Example 11
A regenerative air cooling system is used for an airplane to take 20 tonnes 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 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 the air bled after expansion in the cooling turbine. The cabin is to be maintained at a temperature of 25°C and a pressure of I bar. If the isentropic efficiencies of the compressor and turbine are 90% and 80% respectively, find:
1. Mass of the air bled from cooling turbine to be used for regenerative cooling;
2. Power required for maintaining the cabin at the required condition; and
3. C. O. P. of the system.
Assume the temperature of air leaving to atmosphere from the regenerative heat exchanger as 1 00° C.
Solution
Given :
$Q = 20 TR ; p_1 =0.8 bar ;T_1=10° C =10 + 273 = 283 K ;p_2= 1.2 bar ; $
$p_3=p-4=p_5=4.5bar ;\eta_H=60$%$=0.6 ;T_5= 60° C = 60+273 = 333 K;$
$T_7=25°C=25+273= 298 K ;p_7=p_6=p_6’=1 bar ;\eta_c= 0.9 ;\eta_T= = 80$%$ = 0.8 ;$
$T_8=100° C =100 + 273 = 373 K $
The T-s diagram for the regenerative air cooling system with the given conditions is

Let $T_2$ = Temperature of air at the end of ramming and entering to the, main compressor, $T_3’$= Temperature of air leaving the main compressor. We know that for the isentropic ramming of air ( process 1-2 ),
$\frac{T_2}{T_1}=(\frac{p_2}{p_1})^{\frac{\gamma-1}{\gamma}}=(\frac{1.2}{0.8})^{\frac{1.4-1}{1.4}}=(1.5)^{0.286}=1.123$

Therefore $T_2=T_1\times1.123=283\times1.123=317.8 K$
and for the isentropic compression process 2-3,

$\frac{T_3}{T_2}=(\frac{p_3}{p_2})^{\frac{\gamma-1}{\gamma}}=(\frac{4.5}{1.2})^{\frac{1.4-1}{1.4}}=(3.7)^{0.286}=1.46$

Therefore $T_3=T_2\times1.46=317.8\times1.46=464 K$
Isentropic efficiency of the compressor,

$\eta_{C}=\frac{Isentropic increase in temp}{Actual increase in temp}=\frac{T_3-T_2}{T_3’-T_2}$

$0.9=\frac{464-317.8}{T_3’-317.8}=\frac{146.2}{T_3’-317.8}$

$T_3’=480 K$ We know that effectiveness of the heat exchanger $(\eta_H)$,

$0.6=\frac{T_3’-T_4} {T_3’-T_2}=\frac{480-T_4}{480-317.8}=\frac{480-T_4}{162.2}$

Therefore $T_4=382.7K$
Now for the isentropic cooling in the cooling turbine (process 5-6),
$\frac{T_5}{T_6}=(\frac{p_5}{p_6})^{\frac{\gamma-1}{\gamma}}=(\frac{4.5}{1})^{\frac{1.4-1}{1.4}}=(4.5)^{0.286}=1.54$

Therefore $T_6=\frac{T_5}{1.54}=\frac{333}{1.54}=216 K$
and isentropic efficiency of the cooling turbine,

$\eta_{T}=\frac{ Actual increase in temp }{ Isentropic increase in temp }=\frac{T_5-T_6’}{T_5-T_6}$

$0.8=\frac{333-T_6’}{333-216}=\frac{333-T_6’}{117}$

Therefore $T_6’=239.4 K$ 1. Mass of air bled from the cooling turbine to be used for regenerative cooling
Let
$m_a$= Mass of air bled from the cooling turbine to be used for regenerative cooling,
$m_1$=Total mass of air bled from the main compressor, and
$m_2$= Mass of cold air bled from the cooling turbine for regenerative heat exchanger.
We know that the mass of air supplied to the cabin,

$m_a=m_1-m_2$
$m_a=\frac{210Q}{c_p(T_7-T_6’)}=\frac{210\times20}{1(298-239.4)}=71.7 kg/min$ ..........(i)

$m_2=\frac{m_1(T_4-T_5)}{T_8-T_6’}=\frac{m_1(282.7-333)}{373-239.4}=0.327m_1$ .........(ii) From equation (i), we find that
$m_1-m_2=71.7$ or$m_1-0.372m_1=71.7$

Therefore $m_1=\frac{71.7}{1-0372}=113.4 kg/min$
and $m_2=0.372m_1=0.372\times13.4=42.2 kg/min$ (Ans)
Note:
From equation (ii), $\frac{m_2}{m_1}=0.372$
Therefore we can say that the air bled from the cooling turbine for regenerative cooling is 37.2% of the total air bled from the main compressor.
2. Power required for maintaining the cabin at the required condition
We know that the power required for maintaining the cabin, at the required condition,

$P=\frac{m_1c_p(T_3’-T_2)}{60}=\frac{113.4\times1(480-317.8)}{60}=307 kW$ (Ans)
3. C.O.P. of the system
We know that C.O.P. of the system

$=\frac{210Q}{m_1c_p(T_3’-T_2)}=\frac{210\times20}{113.4\times1(480-317.8)}=0.23$ (Ans)