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

K. Nasr,c:\thermo \module4.doc
• 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
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