12 Energy and Energy Balances

SKF2123-ENERGY BALANCE SESSION 2007/2008

LECTURER : ASSOC. PROF. DR AZEMAN MUSTAFA

7- 1

Energy & Energy Balances Energy & Energy Balances

Azeman Mustafa, PhD

ENERGY BALANCE (SKF 2123) ENERGY BALANCE (SKF 2123)

Chapter 7Chapter 7

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7- 3

7- 4

- What is energy?

- Forms of Energy

- Kinetic energy (KE)

- Potential energy (PE)

- Internal energy (U)

Total Energy , E = KE + PE + U

7.1-7.2 Energy & 1st Law of Thermodynamics

2

21mVKE =

mgzPE =



7- 5

Change in kinetic energy:

Note: ∆ means “change” and is always calculated as “final value minus initial value”

)(21 2

12

212 VVmKEKEKE −=−=∆

)( 1212 zzmgPEPEPE −=−=∆Change in potential energy

Change in potential energy

12UUU −=∆

7- 6

How energy can be transferred between a system and its surroundings?

Heat – energy that flows as a result of temperature difference between a system and its surrounding ; heat is defined positive when it is transferred to the system from the surroundings.

Work – energy that flows in response to any driving force other than a temperature difference. ; work is defined positive when it is done by the system on the surroundings.



7- 7

Types of Work

• Flow work (Wfl) - energy carried across the boundaries of a system with the mass flowing across the boundaries (i.e. internal, kinetic & potential energy)

• Shaft work (Ws) - energy in transition across the boundaries of a system due to a driving force other than temperature, and not associated with mass flow (an example would be mechanical work due to a piston, pump or compressor)

7- 8

ENERGY – CONVERSION UNITS

1 newton (N) = 1 kg. m/s2

1 dyne = 1 g.cm/s2

1 lbf = 32.174 lbm.ft/s2



7- 9

Example 1

1. An automobile weighing 2500 lbm is traveling at 55 miles per hour when the brakes are suddenly applied bringing the vehicle to a stop. After the brakes have cooled to the ambient temperature, how much heat (in Btu) has been transferred from the brakes to the surroundings?

2. Suppose you pour a gallon of water from a height of 10 feet. Howmuch potential energy (ft-Ibf) does the water lose? How fast is the water traveling (m/s) just before the impact?

3. Air at 300oC and 130 kPa flows through a horizontal 7 cm ID pipe at a velocity of 42 cm/sec.

a) Calculate Ek (J/s), assuming ideal gas behaviourb) If the air is heated to 400oC at constant pressure, what is

the rate of change of kinetic energy (J/s) ?

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Example 1 (Solution no. 1)

Q = KE2 – KE1

KE2

KE1



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Suppose you pour a gallon of water from a height of 10 feet. How much potential energy (ft-Ibf) does the water lose?

How fast is the water traveling (m/s) just before the impact?

0 W0,Q 0,∆U s ===

sWQUPEKE −=∆+∆+∆

( )fIbftVm

VVmKE

.4.83)0(21

)(21

222

21

22

−−=−=

−=∆

m = 1 gallon = ? kg∆KE = 83.4 ft.Ibf = ? kg.m2/s2

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a) Calculate Ek (W), assuming ideal gas behaviour

What method is used to calculate m?What is an alternative method to calculate m ?



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b) If the air is heated to 400oC at constant pressure what is ∆Ek (300oC 400oC)?

What method is used to calculate velocity?What is an alternative method to calculate velocity ?

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General Balance Equation

A balance on conserved quantity (i.e. mass, energy, momentum) in a process system may be written as:

Input + generation - output - consumption = accumulation



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7.3 Energy Balance on Closed Systems

• How do you describe a closed system control volume?

• What effect does this have on the mass and energy balances?

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• There is no mass transfer into a closed system • The only way energy can get into or out of a closed

system is by heat transfer or work

a. Heat transfer (Q):b. Work (Ws):

Note: * Work is any boundary interaction that is not heat (mechanical, electrical, magnetic, etc.)

Ws

Q



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First Law of Thermodynamics

Energy can neither be created nor destroyed ; It can only change forms

Input + generation - output - consumption

= accumulation

∴∴ Input Input -- output = accumulationoutput = accumulation

00

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In a closed system, In a closed system,

no mass crosses the boundary, hence the input & no mass crosses the boundary, hence the input & output terms are eliminatedoutput terms are eliminatedenergy can be transferred across the boundary as heat energy can be transferred across the boundary as heat & work, hence the accumulation term may be defined & work, hence the accumulation term may be defined as the change in total energy in the system, i.e.as the change in total energy in the system, i.e.

s W- Q E -

energy system totalin the Change

System in theEnergy Total Initial

System in theEnergy totalFinal

ifE

=

−



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s

if

WQ∆EEE∆E

U PE KE E

−=−=

++=

sWQUPEKEUPEKEE−=∆+∆+∆

∆+∆+∆=∆

Q = heat transferred to the systemWs = work done by the system

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∆E = ∆U + ∆PE + ∆KE = Q – W

Note: (Summation of all heat transfer across system boundary)

(Summation of all work across system boundary)

∑=i

iQQ

∑=i

is,s WW



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For a closed system what is ∆E equal to?

sWQUPEKEE −=∆+∆+∆=∆

• Is it adiabatic? (if yes, Q = 0) • Are there moving parts, e.g. do the walls move? (if

no, Ws = 0) • Is the system moving? (if no, ∆KE = 0) • Is there a change in elevation of the system? (if no,

∆PE = 0 ) • Does temperature, phase, chemical composition

change, or pressure change less than a few atmospheres ? (if no to all, ∆U = 0)

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Example 2

A closed system of mass 5 kg undergoes a process in which there is work of magnitude 9 kJ to the system from the surroundings. Theelevation of the system increases by 700 m during the process. The specific internal energy of the system decreases by 6 kJ/kg and there is no change in kinetic energy of the system. The acceleration of gravity is constant at g=9.6 m/s2. Determine the heat transfer, in kJ.

General energy bal. : ∆U + ∆PE + ∆KE = Q – W

W = - 9 KJ ∆PE = mg∆z = (5kg)(9.6 m/s2)(+700m)(N/1 kg.m/s2)

= 33600 N/m = 33600 J = 33.6 kJ∆U = m∆Û = (- 6 kJ/kg)(5kg) = - 30 kJ∆KE = 0

thus, …….. Q = - 5.4 kJ



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Example 3

5 kg of steam contained within a piston-cylinder assembly undergoes an expansion from state 1, where the specific internal energy isu1=2709.9 kJ/kg, to state 2, where u2=2659.6 kJ/kg. During the process, there is heat transfer to the steam with a magnitude of 80 kJ. Also, a paddle wheel transfers energy to the steam by work in the amount of 18.5 kJ. There is no significant change in the kinetic or potential energy of the steam. Determine the energy transfer bywork from the steam to the piston during the process, in kJ.

m= 5 kg steam

1 2

Q = 80 kJ

Wpw = -18.5 kJ

Wpiston

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Example 3 (Solution)



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Working session 1A cylinder with a movable piston contains 4.00 liters of a gas at 30°C and 5.00 bar. The piston is slowly moved to compress the gas to 8.00 bar.

State 1 : V = 4 L, T = 30oC, P = 5 bar

State 2 : V = ? L, T = ? oC, P = 8 bar

12

Wpiston

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Working session 1 (Cont.)(a) Considering the system to be the gas in the cylinder and

neglecting ∆Ep, write and simplify the closed-system energy balance. Do not assume that the process is isothermal in this part.

(b) Suppose now that the process is carried out isothermally, and the compression work done on the gas equals 7.65 L-bar. If the gas is ideal so that U is a function only of T, how much heat (in joules) is transferred to or from (state which) the surroundings? (Use the gas-constant table in the back of the book to determine the factor needed to convert L-bar to joules.)



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Working session 1 (Cont.)

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Working session 1 (Cont.)

(c) Suppose instead that the process is adiabatic and that U increases as T increases. Is the final system temperature greater than, equal to, or less than 30°C? (Briefly state your reasoning.)

∆U = Q – W

Adiabatic compression,

Q=0 and ∆U = - W

since W has a negative value (work done on the system), ∆U > 0, hence the final temperature should be greater than 30°C



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7.4 Energy Balances on Open Systems

How are open systems control volumes different from closed systemsWhat effect does this have on the energy balance?

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Some common open system steady flow devices

Only one in and one out



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Mixer

Heat Exchanger

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Application of First Law - Conservation of energy for a control volume

+

−

=

m flow massngaccompanyi

volume controlthe intotransfer

energy of ratenet

t timeat work byout dtransferre

being is energywhichat ratenet

t timeat transferheat by

in dtransferrebeing is energywhichat ratenet

t timeat volume control the

within contained energy the of

change of rate time

&

[ ] [ ]outincv

cvcv

PEKEUPEKEUE

EWQtE

++−++=∆

∆+−=∂

∂ &&&



7- 33

General Energy Balances on Open Systems at Steady State

W-Q E

energy system totalin the Change

System theEnteringEnergy Total

System theLeavingEnergy Total

in&&&&

outE

=

−

WQE

EEE inout

&&&

&&&

−=∆

−=∆

WQUPEKEUPEKEE&&

&

−=∆+∆+∆

∆+∆+∆=∆

W - Q E hence, 0 CV&&& =∆=

∂∂tEcv

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( )( )( )

WQgzvUm

mmm

WQPEKEU

if

&&&

&&&

&

&

&

&&

−=

++∆

==−=

−=

−=

−=∆+∆+∆

2ˆ

zzgm∆PE

vvm21∆KE

UUm∆U

2

if

2i

2f

if



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Example 4 (…. Recalling example 1 (no. 3)) Air at 300oC and 130 kPa flows through a horizontal 7 cm ID pipe at a velocity of 42 cm/sec

a) Write and simplify the energy balanceb) Calculate the rate of kinetic energy (W), if the air is heated to 400oC at

constant pressure, assuming ideal gas behaviourc) Why would be correct to say that the rate of heat transfer to the gas equals

the rate of change of kinetic energy? Why?

1 2

Air

T1 =300oCP1=130 kPaV1 = 42 m/s

T2 =400oCP2=130 kPaV2 = ? m/s

Q

7 cm ID

7- 36

Example 4 (Solution)

a) Write and simplify the energy balance

∆E = ∆U + ∆Ek + ∆Ep

b) Calculate Ek (W), assuming ideal gas behaviour

0



7- 37

Example 4 (Solution)c) If the air is heated to 400oC at constant pressure what is ∆Ek (300oC

400oC)?

d) Why would be incorrect to say that the rate of heat transfer to the gas in part (c) must equal the rate of change of kinetic energy?

∆E = ∆U + ∆Ek ….hence ∆E ≠ ∆Ek

7- 38

7.4a Types of Work

Recall …. How energy can be transferred across boundaries of

a closed system ?

an open system?



7- 39

For open systems, two types of work involved

Shaft work , Ws….. Work done by a moving part of a system

Flow works , Wfl (PV) ….. Work done to put a mass of substance into/outside of system boundary

Consider pipe full of flowing fluid (flow due to ∆P where∆P = P1 – P2):

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Consider volume defined by dashed lines “system”, where V = A·L

where P1V1 is work done on system while P2V2 is work done by system on surroundings

/s))(m.....(N/mVPVPW 32ininoutoutfl&&& −=

1

1

2

2



7- 41

7.4b Specific properties and Enthalpy

Total Energy of a flowing fluid (open system)

ininoutoutfl

sfl

VPVPW

WWQPEKEUWQPEKEU

&&&

&&&

&&

−=

+−=∆+∆+∆

−=∆+∆+∆

)(

The fluid possesses an additional form of energy –the flow energy (flow work)

( )...)/,/( ......ˆˆˆ

..) cal (Joule, .......

kgcalkgJEnthalpySpecificVPUH

EnthalpyVPUHWQPEKEH s

+=

∆+∆=∆

−=++∆&

&&

Shaft work

7- 42

Example 5

a. Oxygen at 150 K and 41.64 atm has a tabulated specific volume of 4.684 cm3/g and a specific internal energy of 1706 J/mol. Calculate the specific enthalpy of oxygen this state.



7- 43

Example 6

b. Superheated steam at 5 bar and 200 oC has a tabulated specific internal energy of 2643 kJ/kg. Calculate the specific enthalpy of steam this state assuming ideal gas behaviour.

( )

kgm0.436

1000Lmx

molLx7.865

0.018kgmol

nV

M1

mVV

molL7.865

5bar

K273200x mol.KL.bar0.08314

PRT

nV

nRTPV

33

==

==

=

+==

=

ˆ

gas ideal assuming ..... ? V

VP UH

=

+=ˆ

ˆˆˆ

Compare the specific volume & enthalpy with the values in Table B.5

( )

kgkJ 2862 H

kgkJ 219

1000JkJx

mol.KJ 8.314x

L.bar 0.08314mol.Kx

m

L10xkgm4380x 5bar VP

kgkJ 2643 U

VP UH

3

33

=

=

=

=

+=

ˆ

.ˆ

ˆ

ˆˆˆ

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Working session 2

A horizontal cylinder equipped with a frictionless piston contains 785 cm3 of steam at 400 K and 125 kPa. A total of 83.8 joules of heat is transferred to the steam, causing the steam temperature to rise and the cylinder volume to increase. A constant restraining force is maintained on the piston throughout the expansion, so that the pressure exerted by the piston on the steam remains constant at 125 kPa. The final steam temperature is 480 K. Assuming ideal gas behaviour, calculate

a. the expansion work done (Joule) by the steamb. the change in internal energy (Joule) of the steam



7- 45

Working session 2 (Solution)

State 1 :V1 = 785 cm3 , T1 = 400K, P1 = 125 KPa

State 2 :V2 = ? cm3, T2 = 480K, P2 = 125 kPa

a. the expansion work done (Joule) by the steam

1 2

Q = 83.8 J Wpiston = ? J

7- 46

Working session 2 (Solution)

b. the change in internal energy (Joule) of the steam



7- 47

7.4c Energy balance on an open system at steady state

tEgzVHmWgzVHmQ cv

outout

outoutsinin

inin ∂∂

=

++−−

+++ ∑∑

••••

2

ˆˆ2

ˆˆ22

This work represents everything but the flow work

The flow work is included in the enthalpy term

Input - Output = Accumulation

7- 48

0t

Ecv

outin mm=

∂∂

= ∑∑••

022

22

=

++−−

+++ ∑∑

••••

outout

outoutinin

inin gzVHmsgzVHm WQ

sgzVHmgzVHm WQinin

ininoutout

outout

••••

−=

++−

++ ∑∑ 22

22

Energy Balance on Open Systems at Steady State



7- 49

For an open system what is ∆E equal to?

spk WQE∆E∆ H∆E∆ &&&&&& −=++=

• Is it adiabatic? (if yes, Q = 0) • Are there moving parts, e.g. pump, compressor,

turbine ? (if no, Ws = 0) • Does the average velocity of the fluid change

between the input and the output? ? (if no, ∆KE = 0) • Is there a change in elevation of the system between

the input and the output? ? (if no, ∆PE = 0 ) • Does temperature, phase, chemical composition or

pressure change? (if no to all, ∆H = 0)

7- 50

Single Stream Steady Flow System

NozzlesDiffusersTurbinesCompressorsThrottling Valve

Often the change in kinetic energy of the fluid is small, and the change in potential energy of the fluid is small

( )

−+

−+−=−

•••

inoutinout

inouts zzgVVHHmWQ2

ˆˆ22



7- 51

Nozzles and Diffusers

A nozzle is a device that increases the velocity of a fluid at the expense of pressure

A diffuser is a device that slows a fluid down

nozzle

diffuser

7- 52

Example



7- 53

( )

−+

−+−=−

•••

inoutinout

inouts zzgVVHHmWQ2

ˆˆ22

Is there work in this system? NOIs there heat transfer? Usually it can be ignoredDoes the fluid change elevation?

( )

−

+−=2

ˆˆ022inout

inoutVVHH

enthalpy is converted into kinetic energy

NO

7- 54

Turbines and Compressors

A turbine is a device that produces work at the expense of temperature and pressure

A compressor is a device that increases the pressure of a fluid by adding work to the system

low p

sW&

high p

turbine compressor

sW&

high p

low p



7- 55

( )

−+

−+−=−

→→•••

inoutinout

inouts zzgVVHHmWQ2

ˆˆ22

Is there work in this system? Yes!Is there heat transfer? Usually it can be ignored

Does the fluid change elevation? Usually it can be ignoredDoes the kinetic energy change? Usually it can be ignored

( )inouts HHmW ˆˆ −=− &enthalpy is converted into work

7- 56

Throttling Valve

A throttling valve reduces the fluid pressure

For example, the water that comes into your house goes through a throttling valve, so it doesn’t have excessive pressure in your home.



7- 57

Is there work in this system? NOIs there heat transfer? Usually it can be ignoredDoes the fluid change elevation? NO

Does the fluid change velocity? Usually it can be ignored

( )

−+

−+−=−

→→•••

inoutinout

inouts zzgVVHHmWQ2

ˆˆ22

7- 58

hin = hout

Pin > Pout

For gases that are not ideal, the temperature goes down in a throttling valveFor ideal gases

∆H = Cp ∆TBut ∆H = 0So… ∆T = 0The inlet and outlet temperatures are the same!!!

inout HH ˆˆ0 −=



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Mixing Chamber

Mixing two or more fluids is a common engineering process

Mixing Chamber

7- 60

Mixing Chamber

We no longer have only one inlet and one exit stream

Is there any work done? NoIs there any heat transferred? NoIs there a velocity change? No

Is there an elevation change? No

sgzVHmgzVHm WQinin

ininoutout

outout

••••

−=

++−

++ ∑∑ 22

22

∑∑••

= inoutmm

∑∑ −= ininoutout HmHm ˆˆ0 &



7- 61

Heat Exchanger

A heat exchanger is a device where two moving fluids exchange heat without mixing.

7- 62

Energy balance is the same as a mixing chamber, but…

Two inletsTwo outlets

Material BalanceDivide into two separate streams with equal inlet and outlet flow rates



7- 63

7.5 Tables of Thermodynamic Data

Recall …. Thermodynamics

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7- 65

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• State property – a property of a system component whose value depends only on the state of the system (i.e. temperature, pressure, phase and composition)… e.g. internal energy (U) and hence, enthalpy (H)

• It is impossible to measure the absolute value of state property … but can estimate the change in specific value of U (i.e ∆Û) or H (i.e ∆Ĥ) corresponding to a specified change of state (i.e. temperature, pressure, phase and composition)

• Reference state - specified state (i.e. temperature, pressure or state of aggregation) assigned to measure relative changes in Û or Ĥ… thus, the value of Û or Ĥ of a certain material at a specified state (T,P or phase) is relative to the value of Û or Ĥ of the same material at other specified state (T,P or phase)

Reference states and state properties



7- 67

• change in specific value of U (i.e ∆Û) or H (i.e ∆Ĥ) for the transition from one tabulated state to other, i.e state 1 state 2

As internal energy & enthalpy are state property, reference state is not required…. Just determine the specific value at state 1 and state 2, and calculate the difference, i.e.

∆Û = Û2 – Û1

∆Ĥ = Ĥ2 – Ĥ1

State 2

State 1

7- 68



7- 69

Recall…1. E is always measured relative to reference point!

• Reference plane for PE• Reference frame for KE• Reference state for Û or Ĥ (i.e. usually, but not

necessarily Û or Ĥ = 0)And…1. Changes in E are important, not total values of E2. ∆E depends only on beginning and end states3. Q and W depend on process path (could get to the same

end state with different combinations of Q and W)

7- 70

Example 7

Values of the specific internal energy of Bromine at three conditions are listed here

a. What reference state was used to generate the listed internal energies?b. Calculate ∆Û (kJ/mol) for a process in which Bromine vapor at 300K is

condensed at constant pressure. Calculate also ∆Ĥ (kJ/mol) for the same process.

c. Bromine vapor in a 5 liter-container at 300 K and 0.205 bar is to be heated to 340 K. Calculate the heat (kJ) that must be transferred to the gas to achieve the desired temperature increase, assuming that Û is independent of pressure.

0.00028.2429.62

0.051679.9420.92

0.3100.3101.33

300300340

LiquidVaporvapor

Û(kJ/mol)V (L/mol)P (bar)T(K)State



7- 71

Example 7 (solution)

Values of the specific internal energy of Bromine at three conditions are listed here

a. What reference state was used to generate the listed internal energies?

b. Calculate ∆Û (kJ/mol) for a process in which Bromine vapor at 300K is condensed at constant pressure. Calculate also ∆Ĥ (kJ/mol) for the same process.

0.00028.2429.62

0.051679.9420.92

0.3100.3101.33

300300340

LiquidVaporvapor

Û(kJ/mol)V (L/mol)P (bar)T(K)State

7- 72


∆Û (kJ/mol) = Û2-Û1 =

∆Ĥ (kJ/mol) = ∆Û + ∆(PV) = ∆Û + P∆V

c. Bromine vapor in a 5 liter-container at 300 K and 0.205 bar is to be heated to 340 K. Calculate the heat (kJ) that must be transferred to the gas to achieve the desired temperature increase, assuming that Û is independent of pressure.



7- 73


state 1 : V=5L, T1=300K, P=0.205bar

state 2 : V=5L, T2=340K, P2=? Barand Û # f(P)

General energy bal. for closed system

∆U + ∆PE + ∆KE = Q – Ws ….. ∆PE, ∆KE, Ws = 0Q =∆U = n∆Û∆Û= 29.62 - 28.24 = 1.38 kJ/mol

Assuming ideal gas

kJ 0.0567Un∆Q == ˆ( )( ) mol 0.0411300Kx

mol.KL.bar0.08314

5Lbar 0.205 n

nRTPV

==

=

7- 74

Working session 3

a. Steam at 25oC and 0.0317 bar has a tabulated specific internal energy of 2409.9 kJ/kg. Calculate the specific enthalpy of steamthis state



7- 75

Working session 3

b. The specific enthalpy of steam at 125 kPa varies with temperature approximately as

i) Estimate the internal energy of the steam at 480 Kii) Derive a formula for Ĥ (BTU/Ibm) as a function of T(R)

T(K)5.35 980,34 (J/mol) H +=

7- 76

7.5b Steam Properties

Recall …. Thermodynamics



7- 77

Property of Steam Tables

P – pressure T - temperaturev – specific volumeu – specific internal energyh – specific enthalpy h = u + Pvs – specific entropy

7- 78



7- 79

7- 80

Superheated Properties



7- 81

7.6 Energy Balance Calculation Procedures

1. Select a suitable control volume for analysis, and sketch the system, indicating appropriate boundaries

2. Determining what energy interaction are important, and recognize the sign conventions on such terms

3. Start with the basic 1st law (energy balance) for the chosen system

4. Obtain physical date for the substance under study. Is an equation of state applicable, or must graphical and/or tabular data be employed? What are other property relations for the substance?

7- 82

5. Determine the path of the process between the initial and final states and indicate it in a diagram. Is the process isothermal, isobaric, quasistatic, adiabatic, etc.?

6. What other idealizations or assumptions are necessary to complete the solution? Are kinetic and potential energies negligible, etc.?

7. Draw a suitable diagram for the process, as an aid in picturing the overall problem.

8. Complete the solution for the required item(s) on the basis of the information supplied

Note: * check the units in each equation used!!!



7- 83

Example 8

Air is heated from 25°C to 150°C prior to entering a combustion furnace. The change in specific enthalpy associated with this transition is 3640 J/mol. The flow rate of air at the heater outlet is 1.25 m3/min and the air pressure at this point is 122 kPa

absolute.

Q

HeaterAir

25°C

V=1.25 m3/min T=150°CP=122 kPa

∆Ĥ=3640 J/mol

7- 84


a. Calculate the heat requirement in kW, assuming ideal gas behaviour and that kinetic and potential energy changes from the heater inlet to the outlet are negligible.

General energy bal. for open system

∆H + ∆PE + ∆KE = Q – Ws

and ∆PE=0, ∆KE=0 and W = 0 (why?)

Hence, ∆H = Q and ∆H = 3640 J/molQ = 3640 J/mol but the unit required is kW (kJ/s)

Q = 3640 J/mol x nair mol/s x kJ/1000J …… (kJ/s)



7- 85


Recall mass bal. (SKF1113), for ideal gas,

PV = nRT or use standard ideal gas conversion

thus, Q = 2.631 kW

( )

( ) smol 7230

423Kmol.K

.Pam8.314

60smin

minm1.25Pa 122000

RTPVn

3

3

.=

==

7- 86

Example 9 : Mixing Chamber

Given: showerinlet 1: T1 = 15 oC (cold water)inlet 2: T2 = 60 oC (hot water)

Find: What is the ratio of mass flow rates (hot/cold) in order to get a shower temperature of T3 = 38 oC

Assumptions: (1) steady state steady flow (SSSF)(2) no work, (3) ∆PE=0, ∆KE=0,(4) adiabatic, (5) incompressible water

T2

T1

T3



7- 87

Example 9 : Mixing ChamberBasic equations:(open system)

Solution:

Then

321 mmm &&& =+ 332211 HmHmHm ˆˆˆ &&& =+

32

13

1

2

31

22

1

21

3212211

HHHH

mm

H)mm(1H

mmH

H)mm(HmHm

ˆˆˆˆ

ˆˆˆ

ˆˆˆ

−−

=

+=+

+=+

&

&

&

&

&

&

&&&&

∑∑ =in

inout

out mm &&

∑∑ =in

ininout

outout HmHm ˆˆ &&

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Example 9 : Mixing Chamber (Solution)

kgkJ 159.1C)38(THH

kgkJ 251.1C)60(THH

kgkJ 62.95C)15(THH

o3f3

o2f2

o1f1

===

===

===

ˆˆ

ˆˆ

ˆˆ

12

1

2

mm

1.05

kgkJ159.1)-(251.1

kgkJ62.95)-(159.1

mm

&&

&

&

≈

≅=



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Working session 4 : Heat Exchanger

Saturated steam at 2.00 barg is to be used to heat a stream of ethane at constant pressure. The ethane enters the heat exchanger at 16°C and 1.5 barg at a rate of 795 m3/min and is heated at constant pressure to 93°C. The steam condenses and leaves the exchanger as a liquid at 27°C. The specific enthalpy of ethane at the given pressure is 941 kJ/kg at 16°C and 1073 kJ/kg at 93°C

a. How much heat (kW) must be provided to heat the ethane from 16°C to 93°C?

b. Assuming that all the energy transferred from the steam goes to heat ethane, at what rate in m3/min must steam be supplied to the heat exchanger?

T2

T1

T3

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Example 10

Two hundred kg/min of steam enters a perfectly - insulated steam turbine at 350°C and 40 bar through a 7.5-cm diameter line and exits at 75°C and 5 bar through a 5-cm line. How much energy is transferred to or from the turbine?

State 1

350°C40 bar

ID=7.5-cm

State 2

75°C5 bar ID= 5-cm

Energy bal. for open system

∆H + ∆PE + ∆KE = Q – Ws

Assumptions :

∆PE = 0, Q = 0 (adiabatic)

Hence, ∆H + ∆KE = – Ws



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Hence, Ẃs = 13,460 kW (work produced by turbine)

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Working session 5

A turbine discharges 200 kg/h of saturated steam at 10.0 bar absolute. It is desired to generate steam at 250°C and 10.0 bar by mixing the turbine discharge with a second stream of superheated steam of 300°C and 10.0 bar.

a. If 300 kg/h of the product steam is to be generated, how much heat must be added to the mixer?

b. If instead the mixing is carried out adiabatically, at what rate is the product steam generated

Saturated steam200 kg/h, 10 bar

Superheated steam300oC, 10 bar

Steam250oC, 10 bar



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Working session 5 (solution)

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Example 11

Atmospheric air at 38°C and 97% relative humidity is to be cooled to 18°C and fed to a plant area at the rate of 510 m3/min. Assuming the atmospheric pressure is 760 mm Hg,

(a) calculate the rate (kg/min) at which water condenses(b) calculate the cooling requirement in tons (1 ton of cooling =

12,000 BTU/h), assuming the enthalpy of water vapor is that of saturated steam at the same temperature and the enthalpy of dry air is given by the equation

Ĥ (kJ/mol) = 0.0291 [ T(oC) – 25 ]



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n1 kmol/min38oCRH=97%Yw,1 = ? mol H2O/molYair,1 = ? mol air/mol

cooler

n2 kmol water /min condensed18oC

510 m3/min18oCn3 kmol/min Yw,3 = ? kmol H2O/molYair,3 = ? kmol air/mol

Q&

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7- 97


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Working session 6

A 10.0-m3 tank contains steam at 275°C and 15.0 bar. The tank and its contents are cooled until the pressure drops to 1.2 bar.Some of the steam condenses in the process.

(a) What is the final temperature of the tank contents?(b) How much steam condensed (kg)?(c) How much heat was transferred from the tank?



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7.7 Mechanical Energy Balances for Steady-State Flow Processes

loss friction0mQUF

mW - Fzg

2vP

ρ1

mV V ,VPUH

W- Q zg2vHm

s

2

s

2

≥−∆=

=+∆+∆+ρ

∆

==∆+∆=∆

=

∆+∆+∆

&

&

&

&

&&&Energy balance for an isothermal (constant T) & steady state flow of an incompressible fluid (ρ is constant - for liquids) through a piping system (min = mout)

∆∆ ∆

P v g zρ

+ + =2

20

If there is no shaft work (Ws = 0, i.e., no pump, compressor, etc.), and if the friction losses can be neglected (F = 0), then Bernoulli’s equationresults:

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Example 12

Water from a reservoir passes a dam through a turbine and discharges from a 70 cm- ID pipe at a point 65 m below the reservoir surface. The turbine delivers 0.8 MW. Calculate the required flow rate in m3/min if friction is neglected.



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Working session 6 : Pipe flow

Given: Water is pumped on higher elevation through piping

D1 = 10 cm and D2 = 15 cm; T2 = T1 = Tatm = 20oCp2 = p1 = patm = 101.3 kPa

Find: Pump power

Assumptions:

(1) steady state steady flow (SSSF)(2) adiabatic(3) incompressible water

T2

T1

T3

.



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Problem 1 from Test #1 (session 0506)

Which of the following process requires the most energy. Elaborate and prove your answers

a) Lifting 100 kg of liquid water to a height of 32 feet b) Raising the temperature of 1 kg of liquid water by

10oC without phase changec) Converting 1 kg of liquid water at room

temperature 25 oC and pressure to water vapour.

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Half liter of water in a glass at 64.4oF is to be cooled by adding ice and stirring. The enthalpy of the ice relative to liquid water at 32oF and the specific volume of the ice are -150 BTU/Ibm and 0.01745 ft3/Ibm, respectively. Calculate the mass of ice (Ibm) that must melt to bring the liquid temperature to 41oF, neglecting heat losses to the surroundings.

mi Ibm, Ice32oF (0oC)

ml Ibm½ L water 64.4oF (18oC)

mmix Ibm41 oF (5oC)

Energy Balance for closed system :

∆U + ∆PE + ∆KE = Q – Wpiston∆PE = ∆KE = Q = Wpiston=0



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Problem 3 from Test #1 (session 0506)Hence, ∆U = 0 = Σ (mÛ)in- Σ (mÛ)out= 0

mi Ûi + ml Ûl - mmixÛmix = 0

ml = (0.5L)(35.3145 ft3/L)(62.34 Ibm/ft3) = 1.1023 Ibmmi + ml = mmix

Reference for internal energy @ 32oF (0oC) ….. Table B5

Ûl (liquid water, 18oC) = 75.5 kJ/kg = 32.486 BTU/IbmÛmix (liquid water, 5oC) = 21 kJ/kg = 9.0358 BTU/Ibm

Ûi (0oC) = ? = Ĥ + PV = -150 BTU/Ibm + (1 atm)(0.1745ft3/Ibm)(1.987

BTU/0.7302 ft3.atm)= -150.48 BTU/Ibm

Performing simultaneous mass & energy bal….. mi= 0.1621 Ibm

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At most chemical plants, water is stored in an elevated tank that provides sufficient water pressure for emergency use if the municipal water supply is interrupted. At a petroleum refinery in Kertih, Terangganu, water is stored in an elevated cylindrical storage tank 5 m in diameter and 15 m tall. The elevated tank is vented to the atmosphere. It is painted white to minimize heating by sunlight. It is 25 m from the ground to the bottom of the tank. Water may be withdrawn from the elevated tank at ground level via a 4-inch inside diameter pipe.

a. Estimate the volumetric flow rate (L/min) of water available in an emergency.

b. If the tank is full how long (minutes) that the water supply canlast if the water in the tank is not being replenished during the time of emergency.



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A boiler at a power plant produces 1.33 × 106 kg/hr of steam at 550 °C and 250 bar which feeds a turbine. Steam leaves the turbine at 350 °C and 60 bar. Ten percent of this steam is used to preheat the water going into the boiler; the rest feeds other turbines in the plant. Steam leaves the preheater at 250 °C and 60 bar. Water enters the preheater at 245 °C and 250 bar. After leaving the preheater it goes directly to the boiler.

a. Sketch the diagram of the process and its labels. b. Calculate the work (hp) done by the turbine c. The energy (Btu/s) that must be transferred to the water in

the boiler

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Air at a flow rate of 50 kg/hr is compressed adiabatically from 1 bar and 0 °C to 3 bar and 25 °C. The velocity of the air stream entering the compressor is 5.0 m/s. The velocity of the air stream exiting the compressor is 60.0 m/s. The air stream exits the compressor 3 m higher than it enters.

a. What is the volumetric flow rate (m3/hr) of the air exiting the compressor?

b. What is the inside diameter (inch) of the outlet duct of the compressor?



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A 10000 Liter tank contains steam at 250 ºC and 20.0 bar. The tank and its contents are cooled until the pressure drops to 1.2 bar. Some of the steam condenses in the process.

a. What is the final temperature (ºC) of the tank contents? b. How much steam condensed (kg)? c. How much heat (kJ) was transferred from the tank?

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12 Energy and Energy Balances

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