thermo solutions chapter07

7-1

Chapter 7 ENTROPY

Entropy and the Increase of Entropy Principle

7-1C Yes. Because we used the relation (QH/TH) = (QL/TL) in the proof, which is the defining relation of absolute temperature.

7-2C No. The Q represents the net heat transfer during a cycle, which could be positive.

7-3C Yes.

7-4C No. A system may reject more (or less) heat than it receives during a cycle. The steam in a steampower plant, for example, receives more heat than it rejects during a cycle.

7-5C No. A system may produce more (or less) work than it receives during a cycle. A steam power plant, for example, produces more work than it receives during a cycle, the difference being the net work output.

7-6C The entropy change will be the same for both cases since entropy is a property and it has a fixedvalue at a fixed state.

7-7C No. In general, that integral will have a different value for different processes. However, it will havethe same value for all reversible processes.

7-8C Yes.

7-9C That integral should be performed along a reversible path to determine the entropy change.

7-10C No. An isothermal process can be irreversible. Example: A system that involves paddle-wheel workwhile losing an equivalent amount of heat.

7-11C The value of this integral is always larger for reversible processes.

7-12C No. Because the entropy of the surrounding air increases even more during that process, makingthe total entropy change positive.

7-13C It is possible to create entropy, but it is not possible to destroy it.

7-2

7-14C Sometimes.

7-15C Never.

7-16C Always.

7-17C Increase.

7-18C Increases.

7-19C Decreases.

7-20C Sometimes.

7-21C Yes. This will happen when the system is losing heat, and the decrease in entropy as a result of thisheat loss is equal to the increase in entropy as a result of irreversibilities.

7-22C They are heat transfer, irreversibilities, and entropy transport with mass.

7-23C Greater than.

7-24 A rigid tank contains an ideal gas that is being stirred by a paddle wheel. The temperature of the gasremains constant as a result of heat transfer out. The entropy change of the gas is to be determined.Assumptions The gas in the tank is given to be an ideal gas.Analysis The temperature and the specific volume of the gas remainconstant during this process. Therefore, the initial and the final states of the gas are the same. Then s2 = s1 since entropy is a property.Therefore, 200 kJ

Heat

30 C

IDEAL GAS40 C

Ssys 0

7-3

7-25 Air is compressed steadily by a compressor. The air temperature is maintained constant by heatrejection to the surroundings. The rate of entropy change of air is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 Air is an ideal gas. 4 The process involves no internal irreversibilities such as friction, and thus it is an isothermal, internally reversible process.Properties Noting that h = h(T) for ideal gases, we have h1 = h2 since T1 = T2 = 25 C.Analysis We take the compressor as the system. Noting that the enthalpy of air remains constant, the energy balance for this steady-flow system can be expressed in the rate form as

outin

outin

energiesetc.potential,kinetic,internal,inchangeofRate

(steady)0system

massand work,heat,bynsferenergy tranetofRate

outin 0

QW

EE

EEE P2

AIRT = const.

Q·

12 kW

Therefore,

kW12inout WQP1Noting that the process is assumed to be an isothermal and internally

reversible process, the rate of entropy change of air is determined tobe

kW/K0.0403K298

kW12

sys

airout,air T

QS

7-26 Heat is transferred isothermally from a source to the working fluid of a Carnot engine. The entropychange of the working fluid, the entropy change of the source, and the total entropy change during thisprocess are to be determined.Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is givenby

S QT

Noting that heat transferred from the source is equal to the heat transferred to the working fluid, theentropy changes of the fluid and of the source become

kJ/K1.337K673kJ900

fluid

fluidin,

fluid

fluidfluid T

QTQS

900 kJ

Source400 C(b) kJ/K1.337

K673kJ900

source

sourceout,

source

sourcesource T

QTQS

(c) Thus the total entropy change of the process is0337.1337.1sourcefluidtotalgen SSSS

7-4

7-27 EES Problem 7-26 is reconsidered. The effects of the varying the heat transferred to the working fluidand the source temperature on the entropy change of the working fluid, the entropy change of the source,and the total entropy change for the process as the source temperature varies from 100°C to 1000°C are tobe investigated. The entropy changes of the source and of the working fluid are to be plotted against thesource temperature for heat transfer amounts of 500 kJ, 900 kJ, and1300 kJ. Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"{T_H = 400 [C]}Q_H = 1300 [kJ] T_Sys = T_H "Analysis: (a) & (b) This is a reversible isothermal process, and the entropy change during such a process is given byDELTAS = Q/T""Noting that heat transferred from the source is equal to the heat transferred to the working fluid,the entropy changes of the fluid and of the source become "DELTAS_source = -Q_H/(T_H+273) DELTAS_fluid = +Q_H/(T_Sys+273) "(c) entropy generation for the process:"S_gen = DELTAS_source + DELTAS_fluid

Sfluid

[kJ/K]Ssource

[kJ/K]Sgen

[kJ/K]TH

[C]3.485 -3.485 0 1002.748 -2.748 0 2002.269 -2.269 0 3001.932 -1.932 0 4001.682 -1.682 0 5001.489 -1.489 0 6001.336 -1.336 0 7001.212 -1.212 0 8001.108 -1.108 0 9001.021 -1.021 0 1000

100 200 300 400 500 600 700 800 900 10000

0.5

1

1.5

2

2.5

3

3.5

4

TH [C]

Sfluid

[KJ/K]

QH = 1300 kJ

Ssource = - Sfluid

QH = 500 kJ

QH = 900 kJ

7-5

7-28E Heat is transferred isothermally from the working fluid of a Carnot engine to a heat sink. Theentropy change of the working fluid is given. The amount of heat transfer, the entropy change of the sink,and the total entropy change during the process are to be determined.Analysis (a) This is a reversible isothermal process, and theentropy change during such a process is given by

Heat

SINK95 F

Carnot heat engine

95 FS QT

Noting that heat transferred from the working fluid is equal tothe heat transferred to the sink, the heat transfer become

Btu388.5. Btu5388Btu/R0.7R555 outfluid,fluidfluidfluid QSTQ

(b) The entropy change of the sink is determined from

Btu/R0.7R555Btu388.5

sink

insink,sink T

QS

(c) Thus the total entropy change of the process is07.07.0sinkfluidtotalgen SSSS

This is expected since all processes of the Carnot cycle are reversible processes, and no entropy isgenerated during a reversible process.

7-29 R-134a enters an evaporator as a saturated liquid-vapor at a specified pressure. Heat is transferred tothe refrigerant from the cooled space, and the liquid is vaporized. The entropy change of the refrigerant, the entropy change of the cooled space, and the total entropy change for this process are to be determined.Assumptions 1 Both the refrigerant and the cooled space involve no internal irreversibilities such asfriction. 2 Any temperature change occurs within the wall of the tube, and thus both the refrigerant and thecooled space remain isothermal during this process. Thus it is an isothermal, internally reversible process.Analysis Noting that both the refrigerant and the cooled space undergo reversible isothermal processes, the entropy change for them can be determined from

S QT

(a) The pressure of the refrigerant is maintained constant. Therefore, the temperature of the refrigerant also remains constant at the saturation value,

(Table A-12) K257.4C15.6kPa@160satTT

180 kJ

-5 C

R-134a160 kPa

Then,

kJ/K0.699K257.4

kJ180

trefrigeran

int,refrigerantrefrigeran T

QS

(b) Similarly,

kJ/K0.672K268kJ180

space

outspace,space T

QS

(c) The total entropy change of the process is kJ/K0.027672.0699.0spacetrefrigerantotalgen SSSS

7-6

Entropy Changes of Pure Substances

7-30C Yes, because an internally reversible, adiabatic process involves no irreversibilities or heat transfer.

7-31 The radiator of a steam heating system is initially filled with superheated steam. The valves areclosed, and steam is allowed to cool until the temperature drops to a specified value by transferring heat tothe room. The entropy change of the steam during this process is to be determined.Analysis From the steam tables (Tables A-4 through A-6),

KkJ/kg.949907.68320.04914.57240

04914.0001008.0515.19001008.095986.0C04

KkJ/kg7.2810/kgm95986.0

C015kPa200

22

22

12

2

1

31

1

1

fgf

fg

f

sxss

xvv

T

sTP

v

vv

v

H2O200 kPa 150 C

Q

The mass of the steam is

kg0.02084/kgm0.95986

m0.0203

3

1vVm

Then the entropy change of the steam during this process becomeskJ/K0.132KkJ/kg7.28100.9499kg0.0208412 ssmS

7-7

7-32 A rigid tank is initially filled with a saturated mixture of R-134a. Heat is transferred to the tank from a source until the pressure inside rises to a specified value. The entropy change of the refrigerant, entropychange of the source, and the total entropy change for this process are to be determined.Assumptions 1 The tank is stationary and thus the kinetic and potential energy changes are zero. 2 Thereare no work interactions. Analysis (a) From the refrigerant tables (Tables A-11 through A-13),

KkJ/kg0.781367929.07857.024761.0kJ/kg198.3445.1717857.062.63

7857.00007907.0051201.00007907.004040.0

kPa400

/kgm0.040400.00075330.0998670.40.0007533KkJ/kg0.467878316.04.015457.0

kJ/kg112.7621.1864.028.38

4.0kPa200

22

22

22

12

2

311

11

11

1

1

fgf

fgf

fg

f

fgf

fgf

fgf

sxssuxuuv

xP

xsxss

uxuu

xP

vv

vv

vvv

The mass of the refrigerant is

kg12.38/kgm0.04040

m0.53

3

1vVm

Then the entropy change of the refrigerant becomeskJ/K3.880KkJ/kg0.46780.7813kg12.3812system ssmS

(b) We take the tank as the system. This is a closed system since no mass enters or leaves. Noting that the volume of the system is constant and thus there is no boundary work, the energy balance for this stationaryclosed system can be expressed as

)( 12in

energiesetc.potential,kinetic,internal,inChange

system

massand work,heat,bynsferenergy traNet

outin

uumUQ

EEE

Q Source35 C

R-134a200 kPa

Substituting,kJ1059112.76198.34kg12.3812in uumQ

The heat transfer for the source is equal in magnitude but opposite in direction. Therefore,Qsource, out = - Qtank, in = - 1059 kJ

and

kJ/K3.439K308kJ1059

source

outsource,source T

QS

(c) The total entropy change for this process is kJ/K0.442439.3880.3sourcesystemtotal SSS

7-8

7-33 EES Problem 7-32 is reconsidered. The effects of the source temperature and final pressure on thetotal entropy change for the process as the source temperature varies from 30°C to 210°C, and the finalpressure varies from 250 kPa to 500 kPa are to be investigated. The total entropy change for the process isto be plotted as a function of the source temperature for final pressures of 250 kPa, 400 kPa, and 500 kPa. Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"P_1 = 200 [kPa] x_1 = 0.4 V_sys = 0.5 [m^3]P_2 = 400 [kPa] {T_source = 35 [C]}

"Analysis: "" Treat the rigid tank as a closed system, with no work in, neglect changes in KE and PE of the R134a."E_in - E_out = DELTAE_sysE_out = 0 [kJ] E_in = QDELTAE_sys = m_sys*(u_2 - u_1) u_1 = INTENERGY(R134a,P=P_1,x=x_1) v_1 = volume(R134a,P=P_1,x=x_1)V_sys = m_sys*v_1

"Rigid Tank: The process is constant volume. Then P_2 and v_2 specify state 2."v_2 = v_1 u_2 = INTENERGY(R134a,P=P_2,v=v_2) "Entropy calculations:"s_1 = entropy(R134a,P=P_1,x=x_1)s_2 = entropY(R134a,P=P_2,v=v_2)DELTAS_sys = m_sys*(s_2 - s_1)"Heat is leaving the source, thus:"DELTAS_source = -Q/(T_source + 273)

"Total Entropy Change:"DELTAS_total = DELTAS_source + DELTAS_sys

25 65 105 145 185 2250

0.5

1

1.5

2

2.5

3

Tsource [C]

Stotal

[kJ/K]

P2 = 250 kPa= 400 kPa= 500 kPa

Stotal

[kJ/K]Tsource

[C]0.3848 300.6997 600.9626 901.185 1201.376 1501.542 1801.687 210

7-9

7-34 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. Anelectric heater inside is turned on and kept on until all the liquid vaporized. The entropy change of thewater during this process is to be determined.Analysis From the steam tables (Tables A-4 through A-6)

KkJ/kg6.8649 vaporsat.

KkJ/kg2.81680562.625.03028.1/kgm0.42430.0011.69410.250.001

25.0kPa100

212

11

311

1

1

s

sxssx

xP

fgf

fgf

vv

vvv

H2O2 kg

100 kPa

WeThen the entropy change of the steam becomes

kJ/K8.10KkJ/kg)2.81686.8649)(kg2(12 ssmS

7-35 [Also solved by EES on enclosed CD] A rigid tank is divided into two equal parts by a partition. Onepart is filled with compressed liquid water while the other side is evacuated. The partition is removed andwater expands into the entire tank. The entropy change of the water during this process is to be determined.Analysis The properties of the water are (Table A-4)

KkJ/kg0.75562522.70001018.07549.0

0001018.0001014.002.10

001014.0002034.0

/kgm340020.0

kPa15

/kgm0.002034001017.022 that Noting

KkJ/kg0.8313/kgm0.001017

C60kPa300

22

22

32

2

312

C60@1

3C60@1

1

1

fgf

fg

f

f

f

sxss

xP

ssTP

v

vv

v

vv

vv 1.5 kg compressed

liquid

300 kPa 60 C

Vacuum

Then the entropy change of the water becomeskJ/K0.114KkJ/kg0.83130.7556kg1.512 ssmS

7-10

7-36 EES Problem 7-35 is reconsidered. The entropy generated is to be evaluated and plotted as a functionof surroundings temperature, and the values of the surroundings temperatures that are valid for thisproblem are to be determined. The surrounding temperature is to vary from 0°C to 100°C.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Input Data"P[1]=300 [kPa] T[1]=60 [C] m=1.5 [kg] P[2]=15 [kPa]

Fluid$='Steam_IAPWS'

V[1]=m*spv[1]spv[1]=volume(Fluid$,T=T[1], P=P[1]) "specific volume of steam at state 1, m^3/kg"s[1]=entropy(Fluid$,T=T[1],P=P[1]) "entropy of steam at state 1, kJ/kgK"V[2]=2*V[1] "Steam expands to fill entire volume at state 2"

"State 2 is identified by P[2] and spv[2]"spv[2]=V[2]/m "specific volume of steam at state 2, m^3/kg"s[2]=entropy(Fluid$,P=P[2],v=spv[2]) "entropy of steam at state 2, kJ/kgK"T[2]=temperature(Fluid$,P=P[2],v=spv[2])DELTAS_sys=m*(s[2]-s[1]) "Total entopy change of steam, kJ/K"

"What does the first law tell us about this problem?""Conservation of Energy for the entire, closed system"E_in - E_out = DELTAE_sys"neglecting changes in KE and PE for the system:"DELTAE_sys=m*(intenergy(Fluid$, P=P[2], v=spv[2]) - intenergy(Fluid$,T=T[1],P=P[1]))E_in = 0

"How do you interpert the energy leaving the system, E_out? Recall this is a constant volumesystem."Q_out = E_out "What is the maximum value of the Surroundings temperature?""The maximum possible value for the surroundings temperature occurs when we setS_gen = 0=Delta S_sys+sum(DeltaS_surr)"Q_net_surr=Q_outS_gen = 0 S_gen = DELTAS_sys+Q_net_surr/Tsurr

"Establish a parametric table for the variables S_gen, Q_net_surr, T_surr, and DELTAS_sys. Inthe Parametric Table window select T_surr and insert a range of values. Then place '{' and '}'about the S_gen = 0 line; press F3 to solve the table. The results are shown in Plot Window 1.What values of T_surr are valid for this problem?"

Sgen

[kJ/K]Qnet,surr

[kJ]Tsurr

[K]Ssys

[kJ/K]0.02533 37.44 270 -0.11330.01146 37.44 300 -0.1133

0.0001205 37.44 330 -0.1133-0.009333 37.44 360 -0.1133-0.01733 37.44 390 -0.1133

7-11

260 280 300 320 340 360 380 400-0.020

-0.010

-0.000

0.010

0.020

0.030

Tsurr [K]

Sge

n[k

J/K

]

7-12

7-37E A cylinder is initially filled with R-134a at a specified state. The refrigerant is cooled and condensed at constant pressure. The entropy change of refrigerant during this process is to be determinedAnalysis From the refrigerant tables (Tables A-11E through A-13E),

RBtu/lbm0.06039psia120

F05

RBtu/lbm0.22361F100psia120

F90@22

2

11

1

fssPT

sTP

Q

R-134a120 psia 100 FThen the entropy change of the refrigerant becomes

Btu/R0.3264RBtu/lbm0.223610.06039lbm212 ssmS

7-38 An insulated cylinder is initially filled with saturated liquid water at a specified pressure. The water isheated electrically at constant pressure. The entropy change of the water during this process is to bedetermined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Thecompression or expansion process is quasi-equilibrium.Analysis From the steam tables (Tables A-4 through A-6),

KkJ/kg1.4337kJ/kg467.13

/kgm0.001053

.kPa150

kPa150@1

kPa150@1

3kPa150@1

1

f

f

f

sshh

liquidsatP

vv

Also, kg4.75/kgm0.001053

m0.0053

3

1vVm

We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this stationary closed system can be expressed as

2200 kJ

H2O150 kPa

Sat. liquid

)( 12ine,

outb,ine,


system


outin

hhmWUWW

EEE

since U + Wb = H during a constant pressure quasi-equilibrium process. Solving for h2,

kJ/kg33.930kg4.75kJ220013.467ine,

12 mW

hh

Thus,

KkJ/kg6384.27894.52081.04337.1

2081.00.2226

13.46733.930

kJ/kg33.930kPa150

22

22

2

2

fgf

fg

f

sxssh

hhx

hP

Then the entropy change of the water becomeskJ/K5.72KkJ/kg1.43372.6384kg4.7512 ssmS

7-13

7-39 An insulated cylinder is initially filled with saturated R-134a vapor at a specified pressure. The refrigerant expands in a reversible manner until the pressure drops to a specified value. The finaltemperature in the cylinder and the work done by the refrigerant are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible.Analysis (a) This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the refrigerant tables (Tables A-11 through A-13),

KkJ/kg0.91835kJ/kg246.79

/kgm0.025621

vaporsat.MPa0.8

MPa0.8@1

MPa0.8@1

3MPa0.8@1

1

g

g

g

ssuu

Pvv

Also,

kg952.1/kgm0.025621

m0.053

3

1vVm

and

kJ/kg232.91171.450.987463.62

9874.067929.0

24761.091835.0MPa0.4

22

22

12

2

fgf

fg

f

uxuus

ssx

ssP

R-134a0.8 MPa 0.05 m3

C8.91MPa0.4@sat2 TT

(b) We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this adiabatic closed system can be expressed as

)( 21outb,

outb,


system


outin

uumWUW

EEE

Substituting, the work done during this isentropic process is determined to be kJ27.09kJ/kg)232.91246.79)(kg1.952(21outb, uumW

7-14

7-40 EES Problem 7-39 is reconsidered. The work done by the refrigerant is to be calculated and plotted asa function of final pressure as the pressure varies from 0.8 MPa to 0.4 MPa. The work done for this process is to be compared to one for which the temperature is constant over the same pressure range. Analysis The problem is solved using EES, and the results are tabulated and plotted below.

ProcedureIsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isotherm)T_isotherm=Temperature(R134a,P=P_1,x=x_1)T=T_isothermu_1 = INTENERGY(R134a,P=P_1,x=x_1) v_1 = volume(R134a,P=P_1,x=x_1)s_1 = entropy(R134a,P=P_1,x=x_1) u_2 = INTENERGY(R134a,P=P_2,T=T) s_2 = entropy(R134a,P=P_2,T=T)

"The process is reversible and Isothermal thus the heat transfer is determined by:"Q_isotherm = (T+273)*m_sys*(s_2 - s_1) DELTAE_isotherm = m_sys*(u_2 - u_1) E_in = Q_isothermE_out = DELTAE_isotherm+E_inWork_out_isotherm=E_out

END

"Knowns:"P_1 = 800 [kPa] x_1 = 1.0 V_sys = 0.05[m^3]"P_2 = 400 [kPa]"

"Analysis: "

" Treat the rigid tank as a closed system, with no heat transfer in, neglectchanges in KE and PE of the R134a.""The isentropic work is determined from:"E_in - E_out = DELTAE_sysE_out = Work_out_isenE_in = 0 DELTAE_sys = m_sys*(u_2 - u_1) u_1 = INTENERGY(R134a,P=P_1,x=x_1) v_1 = volume(R134a,P=P_1,x=x_1)s_1 = entropy(R134a,P=P_1,x=x_1) V_sys = m_sys*v_1

"Rigid Tank: The process is reversible and adiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1 u_2 = INTENERGY(R134a,P=P_2,s=s_2) T_2_isen = temperature(R134a,P=P_2,s=s_2)

CallIsothermWork(P_1,x_1,m_sys,P_2:Work_out_Isotherm,Q_isotherm,DELTAE_isotherm,T_isotherm)

7-15

P2

[kPa]Workout,isen

[kJ]Workout,isotherm

[kJ]Qisotherm

[kJ]400 27.09 60.02 47.08500 18.55 43.33 33.29600 11.44 28.2 21.25700 5.347 13.93 10.3800 0 0 0

400 450 500 550 600 650 700 750 8000

10

20

30

40

50

60

P2 [kPa]

Workou

t[kJ]

Isentropic

Isothermal

400 450 500 550 600 650 700 750 8000

10

20

30

40

50

P2 [kPa]

Qisotherm

[kJ]

Qisentropic = 0 kJ

7-16

7-41 Saturated Refrigerant-134a vapor at 160 kPa is compressed steadily by an adiabatic compressor. The minimum power input to the compressor is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis The power input to an adiabatic compressor will be a minimum when the compression process is reversible. For the reversible adiabatic process we have s2 = s1. From the refrigerant tables (Tables A-11 through A-13),

kJ/kg277.06kPa009

KkJ/kg0.9419kJ/kg241.11

/kgm0.12348

vaporsat.kPa160

212

2

kPa160@1

kPa160@1

3kPa160@1

1

hss

P

sshh

P

g

g

gvv 2

R-134a

Also,

kg/s0.27kg/min16.20/kgm0.12348

/minm23

3

1

1

v

Vm 1

There is only one inlet and one exit, and thus m m m1 2 . We take the compressor as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE

( )

W mh mh Q ke pe

W m h hin

in

(since 0)1 2

2 1

Substituting, the minimum power supplied to the compressor is determined to be

W kW9.71kJ/kg241.11277.06kg/s0.27in

7-17

7-42E Steam expands in an adiabatic turbine. The maximum amount of work that can be done by theturbine is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis The work output of an adiabatic turbine is maximum when the expansion process is reversible.For the reversible adiabatic process we have s2 = s1. From the steam tables (Tables A-4E through A-6E),

Btu/lbm1144.2933.690.9725236.14

9725.028448.1

39213.06413.1psia40

RBtu/lbm1.6413Btu/lbm1456.0

F900psia800

22

22

12

2

1

1

1

1

fgf

fg

f

hxhhs

ssx

ssP

sh

TP

1

H2OThere is only one inlet and one exit, and thus m m m1 2 . We take theturbine as the system, which is a control volume since mass crosses theboundary. The energy balance for this steady-flow system can be expressed in the rate form as

2

outin


(steady)0system


outin 0

EE

EEE

( )

mh W mh

W m h h1 2

1 2

out

out

Dividing by mass flow rate and substituting,Btu/lbm311.82.11440.145621out hhw

7-18

7-43E EES Problem 7-42E is reconsidered. The work done by the steam is to be calculated and plotted as a function of final pressure as the pressure varies from 800 psia to 40 psia. Also the effect of varying theturbine inlet temperature from the saturation temperature at 800 psia to 900°F on the turbine work is to be investigated.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"P_1 = 800 [psia] T_1 = 900 [F] P_2 = 40 [psia] T_sat_P_1= temperature(Fluid$,P=P_1,x=1.0)Fluid$='Steam_IAPWS'

"Analysis: "" Treat theturbine as a steady-flow control volume, with no heat transfer in, neglectchanges in KE and PE of the Steam.""The isentropic work is determined from the steady-flow energy equation written per unit mass:"e_in - e_out = DELTAe_sys E_out = Work_out+h_2 "[Btu/lbm]"e_in = h_1 "[Btu/lbm]"DELTAe_sys = 0 "[Btu/lbm]"h_1 = enthalpy(Fluid$,P=P_1,T=T_1) s_1 = entropy(Fluid$,P=P_1,T=T_1)

"The process is reversibleand adiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1 "[Btu/lbm-R]"h_2 = enthalpy(Fluid$,P=P_2,s=s_2)T_2_isen=temperature(Fluid$,P=P_2,s=s_2)

0 100 200 300 400 500 600 700 8000

50

100

150

200

250

300

350

P2 [psia]

Workou

t[Btu/lb

m]

Isentropic ProcessP1 = 800 psia

T1 = 900 F

T1[F]

Workout[Btu/lbm]

520 219.3560 229.6600 239.1650 250.7690 260730 269.4770 279820 291.3860 301.5900 311.9

500 550 600 650 700 750 800 850 900210

232

254

276

298

320

T1 [F]

Workou

t[Btu/lb

m]

Isentropic ProcessP1 = 800 psia

P2 = 40 psia

7-19

7-44 An insulated cylinder is initially filled with superheated steam at a specified state. The steam iscompressed in a reversible manner until the pressure drops to a specified value. The work input during this process is to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible.Analysis This is a reversible adiabatic (i.e., isentropic) process, and thus s2 = s1. From the steam tables (Tables A-4 through A-6),

H2O300 kPa 150 C

kJ/kg2773.8MPa1

KkJ/kg7.0792kJ/kg2571.0

/kgm0.63402

C150kPa300

212

2

1

1

31

1

1

uss

Ps

uTP

v

Also,

kg0.0789/kgm0.63402

m0.053

3

1vVm

We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this adiabatic closed system can be expressed as

)( 12inb,


system


outin

uumUW

EEE

Substituting, the work input during this adiabatic process is determined to be kJ16.0kJ/kg2571.02773.8kg0.078912inb, uumW

7-20

7-45 EES Problem 7-44 is reconsidered. The work done on the steam is to be determined and plotted as afunction of final pressure as the pressure varies from 300 kPa to 1 MPa.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"P_1 = 300 [kPa] T_1 = 150 [C] V_sys = 0.05 [m^3]"P_2 = 1000 [kPa]"

"Analysis: "Fluid$='Steam_IAPWS'

" Treat the piston-cylinder as a closed system, with no heat transfer in, neglectchanges in KE and PE of the Steam. The process is reversible and adiabatic thus isentropic.""The isentropic work is determined from:"E_in - E_out = DELTAE_sysE_out = 0 [kJ] E_in = Work_inDELTAE_sys = m_sys*(u_2 - u_1) u_1 = INTENERGY(Fluid$,P=P_1,T=T_1)v_1 = volume(Fluid$,P=P_1,T=T_1)s_1 = entropy(Fluid$,P=P_1,T=T_1)V_sys = m_sys*v_1

" The process is reversible and adiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1 u_2 = INTENERGY(Fluid$,P=P_2,s=s_2)T_2_isen = temperature(Fluid$,P=P_2,s=s_2)

P2

[kPa]Workin

[kJ]300 0400 3.411500 6.224600 8.638700 10.76800 12.67900 14.4

1000 16

300 400 500 600 700 800 900 10000

2

4

6

8

10

12

14

16

P2 [kPa]

Wor

k in

[kJ]

Work on SteamP1 = 300 kPa

T1 = 150 C

7-21

7-46 A cylinder is initially filled with saturated water vapor at a specified temperature. Heat is transferredto the steam, and it expands in a reversible and isothermal manner until the pressure drops to a specifiedvalue. The heat transfer and the work output for this process are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The cylinder is well-insulatedand thus heat transfer is negligible. 3 The thermal energy stored in the cylinder itself is negligible. 4 Theprocess is stated to be reversible and isothermal.

Q

H2O200 C

sat. vapor T = const

Analysis From the steam tables (Tables A-4 through A-6),

KkJ/kg6.8177kJ/kg2631.1kPa800

KkJ/kg6.4302kJ/kg2594.2

.C200

2

2

12

2

C200@1

C200@11

su

TTP

ssuu

vaporsatT

g

g

The heat transfer for this reversible isothermal process can be determined fromkJ219.9KkJ/kg)6.43026.8177)(kg1.2)(K473(12 ssTmSTQ

We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

)()(

12inoutb,

12outb,in


system


outin

uumQWuumUWQ

EEE

Substituting, the work done during this process is determined to be kJ175.6kJ/kg)2594.22631.1)(kg1.2(kJ9.219outb,W

7-22

7-47 EES Problem 7-46 is reconsidered. The heat transferred to the steam and the work done are to bedetermined and plotted as a function of final pressure as the pressure varies from the initial value to thefinal value of 800 kPa. Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"T_1 = 200 [C] x_1 = 1.0 m_sys = 1.2 [kg]{P_2 = 800"[kPa]"}

"Analysis: "Fluid$='Steam_IAPWS'

" Treat the piston-cylinder as a closed system, neglect changes in KE and PE of the Steam. The process is reversible and isothermal ."T_2 = T_1 E_in - E_out = DELTAE_sysE_in = Q_inE_out = Work_out DELTAE_sys = m_sys*(u_2 - u_1) P_1 = pressure(Fluid$,T=T_1,x=1.0)u_1 = INTENERGY(Fluid$,T=T_1,x=1.0) Kv_1 = volume(Fluid$,T=T_1,x=1.0)s_1 = entropy(Fluid$,T=T_1,x=1.0)V_sys = m_sys*v_1

" The process is reversible and isothermal.W

Then P_2 and T_2 specify state 2."u_2 = INTENERGY(Fluid$,P=P_2,T=T_2)

800 1000 1200 1400 16000

40

80

120

160

200

P2 [kPa]

orkou

t[J]

s_2 = entropy(Fluid$,P=P_2,T=T_2)Q_in= (T_1+273)*m_sys*(s_2-s_1)

P2

[kPa]Qin

[kJ]Workout

[kJ]800 219.9 175.7900 183.7 144.7

1000 150.6 1171100 120 91.841200 91.23 68.851300 64.08 47.651400 38.2 27.981500 13.32 9.6051553 219.9 175.7

800 1000 1200 1400 16000

40

80

120

160

200

P2 [kPa]

Qin

[kJ]

7-23

7-48 A cylinder is initially filled with saturated water vapor mixture at a specified temperature. Steamundergoes a reversible heat addition and an isentropic process. The processes are to be sketched and heat transfer for the first process and work done during the second process are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. 2 The thermal energy stored in thecylinder itself is negligible. 3 Both processes are reversible. Analysis (b) From the steam tables (Tables A-4 through A-6),

Q

H2O100 Cx=0.5

kJ/kg2247.9kPa15

KkJ/kg3542.7kJ/kg2506.0

kJ/kg2675.6

1C100

kJ/kg4.1547)4.2256)(5.0(17.4195.0

C100

323

3

2

2

2

2

2

11

uss

P

suu

hh

xT

xhhhxT

g

g

fgf

We take the contents of the cylinder as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

)( 12outb,in


system


outin

uumUWQ

EEE

For process 1-2, it reduces tokJ5641kg1547.4)kJ/-kg)(2675.65()( 12in12, hhmQ

(c) For process 2-3, it reduces tokJ1291kg2247.9)kJ/-kg)(2506.05()( 32outb,23, uumW

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00

100

200

300

400

500

600

700

s [kJ/kg-K]

T[°C]

101.42 kPa

15 kPa

SteamIAPWS

1 2

3

7-24

7-49 A rigid tank contains saturated water vapor at a specified temperature. Steam is cooled to ambienttemperature. The process is to be sketched and entropy changes for the steam and for the process are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible.Analysis (b) From the steam tables (Tables A-4 through A-6),

KkJ/kg0715.1kJ/kg78.193

0386.0C25

KkJ/kg3542.7kJ/kg2506.0

kJ/kg6720.1

1C100

2

2

2

12

2

1

1

11

suxT

suux

Tg

g

vv

vv

Q

H2O100 Cx = 1

The entropy change of steam is determined fromkJ/K-31.41Kkg7.3542)kJ/-kg)(1.07155()( 12 ssmS w

(c) We take the contents of the tank as the system. This is a closed system since no mass enters or leaves. The energy balance for this closed system can be expressed as

)( 12out


system


outin

uumUQ

EEE

That is, kJ11,511kg193.78)kJ/-kg)(2506.05()( 21out uumQ

The total entropy change for the process is

kJ/K7.39K298kJ11,511kJ/K41.31

surr

outgen T

QSS w

10-3 10-2 10-1 100 101 102 1030

100

200

300

400

500

600

700

v [m3/kg]

T[°C]

101.4 kPa

3.17 kPa

SteamIAPWS

1

2

7-25

7-50 Steam expands in an adiabatic turbine. Steam leaves the turbine at two different pressures. The process is to be sketched on a T-s diagram and the work done by the steam per unit mass of the steam at theinlet are to be determined.Assumptions 1 The kinetic and potential energy changes are negligible. P1 = 6 MPa

T1 = 500 CAnalysis (b) From the steam tables (Tables A-4 through A-6),

P2 = 1 MPa

Turbine

831.0kJ/kg6.2179kPa10

kJ/kg3.2921MPa1

KkJ/kg8826.6kJ/kg1.3423

MPa6C005

3

3

13

3

212

2

1

1

1

1

s

s

s

xh

ssP

hss

Ps

hPT

A mass balance on the control volume gives

321 mmm where 13

12

9.01.0mmmm P3 = 10 kPa

We take the turbine as the system, which is a control volume. The energy balance for thissteady-flow system can be expressed in the rate form as

3121out,11

3322out,11

outin

9.01.0 hmhmWhm

hmhmWhm

EE

s

s

or

kJ/kg3.1169)6.2179)(9.0()3.2921)(1.0(1.34239.01.0

9.01.0

321out,

32out,1

hhhwhhwh

s

s0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00

100

200

300

400

500

600

700

s [kJ/kg-K]

T[°C]

6000 kPa

1000 kPa

10 kPa

SteamIAPWS

1

2

3

The actual work output per unit mass of steam at the inlet is kJ/kg993.9)kJ/kg3.1169)(85.0(out,out sT ww

7-51E An insulated rigid can initially contains R-134a at a specified state. A crack develops, and refrigerant escapes slowly. The final mass in the can is to be determined when the pressure inside drops toa specified value.Assumptions 1 The can is well-insulated and thus heat transfer is negligible. 2 The refrigerant that

remains in the can underwent a reversible adiabatic process. Analysis Noting that for a reversible adiabatic (i.e., isentropic)process, s1 = s2, the properties of the refrigerant in the can are (Tables A-11E through A-13E)

/lbmft0.54530.011822.27720.23550.01182

2355.019962.0

02605.007306.0psia02

RBtu/lbm0.07306F07psia140

322

22

12

2

F07@11

1

fgf

fg

f

f

xs

ssx

ssP

ssTP

vvv

Leak

R-134140 psia

70 F

Thus the final mass of the refrigerant in the can is

lbm2.201/lbmft0.5453

ft1.23

3

2vVm

7-26

Entropy Change of Incompressible Substances

7-52C No, because entropy is not a conserved property.

7-53 A hot copper block is dropped into water in an insulated tank. The final equilibrium temperature of the tank and the total entropy change are to be determined.Assumptions 1 Both the water and the copper block are incompressible substances with constant specificheats at room temperature. 2 The system is stationary and thus the kinetic and potential energies arenegligible. 3 The tank is well-insulated and thus there is no heat transfer.Properties The density and specific heat of water at 25 C are = 997 kg/m3 and cp = 4.18 kJ/kg. C. Thespecific heat of copper at 27 C is cp = 0.386 kJ/kg. C (Table A-3). Analysis We take the entire contents of the tank, water + copper block, as the system. This is a closedsystem since no mass crosses the system boundary during the process. The energy balance for this systemcan be expressed as

120 L

Copper50 kg

WATER

U

EEE

0energiesetc.potential,

kinetic,internal,inChange

system


outin

or,0waterCu UU

0)]([)]([ water12Cu12 TTmcTTmc

where

kg119.6)m0.120)(kg/m997( 33water Vm

Using specific heat values for copper and liquid water at room temperature and substituting, 0C25)(C)kJ/kgkg)(4.18(119.6C80)(C)kJ/kgkg)(0.386(50 22 TT

T2 = 27.0 CThe entropy generated during this process is determined from

kJ/K3.344K298K300.0lnKkJ/kg4.18kg119.6ln

kJ/K3.140K353K300.0lnKkJ/kg0.386kg50ln

1

2avgwater

1

2avgcopper

TTmcS

TTmcS

Thus,kJ/K0.204344.3140.3watercoppertotal SSS

7-27

7-54 A hot iron block is dropped into water in an insulated tank. The total entropy change during thisprocess is to be determined.Assumptions 1 Both the water and the iron block are incompressible substances with constant specificheats at room temperature. 2 The system is stationary and thus the kinetic and potential energies arenegligible. 3 The tank is well-insulated and thus there is no heat transfer. 4 The water that evaporates, condenses back. Properties The specific heat of water at 25 C is cp = 4.18 kJ/kg. C. The specific heat of iron at roomtemperature is cp = 0.45 kJ/kg. C (Table A-3). Analysis We take the entire contents of the tank, water + iron block, as the system. This is a closed systemsince no mass crosses the system boundary during the process. The energy balance for this system can beexpressed as

U

EEE



system


outin

Iron350 C

WATER18 C

or,0wateriron UU

0)]([)]([ water12iron12 TTmcTTmc

Substituting,

C26.72

22 0)C18)(KkJ/kg4.18)(kg100()C350)(KkJ/kg0.45)(kg25(

T

TT

The entropy generated during this process is determined from



1

2avgwater

1

2avgiron

TTmcS

TTmcS

Thus,kJ/K4.08314.12232.8waterirontotalgen SSSS

Discussion The results can be improved somewhat by using specific heats at average temperature.

7-28

7-55 An aluminum block is brought into contact with an iron block in an insulated enclosure. The finalequilibrium temperature and the total entropy change for this process are to be determined.Assumptions 1 Both the aluminum and the iron block are incompressible substances with constant specificheats. 2 The system is stationary and thus the kinetic and potential energies are negligible. 3 The system is well-insulated and thus there is no heat transfer.Properties The specific heat of aluminum at the anticipated average temperature of 450 K is cp = 0.973 kJ/kg. C. The specific heat of iron at room temperature (the only value available in the tables) is cp = 0.45 kJ/kg. C (Table A-3).Analysis We take the iron+aluminum blocks as the system, which is a closed system. The energy balance for this system can be expressed as

U

EEE



system


outin

Aluminum20 kg 200 C

Iron20 kg 100 C

or,

0)]([)]([0

iron12alum12

ironalum

TTmcTTmcUU

Substituting,

K441.4

0)C200)(KkJ/kg973.0)(kg20()C010)(KkJ/kg0.45)(kg20(

2

22

C168.4T

TT

The total entropy change for this process is determined from



1

2avgalum

1

2avgiron

TTmcS

TTmcS

Thus,kJ/K0.169346.1515.1alumirontotal SSS

7-29

7-56 EES Problem 7-55 is reconsidered. The effect of the mass of the iron block on the final equilibriumtemperature and the total entropy change for the process is to be studied. The mass of the iron is to varyfrom 1 to 10 kg. The equilibrium temperature and the total entropy change are to be plotted as a functionof iron mass.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"T_1_iron = 100 [C] {m_iron = 20 [kg]}T_1_al = 200 [C] m_al = 20 [kg]

1 2 3 4 5 6 7 8 9 10180

182

184

186

188

190

192

194

196

198

miron [kg]

T 2C_al = 0.973 [kJ/kg-K] "FromTable A-3 at the anticipated average temperature of 450 K."C_iron= 0.45 [kJ/kg-K] "FromTable A-3 at room temperature, the only valueavailable."

"Analysis: "" Treat the iron plus aluminum as a closed system, with no heat transfer in,no work out, neglect changes in KE and PE of the system. ""The final temperature is found from the energy balance."E_in - E_out = DELTAE_sysE_out = 0 E_in = 0 DELTAE_sys = m_iron*DELTAu_iron + m_al*DELTAu_alDELTAu_iron = C_iron*(T_2_iron - T_1_iron)DELTAu_al = C_al*(T_2_al - T_1_al)

"the iron and aluminum reach thermal equilibrium:"T_2_iron = T_2 T_2_al = T_2 DELTAS_iron = m_iron*C_iron*ln((T_2_iron+273) / (T_1_iron+273))DELTAS_al = m_al*C_al*ln((T_2_al+273) / (T_1_al+273))DELTAS_total = DELTAS_iron + DELTAS_al

Stotal

[kJ/kg]miron

[kg]T2

[C]0.01152 1 197.70.0226 2 195.6

0.03326 3 193.50.04353 4 191.50.05344 5 189.60.06299 6 187.80.07221 7 186.10.08112 8 184.40.08973 9 182.80.09805 10 181.2

1 2 3 4 5 6 7 8 9 100.01

0.02

0.03

0.04

0.05

0.06

0.07

0.08

0.09

0.1

miron [kg]

Sto

tal

[kJ/

kg]

7-30

7-57 An iron block and a copper block are dropped into a large lake. The total amount of entropy changewhen both blocks cool to the lake temperature is to be determined.Assumptions 1 Both the water and the iron block are incompressible substances with constant specificheats at room temperature. 2 Kinetic and potential energies are negligible.Properties The specific heats of iron and copper at room temperature are ciron = 0.45 kJ/kg. C and ccopper = 0.386 kJ/kg. C (Table A-3). Analysis The thermal-energy capacity of the lake is very large, and thus the temperatures of both the ironand the copper blocks will drop to the lake temperature (15 C) when the thermal equilibrium is established.Then the entropy changes of the blocks become

kJ/K1.571K353K288lnKkJ/kg0.386kg20ln

kJ/K4.579K353K288lnKkJ/kg0.45kg50ln

1

2avgcopper

1

2avgiron

TTmcS

TTmcS

We take both the iron and the copper blocks, as thesystem. This is a closed system since no mass crosses the system boundary during the process. The energybalance for this system can be expressed as

Lake15 C

Copper20 kg 80 C

Iron50 kg 80 C

copperironout


system


outin

UUUQ

EEE

or,[Q copper21iron21out )]([)]( TTmcTTmc

Substituting,

kJ1964K288353KkJ/kg0.386kg20K288353KkJ/kg0.45kg50outQ

Thus,

kJ/K6.820K288kJ1964

lake

inlake,lake T

QS

Then the total entropy change for this process is kJ/K0.670820.6571.1579.4lakecopperirontotal SSSS

7-31

7-58 An adiabatic pump is used to compress saturated liquid water in a reversible manner. The work inputis to be determined by different approaches. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible. 3Heat transfer to or from the fluid is negligible.Analysis The properties of water at the inlet and exit of the pump are (Tables A-4 through A-6)

/kgm001004.0kJ/kg90.206MPa15

/kgm001010.0kJ/kg6492.0kJ/kg81.191

0kPa10

32

2

12

2

31

1

1

1

1

v

v

hss

P

sh

xP

pump

15 MPa

10 kPa (a) Using the entropy data from the compressed liquid water tablekJ/kg15.1081.19190.20612P hhw

(b) Using inlet specific volume and pressure values

kJ/kg15.14kPa)100/kg)(15,00m001010.0()( 3121P PPw v

Error = 0.3%(b) Using average specific volume and pressure values

kJ/kg15.10kPa)10(15,000/kgm)001004.0001010.0(2/1)( 312avgP PPw v

Error = 0%Discussion The results show that any of the method may be used to calculate reversible pump work.

7-32

Entropy Changes of Ideal Gases

7-59C For ideal gases, cp = cv + R and

21

12

1

2

1

11

2

22

PTPT

TP

TP

VVVV

Thus,

1

2

1

2

1

2

1

2

1

2

21

12

1

2

1

2

1

212

lnln

lnlnln

lnln

lnln

PPR

TTc

PPR

TTR

TTc

PTPTR

TTc

RTTcss

p

v

v

v VV

7-60C For an ideal gas, dh = cp dT and v = RT/P. From the second Tds relation,

PdPR

TdTc

TdP

PRT

TdPc

TdPv

Tdhds p

p

Integrating,

1

2

1

212 lnln

PP

RTT

css p

Since cp is assumed to be constant.

7-61C No. The entropy of an ideal gas depends on the pressure as well as the temperature.

7-62C Setting s = 0 givespCR

pp P

PTT

PP

cR

TT

PP

RTT

c1

2

1

2

1

2

1

2

1

2

1

2 lnln0lnln

Butkk

pp

p

p PP

TTcck

kk

kccc

cR

1

1

2

1

2,Thus./since111 vv

7-63C The Pr and vr are called relative pressure and relative specific volume, respectively. They are derived for isentropic processes of ideal gases, and thus their use is limited to isentropic processes only.

7-64C The entropy of a gas can change during an isothermal process since entropy of an ideal gas depends on the pressure as well as the temperature.

7-65C The entropy change relations of an ideal gas simplify tos = cp ln(T2/T1) for a constant pressure process

and s = cv ln(T2/T1) for a constant volume process.

Noting that cp > cv, the entropy change will be larger for a constant pressure process.

7-33

7-66 Oxygen gas is compressed from a specified initial state to a specified final state. The entropy changeof oxygen during this process is to be determined for the case of constant specific heats.Assumptions At specified conditions, oxygen can be treated as an ideal gas. Properties The gas constant and molar mass of oxygen are R = 0.2598 kJ/kg.K and M = 32 kg/kmol (TableA-1).Analysis The constant volume specific heat of oxygen at the average temperature is (Table A-2)

KkJ/kg0.690K4292

560298avg,avg vcT

O20.8 m3/kg

25 C

Thus,

KkJ/kg0.105/kgm0.8/kgm0.1lnKkJ/kg0.2598

K298K560lnKkJ/kg0.690

lnln

3

31

2

1

2avg,12 V

VRTTcss v

7-67 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tankstirs the gas, and the pressure and temperature of CO2 rises. The entropy change of CO2 during this process is to be determined using constant specific heats.Assumptions At specified conditions, CO2 can be treated as an ideal gas with constant specific heats at room temperature.Properties The specific heat of CO2 is cv = 0.657 kJ/kg.K (Table A-2).Analysis Using the ideal gas relation, the entropy change is determined to be CO2

1.5 m3

100 kPa 2.7 kg

1.5kPa100kPa150

1

2

1

2

1

1

2

2

PP

TT

TP

TP VV

Thus,

kJ/K0.719

1.5lnKkJ/kg0.657kg2.7

lnlnln1

2avg,

0

1

2

1

2avg,12 T

TmcRTTcmssmS vv V

V

7-34

7-68 An insulated cylinder initially contains air at a specified state. A resistance heater inside the cylinderis turned on, and air is heated for 15 min at constant pressure. The entropy change of air during this process is to be determined for the cases of constant and variable specific heats.Assumptions At specified conditions, air can be treated as an ideal gas. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis The mass of the air and the electrical work done during this process are

We

AIR0.3 m3

120 kPa 17 C

kJ180s6015kJ/s0.2

kg0.4325K290K/kgmkPa0.287

m0.3kPa120

ine,ine,

3

3

1

11

tWWRTPm V

The energy balance for this stationary closed system can be expressed as

)()( 1212ine,outb,ine,


system


outin

TTchhmWUWW

EEE

p

since U + Wb = H during a constant pressure quasi-equilibrium process.(a) Using a constant cp value at the anticipated average temperature of 450 K, the final temperaturebecomes

Thus, K698KkJ/kg1.02kg0.4325

kJ180K290ine,12

pmcW

TT

Then the entropy change becomes

kJ/K0.387K290K698lnKkJ/kg1.020kg0.4325

lnlnln1

2avg,

0

1

2

1

2avg,12sys T

TmcPPR

TTcmssmS pp

(b) Assuming variable specific heats,

kJ/kg706.34kg0.4325

kJ180kJ/kg290.16ine,1212ine, m

WhhhhmW

From the air table (Table A-17, we read = 2.5628 kJ/kg·K corresponding to this hs2 2 value. Then,

kJ/K0.387KkJ/kg1.668022.5628kg0.4325ln 12

0

1

212sys ssm

PPRssmS

7-69 A cylinder contains N2 gas at a specified pressure and temperature. It is compressed polytropicallyuntil the volume is reduced by half. The entropy change of nitrogen during this process is to be determined.Assumptions 1 At specified conditions, N2 can be treated as an ideal gas. 2 Nitrogen has constant specificheats at room temperature.Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The constant volume specificheat of nitrogen at room temperature is cv = 0.743 kJ/kg.K (Table A-2). Analysis From the polytropic relation,

K3.3692K003 13.11

2

112

1

2

1

1

2nn

TTTT

vv

vv

Then the entropy change of nitrogen becomes

kJ/K0.06170.5lnKkJ/kg0.297K300K369.3lnKkJ/kg0.743kg1.2

lnln1

2

1

2avg,2 V

Vv R

TTcmSN

N2

PV1.3 = C

7-35

7-70 EES Problem 7-69 is reconsidered. The effect of varying the polytropic exponent from 1 to 1.4 on theentropy change of the nitrogen is to be investigated, and the processes are to be shown on a common P-vdiagram.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Function BoundWork(P[1],V[1],P[2],V[2],n) "This function returns the Boundary Work for the polytropic process. This function is requiredsince the expression for boundary work depens on whether n=1 or n<>1" If n<>1 then BoundWork:=(P[2]*V[2]-P[1]*V[1])/(1-n)"Use Equation 3-22 when n=1" else BoundWork:= P[1]*V[1]*ln(V[2]/V[1]) "Use Equation 3-20 when n=1" endifend

n=1P[1] = 120 [kPa] T[1] = 27 [C] m = 1.2 [kg] V[2]=V[1]/2Gas$='N2'MM=molarmass(Gas$)R=R_u/MMR_u=8.314 [kJ/kmol-K]"System: The gas enclosed in the piston-cylinder device.""Process: Polytropic expansion or compression, P*V^n = C"P[1]*V[1]=m*R*(T[1]+273)P[2]*V[2]^n=P[1]*V[1]^nW_b = BoundWork(P[1],V[1],P[2],V[2],n) "Find the temperature at state 2 from the pressure and specific volume."T[2]=temperature(gas$,P=P[2],v=V[2]/m)"The entropy at states 1 and 2 is:"s[1]=entropy(gas$,P=P[1],v=V[1]/m)s[2]=entropy(gas$,P=P[2],v=V[2]/m)DELTAS=m*(s[2] - s[1]) "Remove the {} to generate the P-v plot data"{Nsteps = 10 VP[1]=V[1]PP[1]=P[1]Duplicate i=2,Nsteps VP[i]=V[1]-i*(V[1]-V[2])/Nsteps PP[i]=P[1]*(V[1]/VP[i])^nEND }

S [kJ/kg] n Wb [kJ]-0.2469 1 -74.06-0.2159 1.05 -75.36-0.1849 1.1 -76.69-0.1539 1.15 -78.05-0.1229 1.2 -79.44

-0.09191 1.25 -80.86-0.06095 1.3 -82.32-0.02999 1.35 -83.82

0.0009849 1.4 -85.34

7-36

0.4 0.5 0.6 0.7 0.8 0.9 1100

150

200

250

300

350

VP[i]

PP

[i]n = 1

= 1.4

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4-0.25

-0.2

-0.15

-0.1

-0.05

0

0.05

n

S [

kJ/K

]

S = 0 kJ/k

1 1.05 1.1 1.15 1.2 1.25 1.3 1.35 1.4-87.5

-84.5

-81.5

-78.5

-75.5

-72.5

n

Wb

[kJ]

7-37

7-71E A fixed mass of helium undergoes a process from one specified state to another specified state. The entropy change of helium is to be determined for the cases of reversible and irreversible processes.Assumptions 1 At specified conditions, helium can be treated as an ideal gas. 2 Helium has constantspecific heats at room temperature.Properties The gas constant of helium is R = 0.4961 Btu/lbm.R (Table A-1E). The constant volumespecific heat of helium is cv = 0.753 Btu/lbm.R (Table A-2E). Analysis From the ideal-gas entropy change relation,

HeT1 = 540 RT2 = 660 R

Btu/R9.71/lbmft50/lbmft10lnRBtu/lbm0.4961

R540R660lnR)Btu/lbm(0.753)lbm15(

lnln

3

31

2

1

2ave,He v

vv R

TTcmS

The entropy change will be the same for both cases.

7-72 Air is compressed in a piston-cylinder device in a reversible and isothermal manner. The entropychange of air and the work done are to be determined.Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is specified to be reversible.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).Analysis (a) Noting that the temperature remains constant, the entropy change of air is determined from

KkJ/kg0.428kPa90kPa400lnKkJ/kg0.287

lnlnln1

2

1

20

1

2avg,air P

PRPPR

TTcS p

AIR

T = const

QAlso, for a reversible isothermal process,

kJ/kg4.125kJ/kg125.4KkJ/kg0.428K293 outqsTq

(b) The work done during this process is determined from the closed system energy balance,

kJ/kg125.4outin

12outin


system


outin

0)(qw

TTmcUQW

EEE

v

7-38

7-73 Air is compressed steadily by a 5-kW compressor from one specified state to another specified state. The rate of entropy change of air is to be determined.Assumptions At specified conditions, air can be treated as an ideal gas. 2 Air has variable specific heats. Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1).

P2 = 600 kPa T2 = 440 K

Analysis From the air table (Table A-17),

KkJ/kg2.0887kPa600K440

KkJ/kg1.66802kPa100K290

22

2

11

1

sPT

sPT

5 kW

AIRCOMPRESSOR

Then the rate of entropy change of air becomes

kW/K0.00250kPa100kPa600lnKkJ/kg0.2871.668022.0887kg/s1.6/60

ln1

212sys P

PRssmS P1 = 100 kPa T1 = 290 K

7-74 One side of a partitioned insulated rigid tank contains an ideal gas at a specified temperature andpressure while the other side is evacuated. The partition is removed, and the gas fills the entire tank. Thetotal entropy change during this process is to be determined.Assumptions The gas in the tank is given to be an ideal gas, and thus ideal gas relations apply.Analysis Taking the entire rigid tank as the system, the energy balance can be expressed as

)(0

12

12

12


system


outin

TTuu

uumU

EEE

since u = u(T) for an ideal gas. Then the entropy change of the gas becomes

IDEALGAS

5 kmol 40 C

kJ/K28.81

2lnKkJ/kmol8.314kmol5

lnlnln1

2

1

20

1

2avg, V

VVV

v uu NRRTTcNS

This also represents the total entropy change since the tank does not contain anything else, and there are no interactions with the surroundings.

7-39

7-75 Air is compressed in a piston-cylinder device in a reversible and adiabatic manner. The finaltemperature and the work are to be determined for the cases of constant and variable specific heats. Assumptions 1 At specified conditions, air can be treated as an ideal gas. 2 The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air at lowto moderately high temperatures is k = 1.4 (Table A-2). Analysis (a) Assuming constant specific heats, the ideal gas isentropic relations give

K525.3kPa100kPa800K290

1.40.41

1

212

kk

PPTT

AIRReversible

Then,

KkJ/kg0.727K avg,avg 407.7/2525.3290 vcT

We take the air in the cylinder as the system. The energy balance for this stationary closed system can be expressed as

)()( 1212in


system


outin

TTmcuumUW

EEE

v

Thus,kJ/kg171.1K290525.3KkJ/kg0.72712avg,in TTcw v

(b) Assuming variable specific heats, the final temperature can be determined using the relative pressure data (Table A-17),

and

kJ/kg376.16849.91.2311

kPa100kPa800

kJ/kg206.912311.1

K290

2

2

1

2

11

12

1

uT

PPPP

uP

T

rr

r

K522.4

Then the work input becomeskJ/kg169.25kJ/kg206.91376.1612in uuw

7-40

7-76 EES Problem 7-75 is reconsidered. The work done and final temperature during the compressionprocess are to be calculated and plotted as functions of the final pressure for the two cases as the final pressure varies from 100 kPa to 800 kPa. Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Procedure ConstPropSol(P_1,T_1,P_2,Gas$:Work_in_ConstProp,T2_ConstProp)C_P=SPECHEAT(Gas$,T=27)MM=MOLARMASS(Gas$)R_u=8.314 [kJ/kmol-K]R=R_u/MMC_V = C_P - Rk = C_P/C_V T2= (T_1+273)*(P_2/P_1)^((k-1)/k) T2_ConstProp=T2-273 "[C]"DELTAu = C_v*(T2-(T_1+273))Work_in_ConstProp = DELTAuEnd

"Knowns:"P_1 = 100 [kPa] T_1 = 17 [C] P_2 = 800 [kPa]

"Analysis: "" Treat the piston-cylinder as a closed system, with no heat transfer in, neglectchanges in KE and PE of the air. The process is reversible and adiabatic thus isentropic.""The isentropic work is determined from:"e_in - e_out = DELTAe_syse_out = 0 [kJ/kg] e_in = Work_inDELTAE_sys = (u_2 - u_1)u_1 = INTENERGY(air,T=T_1)v_1 = volume(air,P=P_1,T=T_1)s_1 = entropy(air,P=P_1,T=T_1)" The process is reversible and adiabatic or isentropic.Then P_2 and s_2 specify state 2."s_2 = s_1 u_2 = INTENERGY(air,P=P_2,s=s_2)T_2_isen=temperature(air,P=P_2,s=s_2)Gas$ = 'air'Call ConstPropSol(P_1,T_1,P_2,Gas$: Work_in_ConstProp,T2_ConstProp)

100 200 300 400 500 600 700 8000

20406080

100120140160180

P2 [kPa]

Workin[kJ/kg]

P2

[kPa]Workin

[kJ/kg]Workin,ConstProp

[kJ/kg]100 0 0200 45.63 45.6300 76.84 76.77400 101.3 101.2500 121.7 121.5600 139.4 139.1700 155.2 154.8800 169.3 168.9

7-41

7-77 Helium gas is compressed in a piston-cylinder device in a reversible and adiabatic manner. The finaltemperature and the work are to be determined for the cases of the process taking place in a piston-cylinder device and a steady-flow compressor.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is given to be reversibleand adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.Properties The specific heats and the specific heat ratio of helium are cv = 3.1156 kJ/kg.K, cp = 5.1926 kJ/kg.K, and k = 1.667 (Table A-2).

2

HeRev.He

Rev.

Analysis (a) From the ideal gas isentropic relations,

K576.91.6670.6671

1

212 kPa90

kPa450K303kk

PPTT

(a) We take the air in the cylinder as the system. Theenergy balance for this stationary closed system can beexpressed as

)()( 1212in


system


outin

TTmcuumUW

EEE

v

1

Thus, kJ/kg853.4K)303576.9)(KkJ/kg3.1156(12in TTcw v

(b) If the process takes place in a steady-flow device, the final temperature will remain the same but the work done should be determined from an energy balance on this steady-flow device,

)()(

0

1212in

21in

outin


(steady)0system


outin

TTcmhhmW

hmhmW

EE

EEE

p

Thus, kJ/kg1422.3K)303576.9)(KkJ/kg5.1926(12in TTcw p

7-78 An insulated rigid tank contains argon gas at a specified pressure and temperature. A valve is opened, and argon escapes until the pressure drops to a specified value. The final mass in the tank is to bedetermined.Assumptions 1 At specified conditions, argon can be treated as an ideal gas. 2 The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply.Properties The specific heat ratio of argon is k = 1.667 (Table A-2).

ARGON4 kg

450 kPa 30 C

Analysis From the ideal gas isentropic relations,

K0.219kPa450kPa200K303

1.6670.6671

1

212

kk

PP

TT

The final mass in the tank is determined from the ideal gas relation,

kg2.46kg4K219kPa450K303kPa200

121

122

22

11

2

1 mTPTP

mRTmRTm

PPV

V

7-42

7-79 EES Problem 7-78 is reconsidered. The effect of the final pressure on the final mass in the tank is tobe investigated as the pressure varies from 450 kPa to 150 kPa, and the results are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"UNIFORM_FLOW SOLUTION:""Knowns:"C_P = 0.5203"[kJ/kg-K ]"C_V = 0.3122 "[kJ/kg-K ]"R=0.2081 "[kPa-m^3/kg-K]"P_1= 450"[kPa]"T_1 = 30"[C]"m_1 = 4"[kg]"P_2= 150"[kPa]"

"Analysis:We assume the mass that stays in the tank undergoes an isentropic expansionprocess. This allows us to determine the final temperature of that gas at the finalpressure in the tank by using the isentropic relation:"

k = C_P/C_V T_2 = ((T_1+273)*(P_2/P_1)^((k-1)/k)-273)"[C]"V_2 = V_1 P_1*V_1=m_1*R*(T_1+273)P_2*V_2=m_2*R*(T_2+273)

m2[kg]

P2[kPa]

2.069 1502.459 2002.811 2503.136 3003.44 3503.727 400

4 450

150 200 250 300 350 400 4502

2.4

2.8

3.2

3.6

4

P2 [kPa]

m2

[kg]

7-43

7-80E Air is accelerated in an adiabatic nozzle. Disregarding irreversibilities, the exit velocity of air is to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is given to be reversible and adiabatic, and thus isentropic. Therefore, isentropic relations of ideal gases apply. 2 The nozzle operates steadily.Analysis Assuming variable specific heats, the inlet and exit properties are determined to be

and

Btu/lbm152.11R635.9

2.4612.30psia60psia12

Btu/lbm240.9830.12

R1000

2

2

1

2

11

12

1

hT

PPPP

hP

T

rr

r

2AIR1

We take the nozzle as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as

02

/2)V+()2/(

0

21

22

12

222

211

outin


(steady)0system


outin

VVhh

hmVhm

EE

EEE

Therefore,

ft/s2119

ft/s200Btu/lbm1

/sft25,037Btu/lbm152.11240.9822 2

222

1212 VhhV

7-44

7-81 Air is accelerated in an nozzle, and some heat is lost in the process. The exit temperature of air and the total entropy change during the process are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The nozzle operates steadily. Analysis (a) Assuming variable specific heats, the inlet properties are determined to be,

(Table A-17) Th

s11

1350 K

350.49 kJ / kg1.85708 / kJ / kg K 3.2 kJ/s

AIRWe take the nozzle as the system, which is a control volume. The energybalance for this steady-flow system can be expressed in the rate form as 1 2

2+0

+/2)+()2/(

0

21

22

12out

out2

222

11


(steady)0system


outin

VVhhq

QVhmVhm

EE

EEE

outin

Therefore,

kJ/kg297.34/sm1000

kJ/kg12

m/s50m/s3203.2350.492 22

2221

22

out12VVqhh

At this h2 value we read, from Table A-17, T s2 2297.2 K, 1.6924 kJ / kg K

(b) The total entropy change is the sum of the entropy changes of the air and of the surroundings, and isdetermined from

where

KkJ/kg0.1775kPa280

kPa85lnKkJ/kg0.2871.857081.6924ln1

212air

surrairtotal

PPRsss

sss

and

Thus,KkJ/kg0.18840109.01775.0

KkJ/kg0.0109K293

kJ/kg3.2

total

surr

insurr,surr

s

Tq

s

7-45

7-82 EES Problem 7-76 is reconsidered. The effect of varying the surrounding medium temperature from10°C to 40°C on the exit temperature and the total entropy change for this process is to be studied, and theresults are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Function HCal(WorkFluid$, Tx, Px)"Function to calculate the enthalpy of an ideal gas or real gas" If 'Air' = WorkFluid$ then HCal:=ENTHALPY('Air',T=Tx) "Ideal gas equ." else HCal:=ENTHALPY(WorkFluid$,T=Tx, P=Px)"Real gas equ." endifend HCal

"System: control volume for the nozzle""Property relation: Air is an ideal gas""Process: Steady state, steady flow, adiabatic, no work""Knowns - obtain from the input diagram"WorkFluid$ = 'Air'T[1] = 77 [C] P[1] = 280 [kPa] Vel[1] = 50 [m/s]P[2] = 85 [kPa] Vel[2] = 320 [m/s] q_out = 3.2 [kJ/kg] "T_surr = 20 [C]"

"Property Data - since the Enthalpy function has different parametersfor ideal gas and real fluids, a function was used to determine h."h[1]=HCal(WorkFluid$,T[1],P[1])h[2]=HCal(WorkFluid$,T[2],P[2])

"The Volume function has the same form for an ideal gas as for a real fluid."v[1]=volume(workFluid$,T=T[1],p=P[1])v[2]=volume(WorkFluid$,T=T[2],p=P[2])

"If we knew the inlet or exit area, we could calculate the mass flow rate. Since we don't know these areas, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]""Conservation of Energy - SSSF energy balance for neglecting the change in potential energy, no work, but heat transfer out is:"h[1]+Vel[1]^2/2*Convert(m^2/s^2, kJ/kg) = h[2]+Vel[2]^2/2*Convert(m^2/s^2, kJ/kg)+q_outs[1]=entropy(workFluid$,T=T[1],p=P[1])s[2]=entropy(WorkFluid$,T=T[2],p=P[2])

"Entropy change of the air and the surroundings are:"DELTAs_air = s[2] - s[1] q_in_surr = q_out DELTAs_surr = q_in_surr/(T_surr+273)DELTAs_total = DELTAs_air + DELTAs_surr

7-46

stotal

[kJ/kg-K]Tsurr

[C]T2

[C]0.189 10 24.22

0.1888 15 24.220.1886 20 24.220.1884 25 24.220.1882 30 24.220.188 35 24.22

0.1879 40 24.22

10 15 20 25 30 35 400.1878

0.188

0.1882

0.1884

0.1886

0.1888

0.189

Tsurr [C]

s tot

al[k

J/kg

-K]

7-47

7-83 A container is filled with liquid water is placed in a room and heat transfer takes place between thecontainer and the air in the room until the thermal equilibrium is established. The final temperature, theamount of heat transfer between the water and the air, and the entropy generation are to be determined.Assumptions 1 Kinetic and potential energy changes are negligible. 2 Air is an ideal gas with constantspecific heats. 3 The room is well-sealed and there is no heat transfer from the room to the surroundings. 4Sea level atmospheric pressure is assumed. P = 101.3 kPa.Properties The properties of air at room temperature are R = 0.287 kPa.m3/kg.K, cp = 1.005 kJ/kg.K, cv = 0.718 kJ/kg.K. The specific heat of water at room temperature is cw = 4.18 kJ/kg.K (Tables A-2, A-3). Analysis (a) The mass of the air in the room is

kg5.111K)273K)(12/kgmkPa(0.287

)mkPa)(90(101.33

3

1aa RT

Pm V

An energy balance on the system that consists of the water in thecontainer and the air in the room gives the final equilibriumtemperature

C70.2222

1212

)12kJ/kg.K)(kg)(0.7185.111()95kJ/kg.K)(kg)(4.1845(0

)()(0

TTT

TTcmTTcm aawww v

Water45 kg 95 C

Room90 m3

12 C

(b) The heat transfer to the air is kJ4660)120.2kJ/kg.K)(7kg)(0.7185.111()( 12 aa TTcmQ v

(c) The entropy generation associated with this heat transfer process may be obtained by calculating totalentropy change, which is the sum of the entropy changes of water and the air.

kJ/K11.13K273)(95K273)(70.2kJ/kg.K)lnkg)(4.18(45ln

1

2

wwww T

TcmS

kPa122)m(90

K)273K)(70.2/kgmkPakg)(0.287(111.53

32

2 VRTmP a

kJ/K88.14kPa101.3

kPa122kJ/kg.K)ln(0.287K273)(12K273)(70.2

kJ/kg.K)ln(1.005kg)(111.5

lnln1

2

1

2

PP

RTT

cmSa

paa

kJ/K1.7788.1411.13totalgen aw SSSS

7-48

7-84 Air is accelerated in an isentropic nozzle. The maximum velocity at the exit is to be determined.Assumptions 1 Air is an ideal gas with constant specific heats. 2 The nozzle operates steadily. Properties The properties of air at room temperature are cp = 1.005 kJ/kg.K, k = 1.4 (Table A-2a). Analysis The exit temperature is determined from ideal gas isentropic relation to be,

K5.371kPa800kPa100

K2734000.4/1.4/)1(

1

212

kk

PP

TT

We take the nozzle as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

2)(0

200

/2)+()2/(

0

22

12

22

12

222

211

outin


(steady)0system


outin

VTTc

Vhh

VhmVhm

EE

EEE

p

AIR1 2

Therefore,

m/s778.5371.5)K-73kJ/kg.K)(6005.1(2)(2 122 TTc pV

7-49

7-85 An ideal gas is compressed in an isentropic compressor. 10% of gas is compressed to 400 kPa and90% is compressed to 600 kPa. The compression process is to be sketched, and the exit temperatures at thetwo exits, and the mass flow rate into the compressor are to be determined.Assumptions 1 The compressor operates steadily. 2 The process is reversible-adiabatic (isentropic)Properties The properties of ideal gas are given to be cp = 1.1 kJ/kg.K and cv = 0.8 kJ/kg.K.Analysis (b) The specific heat ratio of the gas is

375.18.01.1

vcc

k pP3 = 600 kPa

P2 = 400 kPa

32 kW

COMPRESSOR

The exit temperatures are determined from ideal gas isentropic relations to be,

K437.850.375/1.37/)1(

1

212 kPa100

kPa400K27327

kk

PP

TT

K489.050.375/1.37/)1(

1

313 kPa100

kPa600K27327

kk

PP

TTP1 = 100 kPa T1 = 300 K (c) A mass balance on the control volume gives

321 mmm

s

TP3

P2

P1

where13

12

9.01.0mmmm

We take the compressor as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

3121in11

3322in11

outin


(steady)0system


outin

9.01.0

0

TcmTcmWTcm

hmhmWhm

EE

EEE

ppp

Solving for the inlet mass flow rate, we obtain

kg/s0.158300)-0.9(489.0300)-0.1(437.8K)kJ/kg(1.1

kW32

)(9.0)(1.0 13121 TTTTc

Wmp

in

7-50

7-86 Air contained in a constant-volume tank s cooled to ambient temperature. The entropy changes of theair and the universe due to this process are to be determined and the process is to be sketched on a T-s diagram.Assumptions 1 Air is an ideal gas with constant specific heats.

Air5 kg

327 C100 kPa

Properties The specific heat of air at room temperature iscv = 0.718 kJ/kg.K (Table A-2a). Analysis (a) The entropy change of air is determined from

kJ/K2.488K273)(327K273)(27kJ/kg.K)lnkg)(0.718(5

ln1

2air T

TmcS v

27ºC

327ºC

surr

air

21

2

s

T 1(b) An energy balance on the system gives

kJ1077)2727kJ/kg.K)(3kg)(0.7185(

)( 12out TTmcQ v

The entropy change of the surroundings is

kJ/K3.59K300kJ1077

surr

outsurr T

Qs

The entropy change of universe due to this process iskJ/K1.1059.3488.2surrairtotalgen SSSS

7-51

Reversible Steady-Flow Work

7-87C The work associated with steady-flow devices is proportional to the specific volume of the gas. Cooling a gas during compression will reduce its specific volume, and thus the power consumed by thecompressor.

7-88C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the workoutput of the turbine. Therefore, this is not a good proposal.

7-89C We would not support this proposal since the steady-flow work input to the pump is proportional tothe specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.

7-90 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to thepump is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible.Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5).

2Analysis The power input to the pump can be determined directly from thesteady-flow work relation for a liquid,

45 kg/s

H2O121

2

1

00in PPmpekedPmW vv

Substituting,

kW27433

in mkPa1kJ1kPa)206000)(/kgm0.001017)(kg/s45(W

1

7-91 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the watercan be pumped to is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is assumed to be reversible since we will determine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversiblesteady-flow work relation for a liquid,

P2

Thus,

323

1212

1

00in

mkPa1kJ1Pak)100)(/kgm0.001)(kg/s5(kJ/s25 P

PPmpekedPmW vv

25 kWPUMP

It yields kPa51002P100 kPa

7-52

7-92E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The power input to the compressor is to be determined, and it is also to be compared to the work input for the liquidcase.Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The compressor is adiabatic.Analysis The compression process is reversible and adiabatic, and thus isentropic, s1 = s2. Then the properties of the refrigerant are (Tables A-11E through A-13E)

Btu/lbm115.80psia08


vaporsat.psia15

212

1

1

11

hss

P

shP

R-134a

2

R-134a

2

The work input to this isentropic compressor isdetermined from the steady-flow energy balance to be

)(

0

12in

21in

outin


(steady)0system


outin

hhmW

hmhmW

EE

EEE

11

Thus, Btu/lbm14.899.10080.11512in hhw

If the refrigerant were first condensed at constant pressure before it was compressed, we would use a pumpto compress the liquid. In this case, the pump work input could be determined from the steady-flow workrelation to be

121002

1in PPpekedPw vv

where v3 = vf @ 15 psia = 0.01165 ft3/lbm. Substituting,

Btu/lbm0.1433

in ftpsia5.4039Btu1psia1580)/lbmft0.01165(w

7-53

7-93 A steam power plant operates between the pressure limits of 10 MPa and 20 kPa. The ratio of theturbine work to the pump work is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic. Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5). Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,s1 = s2 and s3 = s4. Then the properties of the steam are

kJ/kg4707.2MPa10

KkJ/kg7.9073kJ/kg2608.9

.kPa20

343

3

kPa20@4

kPa20@44

hss

P

sshh

vaporsatP

g

g

H2O

3

H2O

2

Also, v1 = vf @ 20 kPa = 0.001017 m3/kg.

The work output to this isentropic turbine is determinedfrom the steady-flow energy balance to be 1 4

)(

0

43out

out43

outin


(steady)0system


outin

hhmW

Whmhm

EE

EEE

Substituting,kJ/kg2098.39.26082.470743outturb, hhw

The pump work input is determined from the steady-flow work relation to be

kJ/kg10.15mkPa1

kJ1kPa)2010,000)(/kgm0.001017( 33

121002

1inpump, PPpekedPw vv

Thus,

206.710.152098.3

inpump,

outturb,

ww

7-54

7-94 EES Problem 7-93 is reconsidered. The effect of the quality of the steam at the turbine exit on the network output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to beplotted as a function of this quality.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"System: control volume for the pump and turbine""Property relation: Steam functions""Process: For Pump and Turbine: Steady state, steady flow, adiabatic, reversible or isentropic""Since we don't know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]"

"Knowns:"WorkFluid$ = 'Steam_IAPWS' P[1] = 20 [kPa] x[1] = 0 P[2] = 10000 [kPa] x[4] = 1.0

"Pump Analysis:"T[1]=temperature(WorkFluid$,P=P[1],x=0)v[1]=volume(workFluid$,P=P[1],x=0)h[1]=enthalpy(WorkFluid$,P=P[1],x=0)s[1]=entropy(WorkFluid$,P=P[1],x=0)s[2] = s[1]h[2]=enthalpy(WorkFluid$,P=P[2],s=s[2])T[2]=temperature(WorkFluid$,P=P[2],s=s[2])

"The Volume function has the same form for an ideal gas as for a real fluid."v[2]=volume(WorkFluid$,T=T[2],p=P[2])"Conservation of Energy - SSSF energy balance for pump"" -- neglect the change in potential energy, no heat transfer:"h[1]+W_pump = h[2] "Also the work of pump can be obtained from the incompressible fluid, steady-flow result:"W_pump_incomp = v[1]*(P[2] - P[1])

"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"P[4] = P[1] P[3] = P[2] h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4])s[4]=entropy(WorkFluid$,P=P[4],x=x[4])T[4]=temperature(WorkFluid$,P=P[4],x=x[4])s[3] = s[4]h[3]=enthalpy(WorkFluid$,P=P[3],s=s[3])T[3]=temperature(WorkFluid$,P=P[3],s=s[3])h[3] = h[4] + W_turb W_net_out = W_turb - W_pump

7-55

Wnet,out

[kJ/kg]Wpump

[kJ/kg]Wpump,incomp

[kJ/kg]Wturb

[kJ/kg]x4

557.1 10.13 10.15 567.3 0.5734.7 10.13 10.15 744.8 0.6913.6 10.13 10.15 923.7 0.71146 10.13 10.15 1156 0.81516 10.13 10.15 1527 0.92088 10.13 10.15 2098 1

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00

100200300400500600700800900

10001100

s [kJ/kg-K]

T [°

C]

10000 kPa

20 kPa 0.2 0.6 0.8

Steam

1, 2

3

4

x4 = 1.0= 0.5

0.5 0.6 0.7 0.8 0.9 1500

850

1200

1550

1900

2250

x[4]

Wne

t,out

[kJ/

kg]

7-56

7-95 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highestpossible mass flow rate of water is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kineticenergy changes are negligible, but potential energy changes may be significant. 3 The process is assumed to be reversible since we will determine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.Analysis The highest mass flow rate will be realized when the entireprocess is reversible. Thus it is determined from the reversible steady-flow work relation for a liquid,

Thus,

222

33

12122

1

0in

/sm1000kJ/kg1)m10)(m/s9.8(

mkPa1kJ1kPa)1205000)(/kgm0.001(kJ/s7 m

zzgPPmpekedPmW vvWater

P1 = 120 kPa

P2 = 5 MPa

PUMP

It yieldskg/s1.41m

7-57

7-96E Helium gas is compressed from a specified state to a specified pressure at a specified rate. The power input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, andtwo-stage compression.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R. The specific heat ratio of helium is k = 1.667 (Table A-2E).

2

Analysis The mass flow rate of helium is

lbm/s0.0493R530R/lbmftpsia2.6805

/sft5psia143

3

1

11

RTP

mV

(a) Isentropic compression with k = 1.667:

Btu/s0.7068=hp1since

Btu/s44.11

1psia14psia120

11.667R530RBtu/lbm0.49611.667lbm/s0.0493

11

70.667/1.66

/1

1

21incomp,

hp62.4

kk

PP

kkRTmW

5 ft3/s

W·He

1

(b) Polytropic compression with n = 1.2:


Btu/s33.47

1psia14psia120

11.2R530RBtu/lbm0.49611.2lbm/s0.0493

11

0.2/1.2

/1

1

21incomp,

hp47.3

nn

PP

nnRTmW

(c) Isothermal compression:

hp39.4Btu/s27.83psia14psia120lnR530RBtu/lbm0.4961lbm/s0.0493ln

1

2incomp, P

PRTmW

(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across each stage is the same, and its value is determined from

psia41.0psia120psia1421PPPx

The compressor work across each stage is also the same, thus total compressor work is twice the compression work for a single stage:


Btu/s30.52

1psia14psia41

11.2R530RBtu/lbm0.49611.2lbm/s0.04932

11

22

0.2/1.2

/1

1

1Icomp,incomp,

hp43.2

nnx

PP

nnRTmwmW

7-58

7-97E EES Problem 7-96E is reconsidered. The work of compression and entropy change of the helium isto be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Procedure FuncPoly(m_dot,k, R, T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)

If n =1 then T2=T1W_dot_comp_polytropic= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp) "[hp]"W_dot_comp_2stagePoly = W_dot_comp_polytropic "[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s) "[Btu/s]"Q_dot_Out_2stagePoly = Q_dot_Out_polytropic*convert(hp,Btu/s) "[Btu/s]"

ElseC_P = k*R/(k-1) "[Btu/lbm-R]"T2=(T1+460)*((P2/P1)^((n+1)/n)-460)"[F]"W_dot_comp_polytropic = m_dot*n*R*(T1+460)/(n-1)*((P2/P1)^((n-1)/n) - 1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s)+m_dot*C_P*(T1-T2)"[Btu/s]"

Px=(P1*P2)^0.5T2x=(T1+460)*((Px/P1)^((n+1)/n)-460)"[F]"W_dot_comp_2stagePoly = 2*m_dot*n*R*(T1+460)/(n-1)*((Px/P1)^((n-1)/n) - 1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_2stagePoly=W_dot_comp_2stagePoly*convert(hp,Btu/s)+2*m_dot*C_P*(T1-T2x)"[Btu/s]"

endifEND

R=0.4961[Btu/lbm-R]k=1.667n=1.2P1=14 [psia]T1=70 [F] P2=120 [psia]V_dot = 5 [ft^3/s]P1*V_dot=m_dot*R*(T1+460)*convert(Btu,psia-ft^3)

W_dot_comp_isentropic = m_dot*k*R*(T1+460)/(k-1)*((P2/P1)^((k-1)/k) - 1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isentropic = 0"[Btu/s]"Call FuncPoly(m_dot,k, R,T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)

W_dot_comp_isothermal= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isothermal = W_dot_comp_isothermal*convert(hp,Btu/s)"[Btu/s]"

7-59

n Wcomp2StagePoly

[hp]Wcompisentropic

[hp]Wcompisothermal

[hp]Wcomppolytropic

[hp]1 39.37 62.4 39.37 39.37

1.1 41.36 62.4 39.37 43.481.2 43.12 62.4 39.37 47.351.3 44.68 62.4 39.37 50.971.4 46.09 62.4 39.37 54.361.5 47.35 62.4 39.37 57.54

1.667 49.19 62.4 39.37 62.4

1 1.1 1.2 1.3 1.4 1.5 1.6 1.735

40

45

50

55

60

65

n

Wco

mp,

poly

trop

ic [

hp]

PolytropicIsothermalIsentropic2StagePoly

7-60

7-98 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. The mass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic,polytropic, isothermal, and two-stage compression.Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The specific heat ratio of nitrogen is k = 1.4 (Table A-2).

2Analysis (a) Isentropic compression:

or,

1kPa80kPa48011.4

K300KkJ/kg0.2971.4kJ/s10

11

0.4/1.4

/112

1incomp,

m

PPkkRTmW kk

·N2m 10 kW

It yields1m 0.048 kg / s


or,

1kPa80kPa48011.3

K300KkJ/kg0.2971.3kJ/s10

11

0.3/1.3

/112

1incomp,

m

PPnnRTmW nn

It yieldsm 0.051 kg / s


kPa80kPa480lnK300KkJ/kg0.297kJ/s10ln

2

1incomp, m

PPRTmW


(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each stage is the same, and its value is determined to be

kPa196kPa480kPa8021PPPx


or,

1kPa80kPa19611.3

K300KkJ/kg0.2971.32kJ/s10

11

22

0.3/1.3

/11

1Icomp,incomp,

m

PPnnRTmwmW nn

x

It yieldskg/s0.056m

7-61

7-99 Water mist is to be sprayed into the air stream in the compressor to cool the air as the water evaporates and to reduce the compression power. The reduction in the exit temperature of the compressedair and the compressor power saved are to be determined.Assumptions 1 Air is an ideal gas with variable specific heats. 2 The process is reversible. 3 Kinetic and potential energy changes are negligible. 3 Air is compressed isentropically. 4 Water vaporizes completelybefore leaving the compressor. 4 Air properties can be used for the air-vapor mixture.Properties The gas constant of air is R = 0.287 kJ/kg.K (Table A-1). The specific heat ratio of air is k =1.4. The inlet enthalpies of water and air are (Tables A-4 and A-17)

hw1 = hf@20 C = 83.29 kJ/kg , hfg@20 C = 2453.9 kJ/kg and ha1 = h@300 K =300.19 kJ/kgAnalysis In the case of isentropic operation (thus no cooling or water spray), the exit temperature and thepower input to the compressor are

K610.2=kPa100kPa1200K)300(

4.1/)14.1(

2

/)1(

1

2

1

2 TPP

TT

kk

kW3.6541kPakPa/100120011.4

K300KkJ/kg0.2871.4kg/s)1.2(

11

0.4/1.4

/112

1incomp,

kkPPkkRTmW

When water is sprayed, we first need to check the accuracy of the assumption that the water vaporizes completely in the compressor. In thelimiting case, the compression will be isothermal at the compressor inlettemperature, and the water will be a saturated vapor. To avoid thecomplexity of dealing with two fluid streams and a gas mixture, we disregard water in the air stream (other than the mass flow rate), andassume air is cooled by an amount equal to the enthalpy change of water. Water

20 C

1200 kPa

100 kPa 300 K

W·

1

2

He

The rate of heat absorption of water as it evaporates at the inlet temperature completely is

kW490.8=kJ/kg).9kg/s)(24532.0(C20@maxcooling, fgwhmQ

The minimum power input to the compressor is

kW3.449kPa100kPa1200lnK)K)(300kJ/kg7kg/s)(0.281.2(ln

1

2minin,comp, P

PRTmW

This corresponds to maximum cooling from the air since, at constant temperature, h = 0 and thus, which is close to 490.8 kW. Therefore, the assumption that all the water vaporizes

is approximately valid. Then the reduction in required power input due to water spray becomeskW3.449inout WQ

kW2053.4493.654isothermalcomp,isentropiccomp,incomp, WWW

Discussion (can be ignored): At constant temperature, h = 0 and thus corresponds to maximum cooling from the air, which is less than 490.8 kW. Therefore, the assumption that all the watervaporizes is only roughly valid. As an alternative, we can assume the compression process to be polytropicand the water to be a saturated vapor at the compressor exit temperature, and disregard the remainingliquid. But in this case there is not a unique solution, and we will have to select either the amount of water or the exit temperature or the polytropic exponent to obtain a solution. Of course we can also tabulate theresults for different cases, and then make a selection.

kW3.449inout WQ

7-62

Sample Analysis: We take the compressor exit temperature to be T2 = 200 C = 473 K. Then, hw2 = hg@200 C = 2792.0 kJ/kg and ha2 = h@473 K = 475.3 kJ/kg

Then,

224.1kPa100kPa1200

K300K473 /)1(/)1(

1

2

1

2 nPP

TT nnnn

kW570K)300473(11.224

KkJ/kg0.2871.224kg/s)1.2(

)(1

11 12

/112

1, TT

nnRmPP

nnRT

mW nnincomp

Energy balance:

kW0.202300.19)3kg/s)(475.1.2(kW7.569

)()( 12incomp,out12outincomp, hhmWQhhmQW

Noting that this heat is absorbed by water, the rate at which water evaporates in the compressor becomes

kg/s0746.0kJ/kg)29.830.2792(

kJ/s0.202)(12

waterin,12waterin,airout,

wwwwww hh

QmhhmQQ

Then the reductions in the exit temperature and compressor power input becomeT T T

W W Wcomp in comp comp

2 2 2 610 2 473

654 3 570, ,

, , ,

.

.isentropic water cooled

isentropic water cooled

137.2 C

84.3 kW

Note that selecting a different compressor exit temperature T2 will result in different values.

7-100 A water-injected compressor is used in a gas turbine power plant. It is claimed that the power outputof a gas turbine will increase when water is injected into the compressor because of the increase in the mass flow rate of the gas (air + water vapor) through the turbine. This, however, is not necessarily rightsince the compressed air in this case enters the combustor at a low temperature, and thus it absorbs muchmore heat. In fact, the cooling effect will most likely dominate and cause the cyclic efficiency to drop.

7-63

Isentropic Efficiencies of Steady-Flow Devices

7-101C The ideal process for all three devices is the reversible adiabatic (i.e., isentropic) process. Theadiabatic efficiencies of these devices are defined as

energykineticexitcinsentropienergykineticexitactualand,

inputworkactualinputworkcinsentropi,

outputworkcinsentropioutputworkactual

NCT

7-102C No, because the isentropic process is not the model or ideal process for compressors that are cooled intentionally.

7-103C Yes. Because the entropy of the fluid must increase during an actual adiabatic process as a result of irreversibilities. Therefore, the actual exit state has to be on the right-hand side of the isentropic exitstate

7-104 Steam enters an adiabatic turbine with an isentropic efficiency of 0.90 at a specified state with a specified mass flow rate, and leaves at a specified pressure. The turbine exit temperature and power outputof the turbine are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potential energy changes are negligible. 3The device is adiabatic and thus heat transfer is negligible.Analysis (a) From the steam tables (Tables A-4 through A-6),

kJ/kg2268.32335.30.8475289.27

8475.08234.6

9441.07266.6kPa30

KkJ/kg6.7266kJ/kg3399.5

C500MPa8

22

22

12

2

1

1

1

1

fgsfs

fg

fss

s

s

hxhhs

ssx

ssP

sh

TP

P1 = 8 MPaT1 = 500 C

STEAMTURBINE

T = 90%

P2 = 30 kPaFrom the isentropic efficiency relation,

Thus,

C69.09kPa30@sat22

2

211221

21

kJ/kg2381.4kPa30

kJ/kg2381.42268.33399.50.95.3399

TThP

hhhhhhhh

aa

a

sas

aTT

(b) There is only one inlet and one exit, and thus m m m1 2 . We take the actual turbine as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE

)(

0)peke(since

21outa,

2outa,1

hhmW

QhmWhm

Substituting,kW3054kJ/kg2381.43399.5kg/s3outa,W

7-64

7-105 EES Problem 7-104 is reconsidered. The effect of varying the turbine isentropic efficiency from0.75 to 1.0 on both the work done and the exit temperature of the steam are to be investigated, and theresults are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"System: control volume for turbine""Property relation: Steam functions""Process: Turbine: Steady state, steady flow, adiabatic, reversible or isentropic""Since we don't know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]=m_dot"

"Knowns:"WorkFluid$ = 'Steam_iapws'm_dot = 3 [kg/s] P[1] = 8000 [kPa] T[1] = 500 [C] P[2] = 30 [kPa] "eta_turb = 0.9"

"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potential energy, no heat transfer:"h[1]=enthalpy(WorkFluid$,P=P[1],T=T[1])s[1]=entropy(WorkFluid$,P=P[1],T=T[1]) 0.0 1.0 2.0 3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0

0

100

200

300

400

500

600

700

s [kJ/kg-K]

T [°

C]

8000 kPa

30 kPa

0.2 0.4 0.6

0.8

Steam

1

2s

2

T_s[1] = T[1] s[2] =s[1]s_s[2] = s[1]h_s[2]=enthalpy(WorkFluid$,P=P[2],s=s_s[2])T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2])eta_turb = w_turb/w_turb_s h[1] = h[2] + w_turb h[1] = h_s[2] + w_turb_s T[2]=temperature(WorkFluid$,P=P[2],h=h[2])W_dot_turb = m_dot*w_turb

turb Wturb [kW]0.75 25450.8 2715

0.85 28850.9 3054

0.95 32241 3394

0.75 0.8 0.85 0.9 0.95 12500

2600

2700

2800

2900

3000

3100

3200

3300

3400

turb

Wtu

rb [

kW]

7-65

7-106 Steam enters an adiabatic turbine at a specified state, and leaves at a specified state. The mass flow rate of the steam and the isentropic efficiency are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible.Analysis (a) From the steam tables (Tables A-4 and A-6),

kJ/kg2780.2C150

kPa50

KkJ/kg7.0910kJ/kg3650.6

C600MPa7

22

2

1

1

1

1

ahTP

sh

TP

There is only one inlet and one exit, and thus m m m1 2 . We take the actual turbine as the system, whichis a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

outin


(steady)0system


outin 0

EE

EEE 1

6 MW H2O

2

0)pe(since/2)+()2/(2

12

212outa,

212outa,

211

VVhhmW

QVhmWVhm

2Substituting, the mass flow rate of the steam is determined to be

kg/s6.95m

m 22

22

/sm1000kJ/kg1

2)m/s80()m/s140(3650.62780.2kJ/s6000

(b) The isentropic exit enthalpy of the steam and the power output of the isentropic turbine are

kJ/kg2467.32304.70.9228340.54

0.92286.5019

1.09127.0910kPa50

22

22

12

2

fgsfs

fg

fss

s

s

hxhhs

ssx

ssP

and

kW8174/sm1000

kJ/kg12

)m/s80()m/s140(3650.62467.3kg/s6.95

2/

22

22

outs,

21

2212outs,

W

VVhhmW s

Then the isentropic efficiency of the turbine becomes

73.4%0.734kW8174kW6000

s

a

WW

T

7-66

7-107 Argon enters an adiabatic turbine at a specified state with a specified mass flow rate, and leaves at a specified pressure. The isentropic efficiency of the turbine is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas with constant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constant pressure specific heat of argon is cp = 0.5203 kJ/kg.K (Table A-2). Analysis There is only one inlet and one exit, and thus m m m1 2 . We take the isentropic turbine as thesystem, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed in the rate form as

1

)(

0)peke(since

21outs,

2out,1

outin

s

ss

hhmW

QhmWhm

EE

370 kWAr

TFrom the isentropic relations,

K479kPa1500kPa200K1073

70.667/1.66/1

1

212

kks

s PPTT

Then the power output of the isentropic turbine becomes 2kW14124791073KkJ/kg0.5203kg/min80/6021outs, .sp TTcmW

Then the isentropic efficiency of the turbine is determined from

89.8%898.0kW412.1

kW370

s

a

WW

T

7-108E Combustion gases enter an adiabatic gas turbine with an isentropic efficiency of 82% at a specifiedstate, and leave at a specified pressure. The work output of the turbine is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Combustiongases can be treated as air that is an ideal gas with variable specific heats.Analysis From the air table and isentropic relations, 1

Th

Pr1

1

1174 0

2000 R= 504.71 Btu / lbm

.

AIRT = 82%Btu/lbm417.387.0174.0

psia120psia60

2s1

212

hPPPP rr

There is only one inlet and one exit, and thus m m m1 2 . We take theactual turbine as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as

2

)(

0)peke(since

21outa,

2outa,1

outin

hhmW

QhmWhm

EE

Noting that wa = Tws, the work output of the turbine per unit mass is determined fromBtu/lbm71.7Btu/lbm417.3504.710.82aw

7-67

7-109 [Also solved by EES on enclosed CD] Refrigerant-134a enters an adiabatic compressor with an isentropic efficiency of 0.80 at a specified state with a specified volume flow rate, and leaves at a specifiedpressure. The compressor exit temperature and power input to the compressor are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.

2Analysis (a) From the refrigerant tables (Tables A-11E through A-13E),

0.3 m3/min

R-134aC = 80%

kJ/kg281.21MPa1

/kgm0.16212KkJ/kg0.94779

kJ/kg236.97

vaporsat.kPa120

212

2

3kPa120@1

kPa120@1

kPa120@11

ss

g

g

g

hss

P

sshh

P

vv

From the isentropic efficiency relation, 1

Thus,

C58.9aa

a

saa

s

ThP

hhhhhhhh

CC

22

2

121212

12

kJ/kg292.26MPa1

kJ/kg292.26/0.80236.97281.21236.97/

(b) The mass flow rate of the refrigerant is determined from

kg/s0.0308/kgm0.16212/sm0.3/60

3

3

1

1

v

Vm

There is only one inlet and one exit, and thus m m m1 2 . We take the actual compressor as the system,which is a control volume since mass crosses the boundary. The energy balance for this steady-flow systemcan be expressed as

outin


(steady)0system


outin 0

EE

EEE

)(

0)peke(since

12ina,

21ina,

hhmW

QhmhmW

Substituting, the power input to the compressor becomes,

kW1.70kJ/kg236.97292.26kg/s0.0308ina,W

7-68

7-110 EES Problem 7-109 is reconsidered. The problem is to be solved by considering the kinetic energyand by assuming an inlet-to-exit area ratio of 1.5 for the compressor when the compressor exit pipe insidediameter is 2 cm.Analysis The problem is solved using EES, and the solution is given below.

"Input Data from diagram window"{P[1] = 120 "kPa" P[2] = 1000 "kPa" Vol_dot_1 = 0.3 "m^3/min"Eta_c = 0.80 "Compressor adiabatic efficiency"A_ratio = 1.5 d_2 = 2/100 "m"}

"System: Control volume containing the compressor, see the diagram window.Property Relation: Use the real fluid properties for R134a. Process: Steady-state, steady-flow, adiabatic process."Fluid$='R134a'"Property Data for state 1"T[1]=temperature(Fluid$,P=P[1],x=1)"Real fluid equ. at the sat. vapor state"h[1]=enthalpy(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"s[1]=entropy(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"v[1]=volume(Fluid$, P=P[1], x=1)"Real fluid equ. at the sat. vapor state"

"Property Data for state 2"s_s[1]=s[1]; T_s[1]=T[1] "needed for plot"s_s[2]=s[1] "for the ideal, isentropic process across the compressor"h_s[2]=ENTHALPY(Fluid$, P=P[2], s=s_s[2])"Enthalpy 2 at the isentropic state 2s and pressure P[2]"T_s[2]=Temperature(Fluid$, P=P[2], s=s_s[2])"Temperature of ideal state - needed only for plot.""Steady-state, steady-flow conservation of mass" m_dot_1 = m_dot_2 m_dot_1 = Vol_dot_1/(v[1]*60)Vol_dot_1/v[1]=Vol_dot_2/v[2]Vel[2]=Vol_dot_2/(A[2]*60)A[2] = pi*(d_2)^2/4A_ratio*Vel[1]/v[1] = Vel[2]/v[2] "Mass flow rate: = A*Vel/v, A_ratio = A[1]/A[2]"A_ratio=A[1]/A[2]

"Steady-state, steady-flow conservation of energy, adiabatic compressor, see diagramwindow"m_dot_1*(h[1]+(Vel[1])^2/(2*1000)) + W_dot_c= m_dot_2*(h[2]+(Vel[2])^2/(2*1000))

"Definition of the compressor adiabatic efficiency, Eta_c=W_isen/W_act"Eta_c = (h_s[2]-h[1])/(h[2]-h[1])

"Knowing h[2], the other properties at state 2 can be found."v[2]=volume(Fluid$, P=P[2], h=h[2])"v[2] is found at the actual state 2, knowing P and h."T[2]=temperature(Fluid$, P=P[2],h=h[2])"Real fluid equ. for T at the known outlet h and P."s[2]=entropy(Fluid$, P=P[2], h=h[2]) "Real fluid equ. at the known outlet h and P."T_exit=T[2]"Neglecting the kinetic energies, the work is:"m_dot_1*h[1] + W_dot_c_noke= m_dot_2*h[2]

7-69

SOLUTION

A[1]=0.0004712 [m^2] A[2]=0.0003142 [m^2] A_ratio=1.5d_2=0.02 [m] Eta_c=0.8Fluid$='R134a'h[1]=237 [kJ/kg] h[2]=292.3 [kJ/kg] h_s[2]=281.2 [kJ/kg] m_dot_1=0.03084 [kg/s] m_dot_2=0.03084 [kg/s] P[1]=120.0 [kPa] P[2]=1000.0 [kPa] s[1]=0.9478 [kJ/kg-K] s[2]=0.9816 [kJ/kg-K]

s_s[1]=0.9478 [kJ/kg-K] s_s[2]=0.9478 [kJ/kg-K] T[1]=-22.32 [C] T[2]=58.94 [C] T_exit=58.94 [C] T_s[1]=-22.32 [C] T_s[2]=48.58 [C] Vol_dot_1=0.3 [m^3 /min] Vol_dot_2=0.04244 [m^3 /min]v[1]=0.1621 [m^3/kg] v[2]=0.02294 [m^3/kg] Vel[1]=10.61 [m/s] Vel[2]=2.252 [m/s] W_dot_c=1.704 [kW] W_dot_c_noke=1.706 [kW]

0.0 0.2 0.4 0.6 0.8 1.0 1.2-50

-25

0

25

50

75

100

125

Entropy [kJ/kg-K]

Tem

pera

ture

[C]

1000 kPa

120 kPa

R134a

T-s diagram for real and ideal compressor

Ideal CompressorIdeal Compressor

Real CompressorReal Compressor

7-70

7-111 Air enters an adiabatic compressor with an isentropic efficiency of 84% at a specified state, and leaves at a specified temperature. The exit pressure of air and the power input to the compressor are to bedetermined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats.Properties The gas constant of air is R = 0.287 kPa.m3/kg.K (Table A-1) 2Analysis (a) From the air table (Table A-17),

2.4 m3/s

AIRC = 84%kJ/kg533.98K530

1.2311kJ/kg,290.16K290

22

111

a

r

hT

PhT

From the isentropic efficiency relation12

12

hhhh

a

sC

,

951.7kJ/kg495.0290.16533.980.84290.162

1212

r

as

P

hhhhC 1

Then from the isentropic relation ,

kPa646kPa1001.23117.951

121

2

1

2

1

2 PPP

PPP

PP

r

r

r

r

(b) There is only one inlet and one exit, and thus m m m1 2 . We take the actual compressor as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as

outin


(steady)0system


outin 0

EE

EEE

)(

0)peke(since

12ina,

21ina,

hhmW

QhmhmW

where kg/s884.2)K290)(K/kgmkPa0.287(

)/sm2.4)(kPa100(3

3

1

11

RTP

mV

Then the power input to the compressor is determined to be

kW703kJ/kg)290.16533.98)(kg/s2.884(ina,W

7-71

7-112 Air is compressed by an adiabatic compressor from a specified state to another specified state. The isentropic efficiency of the compressor and the exit temperature of air for the isentropic case are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas with variable specific heats.Analysis (a) From the air table (Table A-17),

2T h

T h

r

a

1 1

2 2

11 386300 K 300.19 kJ / kg,

550 K 554.74 kJ / kg

.P

AIRFrom the isentropic relation,

kJ/kg508.728.7541.386kPa95kPa600

21

212 srr hP

PPP

Then the isentropic efficiency becomes1

C

h hh h

s

a

2 1

2 1

508 72 30019554 74 30019

0 819. .. .

. 81.9%

(b) If the process were isentropic, the exit temperature would be

h Ts s2 2508.72 kJ / kg 505.5 K

7-72

7-113E Argon enters an adiabatic compressor with an isentropic efficiency of 80% at a specified state, and leaves at a specified pressure. The exit temperature of argon and the work input to the compressor are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Argon is an ideal gas withconstant specific heats.Properties The specific heat ratio of argon is k = 1.667. The constantpressure specific heat of argon is cp = 0.1253 Btu/lbm.R (Table A-2E).

2

Analysis (a) The isentropic exit temperature T2s is determined from

R1381.9psia20psia200R550

70.667/1.66/1

1

212

kks

s PPTT

The actual kinetic energy change during this process is

Btu/lbm1.08/sft25,037

Btu/lbm12

ft/s60ft/s2402 22

2221

22 VV

kea

ArC = 80%

1

The effect of kinetic energy on isentropic efficiency is very small. Therefore, we can take the kinetic energy changes for the actual and isentropic cases to be same in efficiency calculations. From the isentropic efficiency relation, including the effect of kinetic energy,

08.15501253.008.15509.13811253.08.0

)()(

212

12

12

12

aaap

ssp

a

s

a

s

TkeTTckeTTc

kehhkehh

ww

C

It yields T2a = 1592 R


outin


(steady)0system


outin 0

EE

EEE

ke2

0)pe(since/2)+()2/(

12ina,

21

22

12ina,

222

211ina,

hhwVVhhmW

QVhmVhmW

Substituting, the work input to the compressor is determined to be Btu/lbm131.6Btu/lbm1.08R5501592RBtu/lbm0.1253ina,w

7-73

7-114 CO2 gas is compressed by an adiabatic compressor from a specified state to another specified state.The isentropic efficiency of the compressor is to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Kinetic and potentialenergy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 CO2 is anideal gas with constant specific heats.Properties At the average temperature of (300 + 450)/2 = 375 K, the constant pressure specificheat and the specific heat ratio of CO2 are k = 1.260 and cp = 0.917 kJ/kg.K (Table A-2). 2Analysis The isentropic exit temperature T2s is

CO21.8 kg/s

K434.2kPa100kPa600K300

00.260/1.26/1

1

212

kks

s PPTT

From the isentropic efficiency relation,

89.5%895.03004503002.434

12

12

12

12

12

12

TTTT

TTcTTc

hhhh

ww

a

s

ap

sp

a

s

a

sC

1

7-115E Air is accelerated in a 90% efficient adiabatic nozzle from low velocity to a specified velocity. The exit temperature and pressure of the air are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Air is an ideal gas withvariable specific heats.Analysis From the air table (Table A-17E),

T h r1 1 153 041480 R 363.89 Btu / lbm, .P

There is only one inlet and one exit, and thus m m m1 2 . We take the nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system can be expressed as

outin


(steady)0system


outin 0

EE

EEE

2AIR

N = 90% 1

2

0)pe(since/2)+()2/(02

12

212

222

211

VVhh

QWVhmVhm

Substituting, the exit temperature of air is determined to be

Btu/lbm351.11/sft25,037

Btu/lbm12

0ft/s800kJ/kg363.89 22

2

2h

From the air table we read T2a = 1431.3 R


12

hhhh

s

aN

,

04.46Btu/lbm349.690.90/363.89351.11363.89/21212 ras Phhh

Nh

Then the exit pressure is determined from the isentropic relation to be

psia52.1psia6053.0446.04

121

2

1

2

1

2 PPP

PPP

PP

r

r

r

r

7-74

7-116E EES Problem 7-115E is reconsidered. The effect of varying the nozzle isentropic efficiency from0.8 to 1.0 on the exit temperature and pressure of the air is to be investigated, and the results are to be plotted.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"Knowns:"WorkFluid$ = 'Air'P[1] = 60 [psia]T[1] = 1020 [F] Vel[2] = 800 [ft/s] Vel[1] = 0 [ft/s]eta_nozzle = 0.9

"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"h[1]=enthalpy(WorkFluid$,T=T[1])s[1]=entropy(WorkFluid$,P=P[1],T=T[1])T_s[1] = T[1] s[2] =s[1]s_s[2] = s[1]h_s[2]=enthalpy(WorkFluid$,T=T_s[2])T_s[2]=temperature(WorkFluid$,P=P[2],s=s_s[2])eta_nozzle = ke[2]/ke_s[2] ke[1] = Vel[1]^2/2 ke[2]=Vel[2]^2/2h[1]+ke[1]*convert(ft^2/s^2,Btu/lbm) = h[2] + ke[2]*convert(ft^2/s^2,Btu/lbm)h[1] +ke[1]*convert(ft^2/s^2,Btu/lbm) = h_s[2] + ke_s[2]*convert(ft^2/s^2,Btu/lbm)T[2]=temperature(WorkFluid$,h=h[2])P_2_answer = P[2] T_2_answer = T[2]

nozzle P2

[psiaT2

[FTs,2

[F]0.8 51.09 971.4 959.2

0.85 51.58 971.4 962.80.9 52.03 971.4 966

0.95 52.42 971.4 968.81 52.79 971.4 971.4

0.8 0.84 0.88 0.92 0.96 1950

960

970

980

nozzle

T s[2

]

T2Ts[2]

7-75

7-117 Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temperature are to be determined.Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changesare negligible. 3 The device is adiabatic and thus heat transfer is negligible. 4 Combustion gases can be treated as air that is an ideal gas with variable specific heats.Analysis From the air table (Table A-17),

4.123kJ/kg,1068.89K1020111 rPhT

From the isentropic relation ,

kJ/kg783.9234.40123.4kPa260

kPa852

1

212 srr hP

PPP

P1 = 260 kPa T1 = 747 CV1 = 80 m/s

AIRN = 92% P2 = 85 kPa

There is only one inlet and one exit, and thus m m m1 2 . We take the nozzle as the system, which is a control volume since mass crosses the boundary. The energy balance for this steady-flow system for theisentropic process can be expressed as

outin


(steady)0system


outin 0

EE

EEE

2

0)pe(since/2)+()2/(2

12

212

222

211

VVhh

QWVhmVhm

ss

ss

Then the isentropic exit velocity becomes

m/s759.2kJ/kg1

/sm1000kJ/kg783.921068.892m/s802

222

212

12 ss hhVV

Therefore,

m/s728.2m/s759.20.9222 sa VVN

The exit temperature of air is determined from the steady-flow energy equation,

kJ/kg806.95/sm1000

kJ/kg12

m/s80m/s728.2kJ/kg1068.89 22

22

2ah

From the air table we read T2a = 786.3 K

7-51

Reversible Steady-Flow Work

7-87C The work associated with steady-flow devices is proportional to the specific volume of the gas. Cooling a gas during compression will reduce its specific volume, and thus the power consumed by thecompressor.

7-88C Cooling the steam as it expands in a turbine will reduce its specific volume, and thus the workoutput of the turbine. Therefore, this is not a good proposal.

7-89C We would not support this proposal since the steady-flow work input to the pump is proportional tothe specific volume of the liquid, and cooling will not affect the specific volume of a liquid significantly.

7-90 Liquid water is pumped reversibly to a specified pressure at a specified rate. The power input to thepump is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible.Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5).

2Analysis The power input to the pump can be determined directly from thesteady-flow work relation for a liquid,

45 kg/s

H2O121

2

1

00in PPmpekedPmW vv

Substituting,

kW27433

in mkPa1kJ1kPa)206000)(/kgm0.001017)(kg/s45(W

1

7-91 Liquid water is to be pumped by a 25-kW pump at a specified rate. The highest pressure the watercan be pumped to is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is assumed to be reversible since we will determine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.Analysis The highest pressure the liquid can have at the pump exit can be determined from the reversiblesteady-flow work relation for a liquid,

P2

Thus,

323

1212

1

00in

mkPa1kJ1Pak)100)(/kgm0.001)(kg/s5(kJ/s25 P

PPmpekedPmW vv

25 kWPUMP

It yields kPa51002P100 kPa

7-52

7-92E Saturated refrigerant-134a vapor is to be compressed reversibly to a specified pressure. The power input to the compressor is to be determined, and it is also to be compared to the work input for the liquidcase.Assumptions 1 Liquid refrigerant is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The compressor is adiabatic.Analysis The compression process is reversible and adiabatic, and thus isentropic, s1 = s2. Then the properties of the refrigerant are (Tables A-11E through A-13E)

Btu/lbm115.80psia08


vaporsat.psia15

212

1

1

11

hss

P

shP

R-134a

2

R-134a

2

The work input to this isentropic compressor isdetermined from the steady-flow energy balance to be

)(

0

12in

21in

outin


(steady)0system


outin

hhmW

hmhmW

EE

EEE

11

Thus, Btu/lbm14.899.10080.11512in hhw

If the refrigerant were first condensed at constant pressure before it was compressed, we would use a pumpto compress the liquid. In this case, the pump work input could be determined from the steady-flow workrelation to be

121002

1in PPpekedPw vv

where v3 = vf @ 15 psia = 0.01165 ft3/lbm. Substituting,

Btu/lbm0.1433

in ftpsia5.4039Btu1psia1580)/lbmft0.01165(w

7-53

7-93 A steam power plant operates between the pressure limits of 10 MPa and 20 kPa. The ratio of theturbine work to the pump work is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kinetic and potential energy changes are negligible. 3 The process is reversible. 4 The pump and the turbine are adiabatic. Properties The specific volume of saturated liquid water at 20 kPa is v1 = vf @ 20 kPa = 0.001017 m3/kg(Table A-5). Analysis Both the compression and expansion processes are reversible and adiabatic, and thus isentropic,s1 = s2 and s3 = s4. Then the properties of the steam are

kJ/kg4707.2MPa10

KkJ/kg7.9073kJ/kg2608.9

.kPa20

343

3

kPa20@4

kPa20@44

hss

P

sshh

vaporsatP

g

g

H2O

3

H2O

2

Also, v1 = vf @ 20 kPa = 0.001017 m3/kg.

The work output to this isentropic turbine is determinedfrom the steady-flow energy balance to be 1 4

)(

0

43out

out43

outin


(steady)0system


outin

hhmW

Whmhm

EE

EEE

Substituting,kJ/kg2098.39.26082.470743outturb, hhw

The pump work input is determined from the steady-flow work relation to be

kJ/kg10.15mkPa1

kJ1kPa)2010,000)(/kgm0.001017( 33

121002

1inpump, PPpekedPw vv

Thus,

206.710.152098.3

inpump,

outturb,

ww

7-54

7-94 EES Problem 7-93 is reconsidered. The effect of the quality of the steam at the turbine exit on the network output is to be investigated as the quality is varied from 0.5 to 1.0, and the net work output us to beplotted as a function of this quality.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

"System: control volume for the pump and turbine""Property relation: Steam functions""Process: For Pump and Turbine: Steady state, steady flow, adiabatic, reversible or isentropic""Since we don't know the mass, we write the conservation of energy per unit mass.""Conservation of mass: m_dot[1]= m_dot[2]"

"Knowns:"WorkFluid$ = 'Steam_IAPWS' P[1] = 20 [kPa] x[1] = 0 P[2] = 10000 [kPa] x[4] = 1.0

"Pump Analysis:"T[1]=temperature(WorkFluid$,P=P[1],x=0)v[1]=volume(workFluid$,P=P[1],x=0)h[1]=enthalpy(WorkFluid$,P=P[1],x=0)s[1]=entropy(WorkFluid$,P=P[1],x=0)s[2] = s[1]h[2]=enthalpy(WorkFluid$,P=P[2],s=s[2])T[2]=temperature(WorkFluid$,P=P[2],s=s[2])

"The Volume function has the same form for an ideal gas as for a real fluid."v[2]=volume(WorkFluid$,T=T[2],p=P[2])"Conservation of Energy - SSSF energy balance for pump"" -- neglect the change in potential energy, no heat transfer:"h[1]+W_pump = h[2] "Also the work of pump can be obtained from the incompressible fluid, steady-flow result:"W_pump_incomp = v[1]*(P[2] - P[1])

"Conservation of Energy - SSSF energy balance for turbine -- neglecting the change in potentialenergy, no heat transfer:"P[4] = P[1] P[3] = P[2] h[4]=enthalpy(WorkFluid$,P=P[4],x=x[4])s[4]=entropy(WorkFluid$,P=P[4],x=x[4])T[4]=temperature(WorkFluid$,P=P[4],x=x[4])s[3] = s[4]h[3]=enthalpy(WorkFluid$,P=P[3],s=s[3])T[3]=temperature(WorkFluid$,P=P[3],s=s[3])h[3] = h[4] + W_turb W_net_out = W_turb - W_pump

7-55

Wnet,out

[kJ/kg]Wpump

[kJ/kg]Wpump,incomp

[kJ/kg]Wturb

[kJ/kg]x4

557.1 10.13 10.15 567.3 0.5734.7 10.13 10.15 744.8 0.6913.6 10.13 10.15 923.7 0.71146 10.13 10.15 1156 0.81516 10.13 10.15 1527 0.92088 10.13 10.15 2098 1

0.0 1.1 2.2 3.3 4.4 5.5 6.6 7.7 8.8 9.9 11.00

100200300400500600700800900

10001100

s [kJ/kg-K]

T [°

C]

10000 kPa

20 kPa 0.2 0.6 0.8

Steam

1, 2

3

4

x4 = 1.0= 0.5

0.5 0.6 0.7 0.8 0.9 1500

850

1200

1550

1900

2250

x[4]

Wne

t,out

[kJ/

kg]

7-56

7-95 Liquid water is pumped by a 70-kW pump to a specified pressure at a specified level. The highestpossible mass flow rate of water is to be determined.Assumptions 1 Liquid water is an incompressible substance. 2 Kineticenergy changes are negligible, but potential energy changes may be significant. 3 The process is assumed to be reversible since we will determine the limiting case.Properties The specific volume of liquid water is given to be v1 = 0.001 m3/kg.Analysis The highest mass flow rate will be realized when the entireprocess is reversible. Thus it is determined from the reversible steady-flow work relation for a liquid,

Thus,

222

33

12122

1

0in

/sm1000kJ/kg1)m10)(m/s9.8(

mkPa1kJ1kPa)1205000)(/kgm0.001(kJ/s7 m

zzgPPmpekedPmW vvWater

P1 = 120 kPa

P2 = 5 MPa

PUMP

It yieldskg/s1.41m

7-57

7-96E Helium gas is compressed from a specified state to a specified pressure at a specified rate. The power input to the compressor is to be determined for the cases of isentropic, polytropic, isothermal, andtwo-stage compression.Assumptions 1 Helium is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of helium is R = 2.6805 psia.ft3/lbm.R = 0.4961 Btu/lbm.R. The specific heat ratio of helium is k = 1.667 (Table A-2E).

2

Analysis The mass flow rate of helium is

lbm/s0.0493R530R/lbmftpsia2.6805

/sft5psia143

3

1

11

RTP

mV

(a) Isentropic compression with k = 1.667:


Btu/s44.11

1psia14psia120

11.667R530RBtu/lbm0.49611.667lbm/s0.0493

11

70.667/1.66

/1

1

21incomp,

hp62.4

kk

PP

kkRTmW

5 ft3/s

W·He

1



Btu/s33.47

1psia14psia120

11.2R530RBtu/lbm0.49611.2lbm/s0.0493

11

0.2/1.2

/1

1

21incomp,

hp47.3

nn

PP

nnRTmW


hp39.4Btu/s27.83psia14psia120lnR530RBtu/lbm0.4961lbm/s0.0493ln

1

2incomp, P

PRTmW

(d) Ideal two-stage compression with intercooling (n = 1.2): In this case, the pressure ratio across each stage is the same, and its value is determined from

psia41.0psia120psia1421PPPx



Btu/s30.52

1psia14psia41

11.2R530RBtu/lbm0.49611.2lbm/s0.04932

11

22

0.2/1.2

/1

1

1Icomp,incomp,

hp43.2

nnx

PP

nnRTmwmW

7-58

7-97E EES Problem 7-96E is reconsidered. The work of compression and entropy change of the helium isto be evaluated and plotted as functions of the polytropic exponent as it varies from 1 to 1.667.Analysis The problem is solved using EES, and the results are tabulated and plotted below.

Procedure FuncPoly(m_dot,k, R, T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)

If n =1 then T2=T1W_dot_comp_polytropic= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp) "[hp]"W_dot_comp_2stagePoly = W_dot_comp_polytropic "[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s) "[Btu/s]"Q_dot_Out_2stagePoly = Q_dot_Out_polytropic*convert(hp,Btu/s) "[Btu/s]"

ElseC_P = k*R/(k-1) "[Btu/lbm-R]"T2=(T1+460)*((P2/P1)^((n+1)/n)-460)"[F]"W_dot_comp_polytropic = m_dot*n*R*(T1+460)/(n-1)*((P2/P1)^((n-1)/n) - 1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_polytropic=W_dot_comp_polytropic*convert(hp,Btu/s)+m_dot*C_P*(T1-T2)"[Btu/s]"

Px=(P1*P2)^0.5T2x=(T1+460)*((Px/P1)^((n+1)/n)-460)"[F]"W_dot_comp_2stagePoly = 2*m_dot*n*R*(T1+460)/(n-1)*((Px/P1)^((n-1)/n) - 1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_2stagePoly=W_dot_comp_2stagePoly*convert(hp,Btu/s)+2*m_dot*C_P*(T1-T2x)"[Btu/s]"

endifEND

R=0.4961[Btu/lbm-R]k=1.667n=1.2P1=14 [psia]T1=70 [F] P2=120 [psia]V_dot = 5 [ft^3/s]P1*V_dot=m_dot*R*(T1+460)*convert(Btu,psia-ft^3)

W_dot_comp_isentropic = m_dot*k*R*(T1+460)/(k-1)*((P2/P1)^((k-1)/k) - 1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isentropic = 0"[Btu/s]"Call FuncPoly(m_dot,k, R,T1,P2,P1,n:W_dot_comp_polytropic,W_dot_comp_2stagePoly,Q_dot_Out_polytropic,Q_dot_Out_2stagePoly)

W_dot_comp_isothermal= m_dot*R*(T1+460)*ln(P2/P1)*convert(Btu/s,hp)"[hp]"Q_dot_Out_isothermal = W_dot_comp_isothermal*convert(hp,Btu/s)"[Btu/s]"

7-59

n Wcomp2StagePoly

[hp]Wcompisentropic

[hp]Wcompisothermal

[hp]Wcomppolytropic

[hp]1 39.37 62.4 39.37 39.37

1.1 41.36 62.4 39.37 43.481.2 43.12 62.4 39.37 47.351.3 44.68 62.4 39.37 50.971.4 46.09 62.4 39.37 54.361.5 47.35 62.4 39.37 57.54

1.667 49.19 62.4 39.37 62.4

1 1.1 1.2 1.3 1.4 1.5 1.6 1.735

40

45

50

55

60

65

n

Wco

mp,

poly

trop

ic [

hp]

PolytropicIsothermalIsentropic2StagePoly

7-60

7-98 Nitrogen gas is compressed by a 10-kW compressor from a specified state to a specified pressure. The mass flow rate of nitrogen through the compressor is to be determined for the cases of isentropic,polytropic, isothermal, and two-stage compression.Assumptions 1 Nitrogen is an ideal gas with constant specific heats. 2 The process is reversible. 3 Kineticand potential energy changes are negligible.Properties The gas constant of nitrogen is R = 0.297 kJ/kg.K (Table A-1). The specific heat ratio of nitrogen is k = 1.4 (Table A-2).

2Analysis (a) Isentropic compression:

or,

1kPa80kPa48011.4

K300KkJ/kg0.2971.4kJ/s10

11

0.4/1.4

/112

1incomp,

m

PPkkRTmW kk

·N2m 10 kW

It yields1m 0.048 kg / s


or,

1kPa80kPa48011.3

K300KkJ/kg0.2971.3kJ/s10

11

0.3/1.3

/112

1incomp,

m

PPnnRTmW nn



kPa80kPa480lnK300KkJ/kg0.297kJ/s10ln

2

1incomp, m

PPRTmW


(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each stage is the same, and its value is determined to be



or,

1kPa80kPa19611.3

K300KkJ/kg0.2971.32kJ/s10

11

22

0.3/1.3

/11

1Icomp,incomp,

m

PPnnRTmwmW nn

x

It yieldskg/s0.056m