Chap.10.Soln

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CANKAYA UNIVERSITY

FACULTY OF ENGINEERING AND ARCHITECTURE MECHANICAL ENGINEERING DEPARTMENT

ME 212 THERMODYNAMICS II

CHAPTER 10 EXAMPLES SOLUTION

1) An ideal vapor-compression refrigerant cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated vapor enters the compressor at -100C, and saturated liquid leaves the condenser at 280C. The mass flow rate of refrigerant is 5 kg/min. Determine

(a) The compressor power, in kW (b) The refrigerating capacity, in tons. (c) The coefficient of performance. Solution:

Known: Refrigerant 134a is the working fluid in an ideal vapor-compression refrigerant cycle. Operating data are known.

2 Analysis: First, fix each of the principle states.

Compressor:

Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

) h h ( m W m m m m m h m h m W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2 c 2 1 e i 2 2 1 1 c e e 2 e e e i i 2 i i i . v . c . v . c v . c                            

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              c W 2 1

3 1 2 2 1 2 2 1 1 cv i e e e i i j j j cv s s m m m s m s m 0 s m s m T Q dt dS          

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        

State 1: T1 = -10oC, saturated vapor  h1 = 241.35 kJ/kg, s1 = 0.9253 kJ/kg.K

State 2: P2 = Psat@28oC=7.2675 bars, s2 = s1  Double interpolation gives: h2 = 267.9 kJ/kg

Expansion valve:

Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

 Neglect work (w=0)   4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 . v . c 0 . v . c v . c h h m m m m m h m h m 0 gz 2 V h m gz 2 V h m W Q dt dE                          

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          

State 3: T3 = 28oC, saturated liquid  h3 = 88.61 kJ/kg State 4: Throttling process  h4 = h3 = 88.61 kJ/kg

4 3

4 Evaporator:

Assumptions:

 steady state steady flow process(SSSF)

 open system  ΔKE = ΔPE =0  Neglect work (w = 0) ) h h ( m Q m m m m m h m h m Q 0 gz 2 V h m gz 2 V h m W Q dt dE 4 1 in 4 1 e i 1 1 4 4 in e e 2 e e e i i 2 i i i 0 . v . c . v . c v . c                            

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               

(a) The compressor power is:

kW 212 . 2 s / kJ 1 kW 1 kg kJ ) 35 . 241 9 . 267 ( s 60 min 1 min kg 5 ) h h ( m W_{c 2 1 }                       

(b) The refrigeration capacity is:

tons 62 . 3 min / kJ 211 ton 1 kg kJ ) 61 . 88 35 . 241 ( min kg 5 ) h h ( m Q_{in 1 4 }                 

(c) The coefficient of performance is:

75 . 5 kg / kJ ) 35 . 241 9 . 267 ( kg / kJ ) 61 . 88 35 . 241 ( h h h h W Q 1 2 4 1 c in          _ in Q 1 4

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2) A vapor-compression refrigeration system circulates Refrigerant 134a at rate of 6 kg/min. The refrigerant enters the compressor at -100C, 1.4 bar, and exits at 7 bar. The isentropic compressor efficiency is 67%. There are no appreciable pressure drops as the refrigerant flows through the condenser and evaporator. The refrigerant leaves the condenser at 7 bar, 240C. Ignoring heat transfer between the compressor and its surroundings, determine

(a) The coefficient of performance. (b) The refrigerating capacity, in tons.

(c) The irreversibility rates of the compressor and expansion valve, each in kW (d) The changes in specific flow availability of the refrigerant passing through the

evaporator and condenser, respectively, each in kJ/kg. Let To = 21oC, po = 1 bar

Solution:

Known: A vapor-compression refrigeration system circulates Refrigerant 134a with a known mass flow rate. Data are given at various locations, the compressor efficiency is specified.

Schematic & Given Data:

6 Compressor:

Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

) h h ( m W m m m m m h m h m W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 s 2 c 2 1 e i s 2 2 1 1 c e e 2 e e e i i 2 i i i . v . c . v . c v . c                            

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              1 s 2 2 1 s 2 2 1 1 cv i e e e i i j j j cv s s m m m s m s m 0 s m s m T Q dt dS          

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         State 1: T1 = -10oC, P1 = 1.4 bars  h1 = 243.40 kJ/kg, s1 = 0.9606 kJ/kg.K State 2: For isentropic compression, P2 = 7 bars, s2s = s1  h2s = 278.06 kJ/kg Using the compressor efficiency,

kg / kJ 13 . 295 67 . 0 40 . 243 06 . 278 40 . 243 h h h h h h h h c 1 s 2 1 2 1 2 1 s 2 C              s2 = 1.0135 kJ/kg.K c W 2 1

7 State 3:

P3 = 7 bars, T = 24oC  h3 = hf@24oC = 82.90 kJ/kg, s3 = 0.3113 kJ/kg.K Expansion valve:

Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

 Neglect work (w=0)   4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 . v . c 0 . v . c v . c h h m m m m m h m h m 0 gz 2 V h m gz 2 V h m W Q dt dE                          

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          

State 4: Throttling process  h4 = h3 = 82.90 kJ/kg, s4 = 0.33011 kJ/kg.K Evaporator: Assumptions: in Q 1 4 4 3

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 steady state steady flow process(SSSF)

 open system  ΔKE = ΔPE =0  Neglect work (w = 0) ) h h ( m Q m m m m m h m h m Q 0 gz 2 V h m gz 2 V h m W Q dt dE 4 1 in 4 1 e i 1 1 4 4 in e e 2 e e e i i 2 i i i 0 . v . c . v . c v . c                            

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               

(a) The coefficient of performance is:

10 . 3 h h h h 1 2 4 1     

(b) The refrigerating capacity is:

tons 564 . 4 min / kJ 211 ton 1 kg kJ ) 90 . 82 40 . 243 ( min kg 6 ) h h ( m Q_{in 1 4 }                 

(a) For the compressor: 1 2 comp

0 J j ) s s ( m T Q 0  _                Thus: kW 55 . 1 s / kJ 1 kW 1 kg kJ ) 9606 . 0 0135 . 1 ( s 60 min 1 min kg 6 ) K 294 ( ) s s ( m T T I_{comp o comp 0 2 1 }                                

Similarly, for the valve;

kW 5530 . 0 s / kJ 1 kW 1 kg kJ ) 3113 . 0 33011 . 0 ( s 60 min 1 min kg 6 ) K 294 ( ) s s ( m T Ivalve o 4 3                              

(d) The change in specific flow availability for refrigerant passing through the evaporator is:

9 kg / kJ 86 . 24 kgK kJ ) 33011 . 0 9606 . 0 ( K ) 294 ( kg kJ ) 90 . 82 40 . 243 ( ) s s ( T ) h h ( e e_f1  _f4  ₁  ₄  _{o 1}  ₄     

For the condenser:

 

kg kJ 78 . 5 kgK kJ ) 0135 . 1 3113 . 0 ( K 294 kgK kJ ) 13 . 295 90 . 82 ( ) s s ( T ) h h ( e e_f3  _f2  ₃ ₂  _{o 3}  ₂     

Comment: Although there is heat transfer to the refrigerant passing through the evaporator, the specific flow availability decreases. This can be explained by noting that the state of the working fluid moves closer to the dead state as it is heated at a temperature below To.

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3) An ideal vapor-compression heat pump cycle with Refrigerant 134a as the working fluid provides 15 kW to maintain a building at 200C when the outside temperature is 50C. Saturated vapor at 2.4 bar leaves the evaporator, and saturated liquid at 8 bar leaves the condenser. Calculate

(a) The power input to the compressor, in kW (b) The coefficient of performance.

(c) The coefficient of performance of a reversible heat pump cycle operating between thermal reservoirs at 20 and 50C

Solution:

Known: An ideal vapor-compression heat pump cycle uses Refrigerant 134a as the working fluid and provides a known energy output to heat a building. Data are known at various locations.

Schematic and Given Data:

Analysis: First, fix each of the principal states Compressor :

11 Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

) h h ( m W m m m m m h m h m W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2 c 2 1 e i 2 2 1 1 c e e 2 e e e i i 2 i i i . v . c . v . c v . c                            

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              2 1 2 1 2 2 1 1 cv i e e e i i j j j cv s s m m m s m s m 0 s m s m T Q dt dS          

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         Condenser: Assumptions:

 steady state steady flow process(SSSF)

 open system  ΔKE = ΔPE =0 out Q 2 3 Condenser c W 2 1

12  Neglect work (w = 0) ) h h ( m Q m m m m m h m h m Q 0 gz 2 V h m gz 2 V h m W Q dt dE 3 2 in 3 2 e i 3 3 2 2 out e e 2 e e e i i 2 i i i 0 . v . c . v . c v . c                             

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               

State 1: p₁ 2.4bars,saturatedvaporh₁ 244.09kJ/kg,s₁ 0.9222kJ/kg.K State 2: p₂ 8bars,s₂ s₁h₂ 268.97 kJ/kg

State 3: p₃ 8bars,saturatedliquid h₃ 93.42kJ/kg Expansion valve:

Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

 Neglect work (w=0)   4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 . v . c 0 . v . c v . c h h m m m m m h m h m 0 gz 2 V h m gz 2 V h m W Q dt dE                          

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           4 3

13 State 4: Throttlingprocessh3 h4 93.42 kJ/kg

(a) To determine the compressor, first find the mass flow rate from

kW 126 . 2 s / kJ 1 kW 1 kg kJ ) 09 . 244 97 . 268 ( s kg 08544 . 0 ) h h ( m W , Thus s / kg 08544 . 0 kW 1 s / kJ 1 kg kJ ) 42 . 93 97 . 268 ( kW 15 h h Q m or ) h h ( m Q 1 2 c 3 2 out 3 2 out                                    

(b) The coefficient of performance is

055 . 7 126 . 2 15 W Q c out     _

(c) For a reversible heat pump operating between reservoirs at

53 . 19 278 293 293 T T T K 278 T and K 293 T C H H max C H        

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4) A vapor-compression heat pump with a heating capacity of 500 kJ/min is driven by a power cycle with a thermal efficiency of 25%. For the heat pump, Refrigerant 134a is compressed from saturated vapor at -100C to the condenser pressure of 10 bar. The isentropic compressor efficiency is 80%. Liquid enters the expansion valve at 9.6 bar, 340C. For the power cycle, 80% of the heat rejected is transferred to the heated space. (a) Determine the power input to the heat pump compressor, in kW.

(b) Evaluate the ratio of the total rate that heat is delivered to the heated space to the rate of heat input to the power cycle. Discuss.

Solution:

Known: Refrigerant 134a is the working fluid in a vapor-compression heat pump driven by a power cycle. Operating data are specified for the heat pump and the power cycle.

Analysis: First, fix each state of the heat pump cycle. Compressor:

15 Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

 

 

 

W m(h h ) ) h h ( m W m m m m m h m h m W 0 gz 2 V h m gz 2 V h m W Q dt dE 1 2 a c 1 s 2 s c 2 1 e i s 2 2 1 1 s c e e 2 e e e i i 2 i i i . v . c . v . c v . c                              

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               

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2 1



1 s 2 a c s c h h h h ) W ( ) W (      _ Evaporator: Assumptions:

 steady state steady flow process(SSSF)

 open system  ΔKE = ΔPE =0  Neglect work (w = 0) in Q 4 1 Evaporator 2 1 net W

16 ) h h ( m Q m m m m m h m h m Q 0 gz 2 V h m gz 2 V h m W Q dt dE 4 1 in 4 1 e i 1 1 4 4 in e e 2 e e e i i 2 i i i 0 . v . c . v . c v . c                            

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                Expansion valve: Assumptions:

 steady state steady flow process(SSSF)

 open system

 ΔKE = ΔPE =0

 Neglect heat transfer (q=0)

 Neglect work (w=0)   4 3 4 3 e i 4 4 3 3 e e 2 e e e i i 2 i i i 0 . v . c 0 . v . c v . c h h m m m m m h m h m 0 gz 2 V h m gz 2 V h m W Q dt dE                          

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           Condenser: 4 3

17 Assumptions:

 steady state steady flow process(SSSF)

 open system  ΔKE = ΔPE =0  Neglect work (w = 0) ) h h ( m Q m m m m m h m h m Q 0 gz 2 V h m gz 2 V h m W Q dt dE 3 2 in 3 2 e i 2 2 3 3 in e e 2 e e e i i 2 i i i 0 . v . c . v . c v . c                            

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               

State 1: T 10 C,saturatedvapor h1 241.34 kJ/kg,s1 0.9253kJ7kg.K 0

1   

State 2: For isentropic compression, p₂ 10 bars,s_2s s₁h_2s 274.63kJ/kg Using the compressor efficiency

kg / kJ 95 . 282 8 . 0 ) kg / kJ 24 . 241 ( ) kg / kJ 63 . 274 ( kg kJ 34 . 241 h h h h h h h h c 1 s 2 1 2 1 2 1 s 2 c             

State 3: p 9.6bars,T 34 C compressedliquid;h3 hf(340C) 97.31kJ/kg 0

3

3     

State 4: Throttling processh4 h3 97.31kJ/kg (a) The mass flow rate of refrigerant is

s / kg 04489 . 0 s 60 min 1 kg / kJ ) 31 . 97 95 . 282 ( min / kJ 500 h h Q m 3 2 1 , out _            

The compressor power becomes in

Q

3 2 Condenser

18 kW 868 . 1 s / kJ 1 kW 1 kg kJ ) 34 . 241 95 . 282 ( s kg 04489 . 0 ) h h ( m Wc 2 1                   

(b) For the power cycle, _c 0.25. With W_powercycle W _heatpump 1.868kW

472 . 7 W Qin,2 cycle     

The total rejected is,

kW 483 . 4 Q ) 8 . 0 ( Q , Thus kW 604 . 5 kW 868 . 1 kW 472 . 7 Q rej 2 , out rej         Finally, 1.715 kW 472 . 7 kW 483 . 4 kJ 1 kW 1 s 60 min 1 min kJ 500 Q Q Q 2 , in 2 , out 1 , out _            

COMMENT: The engine-driven heat pump delivers more energy to the heated space than could be obtained by burning fuel directly.

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5) Air enters the compressor of an ideal Brayton refrigeration cycle at 140 kPa, 270 K, with a volumetric flow rate of 1 m3/s, and is compressed to 420 kPa. The temperature at the turbine inlet is 320 K. Determine

(a) The net power input, in kW (b) The refrigerating capacity, in kW. (c) The coefficient of performance.

(d) The coefficient of performance of a reversible refrigeration cycle operating between reservoirs at TC = 270 K and TH = 320 K.

Solution:

Known: Air is the refrigerant in an ideal Brayton refrigeration cycle. Data are known at various locations and the volumetric flow rate at the compressor inlet is given.

Schematic and Given Data:

Analysis: First, fix each of the principal states

20 State 2: 2.877 h 370.10 kJ/kg kPa 140 kPa 420 9590 . 0 p p p p 2 1 2 1 r 2 r                  State 3: T3 320Kh3 320.29 kJ/kg,pr3 1.7375 State 4: 0.5792 h 233.61kJ/kg kPa 420 kPa 140 7375 . 1 p p p p 4 3 4 3 r 4 r                

(a) The mass flow rate is

s / kg 807 . 1 m . N 10 kJ 1 kPa 1 m / N 10 ) K 270 ( K . kg 97 . 28 kJ 314 . 8 ) kPa 140 )( s / m 1 ( RT p ) A ( v ) A ( m ₃ 2 3 3 1 1 1 1 1                         

Thus, the net power is









kW 05 . 24 s / kJ 1 kW 1 kg kJ ) 61 . 233 29 . 320 ( ) 11 . 270 1 . 370 ( s kg 807 . 1 ) h h ( ) h h ( m W W W W 4 3 1 2 cycle p t cycle                            

(b) The refrigerating capacity is

kW 96 . 65 kg kJ ) 61 . 233 11 . 270 ( s kg 807 . 1 ) h h ( m Q_{in 1 4}             

(c) The coefficient of performance is

743 . 2 05 . 24 96 . 65 W Q cycle in     _

(d) For a reversible cycle operating between thermal reservoirs at 270 K and T_H 320K is 4 . 5 270 320 270 T T T C H C max  _  _  