A regenerative air cooling system is used for an airplane to take 20 TR of refrigeration load. The ambient air at pressure 0.8 bar and temperature 10°C is rammed isentropically till the pressure rises to 1.2 bar. The air bled off the main compressor at 4.5 bar is cooled by the ram air in heat exchanger whose effectiveness is 60%. Air from heat exchanger is further cooled to 60°C in the regenerative heat exchanger with portion of air bled after expansion in cooling turbine. The cabin is to be maintained at a temperature of 25°C and pressure of 1 bar. If isentropic efficiencies of compressor and turbine are 90% and 80% respectively find:

i. Mass of air bled from cooling turbine to be used for regenerative cooling

ii. Power required for maintaining cabin at required condition

iii. C.O.P. of system

Assume temperature of air leaving to atmosphere from regenerative heat exchanger as 100°C.

Mumbai University > Mechanical Engineering > Sem8 > Refrigeration and Air Conditioning

Marks: 12M

Year: Dec 2014

$$\text{Given}:Q=20 TR;p1=0.8 \text{bar};p2=1.2 bar;p3=p4=p5=4.5 \text{bar} ; \ p7=p6=p6'=1 \text{bar};T8=373K \ T1=283K T5=333K T7=298K $$ ![enter image description here][1] ![enter image description here][2] For 1-2, $$\frac{T2}{T1}=\big(\frac{p2}{p1}\big)^{\big(\frac{γ-1}{γ}\big)}$$

Assuming $\gamma$=1.4 we get T2=317.8 K

For 2-3,

$$\frac{T3}{T2}=\big(\frac{p3}{p2}\big)^{\big(\frac{γ-1}{γ}\big)} \ T3=464 K $$ $$\text{Also}, η_c=\frac{(T3-T2)}{(T3'-T2)} \ \text{Since} \ \ η_c=0.9 \ \text{We get}, T3'=480 K$$ We know effectiveness of heat exchanger, $$η_h=0.6 \ 0.6=\frac{(T3'-T4)}{(T3'-T2)}$$ We get, T4=382.7 K For 5-6, $$\frac{T5}{T6}=\big(\frac{p5}{p6}\big)^{\big(\frac{γ-1}{γ}\big)}$$

Assuming $\gamma$=1.4 we get T6=216 K

Also, $η_t=\frac{(T5-T6')}{(T5-T6)}$

Since $η_t=0.8$

We get, T6’=239.4 K

i. Mass of air bled from cooling turbine to be used for regenerative cooling

Let m=mass of air bled from cooling turbine to be used for regenerative cooling

Let m1=total mass of air bled from main compressor

$m= \frac{(210 Q)}{(Cp(T7-T6'))}$=71.7 kg/min

Let m2=mass of cold air bled from cooling turbine for regenerative heat exchanger

So m2= $\frac{m1(T4-T5)}{(T8-T6')}$=0.372 m1

Now m=m1-m2=71.7

Or m1-0.372m1=71.7

Therefore, m1=113.4 kg/min

And m2=42.2 kg/min … ans

ii) Power required for maintaining cabin at required condition

$$P=\frac{m1Cp(T3'-T2)}{60}=307 kW$$ **iii. C.O.P. of system** $$\text{C.O.P.}= \frac{(210 Q)}{(m1Cp(T3'-T2))}=0.23$$