Sol 1014

8/8/2019 Sol 1014

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10-16 A simple ideal Rankine cycle with water as the working fluid operates between the specified pressure limits.

The maximum thermal efficiency of the cycle for a given quality at the turbine exit is to be determined.

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

Analysis For maximum thermal efficiency, the quality at state 4 would be at its minimum of 85% (most closely

approaches the Carnot cycle), and the properties at state 4 would be (Table A-5)

KkJ/kg7440.6)8234.6)(85.0(9441.0kJ/kg3.2274)3.2335)(85.0(27.289

85.0kPa30

44

44

4

4

fg f

fg f

s xssh xhh

xP

Since the expansion in the turbine is isentropic,

kJ/kg5.3115 KkJ/kg7440.6

kPa30003

43

3h

ss

P

Other properties are obtained as follows (Tables A-4, A-5, and A-6),

kJ/kg31.29204.327.289

kJ/kg04.3mkPa1

kJ1 kPa)303000)( /kgm001022.0(

)(

/kgm001022.0

kJ/kg27.289

inp,12

33

121inp,

3kP a30@1

kP a30@1

whh

PPw

hh

f

f

v

v v

Thus,

kJ/kg0.198527.2893.2274

kJ/kg2823.231.2925.3115

14out

23in

hhq

hhq

and the thermal efficiency of the cycle is

0.2972.2823

0.198511

in

outth

q

q

qin

qout

30 kPa

1

3

2

4

3 MPa

s

T

8/8/2019 Sol 1014


10-21E A simple steam Rankine cycle operates between the specified pressure limits. The mass flow rate, the

power produced by the turbine, the rate of heat addition, and the thermal efficiency of the cycle are to be

determined.


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

Btu/lbm18.7746.772.69

Btu/lbm46.7ftpsia5.404

Btu1 psia)12500)( /lbmft01614.0(

)(

/lbmft01614.0

Btu/lbm72.69

inp,12

3

3

121inp,

3psia6@1

psia1@1

whh

PPw

hh

f

f

v

v v

Btu/lbm70.787)7.1035)(6932.0(72.69

6932.0

84495.1

13262.04116.1

psia1

RBtu/lbm4116.1

Btu/lbm0.1302

F800

psia2500

44

44

34

4

3

3

3

3

fgs f s

fg

f

s

h xhhs

ss x

ss

P

s

h

T

P

kJ/kg13.839)70.7870.1302)(90.0(0.1302)( 4s3T3443

43T hhhh

hh

hh

s

Thus,

Btu/lbm39.45541.7698.1224

Btu/lbm41.76972.6913.839

Btu/lbm8.122418.771302.0

outinnet

14out

23in

qqw

hhq

hhq

The mass flow rate of steam in the cycle is determined from

lbm/s2.081kJ1

Btu0.94782

Btu/lbm455.39

kJ/s1000

net

net

netnet w

W

mwmW

The power output from the turbine and the rate of heat addition are

Btu/s2549

kW1016

Btu/lbm)4.8lbm/s)(122081.2(

Btu0.94782

kJ1Btu/lbm)13.8392.0lbm/s)(130081.2()(

inin

43outT ,

qmQ

hhmW

and the thermal efficiency of the cycle is

0.3718kJ1

Btu0.94782

Btu/s2549

kJ/s1000

in

netth

Q

W

qin

qout

1 psia

1

3

2

4

2500 psia

s

T

4s

8/8/2019 Sol 1014


10-24 A solar-pond power plant that operates on a simple ideal Rankine cycle with refrigerant-134a as the working

fluid is considered. The thermal efficiency of the cycle and the power output of the plant are to be determined.


Analysis (a) From the refrigerant tables (Tables A-11, A-12, and A-13),

kJ/kg40.8958.082.88

kJ/kg58.0

mkPa1

kJ1kPa7001400 /kgm 0008331.0

/kgm 0008331.0

kJ/kg 82.88

in,12

3

3

121in,

3MPa7.0@1

MPa7.0@1

p

p

f

f

whh

PPw

hh

v

v v

kJ/kg.2026221.1769839.082.88

9839.058763.0

33230.09105.0MPa7.0

KkJ/kg9105.0

kJ/kg12.276

vaporsat.

MPa4.1

44

4434

4

MP a4.1@3

MP a4.1@33

fg f

fg

f

g

g

h xhh

s

ss x

ssP

ss

hhP

Thus ,

kJ/kg34.1338.17372.186

kJ/kg38.17382.8820.262

kJ/kg72.18640.8912.276

outinnet

14out

23in

qqw

hhq

hhq

and

7.1%kJ/kg186.72

kJ/kg13.34

in

netth

q

w

(b) kW40.02kJ/kg13.34kg/s3netnet wmW

qin

qout

0.7 MPa

1

3

2

4

1.4 MPa

s

T

R-134a

8/8/2019 Sol 1014


10-34 A steam power plant that operates on the ideal reheat Rankine cycle is considered. The turbine work output

and the thermal efficiency of the cycle are to be determined.


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

kJ/kg.5425912.842.251

kJ/kg8.12mkPa1

kJ1kPa208000 /kgm 001017.0

/kgm 700101.0

kJ/kg42.251

in,12

3

3

121in,

3kP a20@1

kP a20@1

p

p

f

f

whh

PPw

hh

v

v v

kJ/kg2.23855.23579051.042.251

9051.00752.7

8320.02359.7kPa20

KkJ/kg2359.7

kJ/kg2.3457

C500

MPa3

kJ/kg1.3105MPa3

KkJ/kg7266.6

kJ/kg5.3399

C500

MPa8

66

66

56

6

5

5

5

5

4

34

4

3

3

3

3

fg f

fg

f

h xhh

s

ss x

ss

P

s

h

T

P

hss

P

s

h

T

P

The turbine work output and the thermal efficiency are determined from

and

kJ/kg0.34921.31052.345754.2595.3399

2.23852.34571.31055.3399

4523in

6543outT ,

hhhhq

hhhhw kJ/kg 1366.4

Thus,

38.9%kJ/kg3492.5

kJ/kg1358.3

kJ/kg3.135812.84.1366

in

netth

in,,net

q

w

www pout T

1

5

2

6s

3

4

8 MPa

20 kPa

8/8/2019 Sol 1014


10-35 EES Problem 10-34 is reconsidered. The problem is to be solved by the diagram window data entry feature of

EES by including the effects of the turbine and pump efficiencies and reheat on the steam quality at the low-pressure

turbine exit Also, the T-s diagram is to be plotted.

Analysis The problem is solved using EES, and the solution is given below.

"Input Data - from diagram window"

{P[6] = 20 [kPa]P[3] = 8000 [kPa]T[3] = 500 [C]P[4] = 3000 [kPa]T[5] = 500 [C]Eta_t = 100/100 "Turbine isentropic efficiency"Eta_p = 100/100 "Pump isentropic efficiency"}

"Pump analysis" function x6$(x6) "this function returns a string to indicate the state of steam at point 6"

x6$=''if (x6>1) then x6$='(superheated)'if (x6<0) then x6$='(subcooled)'

end

Fluid$='Steam_IAPWS'

P[1] = P[6]P[2]=P[3]x[1]=0 "Sat'd liquid" h[1]=enthalpy(Fluid$,P=P[1],x=x[1])v[1]=volume(Fluid$,P=P[1],x=x[1])s[1]=entropy(Fluid$,P=P[1],x=x[1])T[1]=temperature(Fluid$,P=P[1],x=x[1])W_p_s=v[1]*(P[2]-P[1])"SSSF isentropic pump work assuming constant specific volume" W_p=W_p_s/Eta_p

h[2]=h[1]+W_p "SSSF First Law for the pump" v[2]=volume(Fluid$,P=P[2],h=h[2])s[2]=entropy(Fluid$,P=P[2],h=h[2])T[2]=temperature(Fluid$,P=P[2],h=h[2])"High Pressure Turbine analysis" h[3]=enthalpy(Fluid$,T=T[3],P=P[3])s[3]=entropy(Fluid$,T=T[3],P=P[3])v[3]=volume(Fluid$,T=T[3],P=P[3])s_s[4]=s[3]hs[4]=enthalpy(Fluid$,s=s_s[4],P=P[4])Ts[4]=temperature(Fluid$,s=s_s[4],P=P[4])Eta_t=(h[3]-h[4])/(h[3]-hs[4])"Definition of turbine efficiency" T[4]=temperature(Fluid$,P=P[4],h=h[4])s[4]=entropy(Fluid$,T=T[4],P=P[4])v[4]=volume(Fluid$,s=s[4],P=P[4])h[3] =W_t_hp+h[4]"SSSF First Law for the high pressure turbine" "Low Pressure Turbine analysis" P[5]=P[4]s[5]=entropy(Fluid$,T=T[5],P=P[5])h[5]=enthalpy(Fluid$,T=T[5],P=P[5])s_s[6]=s[5]hs[6]=enthalpy(Fluid$,s=s_s[6],P=P[6])Ts[6]=temperature(Fluid$,s=s_s[6],P=P[6])

8/8/2019 Sol 1014


vs[6]=volume(Fluid$,s=s_s[6],P=P[6])Eta_t=(h[5]-h[6])/(h[5]-hs[6])"Definition of turbine efficiency" h[5]=W_t_lp+h[6]"SSSF First Law for the low pressure turbine" x[6]=QUALITY(Fluid$,h=h[6],P=P[6])"Boiler analysis" Q_in + h[2]+h[4]=h[3]+h[5]"SSSF First Law for the Boiler" "Condenser analysis" h[6]=Q_out+h[1]"SSSF First Law for the Condenser" T[6]=temperature(Fluid$,h=h[6],P=P[6])s[6]=entropy(Fluid$,h=h[6],P=P[6])x6s$=x6$(x[6])

"Cycle Statistics" W_net=W_t_hp+W_t_lp-W_pEff=W_net/Q_in

SOLUTION

Eff=0.389 Eta_p=1 Eta_t=1Fluid$='Steam_IAPWS' h[1]=251.4[kJ/kg] h[2]=259.5[kJ/kg]h[3]=3400[kJ/kg] h[4]=3105[kJ/kg] h[5]=3457[kJ/kg]h[6]=2385[kJ/kg] hs[4]=3105[kJ/kg] hs[6]=2385[kJ/kg]P[1]=20[kPa] P[2]=8000[kPa] P[3]=8000[kPa] P[4]=3000[kPa] P[5]=3000[kPa] P[6]=20[kPa]Q_in=3493[kJ/kg] Q_out=2134[kJ/kg] s[1]=0.832[kJ/kg-K]s[2]=0.8321[kJ/kg-K] s[3]=6.727[kJ/kg-K] s[4]=6.727[kJ/kg-K]s[5]=7.236[kJ/kg-K] s[6]=7.236[kJ/kg-K] s_s[4]=6.727[kJ/kg-K]

s_s[6]=7.236[kJ/kg-K] T[1]=60.06[C] T[2]=60.4[C]T[3]=500[C] T[4]=345.2[C] T[5]=500[C] T[6]=60.06[C] Ts[4]=345.2[C] Ts[6]=60.06[C]v[1]=0.001017[m^3/kg] v[2]=0.001014[m^3/kg] v[3]=0.04177[m^3/kg]v[4]=0.08968[m^3/kg] vs[6]=6.922[m^3/kg] W_net=1359[kJ/kg]W_p=8.117[kJ/kg] W_p_s=8.117[kJ/kg] W_t_hp=294.8[kJ/kg]W_t_lp=1072[kJ/kg] x6s$='' x[1]=0x[6]=0.9051

8/8/2019 Sol 1014


13-12 The masses of the constituents of a gas mixture are given. The mass fractions, the mole fractions, the average

molar mass, and gas constant are to be determined.

Properties The molar masses of O2, N2, and CO2 are 32.0, 28.0 and 44.0 kg/kmol, respectively (Table A-1)

Analysis (a) The total mass of the mixture is

kg23kg10kg8kg5222 CONO mmmmm

Then the mass fraction of each component becomes

0.435

0.348

0.217

kg23

kg10mf

kg23

kg8mf

kg23

kg5mf

2

2

2

2

2

2

CO

CO

N

N

O

O

m

m

m

m

m

m

m

m

m

(b) To find the mole fractions, we need to determine the mole numbers of each component first,

kmol0.227kg/kmol44

kg10

kmol0.286kg/kmol28

kg8

kmol0.156kg/kmol32

kg5

2

2

2

2

2

2

2

2

2

CO

CO

CO

N

N

N

O

O

O

M

m N

M

m N

M

m N

Thus,

kmol0.669kmol0.227kmol0.286kmol615.0222 CONO N N N N m

and

0.339

0.428

0.233

kmol0.669

kmol0.227

kmol0.669

kmol0.286

kmol0.699

kmol0.156

2

2

2

2

2

2

CO

CO

N

N

O

O

m

m

m

N

N y

N

N y

N

N y

(c) The average molar mass and gas constant of the mixture are determined from their definitions:

and

KkJ/kg0.242

kg/kmol34.4

kg/kmol34.4

KkJ/kmol8.314

kmol0.669

kg23

m

um

m

mm

M

R R

N

m M

5 kg O2

8 kg N2

10 kg CO2

8/8/2019 Sol 1014


13-35E A mixture is obtained by mixing two gases at constant pressure and temperature. The volume and specific

volume of the mixture are to be determined.

Properties The densities of two gases are given in the problem statement.

Analysis The volume of constituent gas A is

3

3ft1000

lbm/ft0.001

lbm1

A

A A

mV

and the volume of constituent gas B is

3

3ft1000

lbm/ft0.002

lbm2

B

B B

mV

Hence, the volume of the mixture is

3ft200010001000 B A V V V

The specific volume of the mixture will then be

/lbmft666.7 3

lbm2)(1

ft2000 3

m

V v

1 lbm gas A

2 lbm gas B

8/8/2019 Sol 1014


13-40E The mass fractions of components of a gas mixture are given. The mass of 5 ft3

of this mixture and the

partial volumes of the components are to be determined.

Properties The molar masses of N2, O2, and He are 28.0, 32.0, and 4.0 lbm/lbmol, respectively (Table A-1E).

Analysis We consider 100 lbm of this mixture for calculating the molar mass of the mixture. The mole numbers of

each component are

lbmol5.2lbm/lbmol4

lbm10

lbmol9375.0lbm/lbmol32

lbm30

lbmol1429.2lbm/lbmol28

lbm60

He

HeHe

O2

O2O2

N2

N2N2

M

m N

M

m N

M

m N

The mole number of the mixture is

lbmol5804.55.29375.01429.2HeO2N2 N N N N m

The apparent molecular weight of the mixture is

lbm/lbmol92.17lbmol5.5804

lbm100

m

mm

N

m M

Then the mass of this ideal gas mixture is

lbm4.727R)R)(530 /lbmolftpsia(10.73

lbm/lbmol))(17.92ftpsia)(5(3003

3

T R

M Pm

u

mV

The mole fractions are

0.4480

lbmol5.5804

lbmol2.5

0.1680lbmol5.5804

lbmol0.9375

0.3840lbmol5.5804

lbmol2.1429

HeHe

O2O2

N2N2

m

m

m

N

N y

N

N y

N

N y

Noting that volume fractions are equal to mole fractions, the partial volumes are determined from

3

3

3

ft2.24

ft0.84

ft1.92

)ft(5)4480.0(

)ft(5)1680.0(

)ft(5)3840.0(

3HeHe

3O2O2

3N2N2

m

m

m

y

y

y

V V

V V

V V

5 ft3

60% N2

30% O2

10% He

(by mass)

8/8/2019 Sol 1014


14-16 A tank contains dry air and water vapor at specified conditions. The specific humidity, the relative humidity,

and the volume of the tank are to be determined.

Assumptions The air and the water vapor are ideal gases.

Analysis (a) The specific humidity can be determined form its definition,

airdryO/kgHkg0.0143 2

kg21

kg3.0

a

v

m

m

(b) The saturation pressure of water at 30 C is

kPa2469.4C30@satPPg

Then the relative humidity can be determined from

52.9%kPa)2469.4)(0143.0622.0(

kPa)100)(0143.0(

)622.0( gP

P

(c) The volume of the tank can be determined from the ideal gas relation for the dry air,

3

m18.7kPa755.97

K)K)(303kJ/kgkg)(0.28721(

=

kPa755.97245.2100

kPa2.245=kPa)2469.4)(529.0(

a

aa

va

gv

P

T Rm

PPP

PP

V

21 kg dry air

0.3 kg H2O vapor

30 C

100 kPa

8/8/2019 Sol 1014


14-31E A woman drinks a cool canned soda in a room at a specified temperature and relative humidity. It is to be

determined whether the can will sweat.

Assumptions The air and the water vapor are ideal gases.

Analysis The vapor pressure Pv of the air in the house is uniform

throughout, and its value can be determined from

psia254.0psia)50745.0)(50.0(F80@gv PP

The dew-point temperature of the air in the house is

F59.7psia254.0@sat@satdp T T T vP

(from EES)

That is, the moisture in the house air will start condensing when the air temperature drops below 59.7 C. Since the

canned drink is at a lower temperature than the dew-point temperature, some moisture will condense on the can, and

thus it will sweat.

14-41 The pressure, temperature, and relative humidity of air in a room are specified. Using the psychrometric chart,

the specific humidity, the enthalpy, the wet-bulb temperature, the dew-point temperature, and the specific volume of

the air are to be determined.

Analysis From the psychrometric chart (Fig. A-31) we read

(a) airdrykg / OHkg0181.0 2

(b) h 784. kJ / kg dry air

(c) C5.25wbT

(d ) C3.23dpT

(e) airdrykg / m890.0 3v

14-49 The pressure, temperature, and wet-bulb temperature of air are specified. Using the psychrometric chart, the

relative humidity, specific humidity, the enthalpy, the dew-point temperature, and the water vapor pressure are to be

determined.

Analysis From the psychrometric chart in Fig. A-31 or using EES psychrometric functions we obtain

(a) %8.61618.0

(b) airdrykg / OHkg0148.0 2

(c) airdrykJ/kg8.65h

(d ) C4.22wbT

(e) kPa2.34=kPa)780.3)(618.0(C28@satPPP gv

80 F

50% RH

Cola

40 F

Air

1 atm

28 C

T dp=20 C

8/8/2019 Sol 1014


14-50 The pressure, temperature, and wet-bulb temperature of air are specified. The adiabatic saturation temperature

is to be determined.

Analysis For an adiabatic saturation process, we obtained Eq. 14-14 in the text,

21

22121

)(

f g

fg p

hh

hT T c

This requires a trial-error solution for the adiabatic saturationtemperature, T 2. The inlet state properties are

airdrykg / OHkg0148.0 21 (Fig. A-31)

kJ/kg9.2551C28@1 gg hh (Table A-4)

As a first estimate, let us take T 2 =22 C (the inlet wet-bulb temperature). Also, at the exit, the relative humidity is

100% ( 12 ) and the pressure is 1 atm. Other properties at the exit state are

airdrykg / OHkg0167.0 22

4)-A(Table kJ/kg8.2448

4)-A(Table kJ/kg28.92

C22@2

C22@2

fg fg

f f

hh

hh

Substituting,

airdrykg / OHkg0142.028.929.2551

)8.2448)(0167.0()2822)(005.1()(2

21

22121

f g

fg p

hh

hT T c

which is sufficiently close to the inlet specific humidity (0.0148). Therefore, the adiabatic saturation temperature is

T 2 22 C

Discussion This result is not surprising since the wet-bulb and adiabatic saturation temperatures are approximately

equal to each other for air-water mixtures at atmospheric pressure.

14-66E There are a specified number of people in a movie theater in winter. It is to be determined if the theater

needs to be heated or cooled.

Assumptions There is a mix of men, women, and children in the classroom.

Properties The average rate of heat generation from people in a movie theater is 105 W, and 70 W of it is in sensible

form and 35 W in latent form.

Analysis Noting that only the sensible heat from a person contributes to the heating load of a building, the

contribution of people to the heating of the building is

Btu/h119,420W000,35W)(70500people)of (No. sensibleperson,sensiblepeople, QQ

since 1 W = 3.412 Btu/h. The building needs to be heated since the heat gain from people is less than the rate of heat

loss of 130,000 Btu/h from the building.

1 atm

28 C

T dp=20 C

AIR100%

Water

Humidifier

8/8/2019 Sol 1014


14-71 Air enters a cooling section at a specified pressure, temperature, velocity, and relative humidity. The exit

temperature, the exit relative humidity of the air, and the exit velocity are to be determined.

Assumptions 1 This is a steady-flow process and thus the mass flow rate of dry air remains constant during the

entire process ( )m m ma a a1 2 . 2 Dry air and water vapor are ideal gases. 3 The kinetic and potential energy

changes are negligible.

Analysis (a) The amount of moisture in the air remains constant ( 1 = 2) as it flows through the cooling section

since the process involves no humidification or dehumidification. The inlet state of the air is completely specified,

and the total pressure is 1 atm. The properties of the air at the inlet state are determined from the psychrometric

chart (Figure A-31) to be

airdrykg / m877.0

)(airdryO/kgHkg0089.0

airdrykJ/kg0.55

31

221

1

v

h

The mass flow rate of dry air through the cooling

section is

kg/s58.2

)m /40.4m/s)(18(kg) / m877.0(

1

1

223

111

AV mav

From the energy balance on air in the cooling section,

( )

/ . )

.

Q m h h

h

h

aout

kJ /s =(2.58 kg / s)( kJ / kg

kJ / kg dry air

2 1

2

2

1200 60 550

472

The exit state of the air is fixed now since we know both h2 and 2. From the psychrometric chart at this state we

read

(b)airdrykg / m856.0 3

2

2

2

v

46.6%

C24.4T

(c) The exit velocity is determined from the conservation of mass of dry air,

m/s17.6m/s)18(877.0

856.01

1

22

2

2

1

1

2

2

1

121

V V

AV AV mm aa

v

v

v v v

V

v

V

32 C

30%

18 m/s 1 atm

1200

1 2

AIR

8/8/2019 Sol 1014


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Sol 1014