SME
Engineering Thermodynamics · Problem type
Vapor-compression refrigeration cycle
Use this for “R-134a leaves the evaporator at 241.3 kJ/kg, the compressor at 278.2 kJ/kg and the condenser at 101.6 kJ/kg; 0.08 kg/s flows. Find the COP and cooling load.”
Diagram
QH = Q̇L + W = 7.45 + 1.54
QH = 8.99 kW
COP = Q̇L/W = 7.45/1.54 = 4.82
Work moves heat from cold to hot, which never happens by itself.
This is an example. Type your own values to start a new problem.
qL (kJ/kg)
=
h₁ (kJ/kg)
−
h₄ (kJ/kg)
w (kJ/kg)
=
h₂ (kJ/kg)
−
h₁ (kJ/kg)
COP
=
qL (kJ/kg)
÷
w (kJ/kg)
h₃
Enthalpy out of the condenser
example · kJ/kg
ṁ
Refrigerant flow
example · kg/s
Q̇L
Cooling load
calculated · kW
Ẇ
Compressor power
calculated · kW
Formulas
h₄
=
h₃
95.5 = 95.5
qL
=
h₁
−
h₄
149 = 244.5 − 95.5
w
=
h₂
−
h₁
30.9 = 275.4 − 244.5
COP
=
qL
w
4.822
=
149
30.9
Q̇L
=
ṁ
×
qL
7.45 = 0.05 × 149
Ẇ
=
ṁ
×
w
1.545 = 0.05 × 30.9
Assumptions
•
Ideal cycle: saturated vapor into the compressor (1), saturated liquid out of the condenser (3).
•
The expansion valve throttles: no work or heat, so h₄ = h₃.
•
Read h₁, h₂ and h₃ from your refrigerant tables; kinetic and potential energy are neglected.
Step-by-step
WE KNOW
h₁ = 244.5 kJ/kg, h₂ = 275.4 kJ/kg, h₃ = 95.5 kJ/kg, ṁ = 0.05 kg/s
FIND
enthalpy into the evaporator (h₄), heat taken in (qL), compressor work (w), coefficient of performance (COP), cooling load (Q̇L), compressor power (Ẇ)
Step 1 · Find enthalpy into the evaporator (h₄)
Use h₄ = h₃
The valve throttles: the enthalpy passes through unchanged.
h₄
=
95.5
kJ/kg
Step 2 · Find heat taken in (qL)
Use qL = h₁ − h₄
In
the
evaporator
the
refrigerant
boils,
taking
in
heat
from
h₄
up
to
h₁
.
qL = 244.5 − 95.5
qL = 149 kJ/kg
Step 3 · Find compressor work (w)
Use w = h₂ − h₁
The
compressor’s
work
raises
the
enthalpy
from
h₁
to
h₂
.
w
=
275.4
−
244.5
w
=
30.9
kJ/kg
Step 4 · Find coefficient of performance (COP)
Use COP = qL ÷ w
What the refrigerator does (heat taken in) over what it costs (work).
COP
=
149
30.9
COP = 4.822
Step 5 · Find cooling load (Q̇L)
Use Q̇L = ṁ × qL
Each kilogram takes in qL: multiply by the flow (kg/s × kJ/kg = kW).
Q̇L = 0.05 × 149
Q̇L = 7.45 kW
Step 6 · Find compressor power (Ẇ)
Use Ẇ = ṁ × w
The
compressor’s
power:
the
flow
times
w
.
Ẇ
=
0.05
×
30.9
Ẇ
=
1.545
kW
Check
Put the numbers back in. Both sides should match.
95.5 = 95.5 ✓
149 = 244.5 − 95.5 ✓
30.9 = 275.4 − 244.5 ✓
4.822
=
149
30.9
✓
7.45 = 0.05 × 149 ✓
1.545 = 0.05 × 30.9 ✓
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