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Following data is recorded for a reduced ambient air refrigeration system:

Subject:- Refrigeration and Air Conditioning

Topic:- Introduction to Refrigeration

Difficulty:- High


Following data is recorded for a reduced ambient air refrigeration system:

Ram air pressure and temperature 1.1 bar, 293 K
Pressure of air at exit of compressor 3.3 bar
Isentropic efficiency of compressor 80%
Effectiveness of heat exchanger 0.8
First cooling turbine exit pressure of 85% of internal efficiency 0.8 bar
Cabin pressure and temperature 1.01 bar, 25°C
Isentropic efficiency of second cooling turbine 84%
Refrigerating load required 25 tons

Find:

i) Mass flow rate air in kg/min

ii) Compressor power

iii) COP

1 Answer
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Given:p2=110 kPa ;p3=330 kPa ;Pa=0.8 bar;T6=298 K P6=101 kPa T2=293 K

Using,

$\frac{T3i}{T2}=(\frac{p3i}{p2})^\frac{γ-1}{γ}$

We get, T3i=401.04 K

Also, $η_c\frac{(T3i-T2}{(T3-T2}$

Since $η_c$=0.8

We get, T3=428.05 K

Assuming no pressure loss in heat exchanger, p4=p3= 330 kPa

We know effectiveness of heat exchanger,

$η_h$=0.8

0.8=$\frac{T3-T4}{T3-T2}$

We get, T4=320 K

For 4-5, $\frac{T4}{T5i}=(\frac{p4}{p5i})^\frac{γ-1}{γ}$

Since no data is given to find p5 we cannot find T5 and proceed ahead. This could be a correction in the paper. On knowing p5 and then T5 the sum can be solved as follows.

Calculation of Mass flow rate of air in kg/min

$Q ̇_a=m ̇_a Cp(T6-T5)$

Compressor power

$W ̇_c=m ̇_a Cp(T3-T2)$

COP

C.O.P.= $\frac{Q ̇_{a}}{W ̇_{net}}$

=$\frac{3.5*capacity}{W ̇_ram+W ̇_c}$

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