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problem on boot strap evaporative cooling system
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Example 8
The following data refer to a boot strap air cycle evaporative refrigeration system used for an aeroplane to take 20 tonnes of refrigeration load:
Ambient air temperature
Ambient air pressure
Mach number of the flight
Ram efficiency
Pressure of air bled off the main compressor
Pressure of air in the secondary compressor
Isentropic efficiency of the main compressor
Isentropic efficiency of the secondary compressor
Isentropic efficiency of the cooling turbine
Temperature of air leaving the first heat exchanger
Temperature of air leaving the second heat exchange =
Temperature of air leaving the evaporator
Cabin temperature
Cabin pressure
Find:
1. Mass of air required to take the cabin load,
2. Power required for the refrigeration system, and
3. C.O.P. of the system.
Solution
Given:
Q=20 TR ;$T_1=15°C= 15+237=288 K ;p_1=0.8 bar ; M=1.2 ; \eta_R=90% = 0.9;$
$p_3=p_3’=p_4=4 bar; p_5=p_5’=p_5”=p_6=bar ;\eta_{C1}=90%=0.9 ;$
$\eta_{C2}=80% ;\eta_T=80%=0.8 ;T_4=170°C=170+273=443K ;T_5”=155°C=155+273=428K ;$
$T_6=100° C=100+273=373 K ;T_8=25° C=25+273=298 K ;p_8=p_7=p_7”=1 bar$

The T-s diagram for the boot-strap air cycle evaporative refrigeration system, with the given conditions, is shown in the figure below.

Let
$T_2’$= Stagnation temperature of ambient air entering the main compressor,
$p_2$= Pressure of air at the end of isentropic ramming, and
$p_2’=$ Stagnation pressure of ambient air entering the main compressor

We know that
$\frac{T_2’}{T_1}=1+\frac{\gamma-1}{2}\times M_2=1+\frac{1.4-1}{2}\times(1.2)_2=1.288$

Therefore $T_2’=T_1\times1.288=288\times1.288=371 K$

For isentropic process 1-2,
$\frac{p_2}{p_1}=\frac{T_2}{T_1}^{\frac{\gamma}{\gamma-1}}=\frac{371}{288}^{\frac{1.4}{1.4-1}}=(1.288)^{3.5}=2.425$

Therefore $p_2=p_1\times2.425=0.8\times2.428=1.94 bar$

We know that ram efficiency,

$\eta_R=\frac{Actual pressure rise}{Isentropic pressure rise}=\frac{p_2’-p_1}{p2-p_1}$

$0.9=\frac{p_2’-0.8}{1.94-0.8}=\frac{p_2’-0.8}{1.14}$

Therefore $p_2’=0.9\times1.14+0.8=1.826 bar$
Now for the isentropic process 2'-3,
$\frac{T_3}{T_2’}=(\frac{p_3}{p_2’})^{\frac{\gamma-1}{\gamma}}=(\frac{4}{1.826})^{\frac{1.4-1}{1.4}}=(2.19)^{0.286}=1.25$

Therefore $T_3=T_2’\times1.25=371\times1.25=463.8 K$
We know that isentropic efficiency of the main compressor

$\eta_{C1}=\frac{Isentropic increase in temp}{Actual increase in temp}=\frac{T_3-T_2’}{T_3’-T_2’}$

$0.9=\frac{463.8-371}{T_3’-371}=\frac{92.8}{T_3’-371}$

Therefore $T_3’=474 K$
Temperature of air leaving the first heat exchanger,

For the isentropic process 4-5,
$\frac{T_5}{T_4}=(\frac{p_5}{p_4})^{\frac{\gamma-1}{\gamma}}=(\frac{5}{4})^{\frac{1.4-1}{1.4}}=(1.25)^{0.28}=1.066$

Therefore $T_5=T_4\times1.066-443\times1.066=472 K$
Isentropic efficiency of the secondary compressor,
$\eta_{C2}=\frac{T_5-T_4}{T_5’-T_4}$

$0.8=\frac{473-443}{T_5’-443}=\frac{29}{T_5’-443}$

Therefore $T_5’=479 K$ Temperature of air leaving the second heat exchanger,

Temperature of air leaving the evaporator,

Now for the isentropic process 6-7,
$\frac{T_6}{T_7}=(\frac{p_6}{p_7})^{\frac{\gamma-1}{\gamma}}=(\frac{5}{1})^{\frac{1.4-1}{1.4}}=(5)^{0.28}=1.584$
Therefore $T_7=\frac{T_6}{1.586}=\frac{373}{1.584}=235.3 K$
We know that isentropic efficiency of the cooling turbine,
$\eta_T =\frac{Actual increase in temp}{Isentropic increase in temp}=\frac{T_6-T_7’ }{T_6-T_7}$
$0.8=\frac{373-T_7’}{373-235.5}=\frac{373-T_7’}{137.5}$

Therefore $T_7’=263 K$

1. Mass of air required to take the cabin load

$m_a=\frac{210Q}{c_p(T_8-T_7’)}=\frac{210\times20}{1(298-263)}=120 kg/min$ (Ans)

2. Power required for the refrigeration system
We know that power required for the refrigeration system,

$P=\frac{m_ac_p(T_3’-T_2’)}{60}=\frac{120\times1(474-371)}{60}=206 kW$(Ans)

3. C.O.P. of the system
We know that C.O.P. of the system
$=\frac{210Q}{m_ac_p(T_3’-T_2’)}=\frac{210\times20}{120\times1(474-371)}=0.34$ (Ans)