Thermodynamics 091215174947 Phpapp02.Ppt.1

8/3/2019 Thermodynamics 091215174947 Phpapp02.Ppt.1

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Thermodynamics

PREPARED BY: Er.Rashwinder singh


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THERMODYNAMICS It a science that deals with energy transformation, the transformation of heat intowork or vice versa. It was derived from a Greek word therme that means Heat and dynamis that means Strength. SYSTEM : Is that portion in the universe, an atom, a galaxy, a certain quantity ofmatter or a certain volume in space in which one wishes to study. It is a region en-closed by an specified boundary, that may be imaginary, fixed or moving.OPEN SYSTEM : A system open to matter flow or a system in which there is anexchange of mass between the system and the surroundings.CLOSED SYSTEM: A system closed to matter flow or a system in which theres no

exchange of mass between the system and the surroundings.Closed System: Piston in Cylinder

boundary

Open System: Steam turbine

steam in

steam out

Work

boundary


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SURROUNDINGS OR ENVIRONMENT : It is the region all about the system.WORKING SUBSTANCE : A substance responsible for the transformation of energy.

example: steam in a steam turbine, water in a water pumpPURE SUBSTANCE : A substance that is homogeneous in nature and is homoge-neous, or a substance that is not a mixture of different specie, or a substance thatdoes not undergo chemical reaction.PROPERTY : It is a characteristic quality of a certain substance.INTENSIVE PROPERTY : Property that is independent of the mass of a system.EXTENSIVE PROPERTY : Property that is dependent upon the mass of the systemand are total values such as volume and total internal energy.

PROCESS : It is simply a change of state of a substance. If certain property of asubstance is changed, it is said to have undergone a process.CYCLE : It is a series of two or more processes in which the final and the initialstate are the same.ADIABATIC SYSTEM : A system that is impervious to heat. A system (open or closed)in which heat cannot cross its boundary.

PHASES OF A SUBSTANCEA. Solid phaseB. Liquid phaseC. Gaseous or Vapor phase


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SPECIFIC TERMS TO CHARACTERIZED PHASE TRANSITION:1. Vaporization: Change from liquid to vapor2. Condensation: Change from vapor to liquid3. Freezing: Change from liquid to solid

4. Melting: Change from solid to liquid5. Sublimation: Change from solid directly to vapor without passing the liquidstate.

MASS : It is the absolute quantity of matter in it.m mass, kg

VELOCITY: It is the distance per unit time.

secm

td

v

where:v velocity in m/secd distance in meterst time in sec

ACCELERATION: It is the rate of change of velocity with respect to time.

2secm

dtdv

a


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FORCE : Force is the mass multiplied by the acceleration.

KN 1000maF

Newtonor sec

m-kg maF 2

1 Newton = 1 kg-m/sec 2Newton : Is the force required to accelerate 1 kg mass at the rate of 1 m/secper second

WEIGHT : It is the force due to gravity.

KN1000mg

W

N mgW

Where:g gravitational acceleration, m/sec 2

At standard condition (sea level condition)g = 9.81 m/sec 2


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FORCE OF ATTRACTION : From Newtons Law of Gravitation, the force of attraction between two masses is given by the equation

Newtonr

mGmFg 2

21

Where:m1 and m 2 masses in kgr distance apart in metersG gravitational constant in N-m 2 /kg 2 G = 6.670 x 10 -11 N-m2/kg 2

PROPERTIES OF FLUIDS

DENSITY ( ): It is the mass per unit volume

3mkg

Vm

Where; - density in kg/m 3

m mass in kgV volume in m 3


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SPECIFIC VOLUME ( ): It is the volume per unit mass or the reciprocal of itsdensity.

3mKN

1000g

V1000mg

VW

Where:- specific volume in m 3 /kg

SPECIFIC GRAVITY OR RELATIVE DENSITY1. For liquids it is the ratio of its density to that of water at standard

temperature and pressure.2. For gases it is the ratio of its density to that of either air or hydrogen

at some specified temperature and pressure.

SPECIFIC WEIGHT ( ): It is the weight per unit volume.

kg3m1

mV

Where:- specific weight in KN/m 3


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CONVERSION FORMULAS

32C8.1F

8.132F

C460FR

273CK

KPaormKN

AF

P 2

ABSOLUTE SCALE

PRESSURE : Pressure is defined as the normal component of a force perunit area .

If a force dF acts normally on an infinitesimal area dA, the intensity of pres-sure is equal to

dAdFP

where;P pressure, KPaF- force KNA area, m 21 KPa = 1 KN/m 2 1MPa = 1000 KPa


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PASCALS LAW : At any point in a homogeneous fluid at rest the pressures arethe same in all directions.

x

y

z

A

BC

P 1 A1

P 2 A2

P 3 A3

Fx = 0 and Fy = 0P 1A1 P 3A3 sin = 0 1P 2A2 P 3A3cos = 0 2

From Figure:A1 = A3sin 3A

2= A3cos 4

Eq. 3 to Eq. 1P 1 = P 3Eq. 4 to Eq. 2P 2 = P 3Therefore:

P 1 = P 2 = P 3


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Atmospheric Pressure : It is the absolute pressure exerted by the atmosphere.

At Standard ConditionPa = 101.325 KPa

= 1.033 kg/cm2 = 0.101325MPa

= 1.01325 Bar= 760 mm Hg= 76 cm Hg= 14.7 lb/in 2

= 10.33 m of H 2O= 29.921 in of Hg= 33.88 ft of H 2O

Barometer: an instrument used determine the absolute pressure exerted bythe atmosphere.

ABSOLUTE AND GAGE PRESSUREAbsolute Pressure is the pressure measured referred to absolute zero

and using absolute zero as the base.Gage Pressure is the pressure measured referred to the existing atmospheric

pressure and using atmospheric pressure as the base.


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Atmospheric Pressure

Absolute Zero

P gage

P vacuum

P abs

P abs

P abs = P a P vacuum

P abs = P a + P gage


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VARIATION OF PRESSURE WITH ELEVATION

(P + dP) A

(P) A

dh

W

Fx = 0(P + dP)A PA W = 0PA + dPA PA = WdPA = WW = dVdV = AdhdPA = - Adh

dP = - dh

Note: Negative sign is used becausePressure decreases as elevation in-creases and increases as elevationdecreases


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Manometer: it is a device used in measuring gage pressure in length of someliquid column.

Open Type It has an atmospheric surface and is capablein measuring gage pressure.

Differential Type it has no atmospheric surface and is capablein measuring differences of pressure.

open

Open Type

close

close

Differential Type


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FORMS OF ENERGY

Work: It is the force multiplied by the displacement in the direction of the force.W = Fdx+W indicates that work is done by the system-W indicates that work is done on the system

Heat: It is an energy that crosses a systems boundary because of a temperature difference between the system and the surrounding.+Q indicates that heat is added to the system-Q indicates that heat is rejected from the system

Internal Energy: It is the energy acquired due to the overall molecular interaction,or it is the total energy that a molecule has.

U total internal energy, KJu specific internal energy, KJ/kg

Flow Energy or Flow Work: It is the work required in pushing a fluid usually into

the system or out from the system.


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System orControl volume

P 1A1

L1

P 2A2

L2

E f1 = F 1L1 = P 1A1L1A1L1 = V1E f1 = P 1V1

E f2 = F 2L2 = P 2A2L2A2L2 = V2E f2 = P 2V2

E f = E f2 E f1

Ef = P 2V2 P 1V1 KJ

E f = (PV)

E f = E f2 E f1

E f = P 2 2 P 1 1 KJ/kg

E f = (P )

Where:P pressure in KPaV volume in m 3

- specific volume in m 3 /kg


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Kinetic Energy: It is the energy or the work required due to the motion of abody or a system.

kgKJ

)1000(2

vvKE

KJ )1000(2

vvmKE

Joules 2

vvmvdvmKE

dt

dxdvmdx

dt

dvmdxmaKE

FdxKE

21

22

21

22

2

1

21

22

2

1

2

1

2

1

2

1

Where:m mass in kgv velocity in m/sec

m mF

x1 2


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Potential Energy: It is the energy or work required by a system by virtue ofits configuration or elevation.

m

m

1

2

Reference Datum Or Datum LIne

Z

kg

KJ ZZ

1000

gPE

KJ ZZ1000mg

PE

Joules ZZmgPE

ZZmgdZmgdZWPE

12

12

12

2

1 12

2

1


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Enthalpy: It a thermodynamic property that is equal to the sumof the internal energy and the flow energy of a substance.

h = U + PV


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2

22

1

11

222111

21

vAvA

vAvA

mm

LAW OF CONSERVATION OF MASS : Mass is indestructible. In applyingthis law we must except nuclear processes during which mass is convertedinto energy.Verbal Form: Mass Entering Mass Leaving = change of mass stored within

the systemEquation Form: m 1 m2 = mFor an Open System (steady state, steady flow system) the m = 0.

m = 0m1 m2

m 1 m 2 = 0m 1 = m 2

For one dimensional flow, the mass rate of flow entering or leaving a system is

AvAvm where:m mass flow rate, kg/secA cross sectional area, m 2 v- velocity, m/sec

- density, kg/m 3

- specific volume, m3

/kg


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PROPERTIES OF PURE SUBSTANCE

a - sub-cooled liquid b - saturated liquid c - saturated mixture d - saturated vapor e - superheated vapor

Considering that the system is heated at constantpressure where P = 101.325 KPa, the 100C is thesaturation temperature corresponding to 101.325 KPa,and 101.325 KPa is the saturation pressure correspon-ding 100 C.

P P PP

P

Q

30C100C

100C 100CT 100C

(a) (b) (c) (d) (e)

Q Q Q Q


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Saturation Temperature (tsat) - is the highest temperature at a given pressure in whichvaporization takes place.Saturation Pressure (Psat) - is the pressure corresponding to the temperature.Sub-cooled Liquid - is one whose temperature is less than the saturation temperaturecorresponding to the pressure.Compressed Liquid - is one whose pressure is greater than the saturation pressurecorresponding to the temperature.Saturated Liquid - a liquid at the saturation temperatureSaturated Vapor - a vapor at the saturation temperatureSaturated Mixture - a mixture of liquid and vapor at the saturation temperature.Superheated Vapor - a vapor whose temperature is greater than the saturation temperature.

a

b c de

T

F

Saturated Vapor

Saturated Vapor

30C

100C

t 100C

Saturated Mixture

P = C

Critical Point

T- Diagram


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a

b c de

T

S

F

Saturated Vapor

Saturated Vapor

30C

100C

t 100C

Saturated Mixture

P = C

Critical Point

T-S Diagram

F(critical point)- at the critical point the temperature and pressure is unique.For Steam: At Critical Point, P = 22.09 MPa; t = 374.136C


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a

b c de

T

S

F

Saturated Vapor

Saturated Vapor

ta

tsat

te

Saturated Mixture

P = C

Critical Point

T-S Diagram

tsat - saturation temperature corresponding the pressure Pta

- sub-cooled temperature which is less than tsatte - superheated vapor temperature that is greater than tsat


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h-S (Enthalpy-Entropy Diagram)

h

S

t = C (constant temperature curve)

P = C (constant pressure curve)F

I

II

III

I - subcooled or compressed liquid regionII - saturated mixture regionIII - superheated vapor region


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Quality (x):

Lv

v

Lv

v

mmm

mm

mmm

x

Where:mv mass of vapormL mass of liquid

m total massx- qualityThe properties at saturated liquid, saturated vapor, superheatedvapor and sub-cooled or compressed liquid can be determined

from tables. But for the properties at saturated mixture (liquidand vapor) they can be determined by the equationrc = r f + x(r fg)rfg = rg rf

Where: r stands for any property ( , U, h and S)rg property at saturated vapor (from table)rf property at saturated liquid

Note: The properties at siub-cooled or compressed liquid isapproximately equal to the properties at saturated liquid

corresponding the sub-cooled temperature.


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P 1

P 2

1

2

T

S

h = C

T-S Diagram Throttling Process

P1 steam line pressureP2 calorimeter pressure


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KJ )T-(TmC)t-mC(tQ

m,massthegConsiderin

kgKJ

)T-(TC)t-C(tQ

tionintegraby

CdTCdTdQdTdQ

dtdQ

C

1212

1212

Zeroth Law of Thermodynamics:If two bodies are in thermal equilibrium with a third body, they are in thermalequilibrium with each other and hence their temperatures are equal.Specific Heat:

It is the amount of heat required to raise the temperature of a 1 kg mass 1 C or1 K.


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Heat of Transformation: It is the amount of heat that must be transferred whena substance completely undergoes a phase change without a change in

temperature.a. Heat of Vaporization: It is the amount of heat added to vaporize a

liquid or amount of heat removed to condense a vapor or gas.Q = mLwhere: L latent heat of vaporization, KJ/kg

m mass, kg, kg/secb. Heat of Fusion: It is the amount of heat added to melt a solid or the

amount of heat removed to freeze a liquid.Q = mLwhere: L latent heat of fusion, KJ/kg

Sensible Heat: It is the amount of heat that must be transferred (added orremoved) when a substance undergoes a change in temperature withouta change in phase.

Q = mC T = mC t


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THE FIRST LAW OF THERMODYNAMICS (The Law of Conservation of (Energy)Energy can neither be created nor destroyed but can only be

converted from one form to another. Verbal Form:

Energy Entering Energy Leaving = Change of Energy stored in thesystem

Equation Form:E1 E2 = Es

1. First Corollary of the First Law: Application of first Law to a Closed System

U

Q

W For a Closed System (Non FlowSystem),PV, KE and PE are negligible, thereforethe changeof stored energy Es = U

Q W = U 1Q = U + W 2


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By differentiation:dQ = dU + dW 3

where:

dQ Q2 Q1dW W2 W1

Work of a Closed System (NonFlow)

P

V

W = PdV

P

dV

W = FdxF = PAW = PAdxAdx = dVW = PdVdW = PdVFrom Eq. 3dQ = dU + dWdQ = dU + PdV 4


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2. Second Corollary of the First Law: Application of First Law to an Open System

System orControl volume

Datum Line

Q

W

1

2

U1 + P 1V1 + KE 1 + PE 1

U2 + P 2V2 + KE 2 + PE 2

For an Open system (Steady state, Steady Flow system)

Es = 0, thereforeE1 E2 = 0 orE1 = E 2 or Energy Entering = Energy Leaving

Z1

Z2


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U1 + P 1V1 + KE 1 + PE 1 + Q = U 2 + P 2V2 + KE 2 + PE 2 + W 1Q = (U 2 U1) + (P 2V2 P 1V1) + (KE 2 KE1) + (PE 2 PE 1) + W 2Q = U + (PV) + KE + PE + W 3By differentiationdQ = dU + d(PV) + dKE + dPE + dW 4 But dQ Q2 Q1 and dW W2 W1

Enthalpy (h)h = U + PVdh = dU + d(PV) 5 dh = dU + PdV + VdP 6But: dQ = dU + PdV dh = dQ + VdP 7From Eq. 3Q = h + KE + PE + W 8dQ = dh + dKE + dPE + dW 9dQ = dU + PdV + VdP + dKE + dPE + dW 10dQ = dQ + VdP + dKE + dPE + dW0 = VdP + dKE + dPE + dWdW = -VdP - dKE - dPE 11By IntegrationW = - VdP - KE - PE 12


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If KE = 0 and PE = 0Q = h + W 13W = Q - h 14 W = - VdP 15


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IDEAL OR PERFECT GAS

Prepared By:Rashwinder singh


37/98

1. Ideal Gas Equation of StatePV = mRTP = RT

2T

2 V

2P

1T

1 V

1P

CT

PV

RT

P

Where: P absolute pressure in KPaV volume in m 3 m mass in kgR Gas Constant in KJ/kg- KT absolute temperature in K

IDEAL OR PERFECT GAS


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2. Gas Constant

K-m

kg

KJ 8.3143R

K-kg

KJ

M

RR

Where:

R- Gas Constant in KJ/kg-K

Km

kg

KJ constant gas universalR

M Molecular weight kg/kg m

3. Boyles Law If the temperature of a certain quantity ofgas is held constant the volume V is inver-sely proportional to the absolute pressure P.


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C2

V 2

P1

V 1

P

CPV

P

1

C V

P V

4.Charles Law A. At Constant Pressure (P = C)If the pressure of a certain quantity ofgas is held constant, the volume V is directlyproportional to the temperature T during a qua-sistatic change of state


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2

2

1

1T

V

T

V

CT

V T;C V ;T V

B. At Constant Volume (V = C)

If the volume of a certain quantity of gas isheld constant, the pressure P varies directlyas the absolute temperature T.

2

2

1

1

T

P

T

P

CT

P

;TCPTP ;


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5. Avogadros Law All gases at the same temperature and

pressure have the same number of molecules

per unit of volume, and it follows that thespecific weight is directly proportional toits molecular weight M.

M

6.Specific HeatSpecific Heat or Heat Capacity is the amountof heat required to raise the temperature ofa 1 kg mass 1 C or 1 K

A. SPECIFIC HEAT AT CONSTANT PRESSURE (Cp) From: dh = dU + PdV + VdPbut dU + VdP = dQ ; therefore

dh = dQ + VdP 1


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but at P = C ; dP = O; thereforedh = dQ 2

and by integrationQ = h 3

considering m,h = m(h 2 - h 1 ) 4

Q = h = m (h 2 - h 1 ) 5From the definition of specific heat, C = dQ/T

Cp = dQ /dt 6

Cp = dh/dT, thendQ = CpdT 7

and by considering m,dQ = mCpdT 8

then by integration

Q = m Cp T 9

but T = (T 2 - T 1 )Q = m Cp (T 2 - T 1 ) 10


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B SPECIFIC HEAT AT CONSTANT VOLUME (Cv) At V = C, dV = O, and from dQ = dU + PdVdV = 0, therefore

dQ = dU 11then by integration

Q = U 12then the specific heat at constant volumeCv is;

Cv = dQ/dT = dU/dT 13

dQ = CvdT 14and by considering m,

dQ = mCvdT 15and by integration

Q = m U 16

Q = mCv T 17Q = m(U 2 - U 1) 18Q = m Cv(T 2 - T 1 ) 19


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From:h = U + P and P = RTh = U + RT 20

and by differentiation, dh = dU + Rdt 21 but dh =CpdT and dU = CvdT,

therefore CpdT = CvdT + RdT 22

and by dividing both sides of theequation by dT,

Cp = Cv + R 23


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7. Ratio Of Specific Heats

k = Cp/Cv 24 k = dh/du 25 k = h/ U 26

From eq. 32, Cp = kCv 27

substituting eq. 27 to eq. 24 Cv = R/k-1 28

From eq. 24, Cv = Cp/k 29

substituting eq. 29 to eq. 24 Cp = Rk/k-1 30


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8. Entropy Change ( S) Entropy is that property of a substance thatdetermines the amount of randomness and disorder

of a substance. If during a process, an amount ofheat is taken and is by divided by the absolutetemperature at which it is taken, the result iscalled the ENTROPY CHANGE .

dS = dQ/T 31 and by integration

S = dQ/T 32 and from eq. 39

dQ = TdS 33


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GAS MIXTURE

Total Mass of a mixture

inn

mm

x ii

immMass Fraction

Total Moles of a mixture

nn

y ii

Mole Fraction

Where:m total mass of a mixturem i mass of a componentn total moles of a mixturen

i moles of a component

xi mass fraction of a componentyi - mole fraction of a component


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Equation of StateMass Basis

A. For the mixture

iiiii TRmVP

mRTPV

TRnPV

B. For the components

iiiii TRnVP

Mole Basis

A. For the mixture

B. For the components

Where:

R Gas constant of a mixturein KJ/kg- K- universal gas constant in

KJ/kg m- KR

AMAGATS LAW


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AMAGATS LAW The total volume of a mixture V is equal to the volume occupied by eachcomponent at the mixture pressure P and temperature T.

1n1V1

2n2V2

3n3V3

P,T

P = P1

= P2

= P3

T = T 1 = T 2 = T 3


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For the components:

TR

PVn ;

TR

PVn ;

TR

PVn 33

22

11

321

321

321

321

VVVV

P

TR

TR

PV

TR

PV

TR

PV

TR

PV

TRPV

TRPV

TRPV

TRPV

nnnn

The total moles n:

V

Vyi

TRPVTR

PV

y

nn

y

i

i

i

ii

The mole fraction

DALTONS LAW


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The total pressure of a mixture P is equal to the sum of the partial pressurethat each gas would exert at mixture volume V and temperature T.

1n1P1

2n2P2

3n3P3

MIXTURE

n P

T1 = T 2 = T 3 = TV1 = V 2 = V 3 = V

For the mixture

For the components

TR

VPn

TRVPn

TR

VPn

33

22

11

TR

PVn


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321

321

321

321

PPPP

V

TR

TR

VP

TR

VP

TR

VP

TR

PV

TR

VP

TRVP

TRVP

TRPV

nnnn

The total moles n: The mole fraction:

PP

yi

TRPV

TRVP

y

nny

i

i

i

ii

M l l W i ht f i t


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Molecular Weight of a mixture

RRM

MyM ii

MRR

RxR iiGas Constant of a mixture

Specific Heat of a mixture

RCC

CxC

CxC

vp

viiv

piip

Ratio of Specific Heat

uh

C

C

k v

p


54/98

Gravimetric and Volumetric AnalysisGravimetric analysis gives the mass fractions of the components

in the mixture. Volumetric analysis gives the volumetric or molal fractions

of the components in the mixture.

i

i

i

i

i

ii

iii

Mx

Mx

y

MyMy

x


55/98

1. Isobaric Process ( P = C): An Isobaric Process is an internallyreversible constant pressure process.

A. Closed System :(Nonflow)

P

V

21P

dV

Q = U + W 1 any substanceW = PdV 2 any substance

U = m(U 2 - U 1 ) 3 any substanceW = P(V 2 - V 1 ) 4 any substance

Q = h = m(h 2 -h 1 ) 5 any substance

T

S

1

2

dS

TP = C

PROCESSES OF FLUIDS

For Ideal Gas:


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For Ideal Gas:PV = mRTW =mR(T 2 -T 1 ) 5

U = mCv(T

2-T

1) 6

Q = h = mC P (T 2 -T 1 ) 7

Entropy ChangeS = dQ/T 8 any substance

dQ = dhFor Ideal Gasdh = mC PdT

S = dQ/TS = mC P dT/T

S = mC P ln(T 2 /T 1 ) 9B. Open System:Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = 0

Q = h 12


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QW = - KE - PE 13If KE = 0 and PE = 0W = 0 14Q = mC P(T 2 -T 1 ) 15 Ideal Gas

2. Isometric Process (V = C): An Isometric process is internallyreversible constant volume process.

A. Closed System: (Nonflow)

P

V 1

2T

S

T

dS

1

2V = C

b


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Q = U + W 1 any substanceW = PdV at V = C; dV = 0W = 0

Q = U = m(U 2 - U 1 ) 2 any substanceh = m(h 2 -h 1 ) 3 any substance

For Ideal Gas:Q = U = mC v (T 2 -T 1 ) 4

h = mC P(T 2 -T 1 ) 5Entropy Change:

S = dQ/T 6 any substancedQ = dU

dU = mC v dT for ideal gasS = dU/T = mC v dT/TS = mC v ln(T 2 /T 1 ) 6

B Open System:


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B. Open System:Q = h + KE + PE + W 7 any substanceW = - VdP - KE - PE 8 any substance- VdP = -V(P

2-P

1) 9 any substance

Q = U = m(U 2 - U 1 ) 10 any substanceh = m(h 2 -h 1 ) 11 any substance

For Ideal Gas:- VdP = -V(P 2 -P 1 ) = mR(T 1 -T 2 )

Q = U = mC v (T 2 -T 1 ) 12h = mC P(T 2 -T 1 ) 13

If KE = 0 and PE = 0Q = h + W 14 any substance

W = - VdP 15W = - VdP = -V(P 2 -P 1 ) 16 any substanceW = mR(T 1 -T 2 ) 16 ideal gas

h = mC P(T 2 -T 1 ) 17 ideal gas


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3. Isothermal Process(T = C): An Isothermal process is reversibleconstant temperature process.

A. Closed System (Nonflow)

dS

T

S

T1 2

P

V

1

2P

dV

PV = C orT = C


U = m(U 2 - U 1 ) 3 any substanceFor Ideal Gas:dU = mC v dT; at T = C ; dT = 0

Q = W 4

W = PdV ; at PV = C ;


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W PdV ; at PV C ;P 1 V1 = P 2 V2 = C; P = C/VSubstituting P = C/V to W = PdVW = P

1V

1ln(V

2/V

1) 5

Where (V 2 /V 1 ) = P 1 /P 2 W = P 1 V1 ln(P 1 /P 2 ) 6P 1 V1 = mRT 1 Entropy Change:

dS = dQ/T 7S = dQ/T

dQ = TdS ;at T = CQ = T(S 2 -S 1 )

(S 2 -S 1 ) = S = Q/T 8S = Q/T = W/T 9 For Ideal Gas

B Open System (Steady Flow)


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B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance- VdP = -V(P

2-P

1) 12 any substance

h = m(h 2 -h 1 ) 13 any substanceFor Ideal Gas:- VdP = -P 1 V1 ln(P 2 /P 1 ) 14- VdP = P 1 V1 ln(P 1 /P 2 ) 15

P 1 /P 2 = V 2 /V 1 16dh = C PdT; at T = C; dT = 0

h = 0 16If KE = 0 and PE = 0

Q = h + W 17 any substanceW = - VdP = P 1 V1 ln(P 1 /P 2 ) 18For Ideal Gas

h = 0 19Q = W = - VdP = P 1 V1 ln(P 1 /P 2 ) 20

4 Isentropic Process (S = C): An Isentropic Process is an internally


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4. Isentropic Process (S = C): An Isentropic Process is an internallyReversible Adiabatic process in which the entropy remains constant where S = C and PV k = C for an ideal or perfect gas.

For Ideal Gas

1

2

1

1

1

2

1

2

222

1

kkk

k22k1111

k

V

V

P

P

T

T

VPVPand T

VP

T

VP

CPVandCT

PV Using

A Closed System (Nonflow)


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A. Closed System (Nonflow)

T

S

1

2

P

V

1

2

dV

PS = C orPV k = C


U = m(U 2 - U 1 ) 3 any substanceQ = 0 4W = - U = U = -m(U 2 - U 1 ) 5

For Ideal Gas


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For Ideal GasU = mC V(T 2 -T 1 ) 6From PV k = C, P =C/V k , and substituting P

=C/V k to W = PdV, then by integration,

11

11

1

1

1

211

1

1

21

kk

VP

kk

12

1122

PP

kPdV

P

P

k

mRT

k-1

T-TmRPdV

k

VP-VPPdVW 7

8

9

Q = 0

E t Ch


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Entropy ChangeS = 0

S 1 = S 2

B. Open System (Steady Flow)Q = h + KE + PE + W 10 any substanceW = - VdP - KE - PE 11 any substance

h = m(h 2 -h 1 ) 12 any substance

Q = 0W = - h - KE - PE 13From PV k = C ,V =[C/P] 1/k , substituting V to- VdP, then by integration,

PdVkVdP


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11

11

1

1

1

211

1

1

21

kk

kk12

1122

P

P

k

VkPVdP

P

P

k

kmRT

k-1

T-TkmRVdP

k

VP-VPkVdP 14

15

16

If KE = 0 and PE = 00 = h + W 17 any substanceW = - VdP = - h 18 any substance

h = m(h 2 -h 1 ) 19 any substanceQ = 0


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12P

k

k

kk

12

1122

T-TmChW

P

P

k

VkPW

P

P

k

kmRT

k-1

T-TkmRW

k

VP-VPkPdVkVdPW

11

11

1

1

1

211

1

1

21

20

22

21

23


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P

V

dP

V

Area = - VdP

S = C

5. Polytropic Process ( PVn = C): A Polytropic Process is an


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1

2

1

1

1

2

1

2

2

22

1

nnn

n22

n11

11

n

V

V

P

P

T

T

VPVPand T

VP

T

VP

CPVandCT

PV Using

5. Polytropic Process ( PVn C): A Polytropic Process is aninternally reversible process of an Ideal or Perfect Gas in whichPV n = C, where n stands for any constant.

A. Closed System: (Nonflow)


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y ( )

Q = U + W 1

W = PdV 2U = m(U 2 - U 1 ) 3

Q = mC n (T 2 -T 1 ) 4U = m(U 2 - U 1 ) 5

P

V

1

2

dV

P

PV n = C

T

S

2

1

dS

T

PV n = C

From PV n = C, P =C/V n , and substituting


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gP =C/V n to W = PdV, then by integration,

11

11

1

1

1

211

1

1

21

nn

VP

nn

12

1122

P

P

nPdVW

P

P

n

mRT

n-1

T-TmRPdVW

n

VP-VPPdVW

Entropy ChangedS = dQ/TdQ = mC n dT

S = mC n ln(T 2 /T 1 )

6

8

9

10

B. Open System (Steady Flow)


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Q = h + KE + PE + W 11W = - VdP - KE - PE 12

h = m(h 2 -h 1 ) 13Q = mC n (T 2 -T 1 ) 14dQ = mC n dTW = Q - h - KE - PE 15From PV n = C ,V =[C/P] 1/n , substituting V to

- VdP, then by integration,

n

VP-VPnVdP

PdVnVdP

1122

1

16

1n


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11

11

1

1

211

1

21

nn

n12

P

P

n

VnPVdP

P

P

n

nmRT

n-1

T-TnmRVdP

If KE = 0 and PE = 0Q = h + W 19 any substanceW = - VdP = Q - h 20 any substance

h = m(h 2 -h 1 ) 21 any substanceh = mC p (T 2 -T 1 )

Q = mC n (T 2 -T 1 ) 22

17

18

1n


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11

11

1

1

211

1

21

nn

n12

P

P

n

VnPW

P

P

n

nmRT

n-1

T-TnmRW

24

23

6. Isoenthalpic or Throttling Process: It is a steady - state, steady


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p g y , yflow process in which Q = 0; PE = 0; KE = 0; W = 0 and theenthalpy remains constant.

h 1 = h 2 or h = C

Throttling valve

Main steam line

thermometer

Pressure Gauge

Pressure Gauge

To main steam line

Throttling Calorimeter

Irrversible and Paddle Work


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m

W

Q

UWp

Q = U + W - Wp

where: Wp - irreversible or paddle work

Second Law of Thermodynamics:


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Kelvin-Planck statement of the second law:No cyclic process is possible whose sole resultis the flow of heat from a single heat reservoir

and the performance of an equivalent amount of work.For a system undergoing a cycle: The net heat isequal to the network.

Whenever energy is transferred, the level of energy cannot beconserved and some energy must be permanently reduced to a lowerlevel.

When this is combined with the first law of thermodyna-mics,the law of energy conservation, the statementbecomes:Whenever energy is transferred, energy must be conserved,but thelevel of energy cannot be conserved and some energy must bepermanently reduced to a lower level.


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P T Q


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Pm W

1

2

34

T = C

T = C

S = CS = C

VD

V S

3

21

4

TH

TL

QA

QR

Heat Added (T = C) QA = T H( S) 1

Heat Rejected (T = C) Q

R= T

L( S) 2

S = S 2 - S 1 = S 4 - S 3 3 Net Work

W = Q = Q A - Q R 4 W = (T H - T L)( S) 5


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2. Carnot Refrigerator: Reversed Carnot CycleP


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Processes:1 to 2 - Compression (S =C)2 to 3 - Heat Rejection (T = C)3 to 4 - Expansion (S = C)4 to 1 - Heat Addition (T = C)

QR

QA

1

2

4

3

S

T

TH

TL

Heat Added (T=C) Q T ( S) 1


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QA = T L( S) 1 Heat Rejected (T=C)

QR = T H( S) 2 S = S 1 - S 4 = S 2 - S 3 3

Net Work W = Q 4

W = Q R - Q A 5

W = (T H - T L)( S) 6Coefficient of Performance

LH

L

A

TTTCOP

W

QCOP 7

8

LT

9


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1H

L

TCOP 9

Tons of Refrigeration211 KJ/min = 1 TR

3. Carnot Heat Pump:A heat pump uses the samecomponents as the refrigerator but its pur-pose is to reject heat at high energy level.

Performance Factor:

AR

R

R

QQ

QPF

W

QPF 10

11

TH


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1

1

1

COPPF

T

TPF

Q

QPF

TTPF

L

H

A

R

LH

H 12

13

14

15

Carnot Refrigerator


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TH

TL

W

QA

QR

R

Vapor Power Cycle


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RANKINE CYCLEProcesses:

1 to 2 - Expansion (S = C)2 to 3 - Heat Rejection (P = C)3 to 4 - Compression or Pumping (S = C)4 to 1 - Heat Addition (P = C)

Boiler or SteamGenerator

Turbine

Condenser

Pump

WP

QAQR

W t

1

2

34

Major Components of a Rankine Cycle


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j p y1. Steam Generator or Boiler: The working substance absorbs heat

from products of combustion or other sources of heat at constant

pressure which in turn changes the state of the working substance(water or steam) from sub-cooled liquid and finally to superheatedvapor whence at this point it enters the turbine.

2. Steam Turbine: A steady state, steady flow device where steamexpands isentropically to a lower pressure converting some formsof energy (h, KE, PE) to mechanical work that finally be convertedinto electrical energy if the turbine is used to drive an electric gene-rator.

3. Condenser: Steam exiting from the turbine enters this device to re-

ject heat to the cooling medium and changes its state to that of thesaturated liquid at the condenser pressure which occurred at a cons-tant pressure process.

4. Pump: It is also a steady state, steady flow machine where the


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condensate leaving the condenser at lower pressure be pumpedback to the boiler in an isentropic process in order to raise thepressure of the condensate to that of the boiler pressure.

h

S S

T

3

42

1

3

4

1

2

P1

P2

P1

P2

4 2

2

4

Turbine Work ) Id l C l


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a) Ideal CycleW t = (h 1 - h 2) KJ/kgW t = m s(h1 - h 2) KW

b) Actual CycleW t = (h 1 - h 2) KJ/kgW t = m s(h1 - h 2) KW

where: m s - steam flow rate in kg/sec

Turbine Efficiency

100%xhhhh

100%xWW

t

21

2'1t

t

t'

Pump Work ) Id l C l


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a) Ideal CycleWP = (h 4 - h 3) KJ/kgWP = m s(h4 - h 3) KW

b) Actual CycleWP = (h 4 - h 3) KJ/kgWP = m s(h4 - h 3) KW

Pump Efficiency

100%xhh

hh

100%xWW

34'

34

p

p'

pp


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Heat Rejecteda) Ideal Cycle

QR = (h 2 - h 3) KJ/kgQR = m s(h2 - h 3) KWQR = m s(h2 - h 3) KW = mwCpw(two - twi) KW

b) Actual Cycle

QR = (h 2 - h 3) KJ/kgQR = m s(h2 - h 3) KW = m wCpw(two - twi) KW

Where: m w - cooling water flow rate in kg/sectwi - inlet temperature of cooling water in C

two - outlet temperature of cooling water in CCpw - specific heat of water in KJ/kg- C or KJ/kg- KCpw = 4.187 KJ/kg- C or KJ/kg- K

Heat Added:a) Ideal Cycle


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a) Ideal CycleQA = (h 1 - h 4) KJ/kgQA = m s (h1 - h 4) KW

b) Actual CycleQA = (h 1 - h 4) KJ/kgQA = m s (h1 - h 4) KW

Steam Generator or boiler Efficiency

100%x(HV)m

)h(hm

100%xQ

Q

f

41sB

S

AB

Where: Q A - heat absorbed by boiler in KWQS - heat supplied in KWm f - fuel consumption in kg/secHV - heating value of fuel in KJ/kg

Steam Rate


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KW-seckg

ProducedKW

rateFlowSteamSR

Heat Rate

KW-secKJ

ProducedKWSuppliedHeat

HR

Reheat CycleA steam power plant operating on a reheat cycle improves the thermalefficiency of a simple Rankine cycle plant. After partial expansion of the steam in the turbine, the steam flows back to a section in the boiler

which is the re-heater and it will be reheated almost the same to itsinitial temperature and expands finally in the turbine to the con-

denser pressure.


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Reheater

QA

WP

QR

W t

1 kg

1 23

4

56

Regenerative CycleIn a regenerative cycle, after partial expansion of the steam in theturbine, some part of it is extracted for feed-water heating in an open orclose type feed-water heater. The bled steam heats the condensate fromthe condenser or drains from the previous heater causing a decrease inheat absorbed by steam in the boiler which result to an increase in

thermal efficiency of the cycle.

W1 kg


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QA

W P1

QR

W t1

2

3

456

7

W P2

m

Reheat-Regenerative Cycle

For a reheat - regenerative cycle power plant, part of the steam is re-heated in the re-heater and some portion is bled for feed-water heatingto an open or closed type heaters after its partial expansion in theturbine. It will result to a further increase in thermal efficiency of the

plant.

1-m

1-m


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QA

WP1

QR

W t

1 kg

1

2

4

5678

WP2

m

23

For a 1 kg basis of circulating steam, m is the fraction of steamextracted for feed-water heating as shown on the schematic diagramabove where the reheat and bled steam pressure are the same

Thermodynamics 091215174947 Phpapp02.Ppt.1

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Transcript of Thermodynamics 091215174947 Phpapp02.Ppt.1