Module 4. Refrigeration and Heat Pump Systems The Vapor ...

K. Nasr,c:\thermo \module4.doc 

Module 4. Refrigeration and Heat Pump Systems 

Objective: Maintains a certain physical space at a cold temperature for a refrigeration system (cooling) 

and at a high temperature for a heat pump system (heating). 

Concept Support: 

Coefficient of Performance for Refrigerators & Heat Pumps: 

• In a household refrigerator, the freezer compartment where heat is picked up by the refrigerant 

serves as the evaporator, and the coils behind the refrigerator where heat is rejected to the kitchen 

air serves as the condenser. 

• The Refrigerator: Q C → Q H , where W in is input, by maintaining the low T region at a low T: 

β = COP R = Q C /W net,in = Q C /(Q H - Q C ); COP R can be > 1 

• The Heat Pump: Q C → Q H , where W in is input, by maintaining the high T region at a high T: 

γ = COP HP = Q H /W net,in = Q H /(Q H - Q C ); COP HP is always > 1 

• Note that for fixed values of Q H and Q C : COP HP = COP R + 1 or γ = β + 1 

• Most HP’s in operation have an average-season value of COP between 2 and 3. 

The Vapor Refrigeration Cycle: 

• Modeling: The system combines a Compressor, a Condenser, an Expansion Device, and an 

Evaporator. 

Copyright © 2004, K. Nasr. All Rights Reserved Page 1


A Simplified Vapor-Compression Refrigeration Cycle (Carnot - left and Ideal - right) 

• Heat is transferred to the working fluid (refrigerant) in the evaporator and then compressed by the 

compressor. Heat is transferred from the working fluid in the condenser, and then its pressure is 

suddenly reduced in the expansion valve. A refrigeration cycle is used to extract energy from a fluid 

(air) in contact with the evaporator. 

• It’s normally assumed that kinetic and potential energy are negligible, and that the expansion process 

through the valve is a throttling process. 

¯ 

Carnot Operation: 

1. Isentropic Compression 

2. Constant Pressure Heat Rejection 

3. Isentropic Expansion 

4. Constant Pressure Heat Absorption 

• A Carnot cycle is the most efficient, however it is not practical because: 

(I) The compressor would experience wear due to liquid droplets since state (1) is a liquid-vapor 

mixture. 

(II) It is difficult to maintain equilibrium between liquid and vapor if we do compress 

(III) the isentropic expansion device, involving no losses (friction), is expensive to construct. 

⊕ 

Ideal Cycle Operation: Corrects for the drawbacks of Carnot operation by having: 

1. Inlet to compressor is saturated vapor 

2. exit of compressor is superheated vapor 

3. An irreversible throttling process as an expansion process. 

• The performance of the refrigeration cycle is quantified by the coefficient of performance: 

COP = Q evap . / W comp . 

¯ Applications of the Open System 1st and 2nd Laws to the Various Devices: 

• (1st Law): dE . . . 

2 

. 

2 

C.V. 

Vi 

Ve 

= QC.V. 

− WC V. . + ∑ m i(hi 

+ + gz 

i) − ∑ m e(he 

+ + gz 

e 

) 

dt 

2 

2 

• (2nd Law): dS dt 

QCV 

= + σ gen + mi 

si 

− 

T 

C. V. . . 

C. V. 

. 

i 

. . . 

∑ ∑ ∑ 

i 

e 

me 

s 

e 

e 


Assumptions: SSSF, Negligible ∆KE and ∆PE, Single-Inlet, Single-Outlet conditions 


Process Device 1st Law 2nd Law 

. 

. 

1→2: Isentropic Compression Compressor 

WComp . = m(h1 − h 

2) 

s 2 = s 1 

. 

. 

2 →3: p = C. Heat Rejection Condenser 

QCond. 

= m(h3 − h 

2) 

s 2 > s 3 

3 →4: Adiabatic Throttling Throttling Device h 3 = h 4 s 4 > s 3 

(O-Tube, TXV) 

. 

. 

4 →1: p = C. Heat Absorption Evaporator 

Q = m(h − h ) 

s 1 > s 4 

COP = Q Evap . / W Comp . 

Evap. 

1 4 

¯ 

Actual Cycle Operation: Deviates from ideal cycle operation by having: 

1. Pressure drops due to friction in connecting pipes 

2. Heat transfer exists across connecting pipes 

3. Pressure drops occur through the condenser and evaporator tubes 

4. Heat transfer occurs from the compressor 

5. Frictional effects and flow separation occur on the compressor blades 

6. Refrigerant vapor entering the compressor may be slightly superheated 

7. Refrigerant temperature exiting the condenser may be slightly below saturation. 

A desirable effect out of the above list is item 7. Having a subcooled liquid exiting the 

condenser results in having a larger refrigeration effect. 

Problem: 

A vapor-compression refrigeration system for a household refrigerator has a refrigerating capacity of 

1000 Btu/h. The refrigerant, R134a, enters the evaporator at –10 °F and exits at 0 °F. The isentropic 

compressor efficiency is 80%. The refrigerant condenses at 95 °F and exits the condenser subcooled 

at 90 °F. There are no significant pressure drops in the flows through the evaporator and condenser. 

We want to determine the evaporator and condenser pressures, the mass flow rate of the refrigerant, 

the compressor power input, and the coefficient of performance. 

Solution: 

• First, the refrigerating capacity is the rate of heat transfer into the refrigerant flowing in the 

evaporator. It is the cooling effect, that is the rate of heat transfer extracted from air on the outside 

of the evaporator. The refrigerating capacity is normally given in tons of refrigeration for a system 

where one ton of refrigeration is equivalent to 12000 Btu/hr. 

• Second, since the inlet to the evaporator is always a mixture of liquid and vapor, the pressure 

therefore is the saturation pressure at the given temperature. It is also assumed that frictional effects, 

between the refrigerant and the tubes of the evaporator, are neglected so that the pressure at the 

inlet equals the pressure at the exit of the evaporator. 



• Third, since the refrigerant condenses inside the condenser at 95 °F, then the pressure in the 

condenser is the saturation pressure at the given temperature. 

• Let’s determine the properties of R134a at each state to give us a better look at what the system is 

doing. 

lbf 

Interpolatedfrom TableA-12E 

State 1: 

Btu 

1 

= PSat@ 

−10º 

F 

= 16.674 ⇒T1 

= 0º F ⎯⎯ ⎯⎯⎯⎯⎯⎯→h1 

= 102.2 , s 

lbm 1 

= 0. 2281 

2 

Btu 

P 

lbm* 

R 

in 

State 2: 

lbf InterpolatingfromTableA-12E 

Btu 

P Btu 

2 

= PSat@95º 

F 

= 128.62 ⎯⎯ ⎯⎯⎯⎯⎯⎯ →h2s 

= 121.01 , s 

lbm 2s 

= s1 

= 0. 2281 

lbm* 

R 

in2 

Utilizing the concept for the isentropic efficiency of a compressor, being the ratio of the isentropic 

(minimum required) to the actual power, we get the actual enthalpy value of the refrigerant leaving the 

compressor: 

η 

C 

h 

= 

h 

2s 

2a 

− h 

1 

− h 

1 

⇒ h 

2a 

= h 

1 

+ 

h 

2s 

η 

− h 

C 

1 

= 125.7 

Btu 

lbm 

State 3: 

T 40. 72 

3 

= 90 F ( subcooled ) ⇒ h3 

= h f 

+ vf 

(P - Psat 

) = 

Btu 

lbm 

State 4: (3) to (4) is a throttling process, thus from the 1 st Btu 

law: h3 = h4 

= 40. 72 

lbm 

• To determine the mass flow rate, we need to utilize the refrigeration capacity, 

enthalpies at State 4 and 1. 

Q& 

⎛ 1000 Btu ⎞ 

In 

h 

1h 

lbm 

m & = = ⎜ 

⎟ = 0.271 

Btu 

min 

h1 

− h 

4 

( 102.2 40.72) ⎝ − 60 min 

lbm ⎠ 

Q & 

In 

and the specific 

• Now we can calculate the power needed, in horsepower, to run the compressor via the 1 st Law: 

60 min ⎛ 1hp 

⎞ 

W& 

lbm 

Btu ⎛ ⎞ 

C = m& 

( h1 − h2 

) = ( 0.271 

)( 102.2 125.7) min 

− 

lbm ⎜ ⎟⎜ 

⎟ = −0. 

150hp 

Btu 

⎝ 1h 

⎠ 

2545 

⎝ h ⎠ 

• Its performance is measured via Coefficient of Performance (COP). 

• Now we can determine the coefficient of performance for this refrigerator. It has the same notion as 

the efficiency. In this case, “what I get” is cooling off the evaporator and “what I pay for” is 

compressor power. Therefore, 

h 

β = 

h 

1 

2 

− h 

4 

− h 

1 

= 2.62 



Other Questions for Critical Thinking: 

What is the rate of heat rejection on the condenser side? 

Is energy Conserved for this cyclic device? 

What is the entropy change for the refrigerant flowing through the compressor? 

What is the entropy change for the refrigerant flowing through the condenser? 

What is the entropy change for the refrigerant flowing through the expansion valve? 

What is the entropy change for the refrigerant flowing through the evaporator? 



If the air around the condenser experiences a temperature rise of 15 °F, what is the mass flow rate 

of air via a fan across the condenser’s coils? 

Estimate the CFM (ft 3 /min) of air flowing across the condenser coils 

Can you draw a T-s diagram and trace the refrigerant as it goes through from device to device? 

Can you draw an pressure –enthalpy (p-h) diagram as this is what was normally used to get 

enthalpy values prior to the days of computers and tables? 

HW # 13 

Due Tuesday of 10 th Week 

• Generate a concept map for this module and e-mail to me electronically 

• Problem 4.84 of your textbook 

• Problem 10.17 (a), (b) of your textbook 

• Problem 10.41 (a), (b) of your textbook 

You may find examples 10.1, 10.2, 10.3 of your textbook to be quite useful 

Corresponding Coverage in your text book: 

Chapter 10. Refrigeration and Heat pump Systems 

Section 10.1. Vapor Refrigeration Systems 

Section 10.2. Analyzing Vapor-Compression Refrigeration Systems. 

Section 10.3. Refrigerant Properties.

Module 4. Refrigeration and Heat Pump Systems The Vapor ...

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