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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.

iii) COP of system.

Subject:- Refrigeration and Air Conditioning

Topic:- Introduction to Refrigeration

Difficulty:- High

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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