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Determine - i) mass flow rate of air circulated to the cabin ii) net power delivered to the refrigeration system iii) COP of system.

An air cooling system for a jet plane cockpit operates on the simple cycle. The cockpit is to be maintained at 25°C. The ambient air pressure and temperature are 0.35 bar and -15°C respectively. The pressure ratio of the jet compressor is 3. The plane speed is 1000 km/hr. The air is passed through a heat exchanger after compression and cooled to its original condition entering into the air jet. The pressure loss in heat exchanger is 0.1 bar, pressure of air leaving the cooling turbine is 1.013 bar and is also the pressure in the cockpit. The cooling load in cockpit is 70 kW. Determine-

i) mass flow rate of air circulated to the cabin

ii) net power delivered to the refrigeration system

iii) COP of system.


Subject:- Refrigeration and Air Conditioning

Topic:- Introduction to Refrigeration

Difficulty:- High

1 Answer
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Given:$Q ̇_a$=70 kW; p1=35 kPa ; $\frac{p3}{p2}$=3 ;v=1000 kmph=$\frac{1000}{3.6}$ $\frac{m}s=277.78 \frac{m}s$

p3 - p4=10 kPa; P5=101.3 kPa=P6

T1=258K

T6=298K T4=T2

Using,

$$T2=T1+\frac{v^2}{2000C_p} \ \ \ and \ \ \ Cp=1.005 kJ/kgK$$

We get, T2=296.4 K

For isentropic ramming,

$$\frac{T2}{T1}=(\frac{p2}{p1})^{\frac{(γ-1)}γ}$$

Assuming $\gamma$=1.4 we get p2=56.87 kPa

Therefore, p3=170.62 kPa

For 2-3,

$$\frac{T3}{T2}=(\frac{p2}{p1})^{\frac{(γ-1)}γ}$$

Assuming γ=1.4 we get p2=56.87 kPa

Therefore, p3=170.62 kPa

For 2-3,

$$\frac{T3}{T2}=(\frac{p3}{p2})^{\frac{(γ-1)}γ}$$

therefore,T3=405.69 K

P4= p3-10=160.62 kPa

For 4-5,

$$\frac{T4}{T5}=(\frac{p4}{p5})^{\frac{(γ-1)}γ}$$

therefore,T5=259.83 K

Calculation of mass flow rate of air circulated to the cabin

$\dot{Q }_a=\dot{m }_a Cp(T6-T5) $ ∴ $m ̇_a=1.83 $kg/sec

Calculation of net power delivered to the refrigeration system

$\dot{W} _{net}=\dot{W }_{\text{compressor}}+\dot{W }_{\text{ram}}=\dot{m }_a Cp(T3-T2)+m ̇_a Cp(T2-T1)=271.62 kW$

Calculation of COP of system

C.O.P.= $\frac{\dot{Q }_a}{\dot{W }_{\text{net}}}$=0.257

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