tutorial 9 solutions 1

8/3/2019 Tutorial 9 Solutions 1

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MEC2405/CHE2164 – Thermodynamics

Tutorial 9 Solutions

10-67 A cogeneration plant is to generate power and process heat. Part of the steam extracted from theturbine at a relatively high pressure is used for process heating. The net power produced and the utilization

factor of the plant are to be determined.

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

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

( )

( )( )

kJ/kg38.670

kJ/kg41.19260.081.191

kJ/kg0.60mkPa1

kJ1kPa10600 /kgm 0.00101

/kgm 00101.0

kJ/kg81.191

MPa0.6@3

inpI,12

3

3

121inpI,

3kPa10@1

kPa10@1

==

=+=+=

=

⋅

−=

−=

==

==

f

f

f

hh

whh

PPw

hh

Mixing chamber:

332244

outin(steady)0

systemoutin 0

hmhmhmhmhm

E E E E E

eeii

+= → =

= → =∆=−

or,( )( ) ( )( )

( )( )( )

kJ/kg47.31857.690.311

kJ/kg6.57mkPa1

kJ1kPa6007000 /kgm 0.001026

/kgm 001026.0

kJ/kg90.31130

38.67050.741.19250.22

inII,45

3

3

454inII,

3kJ/kg90.311@4

4

33224

=+=+=

=

⋅−=

−=

=≅

=+

=+

=

=

p

p

h f

whh

PPw

m

hmhmh

f

KkJ/kg8000.6

kJ/kg4.3411

C500

MPa7

6

6

6

6

⋅=

=

°=

=

s

h

T

P

( )( ) kJ/kg6.21531.23928201.081.191

8201.0

4996.7

6492.08000.6kPa10

kJ/kg6.2774MPa6.0

88

88

68

8

767

7

=+=+=

=−

=−

=

=

=

=

=

=

fg f

fg

f

h xhhs

ss x

ss

P

hss

P

Then,

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

( )( ) ( )( )

kW 32,866=−=−=

=+=+=

=−+−=

−+−=

6.210077,33

kW210.6kJ/kg6.57kg/s30kJ/kg0.60kg/s22.5

kW077,33kJ/kg6.21536.2774kg/s22.5kJ/kg6.27744.3411kg/s30

inp,outT,net

inpII,4inpI,1inp,

878766outT,

W W W

wmwmW

hhmhhmW

1

2

8

s

T

6

7

7 MPa

10 kPa

0.6 MPa4

5

3

Qout ·

Qin ·

Qproces

·

6

8

1

5

TurbineBoiler

Condenser

Processheater

P IP II

4 2

3

7



Also, ( ) ( )( ) kW782,15kJ/kg38.6706.2774kg/s7.5377process =−=−= hhmQ

( ) ( )( ) kW788,9247.3184.3411kg/s30565in =−=−= hhmQ

and 52.4%=+

=+

=788,92

782,15866,32

in

processnet

Q

QW u

ε

11-13 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, therate 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, therefrigerant 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/kg82.88

kJ/kg82.88liquidsat.

MPa7.0

C95.34kJ/kg50.273MPa7.0

KkJ/kg94779.0

kJ/kg97.236

vaporsat.

kPa120

34

MPa7.0@33

2212

2

kPa120@1

kPa120@11

=≅

==

=

°==

=

=

⋅==

==

=

hh

hhP

T hss

P

ss

hhP

f

g

g

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

and

( ) ( )( )

( ) ( )( ) kW1.83

kW 7.41

=−=−=

=−=−=

kJ/kg236.97273.50kg/s0.05

kJ/kg82.8897.236kg/s0.05

12in

41

hhmW

hhmQ L

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

kW9.23=+=+= 83.141.7inW QQ L H

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

4.06===kW1.83

kW7.41COP

in

RW

Q L

11-17 A commercial refrigerator with refrigerant-134a as the working fluid is considered. The quality of

the refrigerant at the evaporator inlet, the refrigeration load, the COP of the refrigerator, and the theoretical

maximum refrigeration load for the same power input to the compressor are to be determined. Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential energy changes are negligible.

Q H

Q L

0.121

2

3

4

0.7 MPa

s

T

·

W in ·

·

4s



Analysis (a) From refrigerant-134a tables (Tables A-11 through A-13)

0.4795=

=

=

==

=

°=

=

=

°=

=

=

°−=

=

44

4

34

33

3

22

2

1

1

1

kJ/kg23.111

kPa60

kJ/kg23.111

kJ/kg23.111C42

kPa1200

kJ/kg16.295C65

kPa1200

kJ/kg03.230C34

kPa60

xh

P

hh

hT

P

hT

P

hT

P

Using saturated liquid enthalpy at the given

temperature, for water we have (Table A-4)

kJ/kg94.108kJ/kg47.75

C26@2

C18@1

==

==

°

°

f w

f w

hhhh

(b) The mass flow rate of the refrigerant may be determined from an energy balance on the compressor

kg/s0455.0

g75.47)kJ/k 94kg/s)(108.(0.25kJ/kg)23.11116.295(

)()( 1232

= →

−=−

−=−

R

R

www R

m

m

hhmhhm

The waste heat transferred from the refrigerant, the compressor power input, and the refrigeration load are

kW 367.8kJ/kg)23.11116kg/s)(295.0455.0()( 32 =−=−= hhmQ R H

kW513.2kW0.450kJ/kg)03.23016kg/s)(295.0455.0()( in12in =−−=−−= QhhmW R

kW5.404=−−=−−= 450.0513.2367.8inin QW QQ H L

(c) The COP of the refrigerator is determined

from its definition

2.15===513.2

404.5COP

in

L

W

Q

(d ) The reversible COP of the refrigerator for the

same temperature limits is

063.51)27330 /()27318(

1

1 /

1COPmax =

−+−+=

−=

L H T T

Then, the maximum refrigeration load becomes

kW12.72=== kW)513.2)(063.5(inmaxmaxL, W COPQ

11-23 A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression

refrigeration cycle except for the compression process. The mass flow rate of the refrigerant, the condenser

pressure, and the COP of the refrigerator are to be determined.


Q H

Q L 1

2

3

4

s

T

·

·

2

W in ·

60 kPa

-34°C1

23

4

Q H

42°C

Condenser

Evaporator

Compressor

Expansion

valve

1.2 MPa

65°C

Q L

W in

Water

18°C26°C

Qin



Analysis (a) (b) From the refrigerant-134a tables (Tables A-11 through A-13)

kJ/kg97.236vap.)(sat.1

kPa120

kJ/kg87.298C60

kPa8.671

liq.)(sat.0

kJ/kg83.86

kJ/kg83.8630.0

kPa120

11

41

22

2

32

33

3

43

4

4

4

=

=

==

=

°=

=

=

=

=

=

=

=

=

=

h x

PP

hT

P

PP

P x

h

hh

h x

P

kPa671.8

The mass flow rate of the refrigerant is

determined from

kg/s0.00727=−

=−

=kg236.97)kJ/ (298.87

kW45.0

12

in

hh

W m

(c) The refrigeration load and the COP are

kW091.1

kJ/kg)83.8697kg/s)(236.0727.0(

)( 41

=

−=

−= hhmQ L

2.43===kW0.45

kW091.1COP

in

L

W

Q

11-43 A heat pump operating on the ideal vapor-compression refrigeration cycle with refrigerant-134a as

the working fluid is considered. The COP and the rate of heat supplied to the evaporator are to be

determined.


Analysis 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-11, A-12, and A-13),

)throttling( kJ/kg32.107

kJ/kg32.107 liquidsat.

kPa1000

kJ/kg98.277 kPa1000

KkJ/kg93773.0

kJ/kg46.244

vaporsat.

kPa200

34

kPa1000@33

212

2

kPa200@1

kPa200@11

=≅

==

=

=

=

=

⋅==

==

=

hh

hhP

hss

P

ss

hhP

f

g

g

The mass flow rate of the refrigerant is determined from

kg/s0.179kJ/kg)46.244(277.98

kJ/s6)(

12

in12in =

−=

−= → −=

hh

W mhhmW

Then the rate of heat supplied to the evaporator is

kW24.5=−=−= kJ/kg)32.10746kg/s)(244.179.0()( 41 hhmQ L

Q H

Q L

200 kPa

1

2

3

4

1.0 MPa

s

T

·

W in

·

·

4s

Q H

Q L

120 kPa1

2

3

4

s

T

·

W in ·

·

4s

.

QH

60°CCondenser

Evaporator

Compressor

Expansion

valve

120 kPa

x=0.3 QL

W in

1

23

4

.

.



The COP of the heat pump is determined from its definition,

5.09=−

−=

−

−==

46.24498.277

32.10798.277COP

12

32

inHP

hh

hh

w

q H

11-47 An actual heat pump cycle with R-134a as the refrigerant is considered. The isentropic efficiency of

the compressor, the rate of heat supplied to the heated room, the COP of the heat pump, and the COP and

the rate of heat supplied to the heated room if this heat pump operated on the ideal vapor-compression cycle

between the same pressure limits are to be determined.


Analysis (a) The properties of refrigerant-134a are (Tables A-11 through A-13)

26.277kPa800

kJ/kg9506.0

kJ/kg87.247

C)409.10(

kPa200

C09.10

kJ/kg91.87

kJ/kg91.87C)306.29(

kPa750

C06.29

kJ/kg76.291C55

kPa800

212

2

1

1

1

1

kPasat@200

34

33

3

kPasat@7503

22

2

=

=

=

=

=

°+−=

=

°−=

==

=

°−=

=

°==

=

°=

=

shss

P

s

h

T

P

T

hh

hT

P

T T

hT

P

The isentropic efficiency of the compressor is

0.670=−

−

=−

−

= 87.24776.291

87.24726.277

12

12hh

hhsC η

(b) The rate of heat supplied to the room is

kW3.67=−=−= kJ/kg)91.8776kg/s)(291.018.0()( 32 hhmQ H

(c) The power input and the COP are

kW790.0kJ/kg)87.24776kg/s)(291.018.0()( 12in =−=−= hhmW

4.64===790.0

67.3COP

inW

Q H

(d ) The ideal vapor-compression cycle analysis of the cycle is as follows:

kJ/kg.K9377.0

kJ/kg46.244

kPa200@1

kPa200@1==

==

g

gss

hh

kJ/kg25.273kPa800

212

2=

=

=h

ss

P

34

kPa800@3 kJ/kg47.95

hh

hh f

=

==

6.18=−

−=

−

−=

46.24425.273

47.9525.273COP

12

32

hh

hh

kW3.20=−=−= kJ/kg)47.9525kg/s)(273.018.0()( 32 hhmQ H

Q H

Q L

0.2 MPa

1

2

3

4

0.8 MPa

s

T

·

W in ·

·

4s

Q

Q L 1

2

3

4

s

T

·

·

2

W in ·

.

QH

750 kPa

Condenser

Evaporator

Compressor

Expansion

valve

800 kPa

55°C

QL

W in

1

23

4

.

.

tutorial 9 solutions 1

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