Chap.10.Soln

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. Schematic & Given Data:

1

Analysis: First, fix each of the principle states.

Compressor: 2  W c

1 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)

    dE c.v  Ve2 Vi2   i  h i   e  h e   Q c.v.  Wc.v.   m  gz i    m  gz e  dt 2 2 i   e    m  h m  h 0W c

1

 m  m i

1

2

2

e

1 m 2 m  m  m  (h 2  h 1 ) W c

2

 Q dS cv j  isi   m  e s e   cv   m dt j Tj i e  1s 1  m  2s 2 0m 1 m 2 m  m s 2  s1 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: 3

4 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)  Neglect work (w=0)

    dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v . i i i   e e   dt 2 2 i   e   0 0  3h 3  m  4h 4 0m

 m  m i

e

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

3

Evaporator:

4

1

 Q in Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect work (w = 0)     dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v. i i i  e e  i dt 2 2   e   0

 m  4h 4  m  1h 1 0Q in

 m  m i

e

1 m 4 m  m  m  (h 1  h 4 ) Q in (a) The compressor power is:

kg  1 min  kJ  1 kW   m  (h 2  h 1 )   5 (267.9  241.35)    2.212 kW W  c kg  1 kJ / s   min  60 s  (b) The refrigeration capacity is:

 1 ton kg  kJ   m  (h 1  h 4 )   5   3.62 tons Q (241.35  88.61)  in kg  211 kJ / min   min  (c) The coefficient of performance is:



 Q h  h 4 (241.35  88.61)kJ / kg in  1   5.75  Wc h 2  h 1 (267.9  241.35)kJ / kg

4

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) (b) (c) (d)

The coefficient of performance. The refrigerating capacity, in tons. The irreversibility rates of the compressor and expansion valve, each in kW 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:

Analysis: First, fix each of the principal states. 5

Compressor: 2  W c

1 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)

    dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v . i i i   e e dt 2 2 i   e     h m  h 0W m c

1

 m  m i

1

2

2s

e

1 m 2 m  m  m  (h 2s  h 1 ) W c  Q dS cv j  isi   m  e s e   cv   m dt j Tj i e  1s 1  m  2 s 2s 0m 1 m 2 m  m s 2s  s1 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,

C 

h 2s  h 1 h  h1 278.06  243.40  h 2  h 1  2s  243.40   295.13 kJ / kg h 2  h1 c 0.67

s2 = 1.0135 kJ/kg.K 6

State 3: P3 = 7 bars, T = 24oC  h3 = hf@24oC = 82.90 kJ/kg, s3 = 0.3113 kJ/kg.K Expansion valve: 3

4 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)  Neglect work (w=0)

    dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v . i i i   e e   dt 2 2 i   e   0 0  3h 3  m  4h 4 0m

 m  m i

e

3 m 4 m  m h3  h4 State 4: Throttling process  h4 = h3 = 82.90 kJ/kg, s4 = 0.33011 kJ/kg.K Evaporator:

4

1

 Q in Assumptions: 7

   

steady state steady flow process(SSSF) open system ΔKE = ΔPE =0 Neglect work (w = 0)     Ve2 Vi2         Q c.v.  Wc.v.   m i  h i   gz i    m e  h e   gz e   i 2 2   e  

dE c.v dt

0

 m  4h 4  m  1h 1 0Q in

 m  m i

e

1 m 4 m  m  m  (h 1  h 4 ) Q in (a) The coefficient of performance is:



h1  h 4  3.10 h 2  h1

(b) The refrigerating capacity is:

 1 ton kg  kJ   m  (h 1  h 4 )   6   4.564 tons Q (243.40  82.90)  in kg  211 kJ / min   min   Q  (s1  s 2 )   comp (a) For the compressor: 0      m J  T j    0

Thus:

 kJ  1 kW   kg  1 min  I   (1.0135  0.9606)    1.55 kW  comp  To  comp  T0 m(s 2  s1 )  ( 294K ) 6  min  60 s   kg  1 kJ / s  Similarly, for the valve;

 kJ  1 kW   kg  1 min  I  (0.33011  0.3113)    0.5530 kW  valve  To m(s 4  s 3 )  ( 294K ) 6  min  60 s   kg  1 kJ / s 

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

8

e f 1  e f 4  (h 1  h 4 )  To (s1  s 4 )  (243.40  82.90)

kJ kJ  (294)K(0.9606  0.33011)  24.86 kJ / kg kg kgK

For the condenser:

e f 3  e f 2  (h 3  h 2 )  To (s 3  s 2 )  (82.90  295.13)

kJ kJ kJ  294K(0.3113  1.0135)  5.78 kgK kgK kg

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.

9

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 : 10

2

 W c

1 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)

    dE c.v  Ve2 Vi2        Q c.v.  Wc.v.   m i  h i   gz i    m e  h e   gz e  dt 2 2 i   e     h m  h 0W m c

1

 m  m i

1

2

2

e

1 m 2 m  m  m  (h 2  h 1 ) W c  Q dS cv j  isi   m  e s e   cv   m dt j Tj i e  1s 1  m  2s 2 0m 1 m 2 m  m s1  s 2 Condenser:

 Q out 3

Condenser

2

Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0 11



Neglect work (w = 0)     dE c.v  Ve2 Vi2        Q c.v.  Wc.v.   m i  h i   gz i    m e  h e   gz e     dt 2 2 i   e   0  m  h m  h 0  Q out

 m i  m e

2

2

3

3

2 m 3 m  m  m  (h 2  h 3 ) Q in State 1: p1  2.4 bars, saturated vapor  h1  244.09 kJ / kg, s1  0.9222 kJ / kg.K State 2: p 2  8 bars, s 2  s1  h 2  268.97 kJ / kg State 3: p 3  8 bars, saturated liquid  h 3  93.42 kJ / kg Expansion valve: 3

4 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)  Neglect work (w=0)

    dE c.v  Ve2 Vi2    i  h i   e he   Q c.v.  Wc.v.   m  gz i    m  gz e    dt 2 2 i   e   0 0  3h 3  m  4h 4 0m

 m  m i

e

3 m 4 m  m h3  h4 12

State 4: Throttling process  h 3  h 4  93.42 kJ / kg (a) To determine the compressor, first find the mass flow rate from

 m  (h 2  h 3 ) Q out or   m

 Q out  h2  h3

 1 kJ / s     0.08544 kg / s kJ  1 kW  (268.97  93.42) kg 15 kW

Thus, kg kJ  1 kW   m  (h 2  h 1 )   0.08544 (268.97  244.09)   W c s  kg  1 kJ / s    2.126 kW (b) The coefficient of performance is



 Q 15 out   7.055  2.126 Wc

(c) For a reversible heat pump operating between reservoirs at TH  293 K and TC  278 K  max 

TH 293   19.53 TH  TC 293  278

13

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:

14

2

 W net

1 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)

    dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v . i i i   e e dt 2 2 i   e     m  h m  h 0  W c s

1

 m  m i

1

2

2s

e

1 m 2 m  m W c s  m (h 2s  h 1 ) W   m (h  h ) c a



2

1

 ) h  h 1  (W c s  2s  ) h 2  h 1  (W c a

Evaporator:

1

Evaporator

4

 Q in Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect work (w = 0) 15

    dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v. i i i  e e  i dt 2 2   e   0

 m  4h 4  m  1h 1 0Q in

 m  m i

e

1 m 4 m  m  m  (h 1  h 4 ) Q in Expansion valve: 3

4 Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect heat transfer (q=0)  Neglect work (w=0)

    dE c.v  Ve2 Vi2      i hi   ehe   Q c.v.  Wc.v.   m  gz i    m  gz e    dt 2 2 i   e   0 0  3h 3  m  4h 4 0m

 m  m i

e

3 m 4 m  m h3  h4 Condenser:

16

2

Condenser

3

 Q in Assumptions:  steady state steady flow process(SSSF)  open system  ΔKE = ΔPE =0  Neglect work (w = 0)     dE c.v  Ve2 Vi2   m       Q c. v .  W h   gz  m h   gz e  c. v. i i i  e e  i dt 2 2   e   0

 m  3h 3  m  2h 2 0Q in

 m  m i

e

2 m 3 m  m  m  (h 2  h 3 ) Q in State 1: T1  10 0 C, saturated vapor  h1  241.34 kJ / kg, s1  0.9253 kJ7kg.K State 2: For isentropic compression, p 2  10 bars, s 2s  s1  h 2s  274.63 kJ / kg Using the compressor efficiency

c 

h 2s  h 1 h  h1 kJ (274.63 kJ / kg)  (241.24 kJ / kg)  h 2  h 1  2s  241.34   282.95 kJ / kg h 2  h1 c kg 0.8

State 3: p 3  9.6 bars, T3  34 0 C  compressed liquid ; h 3  h f (34 0 C)  97.31 kJ / kg State 4: Throttling process  h 4  h 3  97.31 kJ / kg (a) The mass flow rate of refrigerant is

 Q out,1

 1 min  500 kJ / min    0.04489 kg / s h 2  h 3 (282.95  97.31) kJ / kg  60 s  The compressor power becomes   m



17

kg  kJ  1 kW   m  (h 2  h 1 )   0.04489  W (282.95  241.34)  c s  kg  1 kJ / s    1.868 kW   W  1.868 kW (b) For the power cycle,   0.25. With W c

 Q in , 2 

power cycle

heat pump

 W cycle

 7.472  The total rejected is,

  7.472 kW  1.868 kW  5.604 kW Q rej Thus,   Q  (0.8)Q out, 2

rej

 4.483 kW

 kJ 1 min 1 kW   500   4.483kW   Q out,1  Q out, 2  min 60s 1 kJ    1.715 Finally,  7.472 kW Q in , 2

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

18

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 State 1: T1  270 K  h1  270.11 kJ / kg, p r1  0.9590

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p   420 kPa    2.877  h 2  370.10 kJ / kg State 2: p r 2  p r1  2   0.9590    140 kPa   p1  State 3: T3  320 K  h 3  320.29 kJ / kg, p r 3  1.7375 p   140 kPa    0.5792  h 4  233.61 kJ / kg State 4: p r 4  p r 3  4   1.7375  420 kPa   p3  (a) The mass flow rate is (A)1 (A)1 p1 (1 m 3 / s)(140 kPa)  10 3 N / m 2  1 kJ    3   1.807 kg / s   v1 RT1 1 kPa 10 N . m  8.314 kJ     (270 K )  28 . 97 kg . K   Thus, the net power is   m

   W cycle  Wt  Wp   W cycle  m( h 2  h 1 )  ( h 3  h 4 )  kg  kJ  1 kW     1.807 (370.1  270.11)  (320.29  233.61)  s  kg  1 kJ / s    24.05 kW (b) The refrigerating capacity is kg kJ  m  (h 1  h 4 )  1.807 (270.11  233.61) Q  65.96 kW in s  kg 

(c) The coefficient of performance is



 Q 65.96 in   2.743  24.05 W cycle

(d) For a reversible cycle operating between thermal reservoirs at 270 K and TH  320 K is

 max 

TC 270   5.4 TH  TC 320  270

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