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problem on boot strap cooling system
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Example 7
A boot-strap cooling system of 10 TR capacity is used in an aeroplane. The ambient air temperature and pressure are 20°C and 0.85 bar respectively. The pressure of air increases from 0.85 bar to 1 bar due to ramming action of air. "The pressure of air discharged from the main compressor is 3 bar. The discharge pressure of air from the auxiliary compressor is 4 bar. The isentropic efficiency of each of the compressor is 80%, while that of turbine is 85%. 50% of the enthalpy of air discharged from the main compressor is removed in the first heat exchanger and 30% of the enthalpy of air discharged from the auxiliary compressor is removed in the second heat exchanger using rammed air. Assuming ramming action to be isentropic, the required cabin pressure of 0.9 bar and temperature of the air leaving the cabin not more than 20° C, find :
1. the power required to operate the system, and
2. the C.O.P. of the system.
Draw the schematic and temperature -entropy diagram of the system.
Take $\gamma$=1.4 and $c_p$=1 kJ/kg K.
Solution
Given :
Q = 10 TR ;$T_1 = 20° C = 20 + 273 = 293 K ;p_1 = 0.85 bar ;p_2 = 1 bar ; $ $p_3=p_3’'=p_4=3 bar ;p_5 =p_5’' = p_6 =4 bar ;$ $\eta_{C1}=\eta_{C2}=80$%$ =0.8 ;\eta_T= 85$%$ = 0.85 ; p_7=p_7’=p_8=0.9 bar ;T_8 =200C= 20+273 =293 K ;\gamma= 1.4;c_p= 1 $kJ/kg K
The schematic diagram for a boot-strap cooling system is shown in the figure below. The temperature- entropy (T-s) diagram with the given conditions is shown in the second figure.

We know that for isentropic ramming process. 1-2

$\frac{T_2}{T_1} =(\frac{p_2}{p_1})^\frac{\gamma-1}{\gamma} =(\frac{1}{0.85})^\frac{1.4-1}{1.4}=(1.176)^{0.286} =1.047$
Therefore $T_2=T_1\times1.047=293\times1.047=306.8K=33.8$0C

Now for isentropic process 2-3,

$\frac{T_3}{T_2} =(\frac{p_3}{p_2})^(\frac{\gamma-1}{\gamma}) =(\frac{3}{1})^(\frac{1.4-1}{1.4})=(3)^{0.286} =1.37$

Therefore $T_3=T_2\times1.37=360.8\times1.37=420.3 K=147.3$0C
We know that isentropic efficiency of the compressor,

$\eta_{c1}=\frac{Isentropic increase in temperature}{Actual increase in temperature}=\frac{T_3-T_2}{T_3’-T_2}$

$0.8=\frac{420.3-360.8}{T_3’-360.8}=\frac{113.5}{T_3’-360.8}$

Therefore $T_3’=448.7 $K=175.7 0C
Since 50% of the enthalpy of air discharged, from the main compressor is removed in the first heat exchanger (i.e. during the process 3'-4), therefore temperature of air leaving the-first heat exchanger,
$T_4=0.5\times175.7=87.85$0C=360.85 K
Now for the isentropic process 4-5,

$\frac{T_5}{T_4} =(\frac{p_5}{p_4})^(\frac{\gamma-1}{\gamma}) =(\frac{4 }{3})^(\frac{1.4-1}{1.4})=(1.33)^{0.286} =1.085$

Therefore $T_5=T_4\times1.085=360.85\times1.085=391.5 K=118.5$0C
We know that isentropic efficiency of the auxiliary compressor,
$\eta_{c2}=\frac{T_5-T_4}{T_5’-T_4}$

$0.8=\frac{391.5-360.85}{T_5’-360.85}=\frac{30.65}{T_5’-360.85}$
Therefore $T_5’=399.16 $K=126.16 0C
Since 30% of the enthalpy of air discharged from the auxiliary compressor is removed in the second heat exchanger (i.e. during the process 5'-6), therefore temperature of air leaving the second heat exchanger,

$T_6=0.7\times126.16=88.3$0C =361.3 K
For the isentropic process 6-7,
$\frac{T_7}{T_6} =(\frac{p_7}{p_6})^(\frac{\gamma-1}{\gamma}) =(\frac{0.9 }{4})^(\frac{1.4-1}{1.4})=(0.225)^{0.286} =0.653$

Therefore $T_7=T_6\times0.653=361.3\times0.653=236 K=-37$0C
We know that turbine efficiency

$\ eta_T=\frac{actual increase in temperature}{isentropic increase in temperature}=\frac{T_6-T_7’}{T_6-T_7}$

$0.85=\frac{361.3-T_7’}{361.3-236}=\frac{361.3-T_7’}{125.3}$
Therefore $T_7’=254.8 K=-18.2$0C
1. Power required to operate the system
We know that amount of air required for cooling the cabin,

$m_a=\frac{210Q}{c_p(T_8-T_7’)}=\frac{210\times10}{1(293-254.8)}=55$kg/min
and power required to operate the system,

$P=\frac{m_ac_p(T_3’-T_2)}{60}=\frac{55\times1(448.7-306.8)}{60}=130$kW (Ans)
2. C.O. P. of the system
We know that C.O.P. of the system

$\frac{210Q}{m_ac_p(T_3’-T_2)}=\frac{210\times10}{55\times(448.7-306.8)}=0.269$(Ans)

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