Microsoft Word - Cengel and Boles TOC _2-03-05_.doc

Chapter 11 | 613

EXAMPLE 11–1 The Ideal Vapor-Compression Refrigeration
Cycle

A refrigerator uses refrigerant-134a as the working fluid and operates on an
ideal vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. If the
mass flow rate of the refrigerant is 0.05 kg/s, determine (a) the rate of heat
removal from the refrigerated space and the power input to the compressor,
(b) the rate of heat rejection to the environment, and (c) the COP of the
refrigerator.

Solution A refrigerator operates on an ideal vapor-compression refrigeration
cycle between two specified pressure limits. The rate of refrigeration, the
power input, the rate of heat rejection, and the COP are to be determined.
Assumptions 1 Steady operating conditions exist. 2 Kinetic and potential
energy changes are negligible.
Analysis The T-sdiagram of the refrigeration cycle is shown in Fig. 11–6.
We note that this is an ideal vapor-compression refrigeration cycle, and thus
the compressor is isentropic and the refrigerant leaves the condenser as a
saturated liquid and enters the compressor as saturated vapor. From the
refrigerant-134a tables, the enthalpies of the refrigerant at all four states are
determined as follows:

(a) The rate of heat removal from the refrigerated space and the power input
to the compressor are determined from their definitions:

and

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

It could also be determined from

(c) The coefficient of performance of the refrigerator is

That is, this refrigerator removes about 4 units of thermal energy from the
refrigerated space for each unit of electric energy it consumes.
Discussion It would be interesting to see what happens if the throttling valve
were replaced by an isentropic turbine. The enthalpy at state 4s(the turbine
exit with P 4 s0.14 MPa, and s 4 ss 3 0.35404 kJ/kg · K) is 88.94 kJ/kg,

COPR

Q

#

L

W

#

in



7.18 kW

1.81 kW

3.97

Q

#

HQ

#

LW

#

in7.181.818.99 kW

Q

#

Hm

# 1 h

2 h 32 ^1 0.05 kg>s^231 275.3995.47^2 kJ>kg^4 9.0 kW

W

#

inm

# 1 h

2 h 12 ^1 0.05 kg>s^231 275.39239.16^2 kJ>kg^4 1.81 kW

Q

#

Lm

# 1 h

1 h 42 ^1 0.05 kg>s^231 239.1695.47^2 kJ>kg^4 7.18 kW

h 4
h 3 1 throttling (^2) ¡ h 4 95.47 kJ>kg
P 3 0.8 MPa ¡ h 3 hf @ 0.8 MPa95.47 kJ>kg
P 2 0.8 MPa
s 2 s 1
f¬h 2 275.39 kJ>kg
s 1 sg @ 0.14 MPa0.94456 kJ>kg # K
P 1 0.14 MPa ¡ h 1 hg @ 0.14 MPa239.16 kJ>kg
T
s
QH
4
1
4 s
3
2
0.14 MPa
0.8 MPa W
in
QL
FIGURE 11–6
T-sdiagram of the ideal
vapor-compression refrigeration cycle
described in Example 11–1.