SOLUTIONS: HOMEWORK #6 - University of Colorado...

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8

SOLUTIONS: HOMEWORK #6 Chapter 5 Problems 5-45 A number of brass balls are to be quenched in a water bath at a specified rate. The rate at which heat needs to be removed from the water in order to keep its temperature constant is to be determined.

Assumptions 1 The thermal properties of the balls are constant. 2 The balls are at a uniform temperature before and after quenching. 3 The changes in kinetic and potential energies are negligible.

Properties The density and specific heat of the brass balls are given to be ρ = 8522 kg/m3 and Cp = 0.385 kJ/kg.°C.

Analysis We take a single ball as the system. The energy balance for this closed system can be expressed as

E E E

Q U m u uQ mC T T

in out

out ball

out

− =

− = = −

= −

Net energy transferby heat, work, and mass

system

Change in internal, kinetic, potential, etc. energies

1 24 34 124 34Δ

Δ ( )( )

2 1

1 2

Brass balls, 120°C

Water bath, 5°C

The total amount of heat transfer from a ball is

m V D

Q mC T Tout

= = = =

= − = ° − ° =

ρ ρπ π3

1 2

68522 0 05 0 558

0 558 0 385 120 74 9 88

( ) ( . .

( ) ( . )( . ( ) .

kg / m m)6

kg

kg kJ / kg. C) C kJ / ball

33

Then the rate of heat transfer from the balls to the water becomes

& & ( ( . )Q n Qtotal = = × =ball ball balls /min) kJ / ball100 9 88 988 kJ / min

Therefore, heat must be removed from the water at a rate of 988 kJ/min in order to keep its temperature constant at 50°C since energy input must be equal to energy output for a system whose energy level remains constant. That is, E E Ein out= = when systemΔ 0 .

5-58C It is mostly converted to internal energy as shown by a rise in the fluid temperature. 5-59C The kinetic energy of a fluid increases at the expense of the internal energy as evidenced by a decrease in the fluid temperature.


&

5-61 Air is accelerated in a nozzle from 30 m/s to 180 m/s. The mass flow rate, the exit temperature, and the exit area of the nozzle are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Air is an ideal gas with constant specific heats. 3 Potential energy changes are negligible. 4 The device is adiabatic and thus heat transfer is negligible. 5 There are no work interactions.

Properties The gas constant of air is 0.287 kPa.m3/kg.K (Table A-1). The specific heat of air at the anticipated average temperature of 450 K is Cp = 1.02 kJ/kg.°C (Table A-2).

Analysis (a) There is only one inlet and one exit, and thus & &m m m1 2= = . Using the ideal gas relation, the specific volume and the mass flow rate of air are determined to be

/kgm 0.4525kPa 300

)K 473)(K/kgmkPa 0.287( 33

1

11 =

⋅⋅==

PRT

v

kg/s 0.5304V === )m/s30)(m0.008(/kgm0.4525

11 2311

1A

vm&

(b) We take 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 in the rate form as

& & &

& &

E E E

E E

in out

in out

− =

=

Rate of net energy transfer by heat, work, and mass

system (steady)

Rate of change in internal, kinetic, potential, etc. energies

1 24 34 1 2444 3444Δ 0 0=

( )

20

20

0)peW (since /2)+()2/(2

122

12,

21

22

12

222

211

VVVV

VV

−+−=⎯→⎯

−+−=

≅Δ≅≅=+

TTChh

Qhmhm

avep

&&&&

Substituting,

⎟⎟⎠

⎞⎜⎜⎝

⎛−+−⋅= 22

22

2 /sm 1000kJ/kg 1

2)m/s 30()m/s 180()C200)(KkJ/kg 1.02(0 oT

It yields T2 = 184.6°C (c) The specific volume of air at the nozzle exit is

AIR P2 = 100 kPa V2 = 180 m/s

P1 = 300 kPa T1 = 200°C V1 = 30 m/s A1 = 80 cm2

/kgm 1.313kPa 100

)K 273184.6)(K/kgmkPa 0.287( 33

2

22 =

+⋅⋅==

PRT

v

( )m/s180/kgm1.313

1kg/s0.530412322

2AA

vm =⎯→⎯= V&

A2 = 0.00387 m2 = 38.7 cm2

5-77C Yes. Because energy (in the form of shaft work) is being added to the air.

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 5-79 Steam expands in a turbine. The change in kinetic energy, the power output, and the turbine inlet area are to be determined.

Assumptions 1 This is a steady-flow process since there is no change with time. 2 Potential energy changes are negligible. 3 The device is adiabatic and thus heat transfer is negligible.

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

kJ/kg3240.9

/kgm0.02975C450

MPa10

1

31

1

1

==

⎪⎭

⎪⎬⎫

=

=

hv

T

Po

STEAM m = 12 kg/s ·

P1 = 10 MPa T1 = 450°C V1 = 80 m/s

P2 = 10 kPa x2 = 0.92 V2 = 50 m/s

W·

and

kJ/kg2393.22392.80.92191.8392.0kPa 10

222

2 =×+=+=⎭⎬⎫

==

fgf hxhhxP

Analysis (a) The change in kinetic energy is determined from

( ) kJ/kg1.95

VV−=⎟

⎟⎠

⎞⎜⎜⎝

⎛−=

−=Δ 22

2221

22

/sm1000kJ/kg1

2m/s)(80m/s50

2ke

(b) There is only one inlet and one exit, and thus & &m m m1 2 &= = . We take the 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 in the rate form as

& & &

& &

E E E

E E

in out

in out

− =

=


system (steady)


1 24 34 1 2444 3444Δ 0 0=

⎟⎟⎠

⎞⎜⎜⎝

⎛ −+−−=

≅Δ≅+=+

2

0)peQ (since /2)+()2/(2

122

12out

222out

211

VV

VV

hhmW

hmWhm

&&

&&&&

Then the power output of the turbine is determined by substitution to be

MW 10.2=−−−= kJ/kg)1.953240.92393.2)(kg/s 12(outW&

(c) The inlet area of the turbine is determined from the mass flow rate relation,

2m 0.00446V

V ===⎯→⎯=m/s 80

)/kgm 0.02975)(kg/s 12(1 3

1

1111

1

vmAA

vm

&&


&

5-90 Helium is compressed by a compressor. For a mass flow rate of 90 kg/min, the power input required is 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. 3 Helium is an ideal gas with constant specific heats.

Properties The constant pressure specific heat of helium is Cp = 5.1926 kJ/kg·K (Table A-2a).

Analysis 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 system can be expressed in the rate form as

& & &

& &

E E E

E E

in out

in out

− =

=


system (steady)


1 24 34 1 2444 3444Δ 0 0=

3

He m=90kg/mi ·

P2 = 700 kPa T2 = 430 K

P1 = 120 kPa T1 = 310 K

W·

& & & &

& & & ( ) & ( )

W mh Q mh

W Q m h h mC T Tp

in out

in out

(since ke pe 0)+ = + ≅ ≅

− = − = −1 2

2 1 2 1

Δ Δ

Thus,

( )

kW965=−⋅=

−+=

310)KK)(430kJ/kg26kg/s)(5.19(90/60+kJ/kg) kg/s)(20(90/6012 TTCmQW poutin &&&

5-104 A hot water stream is mixed with a cold water stream. For a specified mixture temperature, the mass flow rate of cold water is to be determined. Assumptions 1 Steady operating conditions exist. 2 The mixing chamber is well-insulated so that heat loss to the surroundings is negligible. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant. 5 There are no work interactions.

Properties Noting that T < Tsat @ 250 kPa = 127.44°C, the water in all three streams exists as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,

h1 ≅ hf @ 80°C = 334.91 kJ/kg h2 ≅ hf @ 20°C = 83.96 kJ/kg h3 ≅ hf @ 42°C = 175.92 kJ/kg

Analysis We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance: & & & & & &m m E m m min out− = = ⎯ →⎯ + =Δ system (steady)0

1 20

Energy balance:

H2O (P = 250 kPa)

T3 = 42°C

T1 = 80°C m1 = 0.5 kg/s

T2 = 20°C m2·

·

0)peke (since

0

332211

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

≅Δ≅Δ===+

=

=Δ=−

WQhmhmhm

EE

EEE

outin

outin

&&&&&

&&

44 344 21&

43421&&


5-104 CONTINUED

Combining the two relations and solving for gives &m2

( ) 3212211 hmmhmhm &&&& +=+

& &m h hh h

m21 3

3 21=

−−

Substituting, the mass flow rate of cold water stream is determined to be

( )( ) ( ) kg/s 0.864=

−−

= kg/s 0.5kJ/kg83.96175.92kJ/kg175.92334.91

2m&

5-106 Feedwater is heated in a chamber by mixing it with superheated steam. If the mixture is saturated liquid, the ratio of the mass flow rates of the feedwater and the superheated vapor is 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. 3 There are no work interactions. 4 The device is adiabatic and thus heat transfer is negligible.

Properties Noting that T < Tsat @ 800 kPa = 170.43°C, the cold water stream and the mixture exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. Thus,

h1 ≅ hf @ 50°C = 209.33 kJ/kg h3 ≅ hf @ 800 kPa = 721.11 kJ/kg and

kJ/kg2839.3C200

kPa8002

2

2 =⎪⎭

⎪⎬⎫

=

=h

T

Po

Analysis We take the mixing chamber as the system, which is a control volume since mass crosses the boundary. The mass and energy balances for this steady-flow system can be expressed in the rate form as

Mass balance: & & & & & & & &m m m m m m m min out in out− = = → = → + =Δ system (steady) 0

1 20 3

Energy balance:

H2O (P = 800 kPa)

Sat. liquid

T1 = 50°C m1

T2 = 200°C m2·

·

& & &

& &

& & & & &

E E E

E E

m h m h m h Q W

in out

in out

− = =

=

+ = ≅ ≅ ≅ ≅


system (steady)


(since ke pe 0)

1 24 34 1 2444 3444Δ

Δ Δ

0

1 1 2 2 3 3

0

Combining the two, ( ) 3212211 hmmhmhm &&&& +=+

Dividing by yields &m2 ( ) 321 1 hyhhy +=+

Solving for y: yh hh h

=−−

3 2

1 3

where is the desired mass flow rate ratio. Substituting, y m m= & / &1 2

4.14=−−

=y 721.11209.332839.3721.11

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 SOLUTIONS: HOMEWORK #7 Chapter 6 Problems 6-2C Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity. 6-3C An electric resistance heater which consumes 5 kWh of electricity and supplies 6 kWh of heat to a room. 6-10C Heat engines are cyclic devices that receive heat from a source, convert some of it to work, and reject the rest to a sink. 6-12C No. Because 100% of the work can be converted to heat. 6-15C No. Such an engine violates the Kelvin-Planck statement of the second law of thermodynamics. 6-20 The power output and fuel consumption rate of a power plant are given. The overall efficiency is to be determined.

Assumptions The plant operates steadily.

Properties The heating value of coal is given to be 30,000 kJ/kg.

sink

HE

60 t/h

coal

FurnaceAnalysis The rate of energy supply (in chemical form) to this power plant is

( )( ) MW 500kJ/h101.8kJ/kg 30,000kg/h 60,000 9 =×=== coalcoalH umQ &&

Then the overall efficiency of the plant becomes

30.0%==== 0.300MW 500MW 150

H

,overall Q

W outnet

&

&η

6-40C The difference between the two devices is one of purpose. The purpose of a refrigerator is to remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a warm medium. 6-47C No. The refrigerator captures energy from a cold medium and carries it to a warm medium. It does not create it.

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 6-54 The COP and the power consumption of a refrigerator are given. The time it will take to cool 5 watermelons is to be determined.

Assumptions 1 The refrigerator operates steadily. 2 The heat gain of the refrigerator through its walls, door, etc. is negligible. 3 The watermelons are the only items in the refrigerator to be cooled.

Properties The specific heat of watermelons is given to be C = 4.2 kJ/kg.°C.

Analysis The total amount of heat that needs to be removed from the watermelons is

( ) ( )( )( ) kJ2520C820CkJ/kg4.2kg105 =−⋅×=Δ= ooswatermelonL TmCQ

cool space

Kitchen air

R COP = 2.5

450 WThe rate at which this refrigerator removes heat is

( )( ) ( )( ) kW1.125kW0.452.5, === innetRL WCOPQ &&

That is, this refrigerator can remove 1.125 kJ of heat per second. Thus the time required to remove 2520 kJ of heat is

min37.3====Δ s2240kJ/s1.125kJ2520

L

L

QQ

t&

This answer is optimistic since the refrigerated space will gain some heat during this process from the surrounding air, which will increase the work load. Thus, in reality, it will take longer to cool the watermelons.

6-81 The sink temperature of a Carnot heat engine and the rates of heat supply and heat rejection are given. The source temperature and the thermal efficiency of the engine are to be determined.

Assumptions The Carnot heat engine operates steadily.

Analysis (a) For reversible cyclic devices we have ⎟⎟⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛

L

H

revL

H

TT

QQ

Thus the temperature of the source TH must be

17°C

source

HE 200 kJ

650 kJ ( ) K 942.5=⎟⎟

⎠

⎞⎜⎜⎝

⎛=⎟⎟

⎠

⎞⎜⎜⎝

⎛= K 290

kJ 200kJ 650

LrevL

HH T

QQ

T

(b) The thermal efficiency of a Carnot heat engine depends on the source and the sink temperatures only, and is determined from

69%or0.69K 942.5

K 29011, =−=−=H

LCth T

Tη

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 6-95 The refrigerated space and the environment temperatures for a refrigerator and the rate of heat removal from the refrigerated space are given. The minimum power input required is to be determined.

Assumptions The refrigerator operates steadily.

Analysis The power input to a refrigerator will be a minimum when the refrigerator operates in a reversible manner. The coefficient of performance of a reversible refrigerator depends on the temperature limits in the cycle only, and is determined from

( ) ( ) ( ) 03.81K2738/K27325

11/

1, =

−+−+=

−=

LHrevR TT

COP

-8°C

25°C

R

300 kJ/min

The power input to this refrigerator is determined from the definition of the coefficient of performance of a refrigerator,

&&

, , min, max

WQ

COPnet inL

R= = = =

300 kJ / min8.03

37.36 kJ / min 0.623 kW

6-103 A heat pump maintains a house at a specified temperature. The rate of heat loss of the house and the power consumption of the heat pump are given. It is to be determined if this heat pump can do the job.

Assumptions The heat pump operates steadily.

Analysis The power input to a heat pump will be a minimum when the heat pump operates in a reversible manner. The coefficient of performance of a reversible heat pump depends on the temperature limits in the cycle only, and is determined from

( ) ( ) ( ) 14.75K27322/K27321

1/1

1, =

++−=

−=

HLrevHP TT

COP

8 kW

House 22°C

HP

110,000 kJ/h

The required power input to this reversible heat pump is determined from the definition of the coefficient of performance to be

kW2.07=⎟⎟⎠

⎞⎜⎜⎝

⎛==

s3600h1

14.75kJ/h110,000

min,,HP

Hinnet COP

QW

&&

This heat pump is powerful enough since 8 kW > 2.07 kW.


6-125 A Carnot heat engine drives a Carnot refrigerator that removes heat from a cold medium at a specified rate. The rate of heat supply to the heat engine and the total rate of heat rejection to the environment are to be determined.

Analysis (a) The coefficient of performance of the Carnot refrigerator is

( ) ( ) ( ) 14.61K258/K300

11/

1, =

−=

−=

LHCR TT

COP

Then power input to the refrigerator becomes

300 K

750 K

HE R

-15°C 400 kJ/min

QH, R·

QH, HE·

QL, HE·

&&

,,

WQ

COPnet inL

R C= = =

400 kJ / min6.14

65.1 kJ / min

which is equal to the power output of the heat engine, . &,Wnet out

The thermal efficiency of the Carnot heat engine is determined from

η th CL

H

TT, .= − = − =1 01 300 K

750 K60

Then the rate of heat input to this heat engine is determined from the definition of thermal efficiency to be

&&

,,

,Q

WH HE

net out

th HE= = =η

65.1 kJ / min0.60

108.5 kJ / min

(b) The total rate of heat rejection to the ambient air is the sum of the heat rejected by the heat engine ( Q ) and the heat discarded by the refrigerator ( ), &

,L HE&

,QH R

kJ/min465.11.65400

kJ/min43.41.655.108

,,,

,,,

=+=+=

=−=−=

innetRLRH

outnetHEHHEL

WQQ

WQQ&&&

&&&

and & & &

, ,Q Q QAmbient L HE H R= + = + =43.4 465.1 508.5 kJ / min

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 Chapter 7 problems 7-10C No. An isothermal process can be irreversible. Example: A system that involves paddle-wheel work while losing an equivalent amount of heat. 7-17C Increase. 7-21C Yes. This will happen when the system is losing heat, and the decrease in entropy as a result of this heat loss is equal to the increase in entropy as a result of irreversibilities. 7-26 Heat is transferred isothermally from a source to the working fluid of a Carnot engine. The entropy change of the working fluid, the entropy change of the source, and the total entropy change during this process are to be determined.

Analysis (a) This is a reversible isothermal process, and the entropy change during such a process is given by

ΔS QT

=

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

ΔSQT

QTfluid

fluid

fluid

in fluid

fluid= = = =, 900 kJ

673 K1.337 kJ / K

Source 400°C

900 kJ

(b) ΔSQT

QTsource

source

source source= = − = − = −out, source 900 kJ

673 K1.337 kJ / K

(c) Thus the total entropy change of the process is

0=−=Δ+Δ=Δ= 337.1337.1total sourcefluidgen SSSS

7-34 An insulated rigid tank contains a saturated liquid-vapor mixture of water at a specified pressure. An electric heater inside is turned on and kept on until all the liquid vaporized. The entropy change of the water during this process is to be determined.

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

( )( )( )( )

KkJ/kg6.8649.

KkJ/kg2.81680568.625.03026.1/kgm0.42430.0011.6940.250.001

25.0kPa100

212

11

311

1

1

⋅=⎭⎬⎫=

⋅=+=+==−+=+=

⎭⎬⎫

==

svaporsat

vv

sxssvxvv

xP

fgf

fgf

H2O 4 kg

100 kPa

WeThen the entropy change of the steam becomes

( ) kJ/K 16.19=⋅−=−=Δ KkJ/kg )2.81686.8649)(kg 4(12 ssmS

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 7-51 An aluminum block is brought into contact with an iron block in an insulated enclosure. The final equilibrium 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 specific heats. 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

Iron 20 kg 100°C

Aluminum 20 kg 200°C

U

EEE outin

Δ=

Δ=−

0energies etc. potential,

kinetic, internal,in Change

system

mass and work,heat,by nsferenergy traNet

4342143421

or,

0)]([)]([0

12alum12

alum

=−+−=Δ+Δ

iron

iron

TTmCTTmCUU

Substituting,

K441.4

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

2

22

==

=−⋅+−⋅

C168.4o

oo

T

TT

The total entropy change for this process is determined from

( )( )

( )( ) kJ/K1.346K473K441.4

lnKkJ/kg0.973kg20ln

kJ/K1.515K373K441.4

lnKkJ/kg0.45kg20ln

1

2

1

2

−=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=Δ

=⎟⎟⎠

⎞⎜⎜⎝

⎛⋅=⎟⎟

⎠

⎞⎜⎜⎝

⎛=Δ

TT

mCS

TT

mCS

avealum

aveiron

Thus, Δ Δ ΔS S Stotal iron alum= + = − =1515 1346. . 0.169 kJ / K


7-62 An insulated tank contains CO2 gas at a specified pressure and volume. A paddle-wheel in the tank stirs 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

CO21.5 m3

100 kPa 1.2 kg

2.1100120

1

2

1

2

1

1

2

2 ===⎯→⎯=kPakPa

PP

TT

TVP

TVP

Thus,

( )

( )( ) ( )kJ/K 0.323=

⋅=

=⎟⎟⎠

⎞⎜⎜⎝

⎛+=−=Δ

1.2lnKkJ/kg0.657kg 2.7

lnlnln1

2,

0

1

2

1

2,12 T

TmC

VV

RTT

CmssmS avevavev


=

7-110 Steam is condensed by cooling water in the condenser of a power plant. The rate of condensation of steam and the rate of entropy generation are to be determined.

Assumptions 1 Steady operating conditions exist. 2 The heat exchanger is well-insulated so that heat loss to the surroundings is negligible and thus heat transfer from the hot fluid is equal to the heat transfer to the cold fluid. 3 Changes in the kinetic and potential energies of fluid streams are negligible. 4 Fluid properties are constant.

Properties The enthalpy and entropy of vaporization of water at 50°C are hfg =2382.7 kJ/kg and sfg= 7.3725 kJ/kg.K (Table A-4). The specific heat of water at room temperature is Cp = 4.18 kJ/kg.°C (Table A-3).

Analysis (a) We take the cold water tubes as the system, which is a control volume. The energy balance for this steady-flow system can be expressed in the rate form as

Steam 50°C

18°C Water

50°C

& & &

& &

& & &

& & ( )

E E E

E E

Q mh mh

Q mC T T

in out

in out

in

in p

− =

=

+ = ≅ ≅

= −


system (steady)


(since ke pe 0)

1 24 34 1 2444 3444Δ

Δ Δ

0

1 2

2 1

0

Then the heat transfer rate to the cooling water in the condenser becomes

kJ/s 3800=C)18CC)(27kJ/kg. kg/s)(4.18 (101

)]([ watercooling

°−°°=

−= inoutp TTCmQ &&

The rate of condensation of steam is determined to be

kg/s 1.595===⎯→⎯=kJ/kg 7.2382kJ/s3800)(

fgsteamsteamfg h

QmhmQ &

&&&

(b) The rate of entropy generation within the condenser during this process can be determined by applying the rate form of the entropy balance on the entire condenser. Noting that the condenser is well-insulated and thus heat transfer is negligible, the entropy balance for this steady-flow system can be expressed as

{

)()(

0

)0 (since 0

34steam12water

4steam2water3steam1water

44223311

entropy of change of Rate

(steady) 0system

generation entropy of Rate

mass andheat by ansferentropy trnet of Rate

ssmssmS

Ssmsmsmsm

QSsmsmsmsm

SSSS

gen

gen

gen

genoutin

−+−=

=+−−+

==+−−+

Δ=+−

&&&

&&&&&

&&&&&

44 344 21&&

43421&&

Noting that water is an incompressible substance and steam changes from saturated vapor to saturated liquid, the rate of entropy generation is determined to be

kW/K 1.100kJ/kg.K) 25kg/s)(7.37 595.1(273+18273+27kJ/kg.K)ln kg/s)(4.18 101(

ln)(ln steam1

2watersteam

1

2water

=−=

−=−+= fgpgfpgen smTT

CmssmTT

CmS &&&&&


3

7-132 Liquid water is heated in a chamber by mixing it with superheated steam. For a specified mixing temperature, the mass flow rate of the steam and the rate of entropy generation 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. 3 There are no work interactions.

Properties Noting that T < Tsat @ 200 kPa = 120.23°C, the cold water and the exit mixture streams exist as a compressed liquid, which can be approximated as a saturated liquid at the given temperature. From tables,

KkJ/kg0.8312kJ/kg251.13

C60

kPa200

KkJ/kg7.8926kJ/kg3071.8

C300

kPa200

KkJ/kg0.2966kJ/kg83.96

C20

kPa200

C60@3

C60@3

3

3

2

2

2

2

C20@1

C20@1

1

1

⋅=≅=≅

⎪⎭

⎪⎬⎫

=

=

⋅==

⎪⎭

⎪⎬⎫

=

=

⋅=≅=≅

⎪⎭

⎪⎬⎫

=

=

o

o

o

o

o

o

o

f

f

f

f

sshh

T

P

sh

T

P

sshh

T

P

Analysis (a) We take the mixing chamber as the system, which is a control volume. The mass and energy balances for this steady-flow system can be expressed in the rate form as

2

MIXING CHAMBER

200 kPa

20°C2.5 kg/s

300°C

1

3 60°C

600 kJ/min

Mass balance: & & & & & &m m E m m min out− = = ⎯ →⎯ + =Δ system (steady)0

1 20

Energy balance:

332211

energies etc. potential, kinetic, internal,in change of Rate

(steady) 0system

mass and work,heat,by nsferenergy tranet of Rate

0

hmQhmhm

EE

EEE

out

outin

outin

&&&&

&&

44 344 21&

43421&&

+=+

=

=Δ=−

Combining the two relations gives ( ) ( ) ( 3223113212211 hhmhhmhmmhmhmQout −+−=+−+= &&&&&&& ) Solving for and substituting, the mass flow rate of the superheated steam is determined to be &m2

( ) ( )( )

( ) kg/s0.152=−

−−=

−−−

=kJ/kg13.2513071.8

kJ/kg13.25183.96kg/s2.5kJ/s)(600/60

32

3112 hh

hhmQm out &&&

Also, & & &m m m3 1 2= + = + =2.5 0.152 2.652 kg / s

(b) The rate of total entropy generation during this process is determined by applying the entropy balance on an extended system that includes the mixing chamber and its immediate surroundings so that the boundary temperature of the extended system is 25°C at all times. It gives

{

0

0

gen,

332211

entropy of change of Rate

0system

generation entropy of Rate

mass andheat by ansferentropy trnet of Rate

=+−−+

=Δ=+−

STQ

smsmsm

SSSS

surrb

out

genoutin

&&

&&&

43421&&

43421&&

Substituting, the rate of entropy generation during this process is determined to be

( )( ) ( )(( )( )

)

K 298kJ/s) 60/600(KkJ/kg0.2966kg/s2.5

KkJ/kg7.8926kg/s0.152KkJ/kg0.8312kg/s2.652,

112233

kW/K0.297=

+⋅−

⋅−⋅=

+−−=surrb

outgen T

QsmsmsmS

&&&&&

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 7-143 Air is compressed in a two-stage ideal compressor with intercooling. For a specified mass flow rate of air, the power input to the compressor is to be determined, and it is to be compared to the power input to a single-stage compressor. √

Assumptions 1 The compressor operates steadily. 2 Kinetic and potential energies are negligible. 3 The compression process is reversible adiabatic, and thus isentropic. 4 Air is an ideal gas with constant specific heats at room temperature.

Properties The gas constant of air is R =0.287 kPa.m3/kg.K (Table A-1). The specific heat ratio of air is k = 1.4 (Table A-2)

Analysis The intermediate pressure between the two stages is

Stage II

900 kPa

Stage I

27°C

100 kPa 27°C

W

Heat

( )( ) kPa300kPa900kPa10021 === PPPx

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

( )( ) ( )( )( )( )( )( )

kJ/kg222.2

1kPa100kPa300

11.4K300KkJ/kg0.2871.42

11

22

0.4/1.4

/11

1,,,

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−⋅

=

−−

== − kkxIincompincomp PP

kkRTww

and ( )( ) kW4.44=== kJ/kg222.2kg/s0.02,incompin wmW &&

The work input to a single-stage compressor operating between the same pressure limits would be

( )( )( ) ( )( )( ) kJ/kg263.21kPa100kPa900

11.4K300KkJ/kg0.2871.41

1

0.4/1.4/1

121

incomp, =⎟⎟

⎠

⎞

⎜⎜

⎝

⎛−⎟⎟

⎠

⎞⎜⎜⎝

⎛−⋅

=−−

= − kkPPkkRTw

and ( )( ) kW5.26 kJ/kg263.2kg/s0.02, === incompin wmW &&

Discussion Note that the power consumption of the compressor decreases significantly by using 2-stage compression with intercooling.

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 SOLUTIONS: HOMEWORK #8 Chapter 8 Problems 8-95 A steady-flow Carnot engine with water as the working fluid operates at specified conditions. The thermal efficiency, the pressure at the turbine inlet, and the net work output are to be determined.

Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Analysis (a) The thermal efficiency is determined from

3

2

4

1

60°C

s

350°C

η th, C60 273 K350 273 K

= − = −++

=1 1TT

L

H46.5%

(b) Note that s2 = s3 = sf + x3sfg

= 0.8312 + 0.891 × 7.0784 = 7.138 kJ/kg·K

Thus ,

MPa1.40KkJ/kg 1387

C 3502

2

2 =⎭⎬⎫

⋅=°=

P.s

T

(c) The net work can be determined by calculating the enclosed area on the T-s diagram,

Thus,

( )( )

( )( ) ( )( ) kJ/kg 1624=−−=−−==

⋅=+=+=

5391138760350Area

KkJ/kg 5391078471083120

43net

44

..ssTTw

....sxss

LH

fgf

Drawing from solution (1-2-3-4) is incorrect. Yellow-shaded area is correct. Water is compressed liquid at the boiler inlet, not a mixture, since s1 < sg at 350 oC. Numerical answers are okay. 8-97C Heat rejected decreases; everything else increases. 8-103C Yes, because the saturation temperature of steam at 10 kPa is 45.81°C, which is much higher than the temperature of the cooling water.

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 8-105 A steam power plant that operates on a simple ideal Rankine cycle is considered. The quality of the steam at the turbine exit, the thermal efficiency of the cycle, and the mass flow rate of the steam are to be determined.


Analysis (a) From the steam tables (Tables A-4, A-5, and A-6),

( )( )( )

kJ/kg 92201091083191kJ/kg0910

mkPa 1kJ 1kPa 1000010/kgm 001010

/kgm 001010

kJ/kg 83191

in12

33

121in

3kPa 10 1

kPa 10 1

...whh.

,.

PPvw

.vv

.hh

,p

,p

@f

@f

=+=+==

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

( )( ) kJ/kg 2089.32392.80.793191.83

0.79350097

0.64936.5966kPa 10

KkJ/kg 6.5966kJ/kg 3373.7

C 500 MPa10

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

fgf

fg

f

hxhh

.sss

xss

P

sh

TP

qin

qout

10 kPa 1

3

2

4

10 MPa

s

(b) q h hq h hw q q

in

out

net in out

kJ / kgkJ / kg

kJ / kg

= − = − == − = − =

= − = − =

3 2

4 1

3373 7 201 92 3171 782089 3 191 83 1897 47

3171 78 1897 47 1274 31

. . .. . .

. . .

and

40.2%===kJ/kg3171.78kJ/kg1274.31

in

netth q

wη

(c) s/kg 165===kJ/kg1274.31kJ/s210,000

net

net

wW

m&

&

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 8-111 A steam power plant operates on a simple ideal Rankine cycle between the specified pressure limits. The thermal efficiency of the cycle, the mass flow rate of the steam, and the temperature rise of the cooling water are to be determined.



( )

( )( )

kJ/kg 198.8906783191

kJ/kg 7.06mkPa 1

kJ 1kPa 107,000/kgm 0.00101

/kgm 001010

kJ/kg .83191

in12

33

121in

3kPa 10 1

kPa 10 1

=+=+=

=

⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

..whh

PPvw

.vv

hh

,p

,p

@f

@f

( )( ) kJ/kg 3.9321582392820083191

820050097

6493079756kPa 10

KkJ/kg 79756kJ/kg 410.53

C500MPa 7

44

44

34

4

3

3

3

3

=+=+=

=−

=−

=⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

...hxhh

..

..s

ssx

ssP

.sh

TP

fgf

fg

f

qin

qout

10 kPa 1

3

2

4

7 MPa

s

Thus, q h hq h hw q q

in

out

net in out

kJ / kgkJ / kg

kJ / kg

= − = − =

= − = − == − = − =

3 2

4 1

3410 5 198 89 3211 612153 93 191 83 196210

3211 61 196210 1249 51

. . .. . .

. . .

and 38.9%===kJ/kg3211.61kJ/kg1249.51

in

net

qw

thη

(b) s/kg0.36===kJ/kg1249.51kJ/s45,000

net

net

wW

m&

&

(c) The rate of heat rejection to the cooling water and its temperature rise are

( )( )

( ) ( )( ) C8.45°=⋅

==Δ

===

CkJ/kg 4.18kg/s 2000kJ/s 70,636

kJ/s ,63670kJ/kg 1962.1kg/s 36.0

watercooling

outwatercooling

outout

o&

&

&&

CmQT

qmQ

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 8-121 A steam power plant that operates on an ideal reheat Rankine cycle between the specified pressure limits is considered. The pressure at which reheating takes place, the total rate of heat input in the boiler, and the thermal efficiency of the cycle are to be determined.



( )( )( )

kJ/kg 9120008983191

kJ/kg 9.08mkPa 1

kJ 1kPa 100009/kgm 001010

/kgm 001010

kJ/kg 38191

in12

33

121in

3kPa 10 @sat1

kPa 10 @sat1

...whh

,.

PPvw

.vv

.hh

,p

,p

=+=+=

=⎟⎟⎠

⎞⎜⎜⎝

⎛

⋅−=

−=

==

==

( )( )( )( )

( )

kJ/kg 52979MPa 1462

kJ/kg 03466pressurereheat theC500

KkJ/kg 400175009790064930

kJ/kg 423458239290083191kPa 10

KkJ/kg 65766kJ/kg .13386

C500 MPa9

434

4

5

5

65

5

66

66

56

6

3

3

3

3

.hss

.P

.hP

ssT

....sxss

....hxhh

ssP

.sh

TP

fgf

fgf

=⎭⎬⎫

==

==

⎭⎬⎫

=°=

⋅=+=+=

=+=+=

⎭⎬⎫

==

⋅==

⎭⎬⎫

°==

MPa 2.146

1

5

2

6s

3

4 9 MPa

10 kPa

(b) The rate of heat supply is

( ) ( )[ ]

( )( ) kJ/s 91,792=−+−=−+−=

kJ/kg5.2979346691.2001.3386kJ/s 254523in hhhhmQ &&

(c) The thermal efficiency is determined from

Thus, ( ) ( )( )

41.3%=−=−=η

=−=−=

kJ/s 91,792kJ/s 53,83911

kJ/s 53,839kJ/kg8319142345kJ/s 25

in

out

16out

QQ

..hhmQ

th &

&

&&

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 8-123 A steady-flow Carnot refrigeration cycle with refrigerant-134a as the working fluid is considered. The coefficient of performance, the amount of heat absorbed from the refrigerated space, and the net work input are to be determined.


Analysis (a) Noting that TH = 30°C = 303 K and TL = Tsat @ 120 kPa = -22.36°C = 250.6 K, the COP of this Carnot refrigerator is determined from

( ) ( ) 4.78=−

=−

=1K 6.250/K 303

11/

1COP CR,LH TT

(b) From the refrigerant tables (Table A-11),

h hh h

g

f

3 30

4 30

263 5091 49

= == =

°

°

@

@

..

C

C

kJ / kgkJ / kg

Thus,

and

( ) kJ/kg 142.3=⎟⎟⎠

⎞⎜⎜⎝

⎛==⎯→⎯=

=−=−=

kJ/kg 172.01K 303K 250.6

kJ/kg 01.17249.9150.26343

HH

LL

L

H

L

H

H

qTT

qTT

qq

hhq

s

QH

QL

120 kPa

1 2

3 430°C

(c) The net work input is determined from

/kgkJ 71.29=−=−= 3.14201.172net LH qqw

8-132 An ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid is considered. The rate of heat removal from the refrigerated space, the power input to the compressor, the rate of heat rejection to the environment, and the COP are to be determined.


Analysis (a) In an ideal vapor-compression refrigeration cycle, the compression process is isentropic, the refrigerant enters the compressor as a saturated vapor at the evaporator pressure, and leaves the condenser as saturated liquid at the condenser pressure. From the refrigerant tables (Tables A-12 and A-13),

( )

( )throttlingkJ/kg 7886

kJ/kg 7886liquidsat.

MPa 70

C634kJ/kg 22270 MPa70

KkJ/kg 93540kJ/kg 86233

vaporsat.kPa 120

34

MPa 70 33

2212

2

kPa 120 1

kPa 120 11

.hh

.hh.P

.T.hss

.P

.ss.hhP

.@f

@g

@g

=≅

==⎭⎬⎫=

°==⎭⎬⎫

==

⋅====

⎭⎬⎫=

QH

QL

0.12 MPa 1

2 3

4

0.7 MPa

s

·

Win·

·

4s

Then the rate of heat removal from the refrigerated space and the power input to the compressor are determined from

and ( ) ( )( )

( ) ( )( ) kW 1.82

kW 7.35

=−=−=

=−=−=

kJ/kg 233.86270.22kg/s 0.05

kJ/kg 788686233kg/s 0.05

12in

41

hhmW

..hhmQL

&&

&&


8-132 CONTINUED

(b) The rate of heat rejection to the environment is determined from

& & & . .Q Q WH L= + = + =in 7 35 182 9.17 kW

(c) The COP of the refrigerator is determined from its definition,

COP 7.35 kW1.82 kWR

in= = =

&

&QW

L 4.04

8-146 A heat pump with refrigerant-134a as the working fluid heats a house by using underground water as the heat source. The power input to the heat pump, the rate of heat absorption from the water, and the increase in electric power input if an electric resistance heater is used instead of a heat pump are to be determined.


Analysis (a) From the refrigerant tables (Tables A-12 and A-13),

( )throttling kJ/kg 4991

kJ/kg 4991C30 MPa01

kJ/kg 36291C60 MPa01

kJ/kg 64247C0kPa 280

34

C30 33

3

22

2

11

1

.hh

.hhT.P

.hT.P

.hTP

@f

=≅

=≅⎭⎬⎫

°==

=⎭⎬⎫

°==

=⎭⎬⎫

°==

°

QH

QL

0.28 MPa 1

3

4

30°C

s

·

·

2

Win·

0°C

House 60°C

Water, 8°C

1 MPa

The mass flow rate of the refrigerant is

( ) kg/s 0.0834kJ/kg 91.49291.36

kJ/s 0060,000/3,6

32=

−=

−==

hhQ

qQ

m H

H

HR

&&&

Then the power input to the compressor becomes

( ) ( )( ) kW 3.65=−=−= kJ/kg 247.64291.36kg/s 0.083412in hhmW &&

(b) The rate of hat absorption from the water is

( ) ( )( ) kW 13.02=−=−= kJ/kg 91.49247.64kg/s 0.083441 hhmQL &&

(c) The electrical power required without the heat pump is

Thus,

kW13.02=−=−=

===

65.367.16

kW16.67kJ/s3600/000,60

inincrease WWW

QW

e

He

&&&

&&

AREN 2110 SOLUTIONS FALL 2006 HOMEWORK ASSIGNMENTS 6, 7 and 8 8-192 A heat pump water heater has a COP of 2.2 and consumes 2 kW when running. It is to be determined if this heat pump can be used to meet the cooling needs of a room by absorbing heat from it.

Assumptions The COP of the heat pump remains constant whether heat is absorbed from the outdoor air or room air.

Analysis The COP of the heat pump is given to be 2.2. Then the COP of the air-conditioning system becomes

COP COPair-cond heat pump= − = − =1 2 2 1 12. .

Then the rate of cooling (heat absorption from the air) becomes

& & ( . )( .Q Wincooling air-cond= COP kW) kW = 8640 kJ / h= =1 2 2 2 4

since 1 kW = 3600 kJ/h. We conclude that this heat pump can meet the cooling needs of the room since its cooling rate is greater than the rate of heat gain of the room.

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