Subject:- Refrigeration and Air Conditioning

Topic:- Introduction to Refrigeration

Difficulty:- High

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 temp 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 the heat exchanger whose effectiveness is 60%. The air from the heat exchanger is further cooled to 60 C in the regenerative heat exchanger with a portion of air bled after expansion in the cooling turbine. The cabin is to be maintained at a temp of 25 C and a pressure of 1 bar. If the isentropic efficiencies of the 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 the cabin at the required condition; and

iii) COP of the system.

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

=20 TR;p1=0.8 bar;p2=1.2 bar;p3=p4=p5=4.5 bar ;

p7=p6=p6^'=1 bar;T8=373K

T1=283K T5=333K T7=298K

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

According $\gamma$=1/4 we get T2=317.8K

For2-3

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

T3=464 K

Also, $η_c=\frac{T3-T2}{T3'-T2}$

Since η_c=0.9

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,

$\frac{T5}{T6}=(\frac{p5}{p6})^\frac{γ-1}{γ}$

Assuming γ=1.4 we get T6=216 K

Also, $η_t=\frac{T5-T6'}{T5-T6}$

Since $η_t=0.8$

We get, $T6’=239.4 K$

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=$\frac{(210 \ Q)}{CP( T7-T6^{\prime})}$ =71.7 kg/min

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

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

Power required for maintaining cabin at required condition

P=$\frac{(m1Cp(T3'-T2))}{60}$=307 kW

C.O.P. of system

C.O.P.= $\frac{(210 Q)}{m1Cp(T3'-T2)}=0.23$