at a temperature of 4.4oC. If the work output of a A is twice that of B, find: i. The intermediate temperature between A and B, ii. The efficiency of each engine, and iii. The heat rejected to the cold sink. [ Apr-05] Given data: TH = 4210C = 421 + 273 = 694 K TL = 4.40C = 4.4 + 273 = 277.4 K Qs1 = 200kJ WA = WB To find: i. The intermediate temperature between A and B, T = ?

ii. The efficiency of each engine, A and B = ? iii. The heat rejected to the cold sink, QR2 = ?



216

Solution: Work output from engine A, WA = QS1 – QR1 = 200 – QR1

For reversible heat engine,

s1H

R1

R1

QT2.47

T Q

694 200So,

T Q

TH TL

QR1 = 0.288T --------(2.48) So, WA = 200 – 0.288 T But WB = 100 – 0.144T --- (2.49) (2WA = WB) and also WB = Qs2 – QR2 = 0.288T – QR2 ------ (2.50) (Qs1 – Qs2) Equating equations (2.49) and (2.50) 100 – 0.144T = 0.288T – QR2 QR2 = 0.432T -100 -------(2.51) Similarly, for reversible engine B,

WA=2WB

A

B

QR1

QS2

WB

QR2



217

2

2

1

2 1

2

2

s

L R

R

s R

R

R

QT

T Q

QTSo, Q Q

277.4 Q

0.288T 0.288T

Q 0.432T 100

T 416.42K

OR T = 143.420C So, QR1 = 0.288 x 416.42 = 119.93 kJ and QR2 = 0.432 x 416.42 – 100 = 79.89 kJ

1

1

R

A

S

QEmergency of engine A, 1

Q

119.93 1 40.04%

200

2

2

2 1

s

B

R

s R

QEfficiency of engine A, 1

Q

79.89 = 1- Q Q

119.93

= 33.39%

Result:

i. The intermediate temperature between A and B, T = 143.420C

ii. The efficiency of each engine, A = 40.04% and B = 33.39% The heat rejected to the cold sink, QR2 = 79.89kJ



218

UNIT – III

PROPERTIES OF PURE SUBSTANCES

PART – A

1. Define latent heat of ice. Total amount of heat added during conversion of ice of 00C into water of 00C. 2. What is pure substance? Pure substance is a substance which has a fixed chemical composition throughout its mass. Examples: Water, nitrogen, carbon dioxide, and helium. A pure substance does not have to be of a single chemical element or compound. A mixture of various chemical elements or compounds also called as a pure substance as long as the mixture is homogeneous. 3. What is saturation temperature and saturation pressure? At a given pressure, the temperature at which a liquid boils is called the saturation temperature. At a given temperature, the pressure at which the liquid boils is called the saturation pressure. It is also called as vapour pressure. 4. Define latent head of vaporization? (MU – Apr‟2000) The amount of heat added during heating of water from boiling point or dry saturated stage is called as latent heat of vaporization or enthalpy of vaporization or latent heat of steam. 5. Find the saturation temp and latent heat of vaporization of steam at Mpa? (MU – Oct‟ 2000) From steam table of 1 MPa or 10 bar Saturation temperature, Tsat = 179.880C Latent heat of vaporization, hfg = 2013.6 kJ/kg 6. Define the terms „ Boiling point‟ and „Melting point‟?(AU – Apr2005) Boiling point: It is the temperature at which the liquid starts to change its state from liquid to vapour. Melting point: It is the temperature at which the solid starts to change its state from solid to liquid. 7. What is meant by super heated steam and indicate its use? (MU – Apr‟2000) If the dry steam is further heated, the process is called superheating and the steam obtained is known as super heated steam.



219

Uses: 1. Superheated steam has more heat energy and more work can be obtained using it. 2. Thermal efficiency increases as the temperature of superheated steam is high. 3. Heat losses be to condensation of steam an cylinder wall is reduced.

8. Define: Sensible heat of water? The amount of heat required to raise the temperature of unit mass of water from 00C to the saturation temperature under a constant pressure. It is denoted by hf. 9. Define the term “Super heat enthalpy”. The heat supplied to the dry steam at saturation temperature to convert it into superheated steam at the temperature Tsip is called superheated or heat of superheat. 10. What are wet and dry steam? The steam which partially evaporated and having water particles in suspension is called wet steam. The steam which fully evaporated state and is not having any water particles is called dry steam. 11. State phase rule of pure substances? The number of independent variables associated with a multi component, multiphase system is given by the phase rule. It is also called as Gibbs phase rule. It is expressed by the equation as

n = C - + 2 Where, n = the number of independent variable. C = the number of components.

= the number of phase present in equilibrium. 12. Define dryness fraction of steam (MU – Apr ‟96, Apr „97, & Apr „ 2001) What is quality if steam? (AU – April 2005) It is defined as the ratio of the mass of the total steam actually present to the mass of the total steam.

Dryness fraction = mass of dry stream

mass of toal mixture



220

13. Explain the terms: Degree of super heat, Degree of subcooling? (AU – April 2005) Degree of superheat: It is the difference between superheated temperatures and saturated temperature at the same pressure. Degree of subcooling: It is the amount of which the water is cooled beyond the saturated temperature at the same pressure. 14. Define triple point and critical point for pure substance? (MU–OCT‟95, OCT ‟96 OCT‟97) Triple point: Triple point is the state at where all the three phases i.e. solid, liquid and vapour to exist in equilibrium. Critical point: It represents the highest pressure and temperature at which the liquid and vapour phases coexist in equilibrium. At the critical point the liquid and vapour phases are indistinguishable i.e. Liquid directly converted into vapour. 15. When saturation pressure increases, what happens to saturation temperature and freezing point?(MU – Apr‟2000) When saturation pressure increases then the saturation temperature is increasing and the freezing point decreasing. 16. Explain the change of phase of water in T-V diagram? 1 – 2 Solid stage 2 – 3 melting stage 3 – 4 liquid stage 4 – 5 Vapourising stage 5 – 6 Superheating stage

Fig T-V diagram of water

Volume

2 1 3

3

4 5

6

2770C

2730C

Tem

per

ature



221

17. Explain the process of steam generation and how the various stages on T-s diagram (MU – OCT ‟95) In the T-s diagram, the region left of the water exists as liquid. In the right of the dry steam line, the water exists as a superheated steam. In between water and dry steam line, the water exists as a set steam. Therefore, the dryness fraction lines are represented in these region. The value of various quantities can be directly read from the diagram. It can be noted from the figure that the water line and steam line are converging with the increase in temperature. At a particular point, the water is directly converted into dry steam without formation of wet steam. This point is called „Critical Point‟. Figure Temperature – Entropy diagram 18. Draw the h-s diagram for steam and show a throttling process on it (MU-OCT‟97)



222

19. Draw a skeleton p-v diagram for water and show an isotherm passing from compressed liquid state to superheated vapor state through vaporization processes. (AU-Nov 2003) Figure p-v diagram water 20. Draw the p-v-T surface of water and also indicate its salient features. (AU-April 2003) Figure The three important thermodynamic properties such as pressure (p), specific volume (v) and temperature(T) are pointed in three dimensional coordinate. This plot is called as p-v-T surface. Here T and v may be viewed as the independent variable ( the vertical surface). All the points on the surface represent equilibrium states. The single phase regions appear as curved surfaces on the p-v-T surface, and the two phase regions as surfaces perpendicular to the p-T plane. 21. Write the formula for calculating entropy change from saturated water to superheat steam condition. (MU-Apr ‟99)

sup

supEntropy of Superheated steam, S logg ps c

s

TS C

T

where Sg – entropy of dry steam



223

Tsup-Super heated temperature Ts-Saturated temperature Cps-Specific heat of super heated steam 22. Determine the condition of steam of 2 bar whose entropy is 6.27 KJ/kg.(MU-APR‟99) From steam Table at 2 bar sg=7.1268 KJ/Kg K Since entropy of given steam of pressure 2 bar is less than entropy of dry steam at that pressure, the steam is in wet condition. 23. Determine specific enthalpy and specific entropy of 1200 C saturated steam?(MU-Apr‟2000) From steam table at 1200C Specific enthalpy, hg=2706kJ/kg Specific entropy, sg =7.1293 kJ/kg K 24. Find the mass of 0.1 m3 of wet steam at a temperature of 1600 and 0.94 dry?(MU-Oct‟98) From steam table at 1600C Vg=0.30676 m3/kg Specific volume of wet steam = x,vg=0.94x0.30676 m3/kg =0.2884 m3/kg

Volume of given west steam 0.1 of steam, m=

volume of wet steam 0.2884

0.35

MassSpecific

M kg

25. One kg of steam at 10 bar has an enthalpy of 2500kJ/kg. Find its quality. H=2500kJ/kg

H=hr+x hfg

At 10 bar from steam tables Hf = 762.6kJ/kg;hfg=2013.6kJ/kg

2500=762.6+x+2013.6 2500 762.6

0.8622013.6

x

26. Determine whether water at the following states is a compressed liquid, a super heated vapor or a mixture of saturated water steam. (a) 18 Mpa, 0.003m3

/kg(b)1300C,200kPa. Case(a0 P=18Mpa V=0.003m3/kg From steam table, corresponding to 18Mpa, read



224

Vg=0.007497 m3/kg Since v=vg, the steam is in wet condition i.e mixture of liquid and steam. Case(b) T=1300C P=200kPa From steam table, corresponding to 200kpa, read Tsat=120.20C Since T< Tsat. the steam is in superheated condition. 27. What is meant by stem power cycles? Thermodynamic cycles which use steam as the working fluid is called steam power cycles. 28. Define the term efficiency ratio? The ratio of actual cycle efficiency to that of ideal cycle efficiency is termed efficiency ratio.

Efficiency ration = Actual cylcle efficiency

Ideal rankine efficiency

29. What is meant by isentropic efficiency? For an expansion process

Isentropic efficiency =Actual work done

Isentropic work done

For a compression process

Isentropic efficiency =Isentropic work done

Actual work done

30. Define specific steam consumption of an ideal Rankine cycle? It is defined as the mass flow of steam required per unit power output.

Specific steam consumption = steam flow in kg/hr

power in kW

31. What is meant by work ratio? What is the importance of work ration in vapour cycles? Work ratio is defined as the ratio of network transfer to the positive work transfer. Work ratio affects the actual cycle efficiency comparing two cycles with the same ideal efficiency, the cycle having smaller work ratio would have smaller actual efficiency. Higher the work ratio, the SSC is lower, resulting in smaller size plant for the given output.



225

32. Name the different process of Ranking Cycle on T –s diagram? Process 1 - 2: isentropic expansion 2 – 3: constant pressure and temperature heat rejection 3 - 4: water is pumped to boiler pressure 4 – 5: constant pressure heat addition in boiler up to saturation Temperature 5 – 1: content pressure and Temperature in boiler 33. What are the effects of condenser pressure on the Rankine cycle? By lowering the condenser pressure, we can increase the cycle efficiency. The main disadvantages lowering the back pressure increase the wetness of steam isentropic compression of a very wet vapour is very difficult. 34. A vapour cycle inherently has two advantages over gas power cycle? What are they? (MU – Oct‟ 95) (i) Isothermal heat transfer (evaporation and condensation) is possible in practice. (ii) The work ratio is high compared to the gas power cycles. 35. What are the limits of maximum and minimum temperatures in a steam power cycles? (MU – Oc ‟95) The limit of maximum temperature of steam is its critical temperatures i.e 374.150C. 36. Mention the improvement made to increase the ideal efficiency of Ranking Cycle. 1. Lowering the condenser pressure 2. Superheated steam is supplied to the turbine. 3. Increasing the boiler pressure to certain limit 4. Implementing reheat and regeneration in the cycle.



226

37. Sketch the flow diagram of Ranking cycle indicating the main components?(MU – Nov 2002) 38. Name the different component in steam power plant working on a Ranking cycle. (MU – Oct‟ 97) Boiler, Turbine, Cooling, Tower or Condenser and Pump 39. Discuss the effects of steam pressure and temperature at inlet to the turbine I a Rankine cycle. (MU-Oct‟95) At the higher pressure the heat rejection is less and this results in increase in efficiency. The cycle efficiency does not increase continuously with boiler pressure up to the critical pressure. The increase in pressure increases the wetness of the steam after expansion, which decreases the adiabatic efficiency, and reason is that latest heat decreases at high pressure. The increase the cost of boiler, and turbine and also it erodes the turbine blades.

Increase in temperature of steam supplied to turbine increases the work done by the turbine and also increases the net cycle efficiency. The efficiency of the superheated cycle continuously increase with pressure. Super heating reduces the specific steam consumption. It also increases the dryness fraction of steam at the end to the expansion. We can reduce the back pressure, which increases the work done. 40. What are the effects o condenser pressure on the Ranking Cycle? By lowering the condenser pressure, we can increase the cycle efficiency. The main disadvantages is lowering the backpressure increases the wetness of steam Isentropic compression has to deal with a non-homogeneous mixture of water and stem. Because of the large specific volume of liquid vapour is very difficult. 41. Why Carnot cycle cannot be realized in practice for vapour power cycles? The main difficulty to attain the cycle in practice is that isothermal condensation is topped before it reaches to saturated liquid condition. Therefore the compressor has to deal with a non-homogeneous mixture of water and stem. Because of the large specific volume of liquid vapour mixture before compression, the compressor size and work input have to be a large. The higher power requirement reduces the plant efficiency as well as work ratio.



227

Boile

r

42. Mention the improvements made to increase the ideal efficiency of Ranking Cycle (MU-Apr ‟98)

1. lowering the condenser pressure 2. Superheated steam is supplied to the turbine 3. Increasing the boiler pressure to certain limit 4. Implementing reheat and regeneration in the cycle.

43. Draw the block diagram of reheat cycle and indicate its use (MU-Oct‟98) Use: The main purpose of reheating is to increase the dryness fraction of steam at exhaust so that blade erosion due to water particles as well as frictional losses reduced. H.P. Turbine L.P.Turbine

` Reheater

Condenser

Feed pump

44. Why reheat cycle is not used for low boiler pressure ?(MU-AP‟95)

At the low reheat pressure the reheat cycle efficiency may be less that the Rankin cycle efficiency. Since the average temperature during heating will then be low.

45. What are the disadvantages of reheating?

The cost of the plant increases due to the reheater and its long connections. It also increases the condenser capacity due to increased dryness fraction.

46. List the advantages of reheat cycle(MU-Oct‟97)

1. Marginal increase in thermal efficiency 2. Increase in work done per kg of steam which results in reduced size of boiler and auxiliaries

for the same output. 3. We can prevent the turbine from erosion.



228

47. Draw the vapour power reheat cycle on T-s diagram and write ideal efficiency of that cycle interms of enthalpy (MU – Apr‟98)

1 3 6 2

5 4 s

1 2 3 4

1 4 3 2

( ) ( ),

( ) ( )reheat

f

h h h hEfficiency

h h h h

48. Draw the regenerative cycle on T-s diagram and show that its efficiency is equal to Carnot cycle efficiency on the diagram 4 1 T 3 3

4 2 2 s The regenerative cycle is represented by 1-2-3-4-1 whereas the Carnot cycle I by 1-2, -4” -1. The area under both the curve is equal and also the area under the heat addition and heat rejection curves for both the cycles are equal, therefore the Regenerative cycle efficiency is equal that of Ranking cycle.

T



229

49. Sketch the schematic diagram of closed and open type feed water heaters in actual regenerative cycle for steam (MU- Apr‟95) Turbine B Condensor Feed water heater Feed pump Feed pump Turbine Condenser Feed heater Feed heater Feed pump 50. What is the function of feed water heaters in the regenerative cycle with bleeding? (MU-Oct ‟99) The main function of feed water heater is to increase the temperature of feed water to the saturation temperature corresponding to the boiler pressure before it enters into the boiler. 51. When will be the efficiency of the regenerative cycle attains maximum? The temperature of the bled steam is approximately halfway between the extreme temperatures of the primary flow cycle.

Boiler

Boil

er



230

52. What are the advantages of bleeding? It increases the thermodynamic efficiency as the heat of the bled steam is not lost in the condenser but it utilized in feed heating. By bleeding the volume flow at the low pressure end is considerably reduced this reduces the design difficulties of blades and also condenser size is reduced. 53. What are the disadvantages of bleeding? Cost of the plant increases and the work done per kg of steam is reduced which results in higher boiler capacity for given output.



231

PART – B 1. Identify the state of H2O for the following conditions.

a) 5 bar, 300C b) 5 bar, 2990C c) 5 bar, 151.90C d) 250 kPa, 5 lit/kg e) 2 Mpa, 3000 kj/kg sp. Enthalpy f) 2 Mpa, 5.88 kj/kg sp. Enthalpy

a) Given p = 5 bar

T = 300C T given is less than T saturation. Hence it is compressed liquid. At 5 bar T saturation is 151.90 C which is greater than the given temperature hence it is compressed liquid.

b) Given temperature is 2000 C which is greater than the saturation temperature and hence it is a superheated vapor.

c) Given temperature is equal to the saturation and hence it may be saturated liquid, saturated vapor or vest steam.

d) For the given pressure (see Appendix A1 – 1 saturated steam – pressure table) V1 = 1.067 lit / kg Vg = 718.7kj/kgk Given sp. Entropy is greater than Sr and less than Sg therefore it should be a wet steam e) For the given pressure (See Appendix A1 – 1 saturated steam pressure table) hf = 1.067 lit /kg hg =7187 kj/kgk Given Sp. Entropy is greater than hg and hence it is a superheated steam. f) For given pressure S1 = 1.7275 kg/kgK Sg = 6.9405 kj/kgK

2. Complete the following table of Properties for 1 kg of H2O

S.l. No. P(bar) T(0C) V(m

2/kg) X h(kg/kg) S(kg/kg) U(kg/kg)

1 1.5 -- 0.5 -- -- -- --

2 2.0 2000C -- -- -- -- --

3 5.0 25 -- -- -- -- --

4 -- 120 -- -- -- 2.0 --

5 6.0 -- -- -- -- -- --



232

Case: 1 At 1.5 bar Vr = 0.001053m3/kg Vg=1.1593m3/kgK Since Vf <Vgiven@ 1.5 bar the given condition is wet steam. For wet steam other properties can be found only if the dryness fractions known. It can be found from the following equation. [email protected] bar

0.5 0.0010530.431

1.1593 0.001053

given g

g f

V VX

V V

x

For wet steam temperature will be the saturation temperature.

T=Tsat @ 1.5 bar =111.370C Specific enthalpy energy =hfxhg @ 1.5 bar =467.7-150x2226.5 =1426.7kJ/kg Specific entropy Sf [email protected] bar =1.4336+0.421(7.2233-1.4336) since Sfg - Sf =3.93 KJ/kg K Specific internal energy =h- pv where p should be in kPa =1426.7 kJ/kg K =1351.7 kj/kg Case2: Given P=2 bar T=h=pv When pressure and temperature are known, comparing the given temperature with the saturation temperature. The state can be identified. Since Sgiven> Tsaturation the state is superheated vapor referring appendix A1 other properties can be found V=1.0803 m2/mkg H=2870.5kJ/kg S=7.5066kJ/KJK U=h-pv

mailto:[email protected]




233

=2870.5-200x1.0803 =2654.44kJ/kg Case 3: Given P=2 bar T=h-pv At 5 bar Tsat=151.860C Since Tgiven< Tsaturation it is a sub cooled liquid

VVf @ Tgiven =0.001m2/kg H=hf@ Tgiven = 104.89 kj/kg

@

151.86 2731.8607 4.186

25 273

0.3760 /

@

s

f p given pt

g

f given

TS S C in

T

in

kj kgk

u u T

=h=pv

=104.89-5000.001 =104.39kj/kg Case 4: Given P=120bar T=2kj/kg At Tgiven Sf =1.5276kj/kg K Sg = 7.1296kj/kg K

0@120 @120given gf C g CS S S S

Hence it is a wet steam S=Sf+XSfg for wet steam

2 1.5276

7.1296 1.5276

0.0843

f

g f

S SX

S s

X

For wet steam pressure will be saturation pressure corresponding to given temperature. i.e, p=1.985 bar h=hf + xhfg

=503.71+0.0843 2188.5 =688.2kj/kg v=vf+xvfg

=0.0010603+0.0843 0.8919 =0.07625 v=vf+xvfg



234

=0.0010603+0.08430.9819 =0.07625m3/kg uu=h-pv

=688.2-198.50.07625 =673.kj/kg Case 5: Given P=60 bar V=0.0m3/kg Vgiven>Vg@60bar therefore it is super heated steam. Referring superheated table for 60 bar at 5400C specific volume is approximately equal to the given value therefore. H=3547.0kj/kg S-6.999kj/kg K U=h-pv

=3517-60000.06 =3157kj/kg S.l. No. P(bar) T(

0C) V(m

2/kg) X h(kg/kg) S(kg/kg) U(kg/kg)

1 1.5 111.37 0.5 0.431 1.4267 3.9289 1351.7

2 2.0 2000C 1.0803 Not applicable 2870.5 7.5066 2654.4

3 5.0 25 0.001 Not applicable 104.89 0.3674 104.39

4 1.985 120 0.07625 Not applicable 688.2 2.0 673.1

5 6.0 540 0.06 Not applicable 3517 6.999 3157

3. A closed vessel of 0.2m3 contains steam at 1 MP and temperature 2500C. If the vessel is cooled so the pressure falls to 350 kPa. Determine the final temperature, heat transfer and change of entropy during the process (Madras University) System :System Working fluid : Steam (ideal gas equation should Not be used) Process : constant volume cooling Known : p1 = 1MPa [10bar] : v1 = 0.2m3 : T1=2500C : p2 = 350 kPa Diagram:



235

To find : T2,Q,S

Analysis : Before solving for any parameter initial and final states must be identified

Initial State : Given P=10 bar T=2500C At 10 bar Tsat@10bar=179.010C Tgiven>Tsat@10bar Therefore it is a super heat stream

Final State : To fix the final state any two independent intensive properties should be known. Given p2 = 3.5 bar. Since the system is of fixed mass and process is of constant volume, final specific volume should be equal to the initial specific volume.

V2 = V1 =V@10bar 2500C =0.2327 M3/kg Knowing pressure and sp. Volume the state can be establish

3

@3.5

3

@3.5

f 2

0.001079

0.5243

Since V it is a wet steam

f bar

f bar

g

mV

kg

mV

kg

V V

a) Final Temperature: Since the final state is steam, final temperature is the saturation

temperature is the saturation temperature at 3.5 bar. Therefore final temperature is 138.880C

b) Heat transfer: Q=W+U from I law for a closed system since volume remains constant W=0 Q=U2-U1 =[H2-p2V2]-[H1-p1V1] =(H2-H1)-(p2V2-p1V1) =m[(h1-h2)-v(p2-p1)] Where

01 @10 250

2 @3.5 2 @3.5

2942.6 /

bar

f bar fg bar

h h

kj kgk

h h x h

tofind

2 @3.5

2

@3.5

0.2327-0.001079 =

0.5243

=0.4418

f bar

g bar

V Vx

V

Substituting x we get, H2 = 584.33+0.4418x2148.1

mailto:Tsat@10bar=179.010C



236

1

1

0.2 =

0.8595

=0.8595kg.

Vm

V

Substituting the values we get Q=0.8595(1533.3-2942.6)+0.2327(350-1000)] =-1341.26Kj Amount of heat removed from the system is 1341.26kj

c) Change in entropy]

S=S2-S1 =m(s2-s1) Where s1=S@10bar,2500C =6.9247 kj/kg K S2 = [email protected]+x2 [email protected]

=1.7275+0.4418x(6.9405-1.7275) =4.0306kj/kgK Substituting the value.

S=m(s1=s2) =0.8596 =0.8595[4.0306-6.9247] =-2.4875 kJ/K

Entropy decrease is due to the removal from the system. 4. Steam at 10 bar and 0.85 dry expands according to the law pv1.2 = c to a final pressure of 1 bar. Determine a) final volume b) final enthalpy System: closed Process: Polytropic expansion (pv1.2 =c) Working fluid: Steam (Ideal Gad education could not be used) Known: p1=10bar n=1.2 X1=0.85 p2 = 1bar

mailto:[email protected]+x2




237

P-v diagram To find: a) Final volume b) Final enthalpy Analysis: a) P2v2

1.2 = P1v11.2

Where Vt = V6+Vfg1 =Vf1+X1Vg1=0.001127+0.85x0.19444 =0.1664m3/kg b) Final enthalpy can be found only if the final state is completely fixed: P2=1 bar Vg2=1.1336m3/kg At 1 bar Vf2 =0.001043m3/kg Since Vf2 <V2<Vg2 it is wet steam



238

2

1.0336 0.001043

1.6940x

=0.669 Hence h2 = hf2+x2hfg2 =417.46 + 0.669 x 2258 1927.2 kj/kg 5. Steam at 15 bar and 3000C is expanded hyperbolically to a pressure of 5 bars. Calculate change in internal energy and work done during the process. System :Closed Process :Hyperbolic(pV=c) Working fluid: Steam (ideal gas eqn. should not be used ) Known: P1 =15bar T1 = 3000C P2 5 bar

To find :a) pu – v diagram :b)W

Analysis : u=u2-u1 =(h2 –p2v2)-(h1-P1v1 =(h2 – h1)-(p2v2-p1v1) (since the processor is hyperbolic]



239

To find the enthalpy values states are to be identified first State1 :Given P=15bar T=3000C At 15 bar Ts = 198.330c Since Tgiven>Tsaturation

h1=3036kj/kg v1=0.17045 m3kg State2: Given p2 = 5 bar To fix the state two independent intensive properties should be known from the process eq. =p1v1=p2v2

1 12

2

3

P VV

P

15 0.17045

5

0.51135m /Kg

knowing p2 & v2 the state can be established

3

f

mAt 5 bar v 0.001093 and

kg

Vg=0.3748m3/kg Since V2 is greater than Vg@5bar is a spearhead steam P2 = 5 bar and V2 =0.51135m3/kg At 5 bar and 300oC V=0.5226m3/kg

Therefore, for 0.511352m

kg

300 275 0.5226 0.51135T 300

0.5226 0.4985

=288oC



240

o o

o

@5bar,300 C @5bar,175 C

2 @5bar,300 C

h hh h 300 288

300 275

3063 30123063 12

25

3038.5kj / kg

Since steam is behaving like an ideal gas in the superheated vapour region, the change in internal energy is very small. c) work done

2

1

21 v

1

pdv

vp v 1n

v

(or)

11 v

2

pd) = p v 1n

p

15 =1500 0.17045 1n

5

kj =280.89

kg

6. 150 kg / s steam at 25 bars and 300oC expands misanthropically in a steam turbine to 0.3 bars. Determine the power output of the turbine. System : Open Process : Study flow – isentropic Working fluid : Steam Known : p1=25bars : T1=300oC : p2 = 0.3bar : m= 150kg/s



241

To find : Power output

Analysis :Q-Wt=m[h+ke+pe] Given the flow is isentropic. Assuming the changes in PE & KE negligible W1=m[h1-h2] State1 : Given P=25bars T=300oC Since Tgiven>Tsat@25bar it is a superheated steam and H1=3008.0kj/kg S1=6.6424kj/kg State2: To fix the state at least two independent bar. From process equation S1 = S2=6.6424 kj/kgK and hence the other properties can be found. Sf@30kpe=0.9441kj/kgK Sg@30kpe since Sf<S2 <Sg, it is a wet steam. To find X2

g3

2 ff2

f

S SX

S

6.6424 0.9441

7.7672 0.9441

2239.6

Substituting these values we get Wt=150(300.0-2239.6) =115.256Mw.

mailto:Sf@30kpe=0.9441kj/kgK



242

7. Steam is throttled from 15 bars to 1 bar, Temperature after throttling is 120oC. Find the initial state if steam. System : Open Process : Throttling process (h=constant) Working fluid : Steam Known : P1=15bars P2=1bar T1=125oC. To find : Initial state Analysis : To fix the initial state two properties must be known Given P1 = 15 bars From the process equation. H2=h1 [email protected] =2726.3

For state 1 P1=15bars H12726.3kJ/kg At 15 bars hn<hg1 it is wet steam




243

n f11

fg1

h hX

h

2726.3844.85

1946.6

0.966

The initial state is wet steam of dry fraction 0.966 8. Steam at 30 bar and 350oC is expanded in a non flow isothermal process to a pressure of 1 bar. The temperature and pressure of the surroundings are 25oC and 100 kpa respectively. Determine the maximum work that can be obtained from this process per kg of steam. Also find the maximum useful work and change in availability. System : Closed Process : Isothermal expansion Working fluid : Steam (ideal gas equ. Should not be used) Known : p1=30bars P2=1bar T1=623 K=T2 T0=298 K P0=100kPa Diagrams: To find :a) Wmaxb) Wmax useful

c) Analysis :a) Wmax=(u1=u2)-T0(s1-s2) To get the property values the states are the be first identified. State: 1 known p1=30 bars T1=350oC At 30 bar Tsaturation =233.9oC



244

Since the given temperature is greater than Tsaturation it is a superheated steam. From steam table V1=0.0090526m3/KG H1=3117.5KJ/Kg S1=6.9246kJ/kgK

u1=h1 =h1-p1v1 =3117.5-3000 x 0.090526 =2845.9kJ/kg State:2 P2=1bar T2=350oC At 1 bar saturation temperature is 99.63oC. Since the given temperature is greater than the saturation temperature it is a superheated steam. From steam table H2=3175.6kJ/kg V2=2.8708m3/kg S2=8.3858kJ/kgK U2=h2-p2v2 =3175.6-100x2.8708 =2888.52kJ/kg Wmax=(2845-2888.52)-298(6.9246-8.3858) =392.82kJ/kg

b) Wmax,useful=(u1-u2)+p0(v1-v2)-To(S1-S2)

=Wmax+po=(V1-V2) =392.82+100(0.090526-2.8708) =114.79KJ/kg

c) Change in availability

=Wmax, useful =114.79kJ/kg

Availability decreases by 114.79kJ/kg.



245

9. Ammonia enters a throttling device at 30oC as saturated liquid. Temperature after throttling is found to be -6oC. What is the condition of ammonia at the exit. System :Open Working fluid :Ammonia Process :Throttling – isenthalpic Known :t1=30oC and saturated liquid T2=-6oC To find : Condition at the exit Diagram Analysis : To fix the final state at least two independent intensive properties must be known, Given t2=-6oC Also H2=h1 Since the process is throttling. H1 can be found from Appendix A2 H1=h1@30oC =322.9kJ/kg =h2 At-6oC Hr=153.5 and Hg=1436.8kJ/kg Since hf@-6

oC<hg@-6

oc the final state is wet.

H2=h12+x2hfg =hf@-6

0C+X2hg@-6

oC



246

1

1 f12

fg1

1 f

fg

h hX

h

h h

h

0.132

The state after throttling is a mixture of dryness fraction 0.132. 10. In refrigeration plant using R134a, vapours enters the condenser at 0.15 MPa, 60oC and leaves as saturated liquid at the same pressure. What is the heart removed per keg of vapour condensed? System : Open system. Working fluid: R134a Process : Condensation at constant pressure Known : 1)p1=0.15MPa 2)t1=6oC 3)p2 = 0.15MPa To find : Heat removed per kg of vapour condensed Diagram: Analysis : By apply SFEE, neglecting changes in kinetic and potential energies.

. . .

Q W m[ h]

On unit mass

Q-W-h Since there is no work interaction Q=h2-h H1=f(p1,t1)from appendix A 4,3 =454.9kJ/kg h2=f(p2)from appendix A4,3



247

=177.3KJ/kg Therefore, q=177.3-454.9 =-277.6KJ/kg 11. One kilogram of water at 300oC expands against a piston in a cylinder until it reaches ambient pressure, 100kPa, at which point the water has a quality of 90%. It may be assumed that the expansion is reversible and adiabatic. What was the initial pressure in the cylinder and how much work is done by the water? System :Closed Process :Reversible adiabatic Known :1) Initial temperature t1=300oC 2) Final Pressure p2=100kPa 3) Final quality x2=0.9 To find :1) Initial pressure (p1) 2) Work done(1W2) Diagram: 12. The neon has a molecular weight of 20.183 and its critical temperature, pressure and

volume are 44.5K,2.73 MPa and 0.0416 3m

kg.mol. Reading from a compressibility chart for a

reduced pressure temperature and reduced volume? System :neon gas at thermo dynamic stage Known :M=20.183 Tcr =44.5K Per=2.73Mpa Vcr=0.0416m3/kg mol Tr=1.3 Pr=2 Z=0.7 To find: a) V b)p c)t d)vr Analysis: since (a) cannot be solved before solving b and C consider (b)



248

u

3 3

r

cr

zRT(a)

P

zR T

pM

0.7 8.314 57.85

5460 20.183

3.05 10 m / gk

b)gives p 2

p 2 p

2 2.73Mpa

5.46Mpa

r

cr

3

r 3

cr

c)given T 1.3

T1.3 T 1.3 44.5

T

57.85K

V 3.05 10d)V

V 2.06 10

1.48

13. It is required to store 32kg of oxygen at 300K and 10MPa. Determine the volume of the tank using ideal gas equation and generalized compressibility char Tcr=154.6K and pcr=5.04MPa. System : oxygen existing in a thermodynamic state M=32.183 Known =t=300K P=10Mpa=10x103KPa To find: volume Analysis =a) using pV=mRT



249

3

3

mRTV

p

8.31432

34

10 10

0.249m

b) using generalized compressibility chart

r 3

cr

r

cr

fromchart

3

3

p 10 10p

p 5.04 10

1.984

TT

T

300 = 1.94

154.6

Z 0.95

0.95 m R TV

p

8.1340.95 32 300

32 =

10 10

V 0.236m

14. Find the pressure of nitrogen gas at T=175 K and V=0.00375 m3/kg. Using.

a) Ideal gas equation b) Vander walls equation of state

Take molecular weight of nitrogen of state

critical

critical

T 126.2K

P 3.39Mpa

system: nitrogen existing in a thermodynamic state known: T= 175k v=0.00375m3/kg Tcr=126.2K Per=3.39MPa



250

To find: pressure

Analysis: a) u sin g p=RT

2

8.314175

Rt 28.013p

0.00375

1385KPa =13.85MPa

ab)Using p v b RT

v

2

2 2 2 2

3

cr

Rt ap

v b v

8.314whereR 0.297KJ/kgK

28.013

27R T 27 0.297 126.2a

64P 64 3.39 10

0.175

cr

3

cr

3

3 2

R T 0.297 126.2b

8 P 8 3.39 10

1.38 10

0.297 175 0.175P

0.00375 1.38 10 0.00375

4986kpa =9.486MPa

15. The normal boiling point of cf3 Br is 58oC critical temperature and pressure and molecular mass Tc=340.9 k=405 Mpa, M=148.9 using R-K equation state, calculate the specific volume of substance these conditions. System: CF3Br is a thermo0 dynamics state NBP=-58.7oC NBP=58.7oC Known: Tc=340,9J Pc=405 Moa M=148.9 To find: Specific value Analysis: the R-K equation can be modified and expressed as:



251

12

0

52 2

c

2 2

6

c

c

3

6

RT a bb

p bPT

R 8314.4where R= 55.84J/kgK

M 148.9

R Ta 0.42748

P

55.84 340.90.42748 706.186

4.05 10

RTb 0.08664

P

55.84 340.90.08664 0.4072 10

4.05 10

T 58.7 273 214.3K

Analysis: From Energy Balance

Mair C(T1-T2)=msteam(hb-ha)

From steam table

Ha=hf@5bar

KJ639.90

kg

ha=hg@5bar

KJ2747.48

kg

Substituting numerical values we get

air

200 2747.48 639.90m

1.005 600 300

1398kg/hr

b) increase in available Energy per unit time



252

0 air air

steam steam b a

steam g f @5bar

t S S

where S m S S

=m S S

2006.8198 1.8602

3600

kw0.275

k

substituting the numerical values We get

UAE = 298[0.275-0.164]

=32.98kw.

16. CALCULATION OF WORK DONE AND HEAT TRANSFER IN FLOW PROCESSES

Final form of equations for various steady flow processes of systems are given in the tabular form.

Systems Work done Heat transfer

Boiler W=0 Q=h2-h1

Turbine W=h2-h1 Q=0

Condenser W=0 Q=h1-h2

Nozzle W=0 Q=0

IMPORTANT FORMULA E For properties of steam:

Properties Wet steam Dry steam Superheated steam

Enthalpy (h) in kJ/kg hwet=hf+xhfg hg=hr+hfg hsup=hg+CP(Tsup-Ts)

Specific volume (v) in m3/kg Vwet=xvg Vdry=Vg g sup

sup

s

T

T

Density () in kg/m3 wet

wet

1

v

g

g

1

V sup

sup

1

v

Work done (W) in kJ/kg Wwet=100Pvwet Wg=100pvg Wsup=100pvsup

Internal energy in UWet=hwet-Wwet Ug=hg-Wg Usup=hsup-Wsup

Specific entropy (s) in KJ/kg K Swet=sr+XSfg Sdry = Sf+Sfg Ssup=Sg+Cpsloge

sup

s

T

T



253

For Non – flow processes:

Processes Work done Heat transfer

Constant volume W=0 Q=(h2-h1)-(p2Vsup2-p1x1vgf)

Constant pressure

W=p(v2-v1) Q=h2-h1

Constant temperature

W=p(v2-v1) Q= h2-h1

Hyperbolic 2

1 1 e

1

VW p v log

V

2

1 1 e 2 1

1

VQ p v log h h

V

Isentropic W=u1-u2 Q=0

Polytropic 1 1 2 2p v p v

Wn 1

1 1 2 2 2 1

nQ p v p v h h

n 1

For Flow Processes:

Systems Work done Heat transfer

Boiler W=0 Q=h2-h1

Turbine W=h2-h1 Q=0

Condenser W=0 Q=h1-h2

Nozzle W=0 Q=0

17. Determine the state of steam at a pressure of 12 bars with its specific volume of 0.175m3/kg? Given data: P=12bar V=0.175m3/kg To find: State of steam (whether dry, wet or super heated) Solution: From steam table of pressure scale at 12 bar, specific volume of dry steam, vg=0.16321m=3/kg, Ts=188oC. Since v>vg, the steam is in super heated condition. Note: V>Vg Super heated V=vg dry steam V<Vg Wet steam



254

We know that

sup

sup g

s

sup

sup s

g

o

sup

T

T

0.175T T 188

0.16321

T 201.58 C

The steam is super heated to 201.58oC

Result: The state of steam is super heated 18. Determine the condition of steam at a temperature of 220oC and enthalpy of 2750kJ/kg. Given data: T=220oC H=2750KJ/Kg To find: State of steam Solution: From steam Table of temperature scale at 220oC Hf=943.7kJ/kg Hfg=1856.2kJ/kg Hg=2799.9kJ/kg Since h<hg, the steam is in wet condition Note: h>hg Super heated steam h=hg dry steam h<hg wet steam

hwet=hf+xhfg

wet f

fg

h hx

h

2750 943.70.973

1856.2



255

Result: The given steam is wet steam of 0.973 dry. 19. Find the specific volume and enthalpy of steam at 9 bar when the condition of steam is (a) wet with dryness fraction 0.98(b) dry saturated and (c) super heated, the temperature of steam 240oC. Given data: P=9bar

(i) x=0.95(ii)x=1 (ii) super heated Tsup=240oC

Solution: From steam table at 9 bar Ts=175.4oC Vg=0.21501m3/kg Hr=742.6kJ/kg Hfg=2029.5kJ/kg Hg=2772.1KJ/kg (a) Wet steam, x=0.95

wet=xg=0.95 0.21501=0.204m3/kg

hwet=hf+xhfg = 742.6+0.952029.5 hwet=2670.625kJ/kg b) Dry steam vg=0.21501m3/kg hg=2772.1KJ/kg c) Super heated steam Tsup= 240oC

sup

sup g

s

3

sup

T

T

2400.21501

175.4

v 0.294m /kg

Corresponding to p=9 bar and T =240oC from the superheated enthalpy table Hsup=2923.29KJ/kg



256

Results: (a) Wet steam v=0.204m3.kg h=2.670.625KJ/kg. (b) Dry saturated v= 0.21501m3/kg h=2772.1kJ/kg (c) Super heated v=0.294m3/kg h=2923.29kJ/kg 20. Find the internal energy of unit mass of steam at a pressure of 7 bar (i) when its quality is 0.8(ii) when its dry and saturated and (iii) super heated, the degree of superheat being 65oC. The specific heat of super heated steam at constant pressure is 2.277kJ/kg. Given data: P=7bar

(i) X=0.80 (ii) Dry saturated and (iii) Superheated, (Tsup-Ts)=65oC

Find : Internal energy Solution: From steam table at 7 bar Ts=164.9oC; Vg=0.27288m3/kg Hr=697.1kJ/kg; hfg=2064.9kJ/kg Hg=2762kJ/kg; sf=1.992kJ/kgK Sfg=4.713kJ/kgK sg=6.705kJ/kgK (i) When x=0.8

hwet=hf+xhfg=697.1+0.82064.9 hwet=2349.02kJ/kg

wet=xg=0.80.27288

wet=0.2183m3/kg Work done, W=100p vwet

W=10070.2183



257

W=152..8kJ/kg Internal energy, U=h-W U=2349.02-152.8 Uwet=2196.22kJ/kg ii) When it is dry and saturated hg=2762kJ/kg

Wdry =100pg

=10070.27288 =191kJ/kg ug=h-W =2762-191 =2570.984kJ/kg iii) When super heated hsup=hg+Cp(Tsup-Tsat) =2762+2.277(65) hsup=2910jH.kg Wsup=100pVsup

=10070.27288229.9 273

164.9 273

sup s

sup s

o

sup

T T 65

T T 65

=164.9+65

T 229.9 C

=219.36kJ/kg Usup=h=W =2910-219.36 Usup=2632.189kJ/kg Results:

(i) When steam is wet, uwet =2196.2kJ/kg. (ii) When stem is dry, Udry=2570.984kJ/kg (iii) When super heated, Usup=2690.6kJ/kg



258

21. Determine the condition of stem whether it is we, dry or super heated for the following cases by using stem tables only.

(i) Steam has a pressure of 10 bar and specific volume 0.22m3/kg (ii) Steam has a pressure 15 bar and temperature 225oC (iii) Steam has a temperature 200oC and enthalpy 2790.9kJ/kg. (iv) Steam has temperature of 120oC and entropy 7kJ/kg K.

Given Data:

(i) p=10bar v=0.22m3/kg (ii) p=15bar T=225oC. (iii) T=200oC h=2790.9kJ/kg (iv) T=120oC s=kJ/kgK

To find: Conditions of steam Solution:

(i) p=1bar;v=0.22m3/kg From S.T. at 10bar, vg the steam is super heated. (ii) P=15bar;t=225oC From S.T. at 15 bar, Ts=198.3oC since Ts<Ts the stem is super heated

. (iii) T=200oC;h=2790.9kJ/kg From S.T. at 200oC,hg=2790.9kJ/kg since h=hg the steam is dry

saturated. (iv) T=120oC; s=7kJ/kgK From S.T. at 120oC sg=7.129kJK since s<sg, the steam is in wet

condition. Results:

(i) Stem is super heated (ii) Steam is super heated (iii) Steam is dry saturated and (iv) Steam is wet.

22. One kg of steam contains 1/3 liquid and 2/3 vapor by volume. The temperature of the steam is 150oC. Find the quality, specific enthalpy of mixture. Given data:

f

g

o

3

m 1kg

1v volume of vessel

3

2v volume of vessel,and

3

T 150 C



259

To find: Quality (x), specific volume, v and specific enthalpy, h Solution: From steam Table at 150oC P=4.76bar;vf=0.001091m3/kg; vg=0.39245m3/kg Hf=632.1kJ/kg; hfg=2113.3kJ/kg.

f

g

f w f

g s g

w

s

w

s

113

2 23

m 1

m 2

m 0.001091 1

m 0.39245 2

m179.85

m

Dryness fraction, x

s

w s w

s

m 1x

m m m1

m

1x 0.005529

179.85 1

Wetness fraction Y=1-x1-0.00552=0.99447 Volume of mixture

=xg+yf

=0.0055290.39245+0.994470.001091 =0.00325m3/kg Enthalpy of mixture



260

fg fh xh yh

0.005529 2113.3 0.99447 632.1

h 640.289KJ/kg

Results:

(i) Quality of steam, x=0.005529 (ii) Specific volume, v=0.00325m3/kg (iii) Enthalpy of mixture, h=640.289kJ/kg.

23. Steam at a bar and 0.7 dry expands of constant volume until 5.5 bar. Find the final condition of steam and the heat absorbed by 1kg of steam. Given data: M = 1kg P1 = 4bar X1 = 0.7dry Process: constant volume P2 =5.5bar Find: Final condition of steam and Q Solution: From steam table at 4 bar Ts-143.6oC; hf1=604.7KJ/kg Hfg1=2132.9kJ/kg; vg1=0.4622m3/kg H1=hf1+xhfg1

=604.7+0.72132.9 h1=hwet=2097.73kJ/kg

1=xg

=0.70.4622

1=0.3235m3/kg From Steam Table at 5.5 bar

g2=0.34298m3/kg; hf2=655.5kJ/kg hfg2=2096.1kJ/kg Volume is constant during the process



261

V2=V1=0.3235m3 Since v2<vg, the steam is we:

2

2 f 2 2 g2

2

0.3235x 0.943dry

0.34298

h h x h

655.5 0.943 2096.1

h 2632.79kJ/kg

For constant volume process, W=0 By first law of thermodynamics,

Q=U

Heat transfer, Q=U2-U1 =(h2-p2v2)-(h1-p1v1) =(h2-h1)-v2(p2-p1) [ v1=v2] =(2632.79-2097.73)-03235(550-400) =486.53kJ Result: Final condition of steam is wet at 0.973 dry Heat absorbed, Q=486.53kJ/kg 24. A closed vessel of 0.2 m3 contains steam at 1 MPa and temperature 250oX. If the vessel is cooled so that pressure falls to 350 kPa. Determine the final temperature, heat transfer, and change of entropy during the process. (MU-Nov‟94&Oct‟95) Given data: V1=02m3 P1=1MPa=10bar T1=250oC P2=350kPa=3.5bar To find:

T2,Q and S. Solution: From steam tables at 10bar, Ts=179.9oC Since T1>Ts, the state would be in the super heated region from steam tables (in super heated region)



262

At 10bar and 250oC Vsup=0.2328m3/kg From steam tables at 3.5bar Vf2=0.001079m3/kg Vg2=0.52397m3/kg T2 =139.9oC Mass of the steam, m

1

sup

v 0.2m

v 0.2328

=0.859kg

Volume of vapour at final stage, V2 Vg2=mx vg2

=0.859 0.524 Vg2=0.45m3 Since the vessel is closed, the cooling process is undergone by constant volume. Therefore, V1=V2 The vapour must be wet in its final stage because Vg2>V1 The dryness fraction, x

1

g2

v 0.2m 0.444

v 0.45

The final temperature of steam is equal to the saturation temperature at 3.5bar. T2 = 139.9oC Final enthalpy of wet steam h2=hf2+hfg2 From steam table at 3.5bar, Hf2=5843kJ/kg Hg2=2147.3kJ/kg

H2=5843+0.4442147.3 H2=1537.7kJ/kg From super heated steam table at 10bar and 250oC H1=2943kJ/kg



263

Heat transfer, Q=m(h2-h2) =0.859(2943-1537.7) =1207.15KJ From S.T. at 10 bar ad 250oC S1=6.926kJ/kg K From S.T. at 3.5 bar Sf2=1.727kJ/kgK Sfg2=5.212kJ/kgK Final entropy, S2=s12+xsfg2

S2=1.727+0.4445.212 S2=4.041kj/kgK Change in Entropy

s=m(s2-s1)=0.859(4.041-6.926) =-249kj/kgK Result: Final Temperature, T2=129.9oC Heat transfer, Q=1207.15kJ/kg

Change in entropy, s=-2.48kJ/kgK 25. Steam is contained in a closed vessel of 30 litres capacity at a pressure of 10 bar with dryness fraction 0.95. Calculate its internal energy. Due to loss by radiation, the pressure of steam falls to 7 bar, calculate the total loss of heat and final quality of team. (MU-Apr;96) Given data: V1=30 lits P1=10bar x1=0.95 P2= 7bar To find: U1,Q and x2 Solution: From steam tables, corresponding to 10 bar, read Vghf & hfg



264

Hf1=726.6kJ/kg Hg2=2013.6kJ/kg Vg1=0.1932m3/kg V1=x1vgt =0.95(0.1932)=0.1847m3/kg h1=hf1+hg1

=7626+0.952013.6 =2675.52kJ/kg

31

1

VMass of steam m= 1000litres 1m

V

30

1000 0.1846

0.163kg

Internal energy, U1=mu1 =m(h1-p1v1)

=0.163(2675.52-10000.1847) U1=406.02kj For closed vessel V1 = v2 From steam tables, corresponding to 7 bar, read Vg2 =0.27268m3/kg

Now, vg2>v2, the steam is in wet condition V2=x2vg2

0.1846=x20.27268 x2=0.677 h2=hf2+x2hg2

=697.1+0.6772064.9 =2095kJ/kg For constant volume process, W=0 By first law of thermodynamics

Q=U

Heat transfer, Q=U2-U1 =m(h2-p2v2)-(h1-p1v1) =m[(h2-h1)-v2(p2-p1)] [ v1=v2] =0.163[(2095-2675.52)-0.1846(1000-700)] =-103.65kJ



265

Result:

1. Internal energy, U1=406.02kJ 2. Final quality of steam, x2=0.677 3. Total Heat loss, Q=-103.65kJ

26. Steam initially at 400kPa and 0.6 dry is heated in a rigid vessel of 0.1m2 volume. The final condition is 600kPa. Find the amount of heat added and mass of steam. (MU-Apr‟97) Given data: V1= 400kPa x1=0.6 v1=0.1m3 P2= 600kPa To find: Q and m Solution: From steam tables, corresponding to 4 bar, read vg,hf & hfg Hf1=604.7kJ/kg Hg2=2132.9kJ/kg Vg1=0.4622m3/kg Specific volume, V1=x1vgt

V1=0.60.4622 =0.277m3/kg h1=hf1+x1hg1

=604.7+0.62132.9 =1884.44kJ/kg Mass of steam,

1

1

vm

v

0.1

0.278

0.36kg

or rigid vessel, V1=v2

v2=0.277m3/kg



266

From steam tables, corresponding to 6 bar, read H2=670.4kJ/kg Hfg2=2085.1kJ/kg Vg2=0.31546m3/kg

Here, vg2>v2, The steam is in wet condition V2=x2vg2

0.277=x20.31546 x2=0.88

h2=hf2+x2hg2

=670.4+0.882085.1 =2505.2kJ/kg BY first law of thermodynamics

Q=W+U For rigid vessel, v1=v2, So, work transfer W=0

Q=U Total heat transfer =m[U2-U1] =m[(h2-p2v2)-(h1-p1v1)] =m[(h2-h1)-v1(p2-p1)] [ v1=v2]

Total heat transfer. Q=0.36[(2505-1884.44)-0.277(600-400)] =-203.46kJ Result:

1. Heat transfer, Q=203.46kJ 2. Mass of steam, m=0.36kg

27. 2kg if steam initially at 5 bar and 0.6 dry is heated at constant pressure until the temperature becomes 350oC. Find the change in entropy and internal energy. (MU-Oct‟97) Given data: m= 2kg p1=5bar x1=0.6 T2= 350oC To find:

S and U



267

Solution: From steam tables, corresponding to 5 bar, read Vg1 hf1 hfg sf1 & sfg Hf1 =640.1kJ/kg; hfg1=2017.4lJ/kg Sf1=1.86kJ/kgK; Sfg1=4.959kJ/kgK Vg1=0.37466m3/kg Specific volume v1=x1vg1

=0.6 0.37466 =0.225 m3/kg Enthalpy,h1=hf1+x1hfg1

=640.1+0.62107.4 =1904.54kJ/kg Entropy,s1=sf1+x1sfg1

=1.86+0.64.959 =4.835kJ/kgK From steam tables, corresponding to 5 bar & 350oC read V2=0.5701m3/kg H2=3168.1kJ/kg S2=7.634kJ/kgK Change in entropy,

S=m(s2-s1) =2(7,634-4.835) =5.598KJ/k Change in internal energy,

U=m(U2-U1) =m(h2-h1)-(p2v2-p1v1) =m[(h2-h1)-p2(v2-v1)] [ p1=p2] =2[(3168.1-1904.54)-500(0.5701-0.225)] =-2182.02kJ Result:

1. Change in entropy, S=5.598kJ/K

2. change in internal energy, U=2182.02kJ



268

28. 2.5kg of steam is heated at constant pressure of 250kPa and 100oc until temperature is 250oC. Find the amount of heat added and change in entropy. Given data: M=2.5kg Process: constant pressure P=250kPa T1=100oC T2=250oC To find:

Q and S Solution: From Mollier diagram at p-250KPa=2.5bar and 100oC h1=2700kJ/kg s1=7.04kJ/kgK Along p=2.5bar line at 250oC H2=2950kJ/kg

S2=7.65kJ/kg K Heat added Q=m(h2-h1)=2.5(2950-2700) =625kJ

change in entropy, S=m(s2-s1) =2.5(7.65-7.04)=1.525kJ/K 29. One kg of steam at a pressure of 700kPa and dry is heated at constant pressure until it becomes saturated. Determine change in internal energy work done. Given data: M=1kg



269

P=700kPa=7bar X1=0.6 Process: Constant pressure process X2=1 To find:

U and W Solution: From steam Table at 7 bar Hf1 =697.1kJ/kg; hfg1=2064.9lJ/kg Hg1=1.86kJ/kgK; vg1=0.27288m3/kg V1=mkvg1

=10.60.27288 =0.164m3 h1=m(hf1+x1hg1)

=1(697.1+0.62064.9)=1936.04kJ Since the final condition of steam is dry and also pressure remains same at 7 bar. V2=vgf=0.27288m3/kg

V2=mVgf=10.27288=0.27288m3

H2=mhg2=12762=2762kJ/kg Change in internal energy

U=u2-u1 =(h2-h1)-(p2v2-p1v1) =[(h2-h1)-p1(v2-v1)] [ p1=p2] =(2762-1936.04)700(0.273-0.164) =749.66kJ Work done, W=p(V2-V1) =700(0.273-0.164) =76.3kJ Result:

Change in internal energy, u=749.66kJ Work done, W=76.3kJ



270

30. Steam at 15 bar and 300oC is expanded hyperbolically to a pressure of 5 bar. Calculate change in internal energy and work done during the process.(MU-Apr‟99) Given data: P1=15bar T1=300oC=573K P2=5bar To find:

U and W Solution: Corresponding to 15 bar & 300oC. read H1=3038.9kJ/kg V1=0.1697m3/kg Pv=C for hyperbolic process P1v1=p2v2

31 12

2

p v 15 0.1697v 0.5091m /kg

p 5

Corresponding to 5 bar, read Vg2=0.37466m3/kg Since v2<vg2, the steam is in super heated condition,

For hyperbolic process T2=T1. From superheated steam tables, corresponding to 5 bar and 300oC, read H2=3064.8kJ/k By first law of thermo dynamics

Q=W+U

For hyperbolic process U=0, Q=W From hyperbolic relation

21 1

1

v 0.5091Q p v In 1500 0.1697 In

V 0.1697

279.65KJ/kg



271

Result:

1. Change in internal energy, U=0 2. Work done, W=279.65kJ/kg

31. A mass of 0.9 kg of steam initially at a pressure 1.5MPa and temperature of 250oC expands to 150Kpa. Assume the process is isentropic. Find the condition of steam and work transfer. Given data: M=0.9kg P1=1.5MPa=15bar T1=250oC P2=150kPa=1.5bar Process: isentropic To find: Condition of steam and work transfer Solution: From steam table, corresponding to p1 =1.5MPa T1=250oC, read V1=0.152m3/kg H1=2923.5kJ/kg S1=6.71kJ/kg For isentropic process. S1=S2

S2=671kJ/kg K From steam table. Corresponding to 150kPa. Read H2 =467.1kJ/kg; hfg2=2226.2kJ/kg Sf2=1.43kJ/kgK; sfg2=5.79kJ/kgK Sg2=7.223kJ/kgK; vg2 =0.159m3/kg

Now, sg2>s2 the steam is in wet condition S2=sf2+x2sfg2

6.71=1.434+x25.79 x2=0.912

h2=hf2+x2hg2

=467.1+0.9122226.2=2497.39kJ/kg



272

v2+x2vg2

=0.9121.159=1.057m3/kg By first law of thermodynamics

Q=W+u For isentropic process, Q=0

W=-u

u=u2-u1 Work transfer/kg =(h1-p1v1)-(h2-p2v2) =[(h1-h2)-(p1v1-p2v2)] [ p1=p2]

=(2923.5-2497.39)-(15000.152-1501.057) =356.66kJ/kg Total work, WTotal-mW =0.9(356.67) =321kJ Result:

(i) The condition of steam is 0.912 dry (ii) Work transfer, W=321kJ

32. A steam at 10bar and 0.85 dry expands according the law of pV1.2=C to final pressure of 1bar. Find the final volume and final enthalpy. Given data: P1=10bar X1=0.85 P2=1bar Expansion according to pV1.2=c Find: Final volume and final enthalpy Solution: From steam Table at 10 bar Vg1=0.1943m3/kg Force the initial condition of steam is wet, initial volume



273

V1=xvg1

=0.850.1943 V1=0.1653m3/kg From polytropic relation P1v2

12- P2v212 [ p1v1

n=p2v2n]

112

12 1

2

112

3

2

pv v

p

0.1653 10

v 1.1263m /kg

From steam Table at 1 bar Hf2 =417.5kJ/kg; hfg2=2256.9kJ/kg Vg2=1.6938m3/kg Since V2<vg2 the condition of steam is wet for steam v2+x2vg2

22

g2

v 1.1262x 0.66

v 1.6938

h2=hf2+x2hg2 h2=417.5+0.66(2256.9) h2=1907.054kJ/kg Result:

(i) Final Volume = 1.1262m3/kg (ii) Final enthalpy=1907.054kJ/kg

33. Two kg of steam at a pressure of 8 bar occupies volume of 0.3m3. If the air expands to a volume 1.5m3 according to the law pV1.35=C. Calculate work done and change in entropy during the process. Given data: M=2kg P1=8bar V1=0.3m3 Expansion:pV1.35=C V2=1.5m3



274

To find:

Work done (W) and Change in entropy (S) Solution:

3

1

0.3v 0.15m /kg

2

From steam Table at 8 bar Vg1=0.24051m3/kg Hf1 =720.9kJ/kg; hg1=2046.5kJ/kg Sf1=2.046kJ/kgK; Sfg1=4.614kJ/kgK Since v1<vg1, the steam is in wet condition after expansion. Dryness fraction,

1

g1

v 0.15x 0.623

v 0.24051

h1hf1xhg1

=720.9+0.6232046.5 h1=1997.25kJ/kg S1=sf1+xsfg1

=2.046+0.6234.614 s1=4.92kJ/kgK P1v2

1.35-P2v21.35

1.35 1.35

12 1

2

2

v 0.3p p 8

v 1.5

p 0.91bar

From steam Table at 0.91 bar Vg2=1.8501m3/kg

3

2

2 g2,

2

1.5Given v 0.75m /kg

2

Sincev v the steam is in wet condition dryness

0.75fraction,x 0.405

1.8501



275

g

g

g

3

int erpolation

At 0.9bar,v 1.8691

At 0.92bar, v 1.8311

1.8691 1.83110.019

2

at0.91bar

v 1.8691 0.019

1.8501m /kg

From steam table at 0.91 bar (by interpolation) Hf2=406.5kJ/kg; hfg2=22.66.4kJ/kg sf2 =1.2735kJ/kgK; sfg2 = 6.1185kJ/kgK h2hf2x2hg2

=406.5+0.4052266.4 h2=1324.39kJ/kg S2=sf2+xsfg2

=1.2735+0.4056.1185 s2=3.7515kJ/kgK

1 1 2 2V VWork done,W=

n 1

800 0.391 1.5

1.35 1

W=295.7kJ Change in entropy

S=m S1=S2

=2(3.7515-4.92)

S=-2.377kJ/K Result: Work transfer, W=295.8kJ

Change in entropy, S=-2.337kJ/K

34. Steam is expanded as for the law V1.15=C from 10 bar and 0.92 dry to 3 bar. Find work done and heat transfer from a non flow system. Given data:

V1.35=C

1=10bar



276

x1=0.92

2=3bar To find:

1. W=? 2. Q=?

Solution: From steam tables, corresponding to 10 bar, read hf1=762.6kJ/kg; hfg1=2013.6kJ/kg3 Vg1=0.1943m3/kg V1=xvg1

=0.920.1943 =0.179m3/kg h1hf1x1hg1

=762.6+0.922013.6 =2615.112kJ/kg From polytropic relation

1v11.15-2v2

1.15

1

1.151

2 1

2

1

15

3

v v

100.179

3

0.51m /kg

1 1 2 2v vWork done,W=

n 1

1000 0.179 300 0.51

1.15 1

W=173.33kJ/kg From steam tables corresponding to 3 bar, read Vg2 =0.60553m3/kg

Here, vg2 >v2,the steam is in wet condition v2=vf2+x2vfg2 0.51=0.001074+x2(0.60553-0.001074) x2=0.842



277

h2 =hf2x2hg2

=561.4+0.8422163.2 =2382.81kJ/kg Heat transferred, Q= h2-h`1 =2382.81-2615.112=-232.3kJ/kg Result:

1. Work done, W=173.33kJ/kg 2. Heat transfer, Q=-232.3kJ/kg 3. (-ve sign indicates that the heat is transferred from the system)

35. One kg of steam at a pressure of 9 bar with the temperature of 200oC throttled to a pressure of 1 bar. Find the quality of steam, and entropy change during the process. Given data: M=1kg

1=9bar T1=200oC

2=1bar Throttling process To find: The quality of steam and enthalpy change Solution: From steam tables at 9 bar Ts=75.4oC Since T1>Ts, the steam is in super heated conditions From steam table at 9 bar and 200oC H1=2832.7kJ.kg S1=6.751kJ/kgK For throttling process, there is no change in enthalpy.

h2=h1=2832.7kJ/kg Corresponding to 1 bar, read Hg2=2675.4kJ/kg and Ts=99.63oC Since h2>hg2. the steam is in super heated condition Corresponding to h2=283.7kJ/kg from the superheated Table at 1 bar, read T2 =178.46oC(By linear interpolation) Degree of super heat, (Tsup-Ts)=178.46-99.63=78.83oC Similarly, read s2=7.798 kJ/kg K



278

Entropy change during the process S2-s1=7.7398-6.75 =0.987 KJ/kgK Result:

1. Final quality of steam or Degree of super heat, Tsup-Tsat=78.83oC

2. Entropy change, s=0.989kJ/kg K Throttling process 36. A sample of steam from a boiler drum at 1.5 Mpa is put through a throttling calorimeter in which the pressure and temperature are found to be 100kpa, 100oC. find the quality of the sample taken from the boiler. Given data: P1=1.5bar T2=100oC P2=100kpa To find: The quality of sample, x=? Solution: From the super heated steam tables, corresponding to 100kpa & 100oC. H2=2676.2kJ/kg H2=hf1+x1hfg1 From steam tables corresponding to 1.5Mpa, read Hf1=844.6kJ/kg Hfg1=1945.3kJ.kg We know that, for throttling process H2=h1 H2=hf1+hfg1 2676.2-844.6+x1(1945.3) x1=0.942 Result: The quality of the sample, x1=0.942



279

37. Steam of 1 Mpa and 0.9 dry is throttled to a pressure of 200kPa. Using steam table, find the quality of steam and change in entropy. Check your answer mollier chart? State whether this process is reversible or irreversible. Given data: P1=1Mpa=10bar X=0.9 Throttling process P2=200kPa=2bar To find:

(i) Quality steam (ii) Change in entropy (iii) During the throttling process, enthalpy remains constant.

H2=h1 From steam tables at 1Mpa or 10bar Hf1=726.6kJ/kg; hfg1=2013.6kJ/kg; Sf1=2.138kJ/kg K` sfg1=4.445kJ/kg K H1=hf1+xhfg1

H1=762.2+0.92013.6 H1=2574.84kJ/kg Sf1=sf,+xsfg Sf1=2.138+0.9x4.445 Sf1=6.1385kJ/kgK From S.T at 200kPa or 2 bar Hf2 =504.7kJ/kg; hfg2=2201.6kJ/kg Sf2=1.53kJ/kgK; sfg2=5.597kJ/kgK H2=hf1+x2hfg

H2=504.7+x22201.6 Since h1=h2

2574.84=504.7+x22201.6 x2=0.94

Quality of steam is wet. S2=1.53+0.94(5.597) S2=6.79kJ/kg K

Change in entropy s=s2-s1=6.79-6.138

s=0.625kJ/kg K.



280

Generally throttling process is an irreversible process. From mollier chart For x1=0.9 at 10bar line note the enthalpy steam S1 =6.1kJ/kg K Since the throttling process is constant enthalpy process, draw horizontal line in the Mollier chart upto 2 bar pressure line as shown in fig. 3.16. Now read entropy of final state of steam. S2=6.76kJ/kgK Change in entropy, (s2-s1)=0.66kJ/kg K

Since s is positive, the process is irreversible. Results:

(i) Quality of steam is 0.94 dry (ii) Change in entropy is 0.66kJ/kg K. (iii) Process is irreversible process.

38. A sample of steam from a boiler drum at 2.0Mpa is passed through a throttling calorimeter after which the pressure and temperature are found to be 1 bar and 150oC.

(a) Find the dryness fraction of steam taken from the boiler. (b) If the calorimeter must show at least 8oCsuperheat for acceptable results, what is the

minimum dryness fraction that could be measured from the given pressures? Given data: P1=2MPa P2=1bar T2=150oC To find :

1. x1 2. minimum dryness fraction for 8oC super heat, xmin



281

Solution: Case a: From steam tables, corresponding to 4 bar 150oC, read H2=2776.3kJ/kg Se know that, for throttling process H2=h1 H1=hf +x1hfg, From steam tables corresponding to 2MPa, read, Hf, =908.6+x1(1888.7) X1=0.989 Case b: From steam tables, corresponding to 1 bar Ts=99.63oC T2=Ts+Tsup =99.63+8 =107.63oC From steam tables, corresponding 1 bar and 107.6 read. H2=2691.48kJ/kg H2=h1=hf,+x1hfg,

2691.48=908.6+x1(1888.7) x1=0.944 Result:

1. Dryness fraction, x1=0.989 2. The minimum dryness fraction when the superheat 8oC

Xmin=0.944 39. A steam initially at a pressure of 15bar and 0.95 dry expands isentropic ally to 7.5 bar and is then throttled until it is dry. Determine per kg of steam:

(i) Change in entropy (ii) Change in enthalpy and (iii) Change in internal energy.

Given data:



282

P1=15bar X1=0.95 Process: isentropic P2=7.5bar Then throttled until it is dry Find:

(i) S(ii) h and (iii) u Solution: From Mollier diagram at pressure 15 bar dryness fraction 0.95 H1=2680kJ/kg; v1=0.1318m3/kg S1=6.2kJ/kg K The steam expands isentropically (vertical line)7.5 bar. Note the following H2=2560kJ/kg x2=0. S2=s1=6.2kJ/kg K Then steam is throttled until it is dry (horizontal line) H3=h2=2560kJ/kg; p3=0.055bar S3=8.33kJ/kg K; v3=32m3/kg Change in entropy

s=s3-s1=8.33-6.2=2.13kJ/kg K change in internal energy,

u=u3-u1 =(h3-h1)-(p3 v3-p1v1)

=(2560-2680)-(5.532-15000.1318) =-141.7kJ/kg Results:

(i) Change in entropy, s=2.13kJ/kg K

(ii) Change in enthalpy, h=120kJ/kg

(iii) Change in internal energy, U=-141.7kJ/kg



283

40. A boiler generates steam at 3 bar and 0.85 dry from water at 45oC; 540 kJ/s heat is added during the evaporation. Calculate the amount of steam generated per hour. Given data: P=3bar X=0.85 T2=45oC Heat added =540kJ/sec. To find: Steam generated per hour Solution: From stream Table at 3 bar Hf=561.4kJ/kg; hfg=2163.3kJ/kg 45oC, hw=hr=188.4kJ/kg Enthalpy of steam,

Hwet = 561.4+0.852163.3 Hwet=2400.2kJ/kg Heat supplied for 1 kg of steam = hwet-hw =2400.2-188.4 =2211.8kJ.kg Total heat supplied, H=540kJ/sec=1944000kJ/hr

19440001944000kJ/hr of heat 878.92kg/hr

2211.8

Result: Amount of steam generated = 878.92kg/hr. 41. Steam enters an adiabatic turbine at 10Mpa 500oC at the rate of 3kg/s and leaves at 50kPa. If power output of turbine is 2MW. Determine velocity of steam at exists of turbine. (MUOct‟97) Given data: P1=10Mpa=100bar T1=500oC M=50kg/s P2=50kg/s



284

W=2MW=2000kw To find: C2 Solution: From steam tables corresponding to 100 bar & 500 read H1=3374.6kJ/kg S1=6.599kJ/kgK In adiabatic process, entropy remains constant. S1=S2

s2=6.599kJ/kg K From steam tables corresponding to 0.5 bar, read Hf2=340.6kJ/kg; hfg2=2305.4kJ/kg; Sf2=1.09kJ/kg K` sfg2=6.504.445kJ/kg K S2=sf2+x2sfg2

6.599=1.091+x2(6.504) x2=0.847

h2=hf2+x2hfg2=

=340.6+0.8472305.4=2293.27kJ/kg BYSFEE

2 2

1 11 1g 2 2g

C CM h z Q m h z w

2 2

Neglect P.E. & C1=0 For adiabatic process, Q=0

The SFEE reduces to

2

11 2

2

1

2

CM h m h w

2

C3 3374.6 3 2293.27 2000

2

C 910.67m/ sec



285

Result: The exit velocity of the turbine, C2=910.67m/sec. 42. 150kg/sec of steam at 25 bar and 300°C expand isentropically in a stream turbine to 0.33bar. Determine the power output of the turbine. (MU-Apr‟99) Given data: M=150 kg/sec P1=25 bar T1=300°C P2=0.33 bar To Find: Power output, W=? Solution: Corresponding to25 bar & 300°C , read H1=3010 4KJ/kg S1=6.648KJ/kg K For isentropic process S1 =S2

S2=6.648 kJ/kg K S2=Sf2+X2 Sfg2 Corresponding to 0.33 bar, read Sf and Sfg

6.648=0.971 +x2x6.766

X2=0.84

h2=hf2+2hfg2

=298.6+0.84 x 2330.6

=2256.304 kJ/kg

Turbine work,W=m(h1-h2)

=150(3010.4-2256.304)

=113113.3kW

=113.11mW Result: The power output of the turbine, W=113.11MW



286

43. A nozzle is supplied with steam of 1Mpa at 200°C with a velocity of 100m/s.The expansion takes place to a pressure of 300kPa. Assuming isentropic efficiency of nozzle to be 90%.Find the final velocity. Given data : P1=1Mpa =10 bar T1=200°C C1=100m/s P2=300kPa=3bar

isentropic=90% To find: C2 Solution: From steam tables at 10 bar,Ts=179.9°C Since T1>Ts, the state would be in the superheated region. From Steam tables(in super heated region), At 10 bar and 200°C H1=2826.8K.kg From steam tables at 3 bar Sf2=1.672kJ/ kgk; hf2=561.2kJ/kg Sfg2=5.319kj/kgk; hfg2=2163.2 KJ/kg S2s=Sf2+x2sSfg2 S1=S2x

6.692=1.672+x2Sfg2 x2=0.94

h2s=hf2+x2hfg2 =561.5+0.94 x2163.2 =2594.91kJ/kg

1 2n

1 2

2

2

efficiency,

2826.8h 0.9=

2826.8 2594.91

h 2625.79 /

s

h hisentropic

h h

K kg

From SFEE

2 2

1 21 1 2 2 .....(1)

2 2




287

The following assumptions are made for the nozzle z1=z2; Q=0 W=0; m=1kg

The equation (1)reduces to

2 2

1 21 2

2

2 2

641.89 / sec

C Ch h

C m

Result : The final velocity of nozzle, C2=641.89 m/sec. 44. A vessel of volume 0.04m3 contains a mixture of saturated water and steam of a

temperature of 250C. The mass of the liquid present is 9kg. find the pressure, mass, specific volume, enthalpy, entropy and internal energy. Given data: V = 0.04m3

T = 250C M1 = 9kg To find: P,m,h,S and U Solution:

From steam tables corresponding to 250C read vf = vl = 0.00125\m3/kg vg = v3 = 0.050037m3/kg p = 39.776bar Total volume occupied by the liquid V1=mlvl = 9x0.001251 = 0.0113m3 Total volume of the vessel, V = volume of liquid + volume of steam = v1+vs 0.04 = 0.0113+Vs Vs = 0.0287m3

Mass of steam, ms = 0.0287

0.5740.050037

s

s

Vkg

V

Mass of mixture of liquid and steam, lm = m1+ ms = 9 + 0.574 = 9.574 kg



288

Total specific volume of the mixture.

30.040.00418 /

9.574

Vv m kg

M

we know that, V=vf + xvfg

0.00418 = 0.001251+ x(0.050037 –0.001251)[vfg = vg – vf] x = 0.06 From steam tables corresponding to 250cC, read hf = 1085.8 J/kg hfg = 1714.6KJ/kg sf = 2.794 KJ/kg sfg = 3.277KJ/kg K Enthalpy of mixture, h = hf + xhfg = 1085.8 + 0.06 x 1714.6 = 1188.67Kj/kg Entropy of mixture, S=sf x sfg = 2.794 + 0.06 x 3.277 = 2.99kJ/kg K internal energy, u=h-pv = 1188.67 – 39.776 x 102 x 0.00418 = 1172 kJ/kg Result: Pressure, P = 39.776 bar Mass of mixture, m=9.574kg Specific volume, v=0.00418m3/kg Enthalpy, h=1188.67 kJ/kg Entropy, s = 2.99kJ/kg K Internal energy, u=1172 kJ/kg

45. In a steam generator compressed water at 10 Mpa, 30C enters a 30 mm diameter tube

at the rate of 3 litres /sev. Steam at 9 Mpa, 400C exit the tube. Find the rate of heat transfer. (November – 2003) Given data : P1 = 10 bar

Tw = 30C D = 30 mm = 0.03 m



289

V1 = 3 litres/sec = 0.003 m3/sec P2 = 9 bar

T2 = 400C To find: Q Solution:

From Steam tables corresponding to 30C, read vf1 = 0.001004 m3/kg hf1 = h1 = 125.7 kJ/kg v1 = mVf1

1

f1

2 24 2

11 4

v 0.003Mass flow rate of steam, m = 2.988 kg/ s

v 0.001004

D 0.03Area of the tube A = 7.068 10 m

4 4

mv 2.988 0.001004Inlet velocity, C 4.24 m/ s

A 7.068 10

From Steam tables corresponding to 9 Mpa and 400C, read v2 = 0.02993 m3 /kg h2 = 3121.2 KJ/kg

22 4

mv 2.988 0.02993Final velocity, C 126.53 m / s.

A 7.068 10

From SFEE

2 2

1 21 1 2 2

C Cm h z g Q m h z g w

2 2

Neglect P.E. & W = 0, then the SFEE reduces to

2 2

1 21 2

C Cm h Q m h

2 2

2 24.246 126.532.988 125.7 Q 2.988 3121.2

2 1000 2 1000

Q = 8974.45 kw

Result: The rate of heat transfer to the water, Q = 8974.45 kw



290

46. Steam at 0.8 Mpa, 250C and flowing at the rate of 1 kg/s passes into a pipe carrying wet steam at 0.8 Mpa, 0.95 dry. After adiabatic mixing the flow rate is 2.3 kg/s. Determine the properties of the steam after mixing. [November – 2004] Given Data: P1 = 0.8 Mpa = 8bar T1 = 2500C M1 = 1 kg/s P2 = 0.8 Mpa = 8 bar X2 = 0.95 M3 = 2.3 kg/s To find: Properties Solution: The sum of mass of the steam before mixing = the sum of mass of the steam after mixing. m1+ m2 = m3 m2 = m3 – m1 = 2.3 – 1 = 1.3 kg/sec The energy balance equation for adiabatic mixing, m1h1 + m2h2 = m3h3 ……….(1)

Corresponding to 8 bar and 250C, read H1 = 2950.4 KJ/kg Corresponding to 8 bar, read h1 & hfg hfg = 720.9 kJ/kg; hfg2 = 2046.5 kJ/kg

2 f 2 2 fg2

3

3

g

h h x h

= 720.1+0.95 2046.5

= 2664.27 kJ/kg

Substituting all the values of the equation (1)

1 2950.4 + 1.3 2664.27 = 2.3 h

h 2788.67 kJ/kg.

Corresponding to 8 bar, read

H 2767.4kJ

/ kg

Since h3 > hg, the steam is in superheated condition. From the Mollier chart, corresponding to 8 bar and h3 = 2788.67 kJ/kg, read



291

Superheated temperature, T3 = 180C Entropy, s3 = 6.645 kJ/kgK Specific volume, v3 = 2.5 m3/kg

The condition of steam after mixing is 0.8 Mpa & 180C Result:

The condition of steam after mixing is 0.8 Mpa & 180C. Enthalpy, h3 = 2788.67 kJ/kg Entropy, s3 = 6.645 kJ/kg K Specific volume, v3 = 2.5 m3/kg

47. Ten kg of water of 45C is heated at a constant pressure of 10 bar until it becomes

superheated vapour at 300C. Find the change in volume, enthalpy internal energy and entropy. [April – 2005] Given Data: M = 10 kg P1 = p2 = 10 bar

T2 = 300C To find :

V, h, S and U Solution:

From steam tables, corresponding to 45C, read vv, & hf V1 = vf1 = 0.001010 m3/ kg. h1hf1 = 188.4 kJ/kg S1=Sf1 = 0.638 kJ/kg K

From steam tables, corresponding to 10 bar and 300C read H2 = 3052.1 kJ/kg. s2 = 7.125 kJ/kg K V2 = 0.258 m3/kg.

Change in volume, V = m (v2 = v1) = 10(0.258 – 0.001010) = 2.5699 m3

Change in volume, V = m (h2 = h1) = 10(3052.1- 188.4) = 28637 KJ



292

Change in entropy, V = m(s2 = s1) = 10 (7.125 – 0.638) = 64.87 KJ/K

Change in internal energy, V = m (u2 = u1)

2 1 2 2 1 1

2 1 1 2 1 1 2

3

m (h h ) (p v p v )

m (h h ) p (v v ) p p

10 (3052.1 -88.4) -1000 (0.258 - 0.001010)

Result:

1. Change in volume, V = 2.5699 m

2. Change in volume, h = 28637 KJ

3. Change in entropy, S =

64.87 KJ/K

4. Change in internal energy, U = 26067.1 kJ



293

UNIT – IV

PART – A

1. State Boyle‟s law? Boyle‟s law states. “The volume of a given mass of gas varies inversely as its absolute pressure, when the temperature remains constant.

1v

p

2. State Charle‟s law. Charle‟s law states, “The volume of a given mass of a gas varies directly as its absolute temperature, when the pressure remains constant.

V T 3. State Joule‟s Law. Joule‟s law states, “The internal energy a given quantity of gas depends only on the temperature”. 4. State Regnault‟s law Regnault‟s law states that CP and Cv of a gas always remains constant. 5. State Avogadro‟s law. Avogadro‟s law states. “Equal volumes of different perfect gases at the same temperature and pressure, contain equal number of molecules”. 6. State Daltons law of partial pressure? Dalton‟s law of partial pressure states. “The total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by individual gases if each one of them occupied separately in the total volume of the mixture at mixture temperature”. P = P1 + P2 + P3 + …… Pk 7. Distinguish between ideal and real gas. An ideal gas is one which strictly follows the gas laws under all conditions of temperature and pressure. In actual practice, there is no real gas which strictly follows the gas laws over the entire range of temperature and pressure. However hydrogen, oxygen, nitrogen and air behave as an ideal gas under certain temperature and pressure limits.



294

8. What are Maxwell relations?

s v

s p

T P

v S

T v

P S

v T

p T

P S

T v

v S

T P

These are known as Maxwell relations 9. Define Joule – Thomson Co – efficient. The Joule – Thomson co – efficient is defined as the change in temperature with change in pressure keeping the enthalpy remains constant. It is denoted by the

h

T

P

10. Define co-efficient of volume expansion and isothermal compressibility. Co-efficient of Volume expansion: The co – efficient of volume expansions is defined as the change in volume with change in

temperature per unit volume keeping the pressure constant. It is denoted by

p

1 v

v T

Isothermal compressibility: It is defined as the change in volume with change in pressure per unit volume keeping the temperature constant. It is denoted by K

1 vK

v P

11. What is compressibility factor? We know that, the perfect gas equation is pv = RT. But for real gas, a correction factor has to be introduced in the perfect gas equation to take into account the deviation of real gas from the perfect gas equation. The factor is known as compressibility factor (z) and is defined by

PvZ

RT



295

12. What is Clasius clapeyron equation? Clapeyron equation which involves relationship between the saturation pressure, saturation temperature, the enthalpy of evaporation and the specific volume of the two phases involved.

fg

fg

hdP

dT Tv

13. State Tds equations Tds Equation are

P

p

v

T

vTds C dT T dp

T

PTds C dT T dv

T

14. State the assumptions made in kinetic theory gases.

1. There is no intermolecular force between particulars. 2. The volume of the molecules is negligible comparison with the gas

15. State Helmholtz function. Helmholtz function is property of a system and given by subtracting the product of absolute temperature (T) and entropy (S) from the internal energy U. 16. State Gibbs function. Gibbs function is property of a system and is given by G = U – TS + pv = H – TS [H = U + pv] Where H – Enthalpy T – Temperature S – Entropy 17. State third law of Thermodynamics. It is states that the entropy of any pure substance in thermodynamics equilibrium approaches zero as the absolute temperature approaches zero.



296

PART – B 1. A Mixture of 2 kg Oxygen (M=32kg/kgmol) and 2kfg argon (m=40 Kg/kmd) is present in an insulated piston in cylinder arrangement at 100kpa, 300K. The piston now compresses the mixture to half of its initial volume. Find the final pressure, temperature and piston work, assume c, for oxygen and argon for oxygen and argon as 0.6618kj/kgk and 0.3122KJ/kgK respectively (Anna University Nov.2003) System : Closed Process : Adiabatic compression Working fluid : Mixture of oxygen and argon.

1. Mo2 = 32kg/kg mol 2. Mar = 40kg/kg mol 3. T1 = 300t\k

Known4. P1 = 100kpa 5. V2 = ½ V1 6. Cv02= 0.6618kj/kgk 7. Cv02= 0.3122kJ/kgK

Diagram: 2KgO2

2kg Ar at 100 kpa, 300K To find: 1. p2; 2. T2 ; 3.1W2

Analysis : a. p1 = 11

1

MRTV

p

Where m=mo2+mAr

=2+2 =4kg

22

2 2

2 8.314 2 8.314 =

2+2 32 2 2 40

kj =0.234

kgk

ARAR

AR AR

mo mR Ro R

mo m mo m

x x



297

1

3

2 2

2

2 2 0.23 300

100

=2.806m

( ) ( )

( )

r ARv

AR

x xV

mo Cvo m CvC

mo m

Diagram:

2 0.6618 2 0.3122

2 2

0.487

0.487 0.234

0.721

p v

kJ

kgK

C C R

0.721

0.487

1.480

p

v

C

C

It is given that

2kg O2

+

2kg Ar

at 100kPa, 300K

11

2

VV

Initial State

Final State



298

12

2

VV

=1.4033 Since the process is adiabatic, assuming the process to be reversible

1

2 1

1 2

1

2 1

1 2

0.48

2

2

2

=300(2)

=418.K

p

T V

T V

T V

T V

mRT

V

2 2 1 11 2

4 0.234 418.4

1.403

279.15

1

279.15 1.403 100 2.806

1.48 1

231.3

kpa

p V pVW

kJ

2.Consider an ideal gas at 300k and 0.86m3/kg. As a result of some disturbance that state of the gas changes to 302K and 0.87m3.kg. Estimate the change in the pressure of the gas as a result of these disturbances. Let p= f(T.v)

p pdp =

TT dv

v

For an ideal gas pv=RT



299

2

2

2

v

T

RTp

v

p RdT

T v

p Rtdvand

v v

RdT Rtdvdp

v v

From the given data dT=2K dv=0.01m3/kg Substituting these values we get

0.287 2 0.287 301 0.01

0.865 0.865

=-0.461kpa

dp

3.Derive the equation.

2

2

P

T p

C vT

p T

and prove that Cp of an ideal gas is a function of T only. (Madras University) From I law of thermodynamics. Dq=dh-vdp

For an ideal gas undergoing constant pressure process Dq=Cp dt Or Tds = Cp dt

v

v

sC T

T

Differentiating w.r.t. v at constant T we get



300

=

v

vT

v

TT

v

C sT

v v v

or

C sT

v T v

p

T T

From Maxwell relation

2

2

v

T v

C pT

v T

For an ideal gas Pv=RT

0

0

v

v

T

RTp

v

p R

T v

p R

T T T v

Cor

v

Cv is a function of temperature alone. 4. Show that the relation p(v-b)=RT satisfies cyclic relation. where b and R are constant. cyclic relation in terms of pa v and T can be expressed as

1T p v

p v T

v T p

where,

2( )T T

p RT RT

v v v b v b



301

1

1

1 =

R/v-b

v-b =

R

v

v

v

T

pp

T

p RT

T v b

2

v

R =

P

-( )

R =

p(v-b)

p p

T p v

u RT RTb Since b

T T p p

p v T RT R v b

v T p v b p R

=-1

5. Verify the validity of the Maxwell relation

pT

s v

p T

for steam at 3000C and 500Kpa.

0

600 400

300600 400

7.3724 7.5662 =

200

Kpa Kpa

T T C

S Ss s

p p

=9.69x10-4

0 0350 250

500/ 500

500

-4

350 250

0.5701 0.4744 =

100

=9.6 10

C C

p kpa kpa

kpa

V Vv V

T T



302

HencepT

s v

p T

6. Determine hfg for water at 1000C from the following data T(0C) p(kpa)

95 84.55 100 101.35 105 120.82

Consider the clausius – clpeyron eqn

1

2 1 2

4

1 1

120.82 1 1

8.31484.55 368 378

18.016

0.357 (1.558 10 )

2291 /

fg

fg

fg

fg

hp

p R T T

hIn

h

h kJ kg

T1=95+273 =368K T2=105+273 =378K p1=84.55kpa p2=120.82kpa 7. Determine the Joule Thomson coefficient for a van der walls gas given by the

equation.2

( )a

p v b RTv

Prove that for large volume (or low pressure), the inversion temperature is equal to za/br. joule-Thomson Coefficient

p

1 =

Cp

T

p

TT v

p

Differentiating



303

2

2 3

2 2 3

2 3

( ) . . . @ tan ,

2( )

2 2

2

p p

p

p

ap v b RTw r t toT cons t Pweget

v

a T a Tp v b R

v p v p

v a a abp R

T v v v

v a abp R

T v v

2 3

2

2 3

2

2 3

1

2

( )1

2

2 3

1 =

2

ph

p

p

T RT

a abp Cp

v v

ap v b

v

a abCp

v v

a abbp

v va abC

pv v

For large value of v at low-pressure.

ph

p

p

2

1 =

C

1 2 =

C

1 2 =

C

abv

T v

p p

ab

pv

ab

RT

When



304

inversion

inversion

0

2athen b=

RT

2or T

h

T

p

a

bR

8. Determine the change of internal energy, enthalpy and entropy for an isothermal process when the gas obeys Vander Wall‟s equation. System : A van der Waal‟s gas under going a process To obtain : An expression for each of the following 1)U2-u1 2)h2-h1

3)s2-s1

We know that from Eqn. 9.25

T v

u pT p

v T

For Van der Waals gas

2

2

( )

v

ap v b RT

v

RT ap

v b v

p R

T v b



305

2

2

2

2 1

1 2

2 1

1 2

Therefore

=

1 1( )

1 1( )

T

v

T T

u RTp

v v b

ap p

v

p a

T v

adu dV

v

u u T aV V

u u aV V

We know that U2=h2-p2v2 U1=h1-p1v1 There fore H2 –h1 = (u2-u1)+(p2v2-p1v1)

2 2 1 1

1 2

1 1( )p v p v a

V V

Equation 9.34 gives

vCv

dT pds dv

T T

But from Equations 9.25

1

v p

p up

T v T

It has already been proved that for Vander Waals gas

2

2

1Therefore

T

v

u a

v v

p ap

T v T

Substituting the above expression in the expression for ds we get



306

s v

2

v

v

dT 1d =C

T

dT 1 =C

T

dT =C

T

ap dv

v T

RTdv

v b T

dvR

v b

Let us Consider an is other malo process

21

1

( )s T

V bs s RIn

V b

9. Prove that Cp-Cv=R for an ideal gas System : Ideal gas undergoing process To prove : Cp-Cv=R Proof : From Equation 9.37

For an ideal gas

pv=RT

p v

p v

v pC C T

T T

Therefore

p

p

p v

v R

T p

v R

T V

R RC C T

p V

RTR

PV

R

10. Calculate the specific volume of dry saturated steam at a pressure of 147 kPa at which the values of temperature T and latent heat L are 110.790C and 2223.3KJ/kg respectively. Further, saturation temperature of steam at pressure of 157 kPa is 112.740C . Neglect the specific volume of water.



307

From Clausius – Clapeyron equation

3

( )

2223.3 112.74 110.79

273 110.79 157 147

1.13 /

fg

fg

fg

g

fg

jdp

dT T V

h dpOr V

T dT

h T

T P

Kg

11. Prove that constant pressure lines in the wet region of a mollier diagram are straight and not parallel and that the slope of a constant pressure line in the superheat region increases with temperature. Let h=f(s.p)

Also dh=Tds+vdp

Comparing the coefficients of dh, we get

p s

p

h hdh ds dp

s p

hT

s

This relation give the slop of the isobars in the Mollier diagram as shown below. h P=C dh ds s In the wet region for a given pressure temperature remains constant and hence it is a

straight line.



308

Higher the pressure, higher will be the saturation temperature and higher will be the slope. there fore isobars slope upward more steely as the pressure increases.

As temperature increases beyond the saturation line, that is, in the superheated vapour region, these lines bend slightly upward in that region.

12. Show that for a perfect gas obeying Pv=RT,

(1) Cv and Cp are independent of pressure (2) Internal energy u is independent of pressure

Let s = f(T,v) and u=f(T,v)

Then ds = .........(1)

du = .........(2)

Also du=Tds-pdv..............(3)

v v

v v

s sdT dv

T v

u udT dv

T T

From (2) and (3)

v

Tds-pdv =

Tds-pdv =c

1.........(4)

v T

T

c

T

u udT dv

T T

udT dv

T

C dT uds p dv

T T T

From equations (1) and (2) by comparing the coefficients

v

2

of dT, weget C

v

v

v

v v

C S

T T

S

T

S ST

T T V

From equation 9.16



309

2 2

2

2

2

=

therefore 0

T v

v

v

T v

v

v

T

s s

v T

s p

v T T

C pT

v T

RTGivenp

V

p R

T v

C

v

Thus Cv is a function of temperature alone or Cv is independent of pressure from equation 9.29

p

2

2 2

2

C =T

From equation 9.18

p

p

T

pT

s

T

C s

p T p

S v

p T

s vT

T p T



310

2

2

2

2

Therefore

For an ideal gas.

RTV=

P

0

0

p

pT

p

p

p

T

C v

p T

V R

T P

v

T

or

C

p

That is Cp is independent of pressure (ii) From equation 9.25

Given pv=RT

0

T v

v

T

T

u pT p

v T

p R

T V

u RT p

V V

up p

V

Hence u is a function of temperature alone and independent of pressure.



311

Derive the following relations.

2

v 2

2

p 2

ai) u=a-T

T

ii) h=g-TT

aiii) C

T

aiv) C =-T

T

) Let a = f(v,T)

v

p

v

p

g

T

i

(1)T v

a ada dv dT

V T

The Hemhotz function „a‟ is given as A = u – Ts Da = du – Tds – SdT Da = - pdv – SdT (4) From (1) and (4), we get

v

as

T

again considering the equation a = u – Ts u = a+Ts

aa T

T

au a T

T

(ii) let g = f (p,T)

1pT

g gdg dp dT

p T

The Gibb‟s function is given as g = h – Ts



312

dg = dh-Tds-sdT (2) Also Dh = dq +vdp Dh = Tds+vdp (3) Substituting (3) in (2) Dg = Tds + vdp – Tds – sdT Dg = vdp – sdT Comparing (1) & (4), we get

5v

gs

T

Again considering the Equation g = h – Ts we get H = g + Ts 6 Substituting (5) in (6), we get

p

p

v

v

gh g T

T

gh g T

T

siii C T

T

it has already been proved that

2

2

2

2

v

v v

v

v

as

T

s a

T T

aC T

T

(iv) We know that

p

p

SC T

T

Therefore



313

2

2

2

2

p p

p

p

gand s

T

s gTherefore

T T

gC T

T

13. AVessel of volume 0.28 m3 contain 10 kg of air at 302K. determine the pressure exerted by the air using

(1) Perfect gas equation (2) Vander Waals equation (3) Generalised compressibility chart.

[Tale Critical temperature of air is 132.8 K; Critical pressure of air is 37.7 bar] Given data: Volume, v = 0.28 m3 Mass, m = 10 kg Temperature T = 302.8K Critical Temperature (Tc) = 132.8K Critical Pressure (pc) = 37.7 bar

= 37.7 100 kN/m2 To find: Pressure (p) Solution: 1. Perfect gas equation:

2

2

pv = mRT

mRT p =

v

10 0.287 302 p = [ R for air is0.287 kJ/kgk]

0.28

= 3095.5 N/m

p = 3095.5 kPa

[ 1 N/m 1 pascal]



314

2. Vander Waals equation:

2

qp (v-b) = RT ....(A)

v

2 2 2 2

c

c

27R (T ) 27 (0.287) (132.8)a =

64p 64p

[Critical Pressure, pc = 37.7 bar = 37.7 100 kN/m2]

2[ 1 bar = 100 kN/m ] 2 227 (0.287) (132.8)

a64 (37.7) (100)

a = 0.162.

We know that

c

c

-3

RT 0.287 132.8b

8p 8 37.7 100

b = 1.26 10

Substituting a, b values in Vander Waals Equation

-3

2

0.162(A) p v-1.26 10 0.287 302

v

Where

V - Specific volume

Volume v 0.28 v=

mass m 10

v = 0.028 m3/kg Substituting Specific volume

-3

2

2

0.162p+ 0.028-1.26 10 0.287 302

(0.028)

p = 3034.7 kN/m



315

3. Generalized compressibility chart:

R

c

R

cR

c

3

R

T 302 T = 2.27K

T 132.8

T 2.27 K

vP 0.028 37.7 100v

RT 0.287 132.8

v 2.76 m /kg

Reduced temperature is 2.27 K and reduced specific volume is 2.76 m3/kg both are intersects at one point. Mark this point on compressibility chart. From chart, corresponding (z) value is 0.99. We know that,

2

pvCompressibility factor (z) =

RT

p 0.028 0.99 =

0.287 302

p = 3064.5 kN/m

Results :

1. Pressure p (By perfect gas equation) = 3095.5 kN/m2 2. Pressure p (By Vander Waals equation) = 3034.7 kN/m2 3. Pressure p (By compressibility chart) = 3064.5 kN/m2

15. Compute the specific volume of steam at 0.75 bar and 570 K using Vander Waals equation. Take critical temperature of steam is 647.3 K and Critical pressure is 220.9 bar. Given data:

Pressure, p = 0.75 bar = 0.75 100kN/m2 = 0.75 100 kPa

2[ 1 bar = 100 kN/m 100kPa]



316

Temperature, T = 570 K Critical Temperature, Tc = 647.3 K

Critical Pressure, pc = 220.9 100 kN/m2 To find Specific volume (v) Solution : We know that Vander Waals equation

2

2 2

c

c

ap (v - b) = RT

v

27R (T )Where, a =

64p

Universal Gas Constant 8.314Where, R = kJ/kgK

Molecular weight of Steam 18

R 0.462 kJ/kgK

2 2

c

c

-3

27 (0.462) (647.3)a =

64 (220.9) 100

a = 1.70

RT 0.462 647.3We know, b =

8p 8 220.9 100

b = 1.69 10

Substituting a, b and pressure and temperature values in Vander Waals equation.

3

2

2

1.700.75 100 v 1.69 10 0.462 570

v

1.7075 (v - 1.69 10-3) = 263.34

v

By trial and error method, we get Specific volume v = 3.58m3/kg Result : Specific volume v = 3.58 m3/kg.



317

16. An ideal gas mixture consisting of 3 kg of air and 7 kg of nitrogen at a temperature of

25C occupies a volume of 1m3. Determine the specific enthalpy, the specific internal energy and specific entropy of the mixture. Assume that air and nitrogen are ideal gases. Take R for air is 0.287 kJ/kgK and for nitrogen is 0.297 kJ/kgK. The other properties of air and nitrogen are given as under.

Name of the gas

Properties

H, kJ/kg U, kJ/kg S, kJ/kg

Nitrogen 309.64 221.11 6.46

Air 298.52 212.90 2.35

Given data :

For air Mass, ma = 3 kg

Temperature, T = 25C + 273 = 298 K Volume, v = 1m3 Gas constant, R = 0.287 kJ/kgK Enthalpy, h = 298.52 kJ/kgK Internal energy, U = 212.90 kJ/kg Entropy, S = 2.35 kJ/kgK For Nitrogen Mass, mn = 7 kg Temperature, T = 298 K Volume, v = 1m3 R = 0.297 kJ/kgK h = 309.64 kJ/kg U = 221.11 kJ/kg S = 6.46 kJ/kgK To find :

1. Specific volume (v) of the mixture 2. Pressure (p) of the mixture 3. Specific Enthalpy (h) of the mixture 4. Specific Internal energy (U) of the mixture 5. Specific Entropy (S) of the mixture

Solution:



318

a

a

3

a

n

n

3

n

volume v 1Specific Volume of air, v

mass m 3

v 0.333m /kg

volume 1 1Specific Volume of nitrogen, v

mass m 7

v 0.142m /kg

a a aa

a

2

a

m R TPressure of air, P pv = mRT

v

3 0.287 298 p

1

p 265.5 kN/m

Pressure of nitrogen,

n n n

n

2

n

m R T 7 0.297 298p

v 1

p 619.5 kN/m

For mixture

a nTotal mass, m = m m 3 7 10kg

m = 10 kg

Temperature, T = 298 K Volume, v = 1m3

m

3

a n

v 11. Specific Volume of mixture, v

m 10

= 0.1m /kg

2. Pressure of mixture, p = p p 265.5 619.5

= 876.07 kN

2

a a n n

/m

m h m h3. Specific Enthalpy of mixture, h =

m

3 298.52+ 7 309.64 =

10

= 306.34 kJ/kg



319

a a n n

a a n n

m U m U4. Internal energy of mixture, U =

m

3 212.90+7 221.11 =

10

= 218.64 kJ/kg

m S m s5. Specific Entropy of mixture, S =

m

3 2.35+7 6.46 =

10

= 5.227 kJ/kg.

Results:

1. Specific Volume of the mixture, vm = 0.1m3/kg 2. Pressure of the mixture, p = 876.07 kN/m2 3. Specific Enthalpy of the mixture, h = 306.34 J/kg 4. Specific Internal energy of the mixture, U = 218.64 kJ/kg 5. Specific Entropy of the mixture, S = 5.227 kJ/kg.

17. A perfect gas of 0.25 kg has a pressure of 298 Kpa, a temperature of 80C, and a volume of 0.08m3. The gas undergoes an irreversible adiabatic process to a final pressure of 350 kPa and find volume of 0.10 m3, work done on the gas is 25 kJ. Find cp, cv. Given data : m = 0.25 kg p1 = 298 Kpa

T1 = 80C + 273 = 353 K v1 = 0.08m3 p2 = 350 Kpa v2 = 0.10m3 W = -25 kJ [Work done on the gas in Negative valve] To find: Cp and Cv Solution: We know that, Perfect gas equation P1v1 = mRT1

1 1

1

p v 298 0.08 R= 0.270

mT 0.25 353



320

Characteristic gas constant, R = 0.270 kJ/kgK Similarly P2v2 = mRT2

2 2

2

2

p v 350 0.10 T

mR 0.25 0.270

T 518.5K

We know that

Heat transfer, Q = W + U

v 2 1 v 2 1

v

v

v

p v

p

Q W mC T T U= m C T T

Q = -25 + 0.25 C (518.5 353)

For adiabatic process, Q = 0

0 = -25 + 0.25 C (518.5 353)

C 0.604 kJ/kgK

We know that, R = C C

0.270 = C 0.604

p C 0.874 kJ/kgK

Results: Cv = 0.604 kJ/kgK Cp = 0.874 kj/kgK

h pP

T 1 vT v

P C T

From equation, we can determine the Joule-Thomson coefficient () in terms of measurable properties such as pressure (p), temperature (T), specific volume (v) and Cp. Let, Enthalpy is a function of pressure and temperature. i.e. h = f(p, T)

pT

h h dh= dp dT

p T

For Throttling process, Enthalpy remains contact H = C dh = 0



321

Substitute dh value in equation.

pT

pT

h h O = dp dT

p T

h h dp dT 0

p T

Divided by dT

h pT

h pT

pT h

p T

p

p T

h p h 0

p T T

h p h

p T T

h 1 h T =

p T p

h 1 h

p p

h 1 h C

p p

The property p

T

h C

p

is known as constant temperature coefficient.

18. Prove that internal energy of an ideal gas is a function of temperature alone? Solution: We know that, ideal gas Equation Pv = RT

RT

p ...(A)v

We know that, Internal energy Equation

v

pdU C dT T dv pdv

T

Divided by dv [From Equation (30)]



322

v

T T v

v

T v T

T

T

U T p C T p

v v T

U p T T p C 0

v T v

U RT T p [From Equation (A)]

v T v

U R T p

v v

= p - p

T

RT p =

v

U 0

v

If the temperature remains constant, there is no change in internal energy with volume. Hence internal energy is a function of temperature alone. 19. Prove that specific heat at constant volume (Cv) of a Vander Waals gas is a function of temperature alone? [May 2005 Anna University] Solution: We know that, Vander Waals equation

2

2

2

v

2

2

v

ap (v - b)= RT

v

a RT p +

v (v b)

RT a p =

(v-b) v

p R

T v b

p 0

T

We know that,



323

2

v

2

T

2

v

2

T

C pT

v T

C p0 0

v T

Thus if the temperature remains constant, there is no change in specific heat with specific volume. Hence specific heat of constant volume is function of temperature alone.

20. Find the value of (h)T for a fluid that obeys the equation of state

2

RT aP

v v

Solution: We know that,

T

v

U p T p

v T

p dU = T p dv ........(A)

T

We know that

2

2

2

2

RT aP (Given)

v v

RT a P -

v v

RT a P

v v

RT a p dv dv

v v



324

2

1

2

1

T 2

v

2v

v

v

1 2

a RT (dU) dv [From equation (A) and P = ]

v v

U = dvv

-1 U = a

v

1 1 U = a .......(B)

v v

We know that,

2 2 1 1 2 2 1 1

1 2

2

12 2

1 2 2 2 1 1

2

1 2 1 2

1 2

a ah U P v P v P v p v

v v

a a RT a RT ah v v

v v v v v v

RT a P = (Given)

v v

a a a a h =

v v v v

2a 2a =

v v

1 2

1 1 ( h)r = 2a .

v v



325

UNIT – V

PART – A

1. Define Psychrometry? The science which deals with the study of behaviour of moist air (mixture of dry air and water vapour) is known as psychrometry. 2. Define dry bulb temperature (DBT). [MU – Oct „97] The temperature which is measured by an ordinary thermometer is known as dry bulb temperature. It is generally denoted by td. 3. Define wet bulb temperature? [MU-Oct. ‟96 & „99] It is the temperature of air measured by a thermometer when its bulb is covered with wet cloth and is exposed to a current rapidly moving air. It is denoted by tw. 4. Define Dew point temperature? [MU – Oct ‟95, April ‟97 & Oct 98] The temperature at which the water vapour present in air begins to condense when the air

is cooled is known as dew point temperature. It is denoted by pdt .

5. Define specific humidity?

It is defined as the mass of water vapour present in one kg of dry air. Specific humidity (W) or Humidity ratio (or) Moisture content

Mass of water vapour

Mass of dry air

6. State saturation ratio? [Mu-Oct. ‟95 & April 2000] It is defined as the ratio of specific humidity of the moist air to the specific humidity of saturation air at the same temperature. Degree saturation (or) Percentage humidity (or) Saturation ratio

Specific humidity of moist air

Specific humidity of saturated air

7. Define relative humidity? [MU-April „98] It is defined as the ratio between mass of water vapour in a given volume and saturated mass of water in same volume and temperature.



326

Relative humidity () Mass of water vapour in a given volume

Saturated mass of water vapour in sazme volume and temperature

8. State Dalton‟s law of Partial pressure? [MU – Oct. „98] The total pressure exerted by air and water vapour mixture is equal to the barometric pressure. i.e. pb = pa + pv Where, pb = Barometric pressure pa = Partial pressure of dry air pv = Partial pressure of water vapour 9. What is meant by Humidification? The addition of water vapour to the air is known as humidification. 10. Define sensible heat factor? The ratio of sensible heat to the total heat is known as sensible heat ratio (or) sensible heat factor

Sensible heatSensible heat factor =

Total heat

11. Define the following: a) Approach b) Range? Approach: The difference in temperature of cooled-water temperature and the wet bulb temperature of the entering the air is known as the approach. Range: The range is the temperature difference between the inlet and exit states of water.



327

PART – B 1. An air conditioning unit receives an air – water vapour mixture at 101 kpa, 35oC and 80% relative humidity.

a. The dew point b. The humidity ratio c. The partial pressure of air d. The mass fraction of water vapour

System: Air water vapour mixture know:

1. DBT = 35oC 2. P = 101kpa

3. = 80% Diagram: To find: 1)Dew point temperature 2) Humidity ratio 3) Partial pressure of air Analysis: Dew point temperature It is the saturation temperature of steam corresponding to the partial pressure Pv. To find Pv consider relative humidity.

v

s

0.8

where s is the saturated pressure corresponding to 35oC as shown in the Ts diagram. Form steam table (appendix)

s = 5.63 kpa. Therefore Pv = 0.8 x Ps = 0.8 x 5.63 = 4.5 kpa



328

Thus the dew point is the saturation at 4.5 kpa. From steam table it is obtained that DPT = 30.93oC

2. Humidity ration () = v

s

0.622

v

s v

0.622

4.50.622

101 4.5

kgofw.v0.029

kgofd.a

3. Partial pressure of air

= - v = 101 – 4.5 = 96.5kpa 4. Mass fraction of water vapour

v

v a

M

M M

Mass associated with each kg of dry air as indicated by humidity ratio is 0.029kg. There fore

0.029

0.029 1.0

= 0.02818 Result 1. Dew point temperature = 30.93oC

2. Humidity ratio = 0.029 kg of w.v

kg of d.a

3. Partial pressure of air = 96.5 kpa 4. Mass function of water vapour = 0.02818

2. Give for an air water vapour mixture that Tmix = 700C and Pmix = 200 kpa and air = 180 kpa, find the dew point humidity ratio and relative humidity? System: Air water vapour mixture Known: 1 Tmix = 700C 2. Pmix = 200 kpa 3. 180 kpa



329

To find: 1. Dew point temperature (DPT)

2. Humidity ratio ()

3. Relative humidity () Diagram: Analysis: 1. Dew point temperature

It is the saturation temperature of steam corresponding to the partial pressure (v) of the water vapour in the mixture. Form Dalton‟s law of partial pressure.

P = a + Pv 200 = 180 + Pv Pv = 20kpa


v

a

0.622

200.622

180

3. Relative humidity ()

v

s

where s is the saturation pressure corresponding to the mixture temperature 700C, from steam table at 70oC, saturation pressure is 31.19kpa

Therefore f = 20

31.19

= 0.641(64.1%) Result:

1. Dew point temperature = 60.060C

2. Humidity ratio = kg of w.v

0.0691kg of d.a

3. Relative humidity () = 64.a%



330

3. The wet bulb and dry bulb temperature of most air mixture at a total pressure of 1 atm are measured with a string psychrometer and are found to be10 and 20oC respectively. Determine the humidity ratio the relative humidity the degree of saturation enthalpy and volume of the mixture per unit mass of dry air. System: Air water vapour mixture Known

1. DBT = 20oC 2. WBT = 100C 3. p = 1 atm = 101.32kpa

To find


2. Relative humidity ()

3. Degree of saturation () 4. Enthalpy (h) 5. Volume (v)

Analysis 1. Humidity ratio () To compute humidity ratio knowing DBT and WBT consider the expression given in equation.

pa 2 1 v2 v2 f

1

v1 f

C t t h h

h h

Where t2 is the WBT = (10oC) T1 is the DBT = (20oC)

2 is the humidity ratio of saturated air at the given WBT given as

s22

s2

0.622

Ps2 is the saturation pressure corresponding to the given WBT. From steam table at 100C Ps2 = 1.2276kps

Therefore 1 = 0.622kg of w.v

kg of d.a

3 kg of w.v7.63 10

kg of d.a

Cpa is the specific heat at constant pressure of air (1.005kj/kgk) hv1 is the enthalpy of water vapour



331

at the given DBT. From steam table at 200C. Hv1 = hg = 2538.1kj/kg hv2 is enthalpy of water vapour at the given WBT from steam table at 100C hv2 = hg =2519.8kj/kg hf is the liquid enthalpy at the given WBT. From steam table at 10oC hf = 42.01kj/kg Substituting the numerical values we get

3

1

1.005 20 20 7.63 10 2519.8 42.01

2538.1 42.01

2. Relative humidity () = vt

st

P

where v1 is the partial pressure of water vapour in the moist air s1 as the saturation pressure

corresponding to the given DBT to find v consider the expression for w1 that is

v11

v1

6.22

Upon rearranging we have

1v1 3

pP

0.622 3.55 10

= 0.575kpa From steam table at 20C ps1 = 2,359kpa Substituting pv1 we have

0.575

2.339

= 0.246(24.6%)

3. Degree of saturation ()

s

s1s

s1

W

W

pwhere W 0.622

p p



332

Ps1 is the saturation pressure corresponding to the given DBT. From steam table at 20oC Ps1 = 2.339kpa

Therefore Ws = 2.339

101.32 2.339

4. Enthalpy of moist air per kg of dry air(h) = Cpat + w (2500 + 1.88t) where Cpa = 1.005KJ/kgk t = DBT 7 = 20oC w = 3.55 x 10-3 Therefore h = 1.005 x 20 + 3.35 x 10-3 (2500 + 1.88 x 10) 5. Specific volume of moist air per kg of dry air(v)

a

a

R T

p

0.287 293

101.32 0.575

4. An air – water vapour mixture at 30oC and me am has a relative humidity of 60 per cent. Obtain

a. The humidity ratio b. The specific volume c. The enthalpy in KJ / kg dry air d. The dew – point

Using psychrometric chart System : Air – water vapour mixture Known : 1) P = 1atm 2) DBT = 300C

3) = 60% To find:

1. The humidity ratio () 2. The specific volume (v) 3. The specific enthalpy (h) 4. The dew – point



333

Analysis : The following steps are to be adopted to obtain the unknown properties.

Step 1: Locate meeting point of 300C DBT with = 60% curve on the psychrometric chart. Step 2: Follow the horizontal line passing through the state located as above in the right hand side and read the y – axis value to obtain the humidity ratio.

kg of w.vW = 0.016

kg of d.a

Step 3: Comparing the relative distance of the point located on either side to the adjacent constant specific volumes lines, the specific volume of given mixture can be computed. V = 0.87m3 / kg of d.a Step 4: Follow the constant enthalpy line passing through the given state up the enthalpy scale and read the corresponding enthalpy value. Step 5: Follow the horizontal line passing through the located point on the left hand side until the saturation line and read the corresponding temperature DPT = 21oC 5. Air at 1 atm, 20oC, and 40 per cent relative humidity enters a heating section at a rate of 10m3/min, and leaves at 30oC, determine (a) the rate of heat transfer in the heating section, and (b) the relative humidity of the air at the exit. Known: 1. P = 1atm 2. DBTin = 200C

3. in = 60% 4. DBTout = 30oC 5. Vin = 10m3/min



334

Diagrams: To find:

1. The rate of heat transfer in the heating section 2. The relative humidity of air at the exit

Analysis: Rate of heat transfer in the heating section Q = ma(h2 – h1) Where ma is the mass flow rate of air that can be obtained as follows:

Pv in = in Ps in

= 0.4 x Psat @ 20oC

The values 1 and 2 are also to be read from the psychometric chart

1

2

kg of w.v0.018

kg of d.a

kg of w.v0.0078

kg of d.a

ma = 115 (0.018 – 0.0078) Therefore = 1.173 kg/min 2. Substituting all the values we get

ma = 15 (29.5 – 78) + 1.173 42



335

= - 5528 KJ

min

Heat load on the cooling coil

KJ5528

min

92 KJ/ s

Tonnage capacity of the cooling coil

92 =

3.5

26.3 tons

Result :

1. Condensate removed = 1.173 kg/min 2. Tonnage capacity of the cooling coil = 26.3 tons.

6. Air is at 37C dry bulbs and 15% relative humidity. The air is to be cooled by evaporation cooling until the relative humidity is 60%. Determine a) The final temperature b) For 0.5 m3/s of air, the water required. System : Air – Water vapour mixture Process : Steady flow process – evaporative cooling Known :

1. t1 = 37C

2. 1 = 15%

3. 2 = 60% 4. V = 0.5 m3/s

Diagrams



336

To find :

1. The final temperature t2 2. Water required for 0.5 m3/s of air

Analysis: 1. To find the final temperature the steps to be followed are: Step 1 : Fix the inlet state on the psychrometric chart Step 2 : Follow a constant enthalpy line passing through state 1 until it reaches 60% relatives humidity line.

Step 3 : Read the corresponding temperature form the x –axis t2 = 24C 2. To compute the water required

w 2 1 a

a

a1

1a1

a1

m ( )m

Vm

V

RTV

P

Where

a1 v1

s1

sat @ 37 C

P P P

= 101.32 - P

= 101.32 - 0.15 P

= 101.32 - 0.15 12.513

= 99.442 Pa

Hence

a1

3

a

0.287 310V

99.44

= 0.895 m /kg

0.5 m

0.895

= 0.56 kg/s

1 and 2 can be read directly from the psychrometric chart corresponding to states 1 and 2



337

1

2

kg of w.v.0.006

kg. of d.a

kg. of w.v.0.011

kg of d.a.

Water required = (0.011 – 0.006) x 0.56 = 2.79 x 10-3kg/s Result

1. The final temperature = 240C 2. Water required = 2.79 x 10-3kg/s

7. An atmospheric air stream at 100C and 40% relative humidity is first heated to 33oC and then passed through an evaporate cooler until the temperature reaches 22oC. Determine

a. The heat added b. Final relative humidity c. Increase in moisture content

System : Air water vapour mixture Process : Steady state – sensible heating followed by

evaporative cooling Known:

1. t1 = 10oC

2. = 40% 3. t2 = 330C 4. t3 = 220C



338

= 0.4 x 2.339 = 0.9356kPa Pain = Pin - Pvin = 101.32 – 0.9356 = 100.4kpa Hence

ainain

ain

RTV

P

0.287 293

100.4

= 0.838m3/kg Therefore

ina

ain

VM

V

10

0.838

= 11.9kg/s

h2 – h1 = [(ha2 – ha1) + (hv2 – hv1)

= [Cps (t2 – t1) + (1.88) (t2 – t1) where

v1

v1

3

0.622

0.93560.622

100.4

kg of w.v5.79610

kg of d.a

Therefore h2 – h1 = [1.005 (38 – 20) + 5.796 x 10-3 x 1.88(30-20)] = 10.158 KJ/kg of d.a 2. Relative humidity at the exit

v22

s2

P

P

Since the humidity ratio remains constant in a sensible heating process

3

v22

v2

kg of w.v5.79610

kg of d.a

P0.622

P



339

The total pressure is also remaining constant

0.622 Pv2 = 2 x p = 2 pv2

pv2 = (0.622 + 2) = 2p

2v2

2

3

3

pP

0.622

5.796 10 101.32

0.622 5.795 10

=0.9354kPa From steam table at 30oC

Ps2 = Psaturation @ 30C = 4.246kpa Therefore

v22

s2

P

P

0.9354 =

4.246

=22.03%

The above problem can be solved by using psychometric chart also. Step 1: Compute the mass flow rate of dry air as given above Step 2: Fix state 1 on the psychometric chart using the inlet condition and read the corresponding in value. h1 = 34.5 KJ/kg

Step 3: Follow the horizontal line from state 1 until it meets the vertical from 30C DBT. This point represents the exit condition on psychometric chart road the corresponding in and

h2 = 45.0KJ/kg

2 = 22% 1) The rate of heat required in the heating section Q = ma (h2 – h1) = 11.93 (45.0 – 34.5) = 125.26 KJ/min



340

2) The exit relative humidy

2 = [Directly read from the chart] Results:

1) The rate of heat required at the heating section = 125.26 KJ/min 2) Relative humidity at the exit = 22%

8. Moist air at 32C and 60% relative humidity enters the cooling coil of a dehumidifier with

a flow rate of 115kg air/min. The air leaves saturated at 10C calculated the condensate removed and the tons of refrigeration required. System : Air – Water vapour mixture Process : Steady – flow dehumidification by cooling. Known.

1) t1 = 32C

2) 1 = 60% 3) ma = 116kg/min

4) t2 = 10C

5) 2 = 100%



341

To find 1) Condensate removed 2) TOR Energy balance across the control volume gives Q = ma h2-ma h1+m1h1 Q = ma (h2 –h1) + m1h1 Where h2 and are to be obtained as follows

1. State 1 is to be fixed on the psychrometric chart that is the inter section point of 32C DBT line with 60% relative humidity line.

2. State 2 is to be fixed on the saturation line corresponding to 10C 3. Specific enthalpy values corresponding to states 1 and 2 are to be read.

h1 = 78KJ/kg h2 = 29.5KJ/kg

and hf is the specific enthalpy of the condensate that is hf at 10C from steam table

h1@10C = 42KJ/kg To find mf the following steps are to be followed. mf = ma 1 2

9. Air has a dry bulb temperature of 25C and bulb temperature of 15C. If the barometer reads 1 bar, calculate

1. Vapour pressure 2. Specific humidity 3. Saturation ratio 4. Relative humidity 5. Drew point temperature 6. Vapour density 7. Enthalpy of mixture

Given data:

Dry bulb temperature, td = 25C

Wet bulb temperature, tw = 15C Barometric pressure, Pb = 1 bar To find:

Vapour pressure(PV) specific humidity (W)

saturation ratio ()

relative humidity () dew point temperature (tdp)



342

vapour density () enthalpy(h)

Solution: 1. Vapour pressure we know that,

Pv = Psw = b sw d w

w

P P t tf

1527.4 1.3t

[from equation (13)] Where Pb = Barometric pressure (or) atmospheric pressure Td = Dry bulb temperature Tw = wet bulb temperature Psw = Saturation pressure corresponding to wet bulb temperature

From steam table, we find that for 15C wet bulb temperature corresponding saturation pressure Psw is 0.01704

v

v

1 0.01704 25 15P 0.01704

1527.4 1.3 15

P 0.0105bar

2. Specific humidity we know that,

W = 0.622b v

P

P P [From Equation (6)]

= 0.622 x 0.0105

1 0.0105

w = 6.6 x 103 kg/kg of dry air 3. Saturation ratio (or) Degree of saturation:

b sv

s s s

P PP

P P P

[From equation (8)]

Where



343

Ps – Saturation pressure corresponding to dry bulb temperature

From steam table, we find that for that for 25C dry bulb temperature, corresponding saturation pressure is 0.03166 bar i.e. Ps = 0.03166bar

0.0105 1 0.03166

0.03166 1 0.0105

0.324

4. Relative humidity:

v

s

P ....... 10

P

0.0105 =

0.03166

= 0.33 = 33%

5. Dew point temperature (tdp): From steam table, we find that for partial pressure (Pv) = 0.005 bar, corresponding specific

volume tdp is 8C

Tdp= 8C

6. Vapour density (v):

From steam table, we find that for 25C dry bulb temperature, corresponding specific volume, (Vg) is 43.40m2/kg Vg= 43.40m2 /kg

Vapour density, v = 1/vg = 1/43.40 = 0.0230

v = 0.0230kg/m3 of saturated steam Hence vapour density at = 33% =0.230 x 33

v = 0.0075kg/m3 7. Enthalpy of mixture: h = CP td + Whg where, cp = Specific heat at constant pressure = 1.005KJ/kgk hg = Specific enthalpy of air corresponding to dry bulb



344

Temperature

From steam table, for 25C dry bulb temperature, corresponding specific enthalpy is 2547.3KJ/kg i.e. hg = 2547.3KJ/kg

3h 1.005 25 6.6 10 2547.3

h 41.9KJ/kg

Results:

1. Vapour pressure, v = 0.0105bar 2. specific humidity, W = 6.6 x 10-3kg/kg of dry air

3. saturation ratio, = 0.324

4. relative humidity, = 0.33

5. dew point temp, (tdp) = 8C

6. vapour density, v = 0.0075kg/m3 7. Enthalpy of mixture, h=41.9KJ/kg

10. An air water mixture at 20C and 1 bar has relative humidity 80% Calculate

1. Partial pressure of the vapour and the air 2. specific humidity 3. saturation ratio 4. dew point temperature 5. density of the air

6. if the mixture is cooled at constant pressure to a temperature of 10C, find the amount of water – vapour condensed pr kg of dry air.

Give data:

Dry bulb temperature, td = 20C Barometric pressure, Pb = 1bar

Relative humidity, = 80% = 0.8 To find:

1. Partial pressure of the vapour and the air (v and pa) 2. specific humidity (W)

3. saturation ratio () 4. dew point temperature (tdp)

5. density of the air (a)

6. amount water vapour condensed (Temperature 10C)



345

Solution:

1. Relative humidity We know that

Relative humidity = = v

s

P ....... 10

P

Where, v-Vapour pressure Ps – Saturation pressure corresponding to dry bulb temperature

From steam table, we find that for 20C dry bulb temperature corresponding saturation pressure is 0.02337 bar i.e. Ps = 0.02337bar

v

v

P

0.02337

P0.8

0.02337

partial pressure of the vapour, v = 0.0186 bar we know that, Barometric pressure = Partial pressure of the vapour + partial pressure of the air

b = v + a = 0.0186 + Pa partial pressure of the air, Pa = 0.981bar 2. Specific humidity:

b v

P 0.0186W 0.622

P P 1 0.0186

W 0.0117kg/kg a dry air

3. Saturation ratio:

b sv

s b v

[from Equation 8 ]

0.0186 1 0.023370.792

0.2337 1 0.0186



346

4. Dew point temperature (tdp)

From steam table, we find that for partial pressure, v = 0.0186, corresponding temperature 16C

So, dew point temperature tdp = 16C

5. Density of the air (a) From gas law, we know that Pava = Ra Ta Specific volume of air

a sa

a

3

R T 0.287 293V

0.9814 100

0.856m /kg

Gas constant R=0.287 KJ/kgk Ta = td + 273 = 20 + 273 = 293

a = 0.9814bar = 0.9814 x 100kN/m3 Density of the air

3

a

3

1 11.16kg/m

V 0.856

1.16kg/m

6. From steam table, for 10C corresponding pressure is 0.01277bar.

i.e. v = a = 0.01227bar Specific humidity

W=0.622v

b v

3

0.012270.622

1 0.01227

7.7 10 kg/kg of dry air

Amount of water vapour condensed =7.7 x 103 kg/kg of dry air Result:

1. Partial pressure of the vapour, v = 0.0186bar

Partial pressure of the air, a = 0.9814bar 2. Specific humidity, W=0.0117 kg/kg of dry air



347

3. Saturation Ratio, = 0.792

4. Dew point temperature, tdp = 16C

5. Density of the air, a = 1.16kg/m3 6. Amount of water vapour condensed= 7.7 x 10-3 kg/kg of dry air

11. The atmospheric air has a dry bulb temperature of 20C and specific humidity of 0.0095 kg/kg of dry air. If the barometer reads 760mm of Hg, Determine

1. Partial pressure of vapour 2. relative humidity 3. dew point temperature

Give data

Dry bulb temperature, td = 20C Specific Humidity, W = 0.0095 kg/kg of dry air Barometric pressure = 760mm of Hg = 1bar To find:

1. Partial pressure of vapour (v) 2. Relative humidity () 3. Dew point temperature (tdp)

Solution: 1. Specific humidity

we know that, W = 0.622 v

b v

v

v

v v

v v

v

0.0951

1 0.0095 0.622

0.0095 0.622 0.0095

0.0095 0.622 0.0095

Partial pressure of vapour v = 0.0152bar 2. Relative humidity

= v

s

Where, s – saturation pressure corresponding to dry bulb temperature

From steam table, for 20C dry bulb temperature corresponding pressure is 0.02337bar



348

i.e. saturation pressure, s = 0.02337bar

0.0152

0.2337

0.65

3. Dew point temperature (tdp)

From steam table, we find that for partial pressure, v = 0.0152bar, corresponding temperature is

14C

So, dew point temperature, (tdp) = 14C Results:

1. Partial pressure of vapour, v = 0.0152bar

2. relative humidity, = 65%

3. dew point temperature, (tdp) = 14C 12. The sing psychrometer is a laboratory test recorded the following reading?

Dry bulb temperature = 25C

We bulb temperature = 15C Barometer reading =760mm of Hg Partial pressure of the vapour = 10mm Hg Determine specific humidity and saturation ratio. Given data:


Wet bulb temperature tw = 15C Barometer pressure = 760mm of Hg

b = 1 bar

Partial pressure of the vapour, v = 10mm of Hg

v = 0.0133bar

[1mm of Hg = 0.00133bar; 10mm of Hg = 0.0133 bar] To find

1. Specific humidity, W

2. Saturation ratio, Solution: 1. Specific humidity



349

w = 0.62 v

b v

0.01330.622

1 0.0133

w = 8.3 x 10-3 kg/kg of dry air 2. Saturation ratio:

b sv

s b v

where

s = Saturation pressure corresponding to dry bulb temperature

From steam table, we find that for 25C dry bulb temperature, saturation pressure is 0.03166bar

i.e. s = 0.03166bar

0.0133 1 0.03166

0.03166 1 0.0133

0.41

Results:

1. Specific humidity, W = 8.3 x 10-3kg/kg of dry air

2. Saturation ratio, = 0.41 13. A mixture of air and water – vapour possesses a volume of 700m3 at 1bar pressure and

temperature 35C. Its relative humidity is 75%. Find the specific humidity, dew point temperature, mass of air and mass of vapour in the mixture. Given data: Volume, v = 700m3

Barometric pressure, b = 1bar


Relative humidity, = 75% = 0.075 To find

1. Specific humidity, W 2. Dew point temperature, tdp 3. mass of air, ma 4. mass of vapour, mv



350

Solution: 1. Relative humidity: we know that,

v

s

where,

v = Partial pressure of the vapour

s = Saturation pressure corresponding to dry bulb temperature

From steam table, we find that for 35C dry bulb temperature, corresponding pressure is 0.05622bar

i.e. s = 0.05622

v

s

v0.750.05622

Vapour pressure, v = 0.0421bar

Specific humidity, W=0.622 v

s v

0.04210.622

1 0.0421

W=0.027kg/kg of dry air From steam table, we find that for vapour pressure,

v = 0.0421bar, corresponding temperature is 30C so, dew point temperature, tdp

From gas law, a V=ma Ra T

ss

a

sa

vm

R T

700m

0.287 308

[Volume, V=700m3 gas constant, Ra = 0.287KJ/kgk

temperature, 35+273 = 308k]



351

b a v

a b v

a

a

s

a

a 700ma

0.287 308

[we know ]

1 0.0421

0.9579bar

0.9579 100kpa 1bar=100kpa

0.9579 100 700m

0.287 308

Mass of air, m 758.55kg

Specific humi

v

a

v

v

mdity,w = From Equation 1

m

m0.027

758.55

Mass of vapour, m 20.48kg

Result: Specific humidity, W= 0.027kg/kg of dry air

Dew point temperature, tdp = 30C Mass of air, ma = 758.5kg Mass of vapour, mv = 20.48 kg

14. Atmospheric air at 76cm of Hg has 25C dry bulb temperature and 15C wet bulb temperature using psychrometric chart, calculate the following. a. Relative humidity b. Humidity ratio c. Dew point temperature d. Enthalpy e. Partial pressure of vapour f. Specific volume of air g. Saturation pressure corresponding to

dry bulb temperature h. Saturation pressure corresponding to

wet bulb temperature Given data: Barometric reading = 76cm of Hg = 760mm of Hg = 1bar

Dry bulb temperature = 25C

Wet bulb temperature = 15C



352

To find:

(a) Relative humidity (b) Humidity ratio (c) Dew point temperature (d) Enthalpy (e) Partial pressure of vapour (f) Specific volume of air (g) Saturation pressure corresponding to dry bulb temperature (h) Saturation pressure corresponding to wet bulb temperature

Solution:

(a) Relative humidity ()

mark point (1) on the psychrometic chart by given dry bulb temperature (25C) and wet bulb

temperature (15C) Reflective humidity passing through point (1) is 35%

So, = 35% (b) Specific humidity (W): From point (1) draw horizontal line to the right till it cuts specific humidity line. From the chart we know that specific humidity. W = 0.007 kg/kg of dry air (c) Dew point temperature (tdp); From point (1) draw a horizontal line to the left till it cuts saturation curve. At this point temperature is 8oC.



353

(d) Enthalpy (h): Draw a inclined line along the constant wet bulb temperature line till it cuts enthalpy line. At this point enthalpy is 43 KJ/kg i.e. h = 43 KJ/kg (c) Dew point temperature (tdp) :

(e) Partial pressure of vapour (v) : Draw a horizontal line to the left till it cuts vapour presence line. At this point vapour

pressure is 8.0 mm of Hg v = 8.0 mm of Hg. (f) Specific volume of air (v): Specific volume line passing through point (1) is 0.85 m3/kg. So v = 0.853/kg

Sp.h

um

idity (

W)

Dry bulb Temperature (td), oC

Wet

bu

lb t

em

p.

15

o C

Saturation Curve

.007 .006

.005

10 15 20 25



354

(g) Saturation pressure corresponding to dry bulb temperature. Draw a horizontal line form point (2) till it cuts partial pressure line. At this point pressure is 12mm of Hg

s (tw) = 12 mm of Hg Results:

a. Reflective humidity, = 35% b. Specific humidity, W = 0.007 kg/kg a dry air c. Dew point temp, tdp = 8oC d. Enthalpy, h = 43 kJ/kg

e. Partial pressure of vapour, v = 8 mm of Hg f. Specific volume, v = 0.85m3/kg g. Saturation pressure corresponding dry bulb temperature = 23 mm of Hg h. Saturation pressure corresponding wet bulb temperature = 12mm of Hg

15. Atmospheric air at 1 bar pressure has 30OC dry bulb temperature and 50% relative humidity. Using psychrometic chart calculate dew point temperature, enthalpy and vapour pressure Given data:

Barometric pressure, b = 1 bar



355

Dry bulb temperature, td = 30oC Relative humidity = 50% To find: 1. Dew point temperature, tdp

2. Enthalphy, (h)

3. vapour pressure, (v) Solution: 1. Dew point temperature: Mark point (1) on the psychrometric chart by given dry bulb temperature (30oC) and relative humidity (50%). From point (1), draw a horizontal line to the left till it cuts saturation curve. At that point temperature is 18o C 2. Enthalpy (h): From point (1) draw a inclined line along the constant wet bulb temperature line till it cuts enthalpy line. At this point enthalpy is 65kJ/kg i.e. h = 65kJ/kg



356

3. Vapour pressure (v): Draw a horizontal line from point (1) till it cuts vapour pressure line. At that point pressure is 17mm of Hg.

v = 17mm of Hg Results:

1. Dew point temperature, tdp = 18oC 2. Enthalpy, h = 65kJ/kg

3. Vapour pressure, v = 17mm of Hg 16. An air - water vapour mixture enters an adiabatic saturator at 30oC and leaves at 20oC, which is the adiabatic saturation temperature. The pressure remains constant at 100kPa. Determine the relative humidity and humidity ratio of the inlet mixture. Given data: Dry bulb temperature, td = 30oC Wet bulb temperature, tw = 20oC

Barometric pressure, b = 100 kPa = 1bar To find

1. Relative humidity, () 2. humidity ratio, W

Solution: 1. Relative humidity: We know that,

v

s

… (A)

[From equation 10]



357

Where, v = Vapour pressure

s = Saturation pressure corresponding to dry bulb temperature We know that,

Partial pressure vapour, b sw d w

v sw

w

t t

1527.4 1.3t

Where [From equation 13]

sw – Saturation pressure corresponding to wet bulb temperature

b – Barometric pressure td – Dry bulb temperature tw – Wet bulb temperature From steam stable, for wet bulb temperature, 25oC, corresponding pressure is 0.02337 bar.

i.e. sw = 0.02337 bar

v = 0.02337 -

1 0.02337 30 20

1527.4 1.3 20

v = 0.0168bar From steam table, for 30oC dry bulb temperature, corresponding temperature is 0.04242bar

i.e. s = 0.04242 bar

Substitute v, s valve in equation (A)

v

a

0.0168A 0.39

0.04242

= 0.39 2. Humidity ratio:

W = 0.622 v

b v

[From equation (6)]

0.0168

0.6221 0.0168

W = 0.0106 kg/kg of dry air



358

Results:

1. Relative humidity = 0.39 2. humidity ratio, W = 0.0106kg/kg of dry air

17. A sling psychrometer reads 40oC dbt and 36oC wbt. Find the humidity ratio, relative humidity, dew point temperature, specific volume of air, density of air, density of water vapour and enthalpy. Given data: Dry bulb temperature, td = 40oC Wet bulb temperature, tw = 36oC To find:

1. Humidity ratio, (W)

2. Relative humidity, () 3. Dew point temperature, tdp 4. Specific volume of (Va)

5. Density of air, (a)

6. Density of water vapour, (v) 7. Enthalpy, (h)

Solution: 1. Humidity ratio:

W = 0.622 x v

b v

…. (A)

We know that, partial vapour pressure,

b sw d w

v sw

w

t t

1527.4 1.3t

Where

sw = Saturation pressure corresponding to wet bulb temperature

b = Barometric pressure = 1 bar From steam table, For 36oC wet bulb temperature, corresponding pressure is 0.05940 bar

i.e. sw = 0.05940 bar



359

v = 0.05940 -

1 0.05940 40 36

1527.4 1.3 36

Partial pressure of vapour, v = 0.0568 bar

Substitute v and b value in equation (A)

(A) W 0.0568

0.6221 0.0568

Humidity ratio, W = 0.037 kg/kg of dry air 2. Relative humidity:

v

s

s

0.0568

Where, s – Saturation pressure From steam table, For 40oC dry bulb temperature, corresponding pressure is 0.07375 bar.

i.e. s = 0.07375 bar

0.0568

0.07375

Relative humidity, = 0.77 3. Dew point temperature (tdp):

From steam table, we find that for partial vapour pressure v = 0.0568 bar, corresponding temperature is 35oC So, dew point temperature (tdp) is 35oC 4. Specific volume (Va) of air: From gas law,

3a aa

a

R T 0.287 316v 0.96m /kg

0.9432 100

Gas content, R = 0.287 kJ/kg



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Ta = td + 273 = 40 + 273 = 313

a = b - v = 1-0.0568 = 0.9432 bar = 0.9432 x 100 kPa =0.9432 x 100 kN/m2

Specific volume of air, va = 0.96m2/kg 5. Density of air:

a

a

1 1

v 0.96

Density of air a = 1.04 kg/m3

6. Density of water vapour (v): From steam table we find that for 40oC dry bulb temperature, corresponding specific volume (vg) is 19.546m3/kg

v

g

1 1

v 19.456

v = 0.05116 kg/m3

Vapour density of relative humidity, = 0.77

Vapour density, v = 0.05116 x 0.77

v = 0.039 kg/m3

7. Total enthalpy: h = Cp td = Whg

where Cp – Specific heat = 1.005kJ/kgk Hg – Specific enthalpy of air corresponding dry bulb temperature. From steam table, For 40oC dry bulb temperature, corresponding specific enthalpy I 2574.4kJ/kg

h = 1.005 (40) + 0.037 (2574.4)

h = 135.45kJ/kg Results:

1. Humidity ratio, W = 0.037 kg/kg of dry air

2. relative humidity, = 0.77 3. dew point temperature, tdp = 35oC 4. Specific volume of dry air, va = 0.96 m3/kg

5. density of air a = 1.04kg/m3



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6. density of water vapour, v = 0.039kg/m3 7. total enthalpy, h = 135.45kJ/kg

18. The atmospheric conditions are td = 25oC and specific humidity = 8.6 x 10-3 kg/kg of dry air. Determine the following a) Partial pressure of vapour b) Relative humidity c) Dew point temperature Given data: Dry bulb temperature, td = 25oC Specific humidity, W = 8.6 x 10-3 kg/kg of dry air To find:

a. Partial pressure of vapour, (v)

b. Relative humidity, () c. Dew point temp, (tdp)

Solution: 1. Specific humidity:

w = 0.622 x v

b v

8.6 x 10-3 = 0.622 x v

b v

b – Barometric pressure = 1 bar

8.6 x 10-3 = 0.622 x v

v1

(1 - v) 8.6 x 10-3 = 0.622 x v

8.6 x 10-3 = 0.622 v + 8.6 x 10-3 v

8.6 x 10-3 = v (0.622 + 8.6 x 10-3)

Vapour pressure, v = 0.0136 bar 2. Relative humidity:

v

s



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Where, v = 0.0136 bar, corresponding temperature is 11oC. So, Dew point temperature, tdp = 11oC Results:

1. Partial pressure of vapour, v = 0.0136 bar

2. Relative humidity, = 43% 3. Dew point temperature, tdp = 11oC

19. Dry bulb and wet temperatures of 1 atmospheric air steam air 40oC and 30oC respectively. Determine (a) The humidity ratio (b) Relative humidity (c) Specific enthalpy 20. The most is at 45oC dry bulb temperature and 30oC wet bulb temperature calculate a. Vapour pressure b. Dew point temperature c. Specific humidity d. Relative humidity e. Degree a saturation f. Vapour density g. Enthalpy of mixture A sling psychrometer reads 40oC DBT and 30oC WBT. Calculate specific humidity, relative humidity, dew point temp, enthalpy and specific volume of mixture. Assume atmospheric air as 1.0132 bar. Note: The procedure for finding the solution for the above problems (No. 4, 5, 6 and 7) are same as problem no. 1 Atmospheric air at a pressure of 1 bar and 25oC gas a relative humidity of 75% find

1. Partial pressure of the water vapour and the air 2. Specific volume 3. Dew point temperature 4. Specific humidity 5. Degree of saturation 6. Density of the mixtures 7. Water vapour condensed per kg of dry air when the mixture is cooled at constant

pressure to a temp of 10oC. [Note1: The procedure for finding the solution for the above problem is same a problem no 2 Note 2: These problems can also be solved by using psychromic chart]



363

21. The atmospheric air at 1 bar, dry bulb temperature 16oC and wet bulb temperature 10oC enters a heating coil whose temperature is 41oC assuming by pass factor of heating coil is 0.5. Determine dry bulb temperature, wet bulb temperature, relative humidity of the air leaving the coil and the sensible heat added to air per kg of dry air. Given data:

Barmetric pressure, b = 1 bar Initial dry bulb temperature, td1 = 16oC Wet bulb temperature, tw1 = 10oC Heating coil temperature, td3 = 41oC By pass factor = 0.5 To find

1. Dry bulb temperature of the air leaving the coil (td2) 2. Wet bulb temperature of the air leaving the coil (tw2)

3. relative humidity of the air leaving the coil (2) Solution: Step1: The initial condition of air i.e. 16oC dry bulb temperature and 10oC wet bulb temperature is marked on the psychrometric chart at point 1. Step 2: Draw a vertical line from 41oC (Heating coil temperature) and draw a horizontal line from point 1. Both are meeting at point 3. Step 3: We know that,



364

By pass factor (BPF) = 3 2

3 2

d d

d d

t t

t t

[From equation (14)]

2d41 t0.5

41 16

td2 = 28.5oC Final dry bulb temperature, td2 = 28.5oC Step 4: Draw a vertical line from 28.5oC 9Final dry bulb temperature) Till it cuts horizontal line mark this point as 2 Step 5: From psychrometric chart, we know Wet bulb temperature line passing through the point 2 is 17oC So, tw2 = 17oC and Relative humidity line passing through the point 2 is 30%



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Step 6: Draw a inclined line from point 1 and point 2 along the constant wet bulb temperature line till it cuts total enthalpy line. From Chart, At point 1, Enthalpy h1 = 30kJ/kg At point 2, Enthalpy h2 = 45kJ/kg

Sensible heat added Q = m(h2 – h1) ( h2 > h1) m = 1 kg

Q = (h2 – h1) = (45-30) Q = 15kJ/kg

Results:

1. Finally dry bulb temperature, td2 = 28.5oC. 2. Final wet bulb temperature, tw2 = 17oC

3. Final relative humidity, 2 = 30% 4. Sensible heat added Q = 15kJ/kg

22. Atmospheric air with dry bulb temperature of 30oC and wet bulb temperature of 18oC without changing its moisture content calculate the following. 1. Original relative humidity 2. Final relative humidity 3. Final wet bulb temperature.



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Given data: Initial dry bulb temperature, td1 = 30oC Initial wet bulb temperature, tw1 = 18oC Final dry bulb temperature, td2 = 16oC To find:

1. Original relative humidity (1)

2. Final relative humidity (2) 3. Final wet bulb temperature (tw2)

Solution: Step 1: The initial condition of air i.e., 30oC dry bulb temperature and 18oC wet bulb temperature is marked on the psychrometic chart at point 1.

From pscyrhometric chart we know that, relative humidity (1) passing through point 1 is 32%.

i.e. 1 = 32% Step 2: Draw a vertical line from 16oC (final dry bulb temperature) and draw a horizontal line from point 1. Both lines intercept at point 2 as shown in figure.



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Step 3: From psychrometric chart,

We know that, relative humidity () passing through point 2 is 70%

i.e. 2 = 70% Draw a inclined line from point 2 along the constant wet bulb temperature line till it cuts saturation curve. At that point temp is 11oC. i.e. tw2 = 11oC Results:

1. Initial relative humidity, 1 = 32%

2. Final relative humidity, 2 = 70% 3. Final wet bulb temperature, tw2 = 11oC



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23. 50m of air at 35oC DBT and 50% RH is called 25oC DBT maintaining its specific humidity constant. Determine 1. Relative humidity of called air 2. Heat removed. Given data: Volume of air, v = 50 m3

Dry bulb temperature, td1 = 35oC

Relative humidity, 1 = 50% Final dry bulb temperature, td2 = 25oC To find:

1. Final relative humidity (2) 2. Heat removed.

Solution: Step 1 :

The initial condition of air i.e. 35C dry bulb temperature and 50% relative humidity is marked on the psychrometric chart at point 1.



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Step 2 :

Draw a vertical line from 25C (Final dry bulb temperature) and draw a horizontal line from point 1. Both lines intercept 2 as shown in figure. Step 3 :

From psychrometric chart, we know relative humidity () passing through the poi9nt 2 is 86%.

So, 2 = 86% Step 4 : Draw a inclined line from point 1 and point 2 along the constant wet bulb temperature line till it cuts total entering line. From chart, At point 1, enthalpy h1 = 79 KJ/kg At point 2, enthalpy h2 = 68 kJ/kg Step 5 : We know that,

Heat added Q = ma (h2 – h1) 1 2 h h

Q = ma (79 – 68) …..(1)

a

a

vm

v …..(2)

Where Ma – Mass of air kg/s v - Volume of air m3/s va - Specific volume of air m3/kg



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From psychrometric chart, we know that, Specific volume of air (va) passing through point 2 is 0.864 So, va = 0.864 Substitute va and v values in Equation (2)

a

a

3

3

3

a 3

a

v 50(2) m 57.8 kg

v 0.864

v unit is m

v unit is m /kg

v mm = kg

v m /kg

aSubstitute m value m equation (1)

Head added, Q = 57.8 (79 - 68)

Q = 636.5 KJ

2 1

h unit is KJ/kg

m unit is kg

Q = m(h h )

kg KJ/kg

KJ

Results:-

1. Final relative humidity, 2 = 86% 2. Heat added Q = 636.5 KJ

24. One kg of air at 24C, 70% RH is mixed adiabatically with 2 kg of air at 16C, 10%RH. Determine final condition of the mixture? [Oct. ‟98 – M.U.] Given data: First steam : Mass of air, m1 = 1 kg

Dry bulb temperature, 1dt 24 C

Relative humidity 1 = 70% Second steam : Mass of air, m2 = 2 kg




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Relative humidity 2 = 10% To find: Final condition of the mixture

i.e. final specific humidity (W3), final enthalpy (h3) final dry bulb temperature 3d(t ) . Final

relative humidity (3). Solution: Step 1 :

The first steam air i.e. 24C dry bulb temperature and 70% RH is marked on the psychrometric chart at point 1. Step 2 :

The second stream of air i.e. 16C dry bulb temperature and 10% RH is marked on the psychrometric chart at point 2. Step 3: Join the points 1 & 2 from the psychrometric chart, we know that, Specific humidity of the first mass of air.



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Specific humidity of the second mass of air W2 = 0.001 kg/kg of dry air We know that,

3 21

2 1 3

3

3

33

3 3

3

W Wm

m W W

W 0.0011

2 0.013 W

0.013-W W 0.001

2

0.013-W 2W 0.002

W 0.005 KJ/kg of dry air

Specific humidity after mixing W3 = 0.005 KJ/Kg of dry air. Step 4 : Draw a horizontal line from W3 = 0.005 till it cuts 1.2 line. None the point 3.



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Step 5: From psychrometric short, we know that,

1. Relative humidity (), passing through point 3 is 40%

i.e. 3 = 40% 2. Draw a vertical lien from point 3. Till it cuts dry bulb temperature line. At this point temperature

is 20C 3. Draw a inclined line from point 3 along the constant wet bulb temperature line till it cuts total enthalpy line. Enthalpy, h3 = 32 KJ/kg Results: Specific humidity after mixing, W3 = 0.005 KJ/kg of dry air

Relative humidity after mixing (3) = 40%

Dry bulb temperature after mixing, 3dt 20 C

Enthalpy after mixing, h3 = 32 KJ/kg

25. If the two steams of air having temperature 15C and 20C pressure of both 1.013 bar, relative humidity 20% and 80%, flow rate 18m3/min and 25 m3/min respectively are mixed adiabatically and the pressure is maintained at 1 bar. Calculate a) specific humidity b) Specific volume. Given data : First steam of air:


Relative humidity 1 = 20% Flow rate, v1 = 18m3/min = 0.33/s



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Second steam of air:


Relative humidity 2 = 80% Flow rate, v2 = 25m3/min = 0.41m3/s To find:

1. Specific humidity 2. Specific volume

Solution: Step 1:

The first steam of air i.e. 15C dry bulb temperature and 20% relative humidity is marked on the psychometric short at point 1. Step 2:

The second steam of air i.e. 20C dry bulb temperature and 80% relative humidity is marked in the psychometric chart at point 2.



375

Step 3 : Chart, we know that, Specific humidity of the first steam of air W1 = 0.002 kg/kg of dry air Specific humidity of the second steam of air W2 = 0.012 kg/kg of dry air Step 4: We know that,

3 21

2 1 3

W Wm .....(1)

m W W

First steam flow rate, v1 = 0.3m3/s (Given)



376

1

11

s

vMass, m ........(2)

v

From psychometric chart, we know specific volume (vs) passing through point 1 is 0.83

1

3

s

1

1

v 0.83 m /kg

0.3 (2) m 0.361 kg/s

0.83

m 0.361 kg/s

Second steam flow rate, v2 = 0.41 m3/s (Given)

2

22

s

vSo mass m .........(3)

v

From psychometric chart, we know specific volume (v) passing through point 2 is 0.85

2

3

s

1

2

2

v 0.85 m /kg

0.4 (3) m 0.482 kg/d

0.85

m 0.482 kg/s

Substitute m1 & m2 value in Equation (1)

3

3

3 3

-4 3

3 3

-3

3

3

3

w 0.120.301(1)

0.482 0.002 w

(0.002 - W ) 0.361 0.482 (W 0.012)

7.22 10 0.361 W = 0.482 W 5.7 10

6.5 10 0.843 W

W 7.7 10 kg/kg of dry air

Specific humidity after mixing. W3 = 0.0077 kg/kg of dry air Step 5: Draw a horizontal line from W3 = 0.0077 till it cuts 1-2 line. Name point 3.



377

From Psychometric chart we know that specific volume passing through point 3 is 0.84 m3/kg

3

3

sv 0.84 m /kg

Results :

1. Specific humidity, W1 = 0.002kg/kg of dry air W2 = 0.012kg/kg of dry air W3 = 0.0077kg/kg of dry air 2. Specific volume,

1sV = 0.83m3/kg

2sV = 0.85m3/kg

3sV = 0.84m3/kg

26. An air conditioning system is designed under the following condition. Outdoor condition -320C DBT and 75% RH Required Indoor condition -220C DBT and 70% RH Amount of fee air calculated -200m3/min Coil dew point temperature - 140C The required condition is achieved by first cooling and dehumidifying and then by heating. Calculate the following.

1. Capacity of cooling coil in tonne 2. Capacity of heating coil in kW 3. Mass of water vapour removed in kg/s

Given data:



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Outdoor condition: Dry bulb temperature,

1dt = 32oC

Relative humidity, 1 = 75% In door conditions Dry bulb temperature,

2dt =220C

Relative humidity, 2 = 75% Amount of free air circulated v = 200m3/min V = 3.33m3/s Coil dew point temperature, i.e. Surface temperature, Ts = 140C To find:

1. Capacity of cooling coil in tones 2. Capacity of heating coil in kW 3. Mass of water vapour removed in kg/s

Solution: Step 1: The out door condition of air i.e. 320C dry bulb temperature and 75% relative humidity is marked on the psychrometic chart at point 1



379

Indoor condition of air i.e. 220C dry bulb temperature and 70% RH is marked on the psychrometic chart at pint 2. Step 2: Draw a vertical line from 140C (Dew point temperature) till it cuts saturation curve. Name the point 4. Join Point 1 and Point 4. Draw a horizontal line from point 2 till it cuts 1- 4 line name the print 3. Step 3: Draw a inclined line from point 1,2,3 and 4 along constant wet bulb temperature line till it cuts total enthalpy line From point At point 1, enthalpy h1 = 82kJ/kg At point 2, enthalpy h2 = 53 kJ/kg At point 3, enthalpy h3 = 48 kJ/kg At point 4, enthalpy h4 = 41 kJ/kg Draw a horizontal line from point 1, 2, 3 and 4 till it cuts specific humidity (W) line From chart we know that, W1 = 0.020kg/kg of dry air W2 = W3 = 0.0175kg/kg of dry air W4 = 0.009kg/kg dry air From Chart we know that, Specific volume (v) passing through point 1 is 0.88 i.e. v1 = 0.88m3/kg



380

Cooling coil capacity = ma (h1 – h3) ….(1) (1 to 3 is cooling process [dry bulb temperature decreases])

a

1

V 3.33m 3.78kg/ s

V 0.88

ma = 3.78kg/s Substitute ma value in (1)

(1) cooling coil capacity = 3.78 (82-48) = 128.52kJ/s = 128.52kW

178.5236.72tonnes

3.5

[1 tonne = 3.5kW] Cooling coil capacity = 36.72tonne We know that, Heating coil capacity = ma x (h2 – h3) = 3.78 x (53-48) = 18.9kJ/s = 18.9kW We know that, Mass of water vapour removed = ma(W1 – W3) = 3.78(0.020 – 0.0115) Mass of water vapour removed = 0.03213kg/s



381

Results:

1. Cooling coil capacity = 36.72tonne 2. Heating coil capacity = 18.9kW 3. Mass of water vapour removed = 0.03213kg/s

27. An air conditioning system is designed under the following conditions. Outdoor conditions 130C DBT and 90C WBT Required conditions 200C DBT and 60% RH Amount of free air circulated 0.30m3/min person. Seating capacity – 60 persons The required condition is achieved first by heating and then by adiabatic humidifying determine the following

1. Capacity of heating coil in kW 2. Capacity of humidifier.

Given data: Outdoor conditions:

Dry bulb temperature, td1 = 13oC Wet bulb temperature, tw1 = 90C

Required indoor conditions: Dry bulb temperature,

2td = 20oC

Relative humidity, 2 = 60% Amount a free air circulated = 0.30m3/min/person V = 5 x 10-1 m3/s/person Seating capacity = 60 person So, Amount of free air circulated = 5 x 10-3 x 60m3/s V = 0.3m3/s. To find:

1. Heating coil capacity in kW 2. Capacity a humidifier

Solution: Step 1:



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Out door conditions of air i.e. 13oC dry bulb temperature 9oC wet bulb temperature is marked on the psychrometic chart at point 1. Required indoor conditions of air 200C dry bulb temperature and 60% relative humidity is marked on the psychrometic chart at point 2. Step 2: Draw a incline line through point 2 along constant wet bulb temperature line till it cuts total enthalpy line Draw a horizontal line from point 2 both the lines intersect at point 3. Step 3: From Psychrometric chart



383

We know that, h1 = 30kJ/kg h2 = h3 = 43kJ/kg W1 = W3 = 0.006kg/kg of dry air W2 = 0.008 kg/kg of dry air Specific volume passing through point 1 is 0.82m3/kg i.e. v1 = 0.82m3/kg

We know that, a

1

V 0.3m 0.36kg/ s

V 0.82

ma = 0.36kg/s Heating coil capacity = ma(h3 – h1) (1-3 line, dry bulb temp increases) = 0.36 (43-30) = 4.68kJ/s = 4.68kW Capacity of the humidifier = ma(W2 – W1) = 0.36(0.008-0.006) = 7.2 x 104kg/s Results: Heating coil capacity = 4.68kW Capacity of the hudifier = 7.2 x 10-4kg/s 28. Water at 30oC flows into cooling tower at the rate of 1.1kg per kg of air. Air enters the tower at a dry bulb temperature of 20oC and a relative humidity of 60% and leaves it a dry bulb temperature of 260C and 90% relative humidity. Make up water is supplied at 20oC. Determine the following



384

i. The temperature of water leaving the tower. ii. The fraction of water evaporated iii. The approach and range of the cooling tower Given data: Inlet water temperature = 30oC Mass of water, mw = 1.1kg

Inlet dry bulb temperature, 1dt =20oC

Relative humidity, 1 = 60% Outlet dry bulb temperature,

2dt = 26oC

Relative humidity, = 90% Make up water temperature = 200C To find:

1. The temperature of water leaving the tower 4wt

2. The fraction of water evaporated 3. The approach and range.

Solution: The initial condition for air, i.e. 20oC dry bulb temperature and 60% relative humidity is marked on the psychormetric chart at point 1. The final dry bulb temperature 260C and relative humidity 90% is marked on the psychrometric chart at point 2.



385

The corresponding wet bulb temperatures are

1

2

o

w

o

w

t 15 C

t 24 C

The corresponding specific humidifier and enthalpies are h1 = 43kJ/kg of dry air h2 = 86kJ/kg of dry air W1 = 0.0088kg water vapour/kg dry air W2 = 0.0218kg water vapour/kg dyr air

Water entering the tower 2wt = 30oC

Enthalpy 3wt = 125.8kJ/kg. [From steam table]

Make up water temperature = 20oC Enthalpy

cwh = 84 kJ/kg [ From steam table]

We know that, Energy balance equation



386

3 4

3 4

W w 2 1 2 1 W

W

w w

mh h h h W W h

m

1

1.1 86 43 0.0218 0.0088 84

h h 38.06kJ/kg

3 4

3 4

4

w w

w w

pW

w

pW

h ht t

C

38.0930 t

C

where,

pC w = Specific heat of water = 4.17kJ/kgk

4

4

w

o

w

38.0930 t

4.17

t 20.86 C

We know that, Approach =

4 1w wt t = 20.86-15 = 5.86oC

Range = 3 4w wt t = 30-20.86 = 9.14oC

Fraction of water evaporated =m(W2 – W1) = 1 (0.0218 – 0.0088) Fraction of water evaporated = 0.013kg/kg dry air Results:

1. Temperature of water leaving the tower, 4wt 20.860C

2. Fraction of water evaporated = 0.013kg/kg of dry air

3. Approach = 5.86oC; Range = 9.140C



387

ME 1201 – ENGINEERING THERMODYNAMICS

(Common to Production Engineering) (Regulation 2004)

Answer ALL questions.

PART – A (10 X 2 = 20 marks)

1. What is heat?

2. Prove that cp – cv = R

3. State zeroth law of thermodynamics. What is its application?

4. Deduce the relation between the COP of heat pump and refrigerator.

5. What is meant by thermodynamic temperature scale? How do you device such scale?

6. What do you understand by pure substance? Give some typical examples.

7. What is critical pint? What are the properties of water at critical point?

8. What are the unique features of van der Waals equation of state?

9. What is compressibility factor? What does it signify? What us its value for van der Waals gas at critical point?

10. What is dew point temperature? How is it related to dry bulb and wet bulb temperature at the saturation condition?

PART – B (5 X 16 = 80 marks)

11. (i) Prove that internal energy is a property. (4)

(ii) 1KG of gas at 1.1 bar, 270 is compressed to 6.6 bar as per, the law pv1.3 = const. Calculate work and heat transfer, if

1. When the gas is ethane (C2H6) with molar mass of 30Kg/k mol and cp of 2.1 k\KJ/Kg.K

2. When the gas is argon (Ar) with molar mass of 40kg/K mol and cp of 0.52 KJ/Kg.K (12)

12. (a) (i) “Two reversible adiabatic lines cannot intersect”. Is this statement true or false? Justify the answer.

(ii) A reversible engine operates between a source at 972oC and two sinks, one at 127oC

The energy rejected is same at both the sinks. What is the ratio of heat supplied to the heat rejected? Also calculate the efficiency. (12)

Or



388

(b) (i) What are the conditions for reversibility? (ii) Differentiate between heat pump and refrigerator. (iii) 50 kg of water is at 313 k enough ice at -5oC is mixed with water in an adiabatic vessel such that at the end of the process all the ice melts and water at 0oC is obtained. Find the mass of ice required and the entropy change of water and ice. Given cp of water = 4.2 KJ/Kg-K, cp of ice =2.1 kg/kg-K and latent heat of ice = 335 kJ/kg. 13. (a) (i) Draw the p-T diagram off a pure substance and label at the phases and phase changes. (ii) What do you understand by dryness fraction? What is its importance? (iii) A rigid tank of 0.03 m3 capacity contains wet vapour at 80 kPa. If the wet vapour mass is 12 kg, calculate the heat added and the quality of the mixture when the pressure inside the tank reaches 7 Mpa.

Or (b) (i) What are the major problems of Carnot vapour cycle? (ii) What are the methods for improving the performance of Ranking cycle? (iii) Stream enters the turbine at 3 Mpa and 400oC and is condensed at 10 kPa. Some quantity of steam leaves the turbine at 0.6 Mpa and enters open feed water heater. Compute the fraction of the steam extracted per kg of steam and cycle thermal efficiency. (14) (a) (i) Write down the Dalton‟s law of partial pressure and explain its importance. (ii) 0.45 kg of CO and 1 kg of air is contained in a vessel of volume 0.4 m3 at 15oC. Air has 23.3% of O2 and 76.7% of N2by mass. Calculate the partial pressure of each constituent and total pressure in the vessel. Molar masses of CO. O2 and N2 are 28.32 and 28 kg/K mol.

Or

(b) (i) What is the use of Clapeyron equation? And write it down for liquid-vapour region. (ii) Explain the flow process of a real gas through a throttle value. Derive the expression for Joule Thomson coefficient and deduce its value for an ideal gas.

(a) (i) Differentiate between Dry bulb temperature and wet bulb temperature. Wet bulb depression and dew point depression.

(ii) Air at 16oC and 25% relative humidity passes through a heater and then through a humidifier to reach final dry bulb temperature of 30oC and 50% relative humidity. Calculate the heat and moisture added to the air. What is the sensible heat factor?



389

Or

(b) (i) In an adiabation mixing of two streams, derive the relationship among the ratio of mass of streams, ratio of enthalpy change and ratio of specific humidity change. (ii) Saturated air at 20oC at a rate of 1.167 m3/sec is mixed adiabatically with the outside air at 35oC and 50% relative humidity at a rate of 0.5 m3/sec. Assuming adiabatic mixing condition at 1 atm, determine specific humidity. Relative humidity dry bulb temperature and volume flow rate of the mixture

NOVEMBER/DECEMBER 2005.


PART A – (10 X 2 = 20 Marks)

1. Heat is form of Energy which is transferred by virtue of temp difference between the bodies. 2. h = u + pv h = u + RT

dh du

RdT dT

(1)

Cp = Cv + R Cp – Cv = R (1) 3. When 2 bodies are in thermal eqm. With the third body individually, then all the three bodies are in thermal eqm. Among themselves.

it is the law of temperature

used for analyzing bodies in thermal eqm. 4.

2

R1 2

2 1 2 1

1 2 1 2

T1 COP 1

T T

T T T T =

T T T T



390

1 + (COP)R = (COP)HP

5. It is scale which is independent of the properties of working substance and has only positive temperatures. It can be devised with the help of heat engines. 6. It is a substance with fixed chemical composition throughout its mass. Eg. Water, helium, nitrogen air etc. 7. Point at which the saturated liquid and saturated vapour states are identical For water, Pc = 22.09 mPa = 221 bar Tc = 647.3 K = 374.15oC

Vc = 0.0568 m3/kmol = 0.0031553m

kg

8. it is a 2 constant real gas equation. It accounts for intermolecular forces and volume occupied by the molecules.

9. It is the ratio of actual vol of real gas to that of ideal gas 1

2

Zc = 0.375. 1

2

10. It is the temp. at which condensation of vapour in air vapour mixture beings, it it is cooled at constant pressure All are equal at saturation condition.

PART B – (5 X 16 = 80 marks) 11. (i) U is a property: Proof (4)

(ii)

1

2 2

1 1

T P

T P

T2=453.6K

(1)R= uR kJ0.277

m kg.K

Cp-Cv=R Cv=1.823kJ

kg.K



391

972+273

127+273 27+273

HE

p

v

C1.152

C

1 2mR(T T )W

n 1

141.8kJ

u

v p

p

v

n Q W

1

Q=+138.1KJ

R kJ(ii) R= 0312

m kg.k

kJ C C R 0.312

kg.K

C = 1.667.

C

-nQ= W

-1

=-58.6HJ

12. (a) (i) true (1) Proof (3)

(ii) T1 Q1

W Q2 Q3 T2 T3 Fig 2Marks

31 2

1 2 3

2 3

QQ Q0

T T T

Q Q



392

1

2

s 1

R z

R

s

Q7.26

Q

Q Q3.63

Q 2Q

Q=1 - 72.46%

Q

Or (b) (i) No free expansion No Friction No loss of equilibrium condition (2) (ii) HP maintains High Temp when R maintains LT. COP of HP=(COP of R)+1 (2) (iii) Heat lost by water =Heat gained by ice. (2)

u pW m w i pi i

i

mp

m1 1 p1

1

m C (T T ) m [C 273 T ) L

m 24kg.

T KJ( S) m C In 28.67

T K

T( S) m C In L

T

KJ30.7

K

13. (a) (i) P.T diagram with labels (4)

(ii) x= sm

m (1)

Indicates quality of steam (1)

3

1 2

Vm /hg

(iii) m 0.0025

.

1 f1 1 fg1

1

1 f1 1 fg1

1 1 1 1

x at 80 kPa

x 0.0007

u u x u =393.1 kJ/kg

Or

u h p 393.1 kJ/kg



393

22 f 2 2 fg

2

2 f 2 2 fg2

2 2 2 2

2 1

u u x at 7mPa.

x 0.0441.

u u x 1316kJ/kg.

(Or)

u h p 1316KJ/kg.

Q= u=m(u u ) 11mJ

Or

(b) (i) Isotropic compression to extremely high pressures. isothermal heat transfer at variable pressures (ii) Increasing the boiler pressure Increasing the steam Outlet temp Decreasing the condenser pressure Reheating Regeneration (2) (iii) h1=191.83KJ/kg h2=192.43KJ/kg h3=670.56 KJ/kg h4=673.2KJ/kg 4 Marks h5=3230.9KJ/kg h6=2829.63KJ/kg x2 =0.836 h2=2192.22 KJ/kg

3 2

6 2

h hy

h h

y 0.181 kg/kg of steam



394

m 5 4

m 2 1

out

m

q (h h ) 255.7kJ/kg.

q (1 y)(h h ) 1.638.36kJ/Kg.

q=1- 35.9%

q

14. (a) (i) = iP (3)

Importance (3)

(ii) m02 = 23.3

0.233100

xl kg

mN2 = 76.7

0.767100

xl kg

(ii) mc2 = 0.45kg (2)

2

2

2

2

o

o

uN

N

uco

co

RuR

M

RR

M

RR

R

(2)

i ii

m RTP

V (2)

P02 = 0.4359 bar PN2 = 1.6399 bar (2) Pc0 = 0.9621 bar

3.04 barP pi (2)

Or (B) (I) For the determination of change in enthalpy associated with a phase change process (or) properties associated with phase change.

2

1 1 2

P 1 1In

P

.

jg

sat

jg jg

sat jg g

h

R T T

Or

h hdP

dT TU T D

(ii)Throttling process is an isenthapic process and fig(2) P-T diagram with h= lines and cooling heating zones inversion curve. (2)



395

Derivation

1

pp

T VV T

P C T

= 0 for an ideal gas. (2) 15. (a)(i)tdb=T recorded when bulb is not affected by the moisture present in air t = T recorded when the bulb is (4)

covered by wet cloth exposed to air WBD = tdb – TWb (4) DPD = tdb - tdP (ii)



396

C 183 KEY h1 = 23kj /Kg hA = 37.5kJ /Kg h2 = 64.5kJ/Kg q = (h2 – h1) q = 41.5kJ/kg (3) W1= 0.0028kJ/kg.

W2 = 0.0134 kJ/kg. Moisture added = W2 = w1=0.0106kJ/kg. (3)

SHF = 1

2 1

0.35Ah h

h h

(2)

Or (b) (i) m1+ m2 = m3 (1) m1h1+m2h2 = m3h3 (1)

m11+m22 = m23 (1) and

3 2 3 21

2 1 3 1 3

h hm

m h h

(5)

(ii) Saturated air h1 = 57.9kJ/Kg

1 =0.015kg/Kg

1= 0.85m3/Kg

2 = 0.0175kg/kg

2 = 0.898 m3/kg



397

ma1= 1

1

1.3725 /V

kg s

(1)

ma2 = 2

2

0.557 /V

kg s

(1)

ma3 = ma1+ma2= 115.76kg/s.

1 2 3 2 2

2 1 1 3 1

a

a

m h h

m h h

(1)

3 = 0.0157 kg/kg h2 = 64.57kJ/kg of dry air (1) From chart

T3 = 26C (1)

3 = 80% (1)

3 = 0.865 m3/kg

V3 = ma3 3 =1.67m3/sec. (1)



398

B.E./B.Tech. DEGREE EXAMINATION, MAY/JUNE 2006.

Third Semester

Mechanical Engineering

ME 1201 – ENGINEERING THERMODYNAMICS (Common to Production Engineering)

(Regulation 2004)

Standard Steam Table, Moilerchart, Psychrometric charts are permitted Answer ALL Questions

PART-A-(10X2=20 MARKS)

1. What is the relationship between a system and its environment when the system is (a)

Adiabatic (b) Isolated? 2. What is meant by enthalpy? 3. State the clausius statement of School Law of Thermodynamics. 4. State few examples of irreversible process. 5. What is a pure substance? Give examples. 6. How evaporation differs from boiling? 7. State Dalton‟s law of partial pressures. On what assumptions this law is based? 8. What is an equation of state? 9. Define the Carnot cycle with P-V and T-S diagram. 10. What is specific humidity? When does it become maximum?

smax

mmW W

G G

11. (i) Derive the general energy equation for a steady flow system and apply the equation to a nozzle and derive an equation for velocity at exit. (8) (ii) In an air compressor, air flows steadily at the rate of 0.5 kg/sec. At entry to the compressor, air has a pressure of 105 kPa and specific volume of 0.86 m3/kg and at exit of the compressor those corresponding values are 705 kPa and 0.16 m3/kg. Neglect Kinetic and Potential energy change. The Internal energy of air leaking the compressor is 95 kJ/kg greater than that of air entering. The cooling water in the compressor absorbs 60 kJ/sec. of heat from the air. Find power required to derive the compressor. (8)

12. (a) Two kg of air at 500 kPa, 800C expands adiabatically in a closed system until its volume is doubled and its temperature becomes equal to that of the surroundings which is at 100 kPa, 5oC for this process, determine

(i) The maximum work (ii) The change in availability and (iii) The irreversibility.

M-2059



399

For air taken, Cv =0.718 kJ/kg K, u=Cv T Where Cv is constant and Pv =mRT where P is pressure in kPa, V volume in m3,‟m‟ mass in kg, R a constant equal to 0.287 kJ/kg K and T temperature in K. (16) (or) (b) Establish the inequality of Clausius and express Entropy change in irreversible process. (16)

13. (a) In a single heater regenerative cycle the steam enters the turbine at 30 bar, 400oC and the exhaust pressure is 0.10 bar. The feed water heater is a direct – contact type which operation at 5 bar. Find (i) The efficiency and the steam rate of the cycle, and

(ii) The increase in mean temperature of heat addition efficiency and steam rate as compared to the Rankine cycle (with out regeneration) Neglect pump work. (16)

(or) (b) One kg of steam is contained in an elastic balloon of spherical shape which supports an internal pressure proportional to its diameter. The initial condition of steam is saturated vapour at 110oC. Heat is transferred to steam until pressure reaches 200Kpa. Determine: (i) Final temperature (ii) Heat transferred. Take Cps = 2.25 kJ/kg K. (16) 14. (a) Entropy is a function of any two properties like P and V, P and T etc., for a pure substances with the help of Maxwell‟s Equation. Prove

(i) Tds=Cv.dT + T [/k].dv

(ii) Tds=CP.dT-V.dp.T

(iii)Tds=[KCv/].dp+[CP/v].dv. (16) (or) (b) Determine change of Internal Energy and change of entropy when the gas obeys Vander Waal‟s equation. (16) 15. (a) The atmospheric air at 30oC DBT and 75% RH enters a cooling coil at the rate of 200 m3 /min. The coil dew point temperature is 14oC and the by pass factor is 0.1 determine

(i) The temperature of air leaving the coil (ii) Capacity of the cooling coil in TR (iii) The amount of water vapour removed (iv) Sensible heat factor for the process. (16)

(or) (b) The volume flow rate of air is 800 m3 / min of re-circulated at 22o C DBT and 10oC dew point temperature is to be mixed with 300 m3/min of fresh air at 30oC DBT and 50% RH. Determine the enthalpy, Specific volume, Humidity ratio and dew point temperature of the mixture. (16)



400

B.E./B.Tech. DEGREE EXAMINATION, Nov/Dec 2006.

Third Semester Mechanical Engineering


(Common to Production Engineering and B.E. (Part-Time)- Second Semester - Regulation 2005)

Standard Steam Table, Moilerchart, Psychrometric charts are permitted

Answer ALL Questions

PART-A-(10X2=20 MARKS)

1. What is the convention for positive and negative work?

2. What are the corollaries to the first law of Thermodynamics?

3. Given Kelvin – Planck statement of the second Law of Thermodynamics.

4. What is a process involved in a Carnot cycle, sketch the same in P-V and T-S diagram.

5. Define critical pressure and temperature for water.

6. Sketch the Rankine cycle on a P-V plane and name the various process.

7. State the Avagodro‟s law and state its significance.

8. Write the Maxwell‟s questions and its significance.

9. Explain the terms (a) Specific humidity (b) Dew point temperature.

10. What is adiabatic mixing and write the equation for that?

PART - B (5 X 16=80 MARKS)

11. (a) In an isentropic flow through nozzle, air flows at the rate of 600kg/hr. At inlet to the nozzle, pressure is 2 MPa and temperature is 127oC. The exit pressure is 0.5 MPa. Initial air velocity is 300 m/s determines (i) Exit velocity of air (ii) Inlet and exit area of nozzle.

(or)

(b) A centrifugal pump delivers 2750 kg of water per minute from initial pressure of 0.8 bar

T 8241



401

absolute to a final pressure of 2.8 bar absolute. The suction is 2m below and the delivery is 5m above the centre of pump. If the suction and delivery pipes are of 15cm and 10cm diameter respectively, make calculation for power required to run the pump.

12. (a) A heat engine operating between two reservoirs at 100 K and 300 K is used to drive heat pump which extracts heat from the reservoir at 300 K is used to drive heat pump which extracts heat from the reservoir at 300 K at a rate twice that at which engine rejects heat to it. If the efficiency of the engine is 40% of the maximum possible and the co-efficient of performance of the heat pump is 50% of the maximum possible, make calculations for the temperature of the reservoir to which the heat pump rejects heat. Also work out the rate of heat rejection from the heat pump if the rate of supply of heat to the engine is 50 kW. (or) (b) One kg of air is contained in a piston cylinder assembly at 10 bar pressure and 500 K temperature. The piston moves outwards and the air expands to 2 bar pressure and 350 K temperature. Determine the maximum work obtainable. Assume the environmental conditions to be I bar and 290 K. Also make calculations for the availability in the initial and final states. 13.(a) 1 kg of steam initially dry saturated at 1.1 MPa expands in a cylinder following the law PV1.13=C. The pressure at the end of expansion is 0.1 MPa. Determine

(i) The final volume (ii) Final dryness fraction (iii) Work done (iv) The change in internal energy (v) The heat transferred.

(b) Steam at a pressure of 2.5 MPa and 500oC is expanded in a steam turbine to condenser pressure of 0.05 MPa. Determine for Rankine cycle:

(i) The thermal efficiency of Rankine cycle (ii) Specific steam consumption.

If the steam pressure is reduced to 1 MPa and the temperature is kept same 500oC. Determine the thermal efficiency and the specific steam consumption. Neglect feed pump work. 14. (a) Derive Tds Equation when

(i) T and V independent (ii) T and P independent (iii) P and V independent

(b) Explain and derive the (i) Jules Thompson co-efficient (ii) Clausius Clapeyron equation.



402

15. (a) A room 7m x 4m x 4m is occupied by an air water vapour mixture at 38oC. The atmospheric pressure is 1 bar and the relative humidity is 70%. Determine humidity ratio, dew point temperature mass of dry air and mass of water vapour. If the mixture of air – water vapour is further cooled at constant pressure until the temperature is 10oC. Find the amount of water vapour condensed. (or) (b) Air at 20oC, 40% RH is mixed adiabatically with air at 40oC 40% RH in the ratio of 1 kg of the former with 2kg of later. Find the final condition of air. Draw the process in chart also as diagram.

UNIT I PART A - Vel Tech HighTech

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