Example 1
Simple evaporative air refrigeration system is used for an aeroplane to take 20 tones of refrigeration load. The ambient air conditions are 20°C and 0.9 bar. The ambient air is rammed isentropically to a pressure of 1bar. The air leaving the main compressor at pressure 3.5 bar is first cooled in the heat exchanger having effectiveness of 0.6 and then in the evaporator where its temperature is reduced by 5°C. The air from the evaporator is passed through the cooling turbine and then it is supplied to the cabin which is to be maintained at a temperature of 25°C and at a pressure of 1.05 bar. If the internal efficiency of the compressor is 80% and that of cooling turbine is 75% determine:
1. Mass of air bled off the main compressor;
2. Power required for the refrigerating system; and
3. C.O.P. of the refrigeration system.
Solution
Given:
Q=20 TR ; $T_1= 20° C = 20 + 2 73 = 293 K ;p_1 = 0.9 bar ; p_2 = 1 bar ;p_3=p_3’= 3.5 bar ;\eta_H=0.6 ;$
$T_6= 25° C = 25 + 273 = 298 K ;p_6= 1.05 bar ;\eta_C= 80$%= 0.8 ;$\eta_T= 75$%$=0.75$
The T-s diagram for the simple evaporative air refrigeration system with the given conditions is shown below
Let
$T_2$ = Temperature of air entering the main compressor,
$T_3$ = Temperature of air after isentropic compression in the main compressor
$T_3’$ = Actual temperature of air leaving the main compressor; and
$T_4$ = Temperature of air entering the evaporator. We know that for an isentropic ramming process 1-2,

$\frac{T_2}{T_1} =(\frac{p_2}{p_1})^\frac{\gamma-1}{\gamma} =(\frac{1 }{0.9})^\frac{1.4-1}{1.4}=(1.11)^{0.286} =1.03$ ………….(taking$ \gamma$=1.4)

Therefore $T_2=T_1\text1.03=293\text=301.8 K$
Now for the isentropic compression process 2-3,

$\frac{T_3}{T_2} =(\frac{p_3}{p_2})^\frac{\gamma-1}{\gamma} =(\frac{3.5 }{1})^\frac{1.4-1}{1.4}=(3.5)^{0.286} =1.43$

Therefore $T_3=T_2\text1.43=301.8\text1.43=431.6 K$
We know that efficiency of the compressor, $\eta_C=\frac{isentropic increase in temperature}{actual increase in temperature}=\frac{T_3-T_2}{T_3’-T_2}$

$0.8=\frac{431.6-301.8}{T_3’-301.8}=\frac{129.8}{T_3’-301.8}$

Therefore $T_3’=301.8+\frac{129.8}{0.8}=464 K$
Effectiveness of the heat exchanger ($\eta_H$),

$0.6=\frac{T_3’-T_4}{T_3’-T_2’}=\frac{646 -T_4}{464-301.8}=\frac{464-T_4}{162.2} …..(T_2’=T_2) $
Therefore $T_4=646-0.6\text162.2=3667 K=39.7$° C
Since the temperature of air in the evaporator is reduced by 5° C, therefore the temperature of air leaving the evaporator and entering the cooling turbine,

$T_4’=T_4-5=93.7-5=88.7$° C=361.7 K
Now for the isentropic expansion process 4-5,
$\frac{T_4’}{T_5} =(\frac{p_3}{p_6})^\frac{\gamma-1}{\gamma} =(\frac{3.5 }{1.05})^(\frac{1.4-1}{1.4})=(3.33)^{0.286} =1.41$

Therefore $T_5=\frac{T_4’}{1.41}=\frac{361.7}{1.41}=256.5 K$
Efficiency of the cooling turbine,
$\ eta_T=\frac{actual increase in temperature}{isentropic increase in temperature}=\frac{T_4’-T_5’}{T_4’-T_5}$

$0.75=\frac{361.7-T_5’}{361.7-256.5}=\frac{361.7-T_5’}{105.2}$
Therefore $T_5’=361.7-0.75\text105.2=282.8 K$
1. Mass of air bled off the main compressor
We know that mass of air bled off the main. compressor,

$m_a=\frac{210Q}{c_p(T_6-T_5’)}=\frac{210\text20}{1(298-282.8)}=276kg/min $ (Ans)
2. Power required for the refrigerating system, We know that power required for the refrigerating system,

$p=\frac{m_ac_p(T_3’-T_2’)}{60}=\frac{276\text1(464-301.8)}{60}=746kW$ (Ans)
3. C.O.P. of the refrigerating system
We know that C.O.P. of the refrigerating system

$\frac{210Q}{P\text60}=\frac{210\text20}{746\text60}=0.094$ (Ans)