advanced thermodynamics: lecture 7

Advanced Thermodynamics: Lecture 7

Shivasubramanian [email protected]

Shivasubramanian Gopalakrishnan [email protected] ME 661

Entropy change for an Ideal gas

ds =du

T+

Pdv

T

substituting du = cv

dT and P = RT/v

ds = cv

dT

T+ R

dv

v

The entropy change for a process is obtained by integrating thisrelation between the end states

s2

� s1

=

Z2

1

cv

(T )dT

T+ Rln

v2

v1

s2

� s1

=

Z2

1

cp

(T )dT

T� Rln

P2

P1


Constant Specific heat analysis (approximate)

s2

� s1

= cv ,avg ln

T2

T1

+ Rlnv2

v1

(kJ/kg · K )

s2

� s1

= cp,avg ln

T2

T1

� RlnP2

P1

(kJ/kg · K )


Variable Specific heat analysis (exact)

It is convenient to perform an integral with absolute zero as areference once and tabulate the results for di↵erent temperaturesfor a function defined as

so =

ZT

0

cp

(T )dT

T

so is a function of temperature alone, and its value is zero atabsolute zero temperature.

Z2

1

cp

(T )dT

T= so

2

� so1

s2

� s1

= so2

� s01

� RlnP2

P1

(kJ/kg · K )


Entropy Change of an Ideal Gas

Air is compressed from an initial state of 100 kPa and 17C to afinal state of 600 kPa and 57C. Determine the entropy change ofair during this compression process by using (a) property valuesfrom the air table and (b) average specific heats.


Isentropic compression of an Ideal gas

lnT2

T1

= � R

cv

lnv2

v1

✓T2

T1

◆

s=const

=

✓v2

v1

◆�R

c

v

since R = cp

� cv

, k = cp

/cv

, then R/cv

= k � 1

✓T2

T1

◆

s=const

=

✓v2

v1

◆1�k

✓T2

T1

◆

s=const

=

✓P2

P1

◆ k�1

k

✓P2

P1

◆

s=const

=

✓v2

v1

◆k


Reversible Steady Flow WorkReversible (quasi-equilibrium) moving boundary work associated with closedsystems is expressed in terms of the fluid properties as

Wb

=

Z2

1

PdV

The energy balance for a steady-flow device undergoing an internally reversibleprocess can be expressed in di↵erential form as

�qrev

� �wrev

= dh + dke + dpe

�qrev

= Tds Tds = dh � vdP

Therefore�q

rev

= dh � vdP

Hence

��wrev

= �Z

2

1

vdP ��ke ��pe

Work input to steady-flow devices such as compressors and pumps as

�wrev,in =

Z2

1

vdP +�ke +�pe


Compressing a Substance in the Liquid versus Gas Phases

Determine the compressor work input required to compress steamisentropically from 100 kPa to 1 MPa, assuming that the steamexists as (a) saturated liquid and (b) saturated vapor at the inletstate.


Proof that Steady-Flow Devices Deliver the Most andConsume the Least Work when the Process Is ReversibleConsider two steady-flow devices, one reversible and the other irreversible,operating between the same inlet and exit states. The energy balance for eachof these devices can be expressed in the di↵erential form as

�qrev

� �wrev

= dh + dke + dpe

�qact

� �wact

= dh + dke + dpe

The right-hand sides of these two equations are identical since both devices areoperating between the same end states. Thus

�wrev

� �wact

= �qrev

� �qact

However�q

rev

= Tds

�wrev

� �wact

T=

�qrev

� �qact

T� 0

Since

ds � �qact

T

) �wrev

� �wact


Isentropic (Pv k ! constant):

(7–57a)

Polytropic (Pvn ! constant):

(7–57b)

Isothermal (Pv ! constant):

(7–57c)

The three processes are plotted on a P-v diagram in Fig. 7–45 for thesame inlet state and exit pressure. On a P-v diagram, the area to the left ofthe process curve is the integral of v dP. Thus it is a measure of the steady-flow compression work. It is interesting to observe from this diagram that ofthe three internally reversible cases considered, the adiabatic compression(Pv k ! constant) requires the maximum work and the isothermal compres-sion (T ! constant or Pv ! constant) requires the minimum. The workinput requirement for the polytropic case (Pvn ! constant) is between thesetwo and decreases as the polytropic exponent n is decreased, by increasingthe heat rejection during the compression process. If sufficient heat isremoved, the value of n approaches unity and the process becomes isother-mal. One common way of cooling the gas during compression is to usecooling jackets around the casing of the compressors.

Multistage Compression with IntercoolingIt is clear from these arguments that cooling a gas as it is compressed is desir-able since this reduces the required work input to the compressor. However,often it is not possible to have adequate cooling through the casing of thecompressor, and it becomes necessary to use other techniques to achieveeffective cooling. One such technique is multistage compression with inter-cooling, where the gas is compressed in stages and cooled between each stageby passing it through a heat exchanger called an intercooler. Ideally, the cool-ing process takes place at constant pressure, and the gas is cooled to the initialtemperature T1 at each intercooler. Multistage compression with intercoolingis especially attractive when a gas is to be compressed to very high pressures.

The effect of intercooling on compressor work is graphically illustrated onP-v and T-s diagrams in Fig. 7–46 for a two-stage compressor. The gas iscompressed in the first stage from P1 to an intermediate pressure Px, cooled atconstant pressure to the initial temperature T1, and compressed in the secondstage to the final pressure P2. The compression processes, in general, can bemodeled as polytropic (Pv n ! constant) where the value of n varies betweenk and 1. The colored area on the P-v diagram represents the work saved as aresult of two-stage compression with intercooling. The process paths for single-stage isothermal and polytropic processes are also shown for comparison.

wcomp,in ! RT ln¬P2

P1

wcomp,in !nR 1T2 " T1 2

n " 1!

nRT1

n " 1c a P2

P1b 1n"12>n

" 1 d

wcomp,in !kR 1T2 " T1 2

k " 1!

kRT1

k " 1c a P2

P1b 1k"12>k

" 1 dChapter 7 | 367

P

1

P2

P1

Isentropic (n = k)Polytropic (1 < n < k)Isothermal (n = 1)

v

FIGURE 7–45P-v diagrams of isentropic, polytropic,and isothermal compression processesbetween the same pressure limits.

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Isentropic (Pv k = C)

wcomp,in =

kR(T2

� T1

)

k � 1=

kRT1

k � 1

"✓P2

P1

◆(k�1)/k

� 1

#

Polytropic (Pvn = C)

wcomp,in =

nR(T2

� T1

)

n � 1=

nRT1

n � 1

"✓P2

P1

◆(n�1)/n

� 1

#

Isothermal (Pv = C)

wcomp,in = RTln

P2

P1

Image source: Thermodynamics An engineering approach by Cengel and Boles 7th Edition


Multistage Compression with Intercooling

The size of the colored area (the saved work input) varies with the valueof the intermediate pressure Px, and it is of practical interest to determinethe conditions under which this area is maximized. The total work input fora two-stage compressor is the sum of the work inputs for each stage of com-pression, as determined from Eq. 7–57b:

(7–58)

The only variable in this equation is Px. The Px value that minimizes thetotal work is determined by differentiating this expression with respect to Pxand setting the resulting expression equal to zero. It yields

(7–59)

That is, to minimize compression work during two-stage compression, thepressure ratio across each stage of the compressor must be the same. Whenthis condition is satisfied, the compression work at each stage becomesidentical, that is, wcomp I,in ! wcomp II,in.

Px ! 1P1P2 2 1>2¬or¬Px

P1!

P2

Px

!nRT1

n " 1c a Px

P1b 1n"12>n

" 1 d #nRT1

n " 1c a P2

Pxb 1n"12>n

" 1 d wcomp,in ! wcomp I,in # wcomp II,in

368 | Thermodynamics

P

1

P2

P1

Isothermal

Px

Polytropic

Work saved

Intercooling

2

T

s

1T1

Intercooling

2

P2

P1

Px

v

FIGURE 7–46P-v and T-s diagrams for a two-stagesteady-flow compression process.

EXAMPLE 7–13 Work Input for Various Compression Processes

Air is compressed steadily by a reversible compressor from an inlet state of100 kPa and 300 K to an exit pressure of 900 kPa. Determine the compressorwork per unit mass for (a) isentropic compression with k ! 1.4, (b) polytropiccompression with n ! 1.3, (c) isothermal compression, and (d) ideal two-stage compression with intercooling with a polytropic exponent of 1.3.

Solution Air is compressed reversibly from a specified state to a specifiedpressure. The compressor work is to be determined for the cases ofisentropic, polytropic, isothermal, and two-stage compression.

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Multistage Compression with Intercooling

The size of the colored area (the saved work input) varies with the valueof the intermediate pressure Px, and it is of practical interest to determinethe conditions under which this area is maximized. The total work input fora two-stage compressor is the sum of the work inputs for each stage of com-pression, as determined from Eq. 7–57b:

(7–58)

The only variable in this equation is Px. The Px value that minimizes thetotal work is determined by differentiating this expression with respect to Pxand setting the resulting expression equal to zero. It yields

(7–59)

That is, to minimize compression work during two-stage compression, thepressure ratio across each stage of the compressor must be the same. Whenthis condition is satisfied, the compression work at each stage becomesidentical, that is, wcomp I,in ! wcomp II,in.

Px ! 1P1P2 2 1>2¬or¬Px

P1!

P2

Px

!nRT1

n " 1c a Px

P1b 1n"12>n

" 1 d #nRT1

n " 1c a P2

Pxb 1n"12>n

" 1 d wcomp,in ! wcomp I,in # wcomp II,in

368 | Thermodynamics

P

1

P2

P1

Isothermal

Px

Polytropic

Work saved

Intercooling

2

T

s

1T1

Intercooling

2

P2

P1

Px

v

FIGURE 7–46P-v and T-s diagrams for a two-stagesteady-flow compression process.

EXAMPLE 7–13 Work Input for Various Compression Processes

Air is compressed steadily by a reversible compressor from an inlet state of100 kPa and 300 K to an exit pressure of 900 kPa. Determine the compressorwork per unit mass for (a) isentropic compression with k ! 1.4, (b) polytropiccompression with n ! 1.3, (c) isothermal compression, and (d) ideal two-stage compression with intercooling with a polytropic exponent of 1.3.

Solution Air is compressed reversibly from a specified state to a specifiedpressure. The compressor work is to be determined for the cases ofisentropic, polytropic, isothermal, and two-stage compression.

cen84959_ch07.qxd 3/31/05 4:25 PM Page 368

wcomp,in = w

comp,I + wcomp,II

wcomp,in =

nRT1

n � 1

"✓Px

P1

◆(n�1)/n

� 1

#+

nRT1

n � 1

"✓P2

Px

◆(n�1)/n

� 1

#

Px

=pP1

P2



Work Input for Various Compression Processes

Air is compressed steadily by a reversible compressor from an inletstate of 100 kPa and 300 K to an exit pressure of 900 kPa.Determine the compressor work per unit mass for (a) isentropiccompression with k = 1.4, (b) polytropic compression withn = 1.3, (c) isothermal compression, and (d) ideal two- stagecompression with intercooling with a polytropic exponent of 1.3.


Isentropic E�ciency of Turbines

The isentropic e�ciency of a turbine is defined as the ratio of theactual work output of the turbine to the work output that wouldbe achieved if the process between the inlet state and the exitpressure were isentropic

⌘turbine

⇡ h1

� h2a

h1

� h2s

Isentropic efficiencies are defined differently for different devices sinceeach device is set up to perform different tasks. Next we define the isen-tropic efficiencies of turbines, compressors, and nozzles by comparing theactual performance of these devices to their performance under isentropicconditions for the same inlet state and exit pressure.

Isentropic Efficiency of TurbinesFor a turbine under steady operation, the inlet state of the working fluid andthe exhaust pressure are fixed. Therefore, the ideal process for an adiabaticturbine is an isentropic process between the inlet state and the exhaust pres-sure. The desired output of a turbine is the work produced, and the isen-tropic efficiency of a turbine is defined as the ratio of the actual workoutput of the turbine to the work output that would be achieved if theprocess between the inlet state and the exit pressure were isentropic:

(7–60)

Usually the changes in kinetic and potential energies associated with a fluidstream flowing through a turbine are small relative to the change in enthalpyand can be neglected. Then the work output of an adiabatic turbine simplybecomes the change in enthalpy, and Eq. 7–60 becomes

(7–61)

where h2a and h2s are the enthalpy values at the exit state for actual andisentropic processes, respectively (Fig. 7–49).

The value of hT greatly depends on the design of the individual compo-nents that make up the turbine. Well-designed, large turbines have isentropicefficiencies above 90 percent. For small turbines, however, it may drop evenbelow 70 percent. The value of the isentropic efficiency of a turbine isdetermined by measuring the actual work output of the turbine and by cal-culating the isentropic work output for the measured inlet conditions and theexit pressure. This value can then be used conveniently in the design ofpower plants.

hT !h1 ! h2a

h1 ! h2s

hT "Actual turbine work

Isentropic turbine work"

wa

ws

Chapter 7 | 371

h

s

1

h2s

h2a

h1

2s2a

P2

P1Inlet state

Actual process

Isentropic process

Exit

pressure

s2s = s1

wa ws

FIGURE 7–49The h-s diagram for the actual andisentropic processes of an adiabaticturbine.

EXAMPLE 7–14 Isentropic Efficiency of a Steam Turbine

Steam enters an adiabatic turbine steadily at 3 MPa and 400°C and leaves at50 kPa and 100°C. If the power output of the turbine is 2 MW, determine (a)the isentropic efficiency of the turbine and (b) the mass flow rate of thesteam flowing through the turbine.

Solution Steam flows steadily in a turbine between inlet and exit states. Fora specified power output, the isentropic efficiency and the mass flow rate areto be determined.Assumptions 1 Steady operating conditions exist. 2 The changes in kineticand potential energies are negligible.

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Isentropic E�ciency of Turbines

Steam enters an adiabatic turbine steadily at 3 MPa and 400C andleaves at 50 kPa and 100C. If the power output of the turbine is 2MW, determine (a) the isentropic e�ciency of the turbine and (b)the mass flow rate of the steam flowing through the turbine.


Isentropic E�ciency of Compressors and Pumps

The isentropic e�ciency of a compressor is defined as the ratio ofthe work input required to raise the pressure of a gas to a specifiedvalue in an isentropic manner to the actual work input

⌘compressor

⇡ h2s

� h1

h2a

� h1

Isentropic Efficiencies of Compressors and PumpsThe isentropic efficiency of a compressor is defined as the ratio of thework input required to raise the pressure of a gas to a specified value in anisentropic manner to the actual work input:

(7–62)

Notice that the isentropic compressor efficiency is defined with the isen-tropic work input in the numerator instead of in the denominator. This isbecause ws is a smaller quantity than wa, and this definition prevents hCfrom becoming greater than 100 percent, which would falsely imply that theactual compressors performed better than the isentropic ones. Also noticethat the inlet conditions and the exit pressure of the gas are the same forboth the actual and the isentropic compressor.

When the changes in kinetic and potential energies of the gas being com-pressed are negligible, the work input to an adiabatic compressor becomesequal to the change in enthalpy, and Eq. 7–62 for this case becomes

(7–63)

where h2a and h2s are the enthalpy values at the exit state for actual andisentropic compression processes, respectively, as illustrated in Fig. 7–51.Again, the value of hC greatly depends on the design of the compressor.Well-designed compressors have isentropic efficiencies that range from 80to 90 percent.

When the changes in potential and kinetic energies of a liquid are negligi-ble, the isentropic efficiency of a pump is defined similarly as

(7–64)

When no attempt is made to cool the gas as it is compressed, the actualcompression process is nearly adiabatic and the reversible adiabatic (i.e.,isentropic) process serves well as the ideal process. However, sometimescompressors are cooled intentionally by utilizing fins or a water jacketplaced around the casing to reduce the work input requirements (Fig. 7–52).In this case, the isentropic process is not suitable as the model process sincethe device is no longer adiabatic and the isentropic compressor efficiencydefined above is meaningless. A realistic model process for compressorsthat are intentionally cooled during the compression process is thereversible isothermal process. Then we can conveniently define an isother-mal efficiency for such cases by comparing the actual process to areversible isothermal one:

(7–65)

where wt and wa are the required work inputs to the compressor for thereversible isothermal and actual cases, respectively.

hC !wt

wa

hP !ws

wa!

v 1P2 " P1 2h2a " h1

h C !h2s " h1

h2a " h1

h C !Isentropic compressor work

Actual compressor work!

ws

wa

Chapter 7 | 373

h2s

h

s

1h1

h2a

2s

2a

P2

P1

Inlet state

s2s = s1

wa

Exit

pressure

ws

Actual process

Isentropic process

FIGURE 7–51The h-s diagram of the actual andisentropic processes of an adiabaticcompressor.

Air

COMPRESSOR

Coolingwater

FIGURE 7–52Compressors are sometimesintentionally cooled to minimize thework input.

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E↵ect of E�ciency on Compressor Power Input

Air is compressed by an adiabatic compressor from 100 kPa and12C to a pressure of 800 kPa at a steady rate of 0.2 kg/s. If theisentropic e�ciency of the compressor is 80 percent, determine (a)the exit temperature of air and (b) the required power input to thecompressor.


Isentropic E�ciency of Nozzles

The isentropic e�ciency of a nozzle is defined as the ratio of theactual kinetic energy of the fluid at the nozzle exit to the kineticenergy value at the exit of an isentropic nozzle for the same inletstate and exit pressure.

⌘compressor

⇡ h1

� h2a

h1

� h2s

Isentropic Efficiency of NozzlesNozzles are essentially adiabatic devices and are used to accelerate a fluid.Therefore, the isentropic process serves as a suitable model for nozzles. Theisentropic efficiency of a nozzle is defined as the ratio of the actual kineticenergy of the fluid at the nozzle exit to the kinetic energy value at the exit ofan isentropic nozzle for the same inlet state and exit pressure. That is,

(7–66)

Note that the exit pressure is the same for both the actual and isentropicprocesses, but the exit state is different.

Nozzles involve no work interactions, and the fluid experiences little orno change in its potential energy as it flows through the device. If, in addi-tion, the inlet velocity of the fluid is small relative to the exit velocity, theenergy balance for this steady-flow device reduces to

Then the isentropic efficiency of the nozzle can be expressed in terms ofenthalpies as

(7–67)

where h2a and h2s are the enthalpy values at the nozzle exit for the actualand isentropic processes, respectively (Fig. 7–54). Isentropic efficiencies ofnozzles are typically above 90 percent, and nozzle efficiencies above 95percent are not uncommon.

hN !h1 ! h2a

h1 ! h2s

h1 " h2a #V 2a¬¬

2

2

hN "Actual KE at nozzle exit

Isentropic KE at nozzle exit"

V 2a2

V 2s2

Chapter 7 | 375

(b) The required power input to the compressor is determined from the energybalance for steady-flow devices,

Discussion Notice that in determining the power input to the compressor, weused h2a instead of h2s since h2a is the actual enthalpy of the air as it exitsthe compressor. The quantity h2s is a hypothetical enthalpy value that the airwould have if the process were isentropic.

" 58.0 kW

" 10.2 kg>s 2 3 1575.03 ! 285.14 2 kJ>kg 4 W#

a,in " m# 1h2a ! h1 2 m

#h1 # W

#a,in " m

#h2a

E#

in " E#

out

EXAMPLE 7–16 Effect of Efficiency on Nozzle Exit Velocity

Air at 200 kPa and 950 K enters an adiabatic nozzle at low velocity and isdischarged at a pressure of 80 kPa. If the isentropic efficiency of the nozzleis 92 percent, determine (a) the maximum possible exit velocity, (b) the exittemperature, and (c) the actual exit velocity of the air. Assume constantspecific heats for air.

h2s

h

s

1h1

h2a

2s2a

P2

P1

Inlet state

Actual process

Isentropic process

s2s = s1

Exit

pressure

V22a

2V 2

2s2

FIGURE 7–54The h-s diagram of the actual andisentropic processes of an adiabaticnozzle.

cen84959_ch07.qxd 3/31/05 4:25 PM Page 375



E↵ect of E�ciency on Nozzle Velocity Exit

Air at 200 kPa and 950 K enters an adiabatic nozzle at lowvelocity and is discharged at a pressure of 80 kPa. If the isentropice�ciency of the nozzle is 92 percent, determine (a) the maximumpossible exit velocity, (b) the exit temperature, and (c) the actualexit velocity of the air. Assume constant specific heats for air.


Entropy Balance

The entropy change of a system during a process is equal to thenet entropy transfer through the system boundary and the entropygenerated within the system.

Sin

� Sout

+ Sgen

= �Ssystem

The second law of thermodynamics states that entropy can becreated but it cannot be destroyed. Entropy can be transferred bythe following mechanisms

Heat transfer

Mass flow


Entropy Generation in a Wall

Consider steady heat transfer through a 5-m ⇥ 7-m brick wall of ahouse of thickness 30 cm. On a day when the temperature of theoutdoors is 0C, the house is maintained at 27C. The temperaturesof the inner and outer surfaces of the brick wall are measured to be20C and 5C, respectively, and the rate of heat transfer through thewall is 1035 W. Determine the rate of entropy generation in thewall, and the rate of total entropy generation associated with thisheat transfer process.


Entropy Generation during a Throttling Process

Steam at 7 MPa and 450C is throttled in a valve to a pressure of 3MPa during a steady-flow process. Determine the entropygenerated during this process and check if the increase of entropyprinciple is satisfied.


Entropy Generated when a Hot Block Is Dropped in a Lake

A 50-kg block of iron casting at 500 K is thrown into a large lakethat is at a temperature of 285 K. The iron block eventuallyreaches thermal equilibrium with the lake water. Assuming anaverage specific heat of 0.45 kJ/kg K for the iron, determine (a)the entropy change of the iron block, (b) the entropy change ofthe lake water, and (c) the entropy generated during this process.


Entropy Generation in a Mixing Chamber

Water at 20 psia and 50F enters a mixing chamber at a rate of 300lbm/min where it is mixed steadily with steam entering at 20 psiaand 240F. The mixture leaves the chamber at 20 psia and 130F,and heat is lost to the surrounding air at 70F at a rate of 180Btu/min. Neglecting the changes in kinetic and potential energies,determine the rate of entropy generation during this process.


advanced thermodynamics: lecture 7

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