THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I
8th Edition
ISBN: 9781307434316
Author: CENGEL
Publisher: INTER MCG
bartleby

Videos

Question
Book Icon
Chapter 11.10, Problem 117RP
To determine

The greatest amount of exergy destroyed among the processes of system.

Expert Solution & Answer
Check Mark

Answer to Problem 117RP

The greatest amount of exergy destroyed among the processes of system is 30.94kW.

Explanation of Solution

Sketch the schematic layout of the two-stage compression refrigeration system as in Figure (1).

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I, Chapter 11.10, Problem 117RP , additional homework tip  1

Sketch the Ts diagram of the two-stage compression refrigeration as in Figure (2).

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I, Chapter 11.10, Problem 117RP , additional homework tip  2

Write the relation between the specific enthalpies at the inlet and exit of throttling.

hi=he (I)

Here, specific enthalpy of the refrigerant at the inlet of throttling is hi and specific enthalpy of the refrigerant at the exit of throttling is he.

Write the energy balance equation for the separator.

m˙6(h8h5)=m˙2(h1h4)

m˙6=m˙2h1h4h8h5 (II)

Here, mass flow rate of the refrigerant at the end of high pressure compressor is m˙2, mass flow rate at the inlet of evaporator is m˙6, specific enthalpy of refrigerant at the compressor inlet of high pressure cycle is h1, specific enthalpy at the exit of expansion valve at high pressure cycle is h4, specific enthalpy at the exit of compressor at the low pressure cycle is h8, and the specific enthalpy at the inlet of expansion valve at the low pressure cycle is h5.

Write the expression to calculate the quality of the refrigerant (x) from specific enthalpy of saturated refrigerant.

x=hhfhfg (III)

Here, specific enthalpy of the saturated liquid is hf and the specific enthalpy of the saturated refrigerant is hfg.

Write the expression to calculate the specific entropy for saturated refrigerant (s) from quality.

s=sf+xhfg (IV)

Here, specific entropy of the saturated liquid is sf and the specific entropy of the saturated refrigerant is sfg.

Write the expression to calculate the rate of cooling produced (qL) by the two-stage compression refrigeration.

qL=h7h6 (V)

Here, specific enthalpy at the inlet of low pressure compressor is h7 and the specific enthalpy at the inlet evaporator is h6.

Write the expression to calculate the rate of heat rejected from the system (qH).

qH=h2h3 (VI)

Here, specific enthalpy at the exit of high pressure compressor is h2, and specific enthalpy at high pressure throttle inlet is h3.

Write the general expression to calculate the exergy destruction (X˙dest) during a process.

X˙dest=m˙T0sgen

X˙dest=m˙T0(sesiqinTsource+qoutTsink) (VII)

Here, mass flow rate is m˙, surrounding temperature is T0, specific entropy generated is sgen, specific entropy at the exit is se, specific entropy at the inlet is si, heat input to the process is qin, source temperature is Tsource, heat output from the process is qout, and the sink temperature is Tsink.

Conclusion:

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at high pressure compressor inlet temperature (T1) of 8.9°C as follows:

h1=hg=255.60kJ/kgs1=sg=0.92711kJ/kgK

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of high pressure compression exit at pressure (P2) of 1400kPa and specific entropy (s2) of 0.92711kJ/kgK as 281.56kJ/kg.

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle inlet pressure of 1400 kPa as follows:

h3=hf=127.25kJ/kgs3=sf=0.45325kJ/kgK

Substitute 127.25kJ/kg for h3 in Equation (I).

h4=127.25kJ/kg

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at high pressure cycle throttle exit temperature of 8.9°C as follows:

hf=63.85kJ/kghfg=191.69kJ/kgsf=0.24751kJ/kgKsfg=0.6796kJ/kgK

Substitute 63.85kJ/kg for hf, 191.69kJ/kg for hfg, and 127.25kJ/kg for h4 in Equation (III).

x4=127.25kJ/kg63.85kJ/kg191.69kJ/kgx4=0.3307

Substitute 0.24751kJ/kgK for sf, 0.6796kJ/kgK for sfg, and 0.3307 for x4 in Equation (IV).

s4=0.24751kJ/kgK+0.3307(0.6796kJ/kgK)s4=0.47225kJ/kgK

From the Table A-12 of “Saturated refrigerant R-134a: Pressure”, obtain the properties of refrigerant at low pressure cycle throttle inlet temperature of 8.9°C as follows:

P5=Psat=400kPah5=hf=63.91kJ/kgs5=sf=0.24752kJ/kgK

Here, pressure of refrigerant at the separator is P5.

Substitute 63.91kJ/kg for h5 in Equation (I).

h6=63.91kJ/kg

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure expansion valve exit temperature (T6) of 32°C as follows:

hf=10.09kJ/kghfg=220.83kJ/kgsf=0.04249kJ/kgKsfg=0.9157kJ/kgK

Substitute 63.91kJ/kg for h6, 10.09kJ/kg for hf, and 220.83kJ/kg for hfg in Equation (III).

x6=63.9110.09220.83=0.2437

Substitute 0.2437 for x6, 0.04249kJ/kgK for sf, and 0.95819kJ/kgK for sg in Equation (IV).

s6=0.04249kJ/kgK+(0.2437)0.9157kJ/kgK=0.26559kJ/kgK

From the Table A-11 of “Saturated refrigerant R-134a: Temperature”, obtain the properties of refrigerant at low pressure compressor inlet temperature (T7) of 32°C as follows:

h7=hg=230.93kJ/kgs7=sg=0.95819kJ/kgK

From the Table A-13 of “Superheated refrigerant R-134a”, obtain the specific enthalpy of low pressure compression exit at pressure (P8) of 400kPa and specific entropy (s8)  of 0.95819kJ/kgK as 264.51kJ/kg.

Substitute 2kg/s for m˙2, 255.60kJ/kg for h1, 127.25kJ/kg for h4, 264.51kJ/kg for h8, and 63.91 kJ/kg for h5 in Equation (II).

m˙6=2kg/s×255.60kJ/kg127.25kJ/kg264.51kJ/kg63.91 kJ/kgm˙6=1.280kg/s

Substitute 230.93kJ/kg for h7, and 63.91kJ/kg for h6 in Equation (V).

qL=230.93kJ/kg63.91kJ/kgqL=167.02kJ/kg

Substitute 127.25kJ/kg for h3 and 281.56kJ/kg for h2 in Equation (VI).

qH=281.56kJ/kg127.25kJ/kgqH=154.31kJ/kg

Rewrite the Equation (VII) for the process 2 to 3.

X˙dest,2-3=m˙2T0(s3s2+qHTH) (VIII)

Here, the temperature of the high temperature reservoir is TH.

Substitute 2kg/s for m˙2, 20°C for T0, 0.45325kJ/kgK for s3, 0.92711kJ/kgK for s2, 154.31kJ/kg for qH, and 20°C for TH in Equation (VIII).

X˙dest,2-3=2kg/s×20°C×(0.45325kJ/kgK0.92711kJ/kgK+154.31kJ/kg20°C)=2kg/s×(20+273)K×(0.45325kJ/kgK0.92711kJ/kgK+154.31kJ/kg(20+273)K)=30.94kW

Rewrite the Equation (VII) for the process 3 to 4.

X˙dest,3-4=m˙2T0(s4s3) (IX)

Substitute 2kg/s for m˙2, 20°C for T0, 0.45325kJ/kgK for s3, and 0.47225kJ/kgK for s4 in Equation (IX).

X˙dest,3-4=2kg/s×20°C×(0.47225kJ/kgK0.45325kJ/kgK)=2kg/s×(20+273)K×(0.47225kJ/kgK0.45325kJ/kgK)=11.03kW

Rewrite the Equation (VII) for the process 5 to 6.

X˙dest,5-6=m˙6T0(s6s5) (X)

Substitute 1.280kg/s for m˙6, 20°C for T0, 0.26559kJ/kgK for s6, and 0.24752kJ/kgK for s5 in Equation (X).

X˙dest,5-6=1.280kg/s×20°C×(0.26559kJ/kgK0.24752kJ/kgK)=1.280kg/s×(20+273)K×(0.26559kJ/kgK0.24752kJ/kgK)=6.793kW

Rewrite the Equation (VII) for the process 6 to 7.

X˙dest,6-7=m˙6T0(s7s6qLTL) (XI)

Substitute 1.280kg/s for m˙6, 20°C for T0, 0.26559kJ/kgK for s6, and 0.95819kJ/kgK for s7, 167.02kJ/kg for qL, and 22°C for TL in Equation (XI).

X˙dest,6-7=1.280kg/s×20°C×(0.95819kJ/kgK0.26559kJ/kgK167.02kJ/kg(22°C))=1.280kg/s×20°C×(0.95819kJ/kgK0.26559kJ/kgK167.02kJ/kg(22+273)K)=10.18kW

Rewrite the Equation (VII) for the separator process.

X˙dest,separator=T0[m˙6(s5s8)m˙2(s1s4)] (XII)

Substitute 1.280kg/s for m˙6, 20°C for T0, 0.24752kJ/kgK for s5, 0.95819kJ/kg for s8, 2kg/s for m˙2, 0.47225kJ/kgK for s4, and 0.92711kJ/kgK for s1 in Equation (XII).

X˙dest,separator=20°C[1.280kg/s(0.24752kJ/kgK0.95819kJ/kg)2kg/s(0.92711kJ/kgK0.47225kJ/kgK)]=(20+273)K[1.280kg/s(0.24752kJ/kgK0.95819kJ/kg)2kg/s(0.92711kJ/kgK0.47225kJ/kgK)]=0.197kW

The exergy destruction for the isentropic processes is zero. Hence,

X˙dest,1-2=0X˙dest,7-8=0

Thus, the greatest amount of exergy destroyed among the processes of system is 30.94kW.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Chapter 11 Solutions

THERMODYNAMICS(SI UNITS,INTL.ED)EBOOK>I

Ch. 11.10 - An ice-making machine operates on the ideal...Ch. 11.10 - A 10-kW cooling load is to be served by operating...Ch. 11.10 - 11–13 An ideal vapor-compression refrigeration...Ch. 11.10 - 11–14 Consider a 300 kJ/min refrigeration system...Ch. 11.10 - 11–16 Repeat Prob. 11–14 assuming an isentropic...Ch. 11.10 - 11–17 Refrigerant-134a enters the compressor of a...Ch. 11.10 - A commercial refrigerator with refrigerant-134a as...Ch. 11.10 - 11–19 Refrigcrant-134a enters the compressor of a...Ch. 11.10 - A refrigerator uses refrigerant-134a as the...Ch. 11.10 - The manufacturer of an air conditioner claims a...Ch. 11.10 - Prob. 23PCh. 11.10 - How is the second-law efficiency of a refrigerator...Ch. 11.10 - Prob. 25PCh. 11.10 - Prob. 26PCh. 11.10 - Prob. 27PCh. 11.10 - 11–28 Bananas are to be cooled from 28°C to 12°C...Ch. 11.10 - A vapor-compression refrigeration system absorbs...Ch. 11.10 - A refrigerator operating on the vapor-compression...Ch. 11.10 - A room is kept at 5C by a vapor-compression...Ch. 11.10 - Prob. 32PCh. 11.10 - 11–33 A refrigeration system operates on the ideal...Ch. 11.10 - When selecting a refrigerant for a certain...Ch. 11.10 - Consider a refrigeration system using...Ch. 11.10 - A refrigerant-134a refrigerator is to maintain the...Ch. 11.10 - A refrigerator that operates on the ideal...Ch. 11.10 - A heat pump that operates on the ideal...Ch. 11.10 - Do you think a heat pump system will be more...Ch. 11.10 - What is a water-source heat pump? How does the COP...Ch. 11.10 - Prob. 42PCh. 11.10 - Refrigerant-134a enters the condenser of a...Ch. 11.10 - Prob. 45PCh. 11.10 - A heat pump using refrigerant-134a heats a house...Ch. 11.10 - How does the COP of a cascade refrigeration system...Ch. 11.10 - A certain application requires maintaining the...Ch. 11.10 - Consider a two-stage cascade refrigeration cycle...Ch. 11.10 - Can a vapor-compression refrigeration system with...Ch. 11.10 - Prob. 52PCh. 11.10 - Prob. 53PCh. 11.10 - Repeat Prob. 1156 for a flash chamber pressure of...Ch. 11.10 - Prob. 56PCh. 11.10 - Prob. 57PCh. 11.10 - 11–58 Consider a two-stage cascade refrigeration...Ch. 11.10 - Prob. 59PCh. 11.10 - A two-evaporator compression refrigeration system...Ch. 11.10 - A two-evaporator compression refrigeration system...Ch. 11.10 - Repeat Prob. 1163E if the 30 psia evaporator is to...Ch. 11.10 - How does the ideal gas refrigeration cycle differ...Ch. 11.10 - Devise a refrigeration cycle that works on the...Ch. 11.10 - How is the ideal gas refrigeration cycle modified...Ch. 11.10 - Prob. 66PCh. 11.10 - How do we achieve very low temperatures with gas...Ch. 11.10 - 11–68E Air enters the compressor of an ideal gas...Ch. 11.10 - Prob. 69PCh. 11.10 - Air enters the compressor of an ideal gas...Ch. 11.10 - Repeat Prob. 1173 for a compressor isentropic...Ch. 11.10 - Prob. 73PCh. 11.10 - Prob. 74PCh. 11.10 - Prob. 75PCh. 11.10 - A gas refrigeration system using air as the...Ch. 11.10 - An ideal gas refrigeration system with two stages...Ch. 11.10 - Prob. 78PCh. 11.10 - Prob. 79PCh. 11.10 - What are the advantages and disadvantages of...Ch. 11.10 - Prob. 81PCh. 11.10 - Prob. 82PCh. 11.10 - An absorption refrigeration system that receives...Ch. 11.10 - An absorption refrigeration system receives heat...Ch. 11.10 - Heat is supplied to an absorption refrigeration...Ch. 11.10 - Prob. 86PCh. 11.10 - Prob. 87PCh. 11.10 - Prob. 88PCh. 11.10 - Prob. 89PCh. 11.10 - Consider a circular copper wire formed by...Ch. 11.10 - An iron wire and a constantan wire are formed into...Ch. 11.10 - Prob. 92PCh. 11.10 - Prob. 93PCh. 11.10 - Prob. 94PCh. 11.10 - Prob. 95PCh. 11.10 - Prob. 96PCh. 11.10 - Prob. 97PCh. 11.10 - Prob. 98PCh. 11.10 - A thermoelectric cooler has a COP of 0.18, and the...Ch. 11.10 - Prob. 100PCh. 11.10 - Prob. 101PCh. 11.10 - Prob. 102PCh. 11.10 - Prob. 103RPCh. 11.10 - Prob. 104RPCh. 11.10 - Prob. 105RPCh. 11.10 - A heat pump that operates on the ideal...Ch. 11.10 - A large refrigeration plant is to be maintained at...Ch. 11.10 - Repeat Prob. 11112 assuming the compressor has an...Ch. 11.10 - A heat pump operates on the ideal...Ch. 11.10 - An air conditioner with refrigerant-134a as the...Ch. 11.10 - An air conditioner operates on the...Ch. 11.10 - Consider a two-stage compression refrigeration...Ch. 11.10 - A two-evaporator compression refrigeration system...Ch. 11.10 - Prob. 116RPCh. 11.10 - Prob. 117RPCh. 11.10 - Prob. 118RPCh. 11.10 - Consider a regenerative gas refrigeration cycle...Ch. 11.10 - Prob. 120RPCh. 11.10 - The refrigeration system of Fig. P11122 is another...Ch. 11.10 - Repeat Prob. 11122 if the heat exchanger provides...Ch. 11.10 - An ideal gas refrigeration system with three...Ch. 11.10 - Derive a relation for the COP of the two-stage...Ch. 11.10 - Prob. 129FEPCh. 11.10 - Prob. 130FEPCh. 11.10 - Prob. 131FEPCh. 11.10 - Prob. 132FEPCh. 11.10 - An ideal vapor-compression refrigeration cycle...Ch. 11.10 - Prob. 134FEPCh. 11.10 - An ideal gas refrigeration cycle using air as the...Ch. 11.10 - Prob. 136FEPCh. 11.10 - Prob. 137FEPCh. 11.10 - Prob. 138FEP
Knowledge Booster
Background pattern image
Mechanical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, mechanical-engineering and related others by exploring similar questions and additional content below.
Recommended textbooks for you
Text book image
Elements Of Electromagnetics
Mechanical Engineering
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Oxford University Press
Text book image
Mechanics of Materials (10th Edition)
Mechanical Engineering
ISBN:9780134319650
Author:Russell C. Hibbeler
Publisher:PEARSON
Text book image
Thermodynamics: An Engineering Approach
Mechanical Engineering
ISBN:9781259822674
Author:Yunus A. Cengel Dr., Michael A. Boles
Publisher:McGraw-Hill Education
Text book image
Control Systems Engineering
Mechanical Engineering
ISBN:9781118170519
Author:Norman S. Nise
Publisher:WILEY
Text book image
Mechanics of Materials (MindTap Course List)
Mechanical Engineering
ISBN:9781337093347
Author:Barry J. Goodno, James M. Gere
Publisher:Cengage Learning
Text book image
Engineering Mechanics: Statics
Mechanical Engineering
ISBN:9781118807330
Author:James L. Meriam, L. G. Kraige, J. N. Bolton
Publisher:WILEY
Power Plant Explained | Working Principles; Author: RealPars;https://www.youtube.com/watch?v=HGVDu1z5YQ8;License: Standard YouTube License, CC-BY