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problems on aircraft cooling systems - Simple Refrigeration Cycle
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Example 1
A simple air cooled system is used for an aeroplane having a load of 10 tones. The atmospheric pressure and temperature are 0.9 bar and 10°C respectively. The pressure increases to 1.013 bar due to ramming. The temperature of the air is reduced by 50°C in the heat exchanger. The pressure in the cabin is 1.01 bar and the temperature of air leaving the cabin is 25°C. Determine :
1 Power required to take the load of cooling in the cabin; and
2. C.O.P. of the system.
Assume that all the expansions and compressions are isentropic. The pressure of the compressed air is 3.5 bar.

Solution

Given :
Q = 10 TR ;$p_1= 0.9 bar ;T_1= 10°C = 10 + 273 = 283 K$ ;
$p_2= 1.013 bar ;p_5=p_6= 1.01 bar;$
$T_6=25°C=25+273=298 K;p_3=3.5$ bar

1. Power required taking the load of cooling in the cabin

First of all, let us find the mass of air ($m_a$) required for the refrigeration purpose. Since the compressions and expansions are isentropic, therefore the various processes on the T-s diagram are as shown in diagram below

Let
$T_2$ = Temperature of air at the end of ramming or entering the main compressor,
$T_3$ = Temperature of air leaving the main compressor after isentropic compression,
$T_4$ = Temperature of air leaving the heat exchanger, and
$T_5$ = Temperature of air leaving the cooling turbine We know that
$\frac{T_2}{T_1}=(\frac{p_2}{p_1})^\frac{\gamma-1}{\gamma}=(\frac{1.013}{0.9})^\frac{1.4-1}{1.4}=(1.125)^{0.286}=1.1034$

Therefore $T_2=T_1\times1.034=283\times1.034=292.6$K

Similarly $\frac{T_3}{T_2}=(\frac{p_3}{p_2})^\frac{\gamma-1}{\gamma}=(\frac{3.5}{1.013})^\frac{1.4-1}{1.4}=(3.45)^{0.286}=1.425$

Therefore $T_3=T_2\times1.425=283\times1.425=417K =144°C$

Since the temperature of air is reduced by 50° C in the heat exchanger, therefore temperature of air leaving the heat exchanger,
$T_4=144-50=94°C =367K$

We know that
$\frac{T_5}{T_4}=(\frac{p_5}{p_4})^\frac{\gamma-1}{\gamma}=(\frac{1.01}{3.5})^\frac{1.4-1}{1.4}=(0.288)^{0.286}=0.7$

Therefore $T_5=T_4\times0.7=367\times0.7=257K$

We know that mass of air required for the refrigeration purpose,
$m_a=\frac{210Q}{c_p(T_6-T_5)}=\frac{210\times10}{1(298-257)}=512$kg/min …….(Taking $c_p$ for air = 1 kJ/kg K)
Therefore Power required to take the load of cooling in the cabin,
$p=\frac{m_ac_p(T_3-T_2)}{60}=\frac{51.2\times1(417-292.6)}{60}=106 kW$(Ans)
2. C.O.P. of the system
We know that C.O.P, of the system
=$\frac{210Q}{P\times60}=\frac{210\times10}{106\times60}=0.33$(Ans)

Example 2
An aircraft refrigeration plant has to handle a cabin load of 30 tones. The atmospheric temperature is 17°C. The atmospheric air is compressed to a pressure of 0.95 bar and temperature of 30° C due to ram action. This air is then further compressed in a compressor to 4.75 bar, cooled in a heat exchanger to 67°C, expanded in a turbine to I bar pressure and supplied to the cabin. The air leaves the cabin at a temperature of 27°C. The isentropic efficiencies of both compressor and turbine are 0.9. Calculate the mass of air circulated per minute and the C.O.P. For air, $c_p$= 1.004 kJ/kg K and $\frac{c_p}{c_v}=\gamma=1.4$.

Solution

Given:
Q=30 TR ;$T_1=17° C=17+237=290 K ;p_2=0.95 bar ;$
$T_2=30° C=30+273=303 K ;p_3=p_3’=4.75 bar$;
$T_4=67° C=67+273=340 ;p_s=p_s’=1 bar ;T_6=27° C=27+273=300 K ;$
$\eta_c=\eta_T=0.9 ;c_p=1.004 kJ/kg K; \frac{c_p}{c_v}=\gamma =1.4$
The T-s diagram for the simple air refrigeration cycle with the given conditions is shown in diagram below.

Let
$T_3$= Temperature of the air after isentropic compression in the compressor,
$T_3’$= Actual temperature of the air leaving the compressor,
$T_5$= Temperature of the air leaving the turbine after isentropic expansion,
$T_5’$= Actual temperature of the air leaving the turbine.
We know that for isentropic compression process 2-3,
$\frac{T_3}{T_2}=(\frac{p_3}{p_2})^\frac{\gamma-1}{\gamma}=(\frac{4.75}{0.95})^\frac{1.4 -1}{1.4}=5^{0.286}=1.584$
Therefore $T_3=T_2\times1.584=303\times1.584=480$K

and isentropic efficiency of the compressor,
$\eta_C=\frac{Isentropic temperature rise}{Actual temperature rise}=\frac{T_3-T_2}{T_3’-T_2}$
$0.9=\frac{480-303}{T_3’-303}=\frac{177}{T_3’-303}$
Therefore
$T_3’-303=\frac{177}{0.9}=16.7 (or) T_3’=303+16.7=499.7K$ Now for the isentropic expansion process 4-5
$\frac{T_4}{T_5}=(\frac{p_4}{p_5})^\frac{\gamma-1}{\gamma}=(\frac{4.75}{1})^\frac{1.4 -1}{1.4}=4.75^{0.286}=1.561$
Therefore $T_5=\frac{T_4}{1.561}=\frac{P340}{1.561}=217.8$K

and isentropic efficiency of the turbine,
$\eta_T=\frac{Actual temperature rise}{Isentropic temperature rise}=\frac{T_4-T_5’}{T_4-T_5}$
$0.9=\frac{340-T_5’}{340-217.8}=\frac{340-T_5’}{122.2}$
Therefore $T_5’=340-0.9\times122.2=230$ K
Mass of air circulated per minute

$m_a=\frac{210Q}{c_p(T_6-T_5’)}=\frac{210\times10}{1.004(300-230)}=89.64$kg/min (Ans)
C.O.P
=$\frac{210Q}{m_ac_p(T_3’-T_2)} =\frac{210\times30}{89.64\times1.004(499.7-303)} =0.356$(Ans)

Example 3
An aircraft moving with speed of 1000 km/h uses simple gas refrigeration cycle for air-conditioning. The ambient pressure and temperature are 0.35 bar and -10° C respectively. The pressure ratio of compressor is 4.5. The heat exchanger effectiveness is 0.95. The isentropic efficiencies of compressor and expander are 0.8 each. The cabin pressure and temperature is 1.06 bar and 25° C. Determine temperatures and pressures at all points of the cycle. Also find the volume flow rate through compressor inlet and expander outlet for 100 TR. Take $c_p$= 1.005 kJ/kg K ; R = 0.287 kJ/kg K and $\frac{c_p}{c_v}$=1.4 = for air.

Solution

Given :
$V = 1000 km / h = 277.8 km/s ;p_1=0.35 bar ;T_1 = C= 1 0 + 273 = 263 K$;
$\frac{p_3}{p_2} = 4.5 ;\eta_E = 0.95 ;\eta_C=\eta_T=0.8 ;$
$p_5=p_5’= 1.06 bar ; T_6 = 25° C = 25 + 273 = 298 K ;$
$Q = 100 TR ; c_p = 1.005 kJ/kg K ; R = 287 kJ/kg K = 287 J/kg K ; \frac{c_p}{c_v}=\gamma=$ 1.4 The T-s diagram for the simple gas refrigeration cycle with the given conditions is shown above.

Let

$T_2$ and $p_2$= Stagnation temperature and pressure of the ambient air entering the compressor,

$T_3$ and $p_3$=Temperature and pressure of the air leaving the compressor after isentropic compression,
$T_3’$ = Actual temperature of the air leaving the compressor,
$T_4$= Temperature of the air leaving the heat exchanger or entering the expander,
$p_4$ = Pressure of the air leaving the heat exchanger or entering the expander
$T_5$5 = Temperature of the air leaving the expander after isentropic expansion,
$T_5’$= Actual temperature of the air leaving the expander.

We know that
$T_2=T_1+\frac{v^2}{2000c_p}=263+\frac{277.8^2}{2000\times1.005}$
$=263+38.4=301.4 K$(Ans)

and $\frac{p_2}{p_1}=(\frac{T_2}{T_1})^\frac{\gamma}{\gamma-1}=(\frac{301.4}{263})^\frac{1.4}{1.4-1}=(1.146)^{3.5}=1.611$

Therefore $p_2=p_1\times1.611=0.564 bar$(Ans)

Since $\frac{p_3}{p_2}=4.5$ (Given),

therefore $p_3=p_2\times4.5=0.564\times4.5=2.54 bar$(Ans)

We know that for isentropic compression process 2-3,

$\frac{T_3}{T_2}=(\frac{p_3}{p_2})^\frac{\gamma-1}{\gamma}=(4.5)^\frac{1.4 -1}{1.4}=(4.5)^{0.286}=1.537$
Therefore $T_3=T_2\times1.537=301.4\times1.537=463.3$K

We also know that isentropic efficiency of the compressor,
$\eta_C=\frac{Isentropic temperature rise}{Actual temperature rise}=\frac{T_3-T_2}{T_3’-T_2}$
$0.8=\frac{463.3-301.4}{T_3’-301.4}=\frac{161.9}{T_3’-301.4}$

Therefore $T_3’-301.4=\frac{161.9}{0.8}=202.4$

Therefore $T_3’=301.4+202.4=503.8 K$(Ans)

Effectiveness of the heat exchanger($\eta_H$),

$0.95=\frac{T_3’-T_4}{T_3’-T_2}=\frac{503.8-T_4}{503.8-301.4}=\frac{503.8-T_4}{202.4}$
Therefore $T_4=503.8-0.95\times202.4=311.5 K$(Ans)

and $p_4=p_3=2.54$bar (Ans)

Now for isentropic efficiency of the expander,

$\eta_E=\frac{Actual temperature rise}{Isentropic temperature rise}=\frac{T_4-T_5’}{T_4-T_5}$

$0.8=\frac{311.5-T_5’}{311.5-243}=\frac{311.5-T_5’}{68.5}$
Therefore $T_5’=311.5-0.8\times68.5=256.7$ K (Ans)

Volume flow rate
Let $v_2$ = Volume flow rate through the compressor inlet, and
$V_5’$= Volume flow rate through the expander outlet.

We know that mass flow rate of air,
$m_a=\frac{210Q}{c_p(T_6-T_5’)}=\frac{210\times100}{1.005(298-256.7)}=506$kg/min (Ans)
and $p_2v_2=m_aRT_2$

therefore
$v_2=\frac{m_aRT_2}{p_2}=\frac{506\times287\times301.4}{0.564\times10^5}=776m^3/min$(Ans) …(R is taken in J/kg K and $p_2$ is taken in N/m2)

Similarly $p_5’v_5’=m_aRT_5’$

$v_5’=\frac{m_aRT_5’}{p_5’}=\frac{506\times287\times256.7}{1.06\times10^5}=351.7 m^3/min$ (Ans)

Example 4
The cock pit of a jet plane flying at a speed of 1200 km/h is to be cooled by a simple air cooling system. The cock pit is to be maintained at 25°C and the pressure in the cock pit is 1 bar. The ambient air pressure and temperature are 0.85 bar and 30°C. The other data available is as follows: Cock-pit cooling load = 10 TR ; Main compressor pressure ratio = 4 ; Ram efficiency = 90% ; Temperature of air leaving the heat exchanger and entering the cooling turbine = 60° C ; Pressure drop in the heat exchanger = 0.5 bar ; Pressure loss between the cooler turbine and cockpit = 0.2 bar. Assuming the isentropic efficiencies of main compressor and cooler turbine as 80%, find the quantity of air passed through the cooling turbine and C.O.P. of the system. Take $\gamma$=1.4 and $c_p$= 1 kJ/kg K.

Solution

Given : V = 1200 km/h = 333.3 m/s ;$T_6= 25 ° C = 25 + 273 = 298 K ;p_6 = 1 bar ;p_1= 0.85 bar ;T_1 = 30° C = 30 + 273 = 303 K : Q=10 TR ;\frac{p_3}{p_2}= 4 ;\eta_R=90% =0.9$ ;
$T_4=60°C=60+273=333K;p_4=(p_3’-0.5) bar ;p_5=p_5’=p_6+0.2=1+0.2=1.2 bar ; \eta_C=\eta_T= 80% = 0.8$
The T-s diagram for the simple air cooling system with the given conditions is shown in diagram below.

Let

$T_2’$ = Stagnation temperature of the ambient air entering the main compressor = $T_2$ ,
$p_2$ = Pressure of air after isentropic ramming, and
$p_2’$ = Stagnation pressure of air entering the main compressor. We know that $T_2=T_2’=T_1+\frac{v^2}{2000c_p}=303+\frac{333.3^2}{2000\times1}$ $=303+55.5=358.5 K$
and
$\frac{p_2}{p_1}=(\frac{T_2}{T_1})^\frac{\gamma}{\gamma-1}=(\frac{358.5}{303})^\frac{1.4}{1.4-1}=(1.183)^{3.5}=1.8$
Therefore $p_2=p_1\times1.8=1.53 bar$
We know that ram efficiency,
$\eta_R =\frac{Actual pressure rise}{Isentropic pressure rise}=\frac{p_2’-p_1}{p_2-p_1}$
$0.9=\frac{p_2’-0.85}{1.53-0.85}=\frac{p_2’-0.85}{0.68}$

Therefore $p_2’=0.9\times0.68+0.85=1.46 bar$

Now for the isentropic process 2'-3 ,

$\frac{T_3}{T_2’}=(\frac{p_3}{p_2’})^\frac{\gamma-1}{\gamma}=(4)^\frac{1.4 -1}{1.4}=(4)^{0.286}=1.486$

Therefore $T_3=T_2’\times1.486=358.5\times1.486=532.7$K

and isentropic efficiency of the compressor,
$\eta_C=\frac{Isentropic temperature rise}{Actual temperature rise}=\frac{T_3-T_2’}{T_3’-T_2’}$

$0.8=\frac{532.7-358.5}{T_3’-358.5}=\frac{174.2}{T_3’-358.5}$

Therefore $T_3’=576 K$

Since the pressure ratio of the main compressor$(\frac{p_3}{p_2’})$ is 4, therefore pressure of air leaving the main compressor,
$p_3=p_3’=4p_2’=2\times1.46=5.84$bar

Therefore pressure drop in the heat exchanger
= 0.5 bar

Pressure of air after passing through the heat exchanger or at entrance to the cooling turbine, $p_4=p_3’-0.5=5.84-0.5=5.34$ bar

Also there is a pressure loss of 0.2 bar between the cooling turbine and the cock pit. Therefore pressure of air leaving the cooling turbine,

$p_5=p_5’=p_6+0.2=1+0.2=1.2$ bar

Now for the isentropic process 4-5,
$\frac{T_4}{T_5}=(\frac{p_4}{p_5})^\frac{\gamma-1}{\gamma}=(\frac{5.34}{1.2})^\frac{1.4 -1}{1.4}=4.45^{0.286}=1.53$
Therefore $T_5=\frac{T_4}{1.53}=\frac{333}{1.53}=217.6 K$

We know that isentropic efficiency of the cooling turbine,
$\eta_T=\frac{Actual temperature rise}{Isentropic temperature rise}=\frac{T_4-T_5’}{T_4-T_5}$
$0.8=\frac{333-T_5’}{333-217.6}=\frac{333-T_5’}{15.4}$

Therefore $T_5’=240.7$ K
Quantity of air passed through the cooling turbine

We know that quantity of air passed through the cooling turbine,
$m_a=\frac{210Q}{c_p(T_6-T_5’)}=\frac{210\times10}{1 (298-240.7)}=36.6$kg/min (Ans)

C.O.P. of the system
We know that C.O.P. of the system
=$\frac{210Q}{m_ac_p(T_3’-T_2’)}=\frac{210\times10}{36.5\times1(576-358.5)}=0.264$ (Ans)

Example 5
In an aeroplane simple air refrigeration is used. The main compressor delivers the air at 5 bar and 200°C. The bled air taken from compressor is passed through a heat exchanger, cooled with the help of ram air so that the temperature of air leaving the heat exchanger is 45°C and the pressure is 4.5 bar. The cooling turbine drives the exhaust fan which is needed to force the ram air through the heat exchanger. The air leaving the heat exchanger passes through the cooling turbine and then supplied to cabin at 1 bar. The pressure loss between the cooling turbine and cabin is 0.2 bar. If the rate of flow of air through the cooling turbine is 20 kg/min, determine the following:
1. The temperature of the air leaving the expander;
2.The power delivered to the ram air which is passed through the heat exchanger, and
3. The refrigeration load in tones when the temperature of the air / leaving the cabin is limited to 25°C.

Assume that the isentropic efficiency of the cooling turbine is 75% and no loss of heat from air between the cooling turbine and cabin. Take $\gamma$= 1.4 and $c_p=1$kJ/kg K.

Solution

Given:
$p_3= 5 bar ;T_3=200° C=200+273=473K;T_4=45° C = 45+273 = 318 K ;$
$p_4 = 4.5 bar ;m_a= 20 kg/min ;T_6=25° C=25+273=298 K ;\eta_T= 75%. =0.75 ;\gamma=1.4 ;c_p=1$ kJ/kg K

The various processes on the T-s diagram are shown below. The point 3 represents the air delivered from the compressor to heat exchanger and the point 4 shows the condition of air leaving the heat exchanger. The vertical line 4-5 represents the isentropic expansion of air in the cooling turbine and the curve 4-5’ shows the actual expansion of air in the cooling turbine due to internal friction. The line 5'-6 represents the refrigeration process.

1. Temperature of air leaving the expander

Let $T_5$= Temperature of air at the end of isentropic expansion in the cooling turbine or expander, and
$T_5’$= Actual temperature of air leaving the cooling turbine or expander.

We know that
$\frac{T_5}{T_4}=(\frac{p_5}{p_4})^\frac{\gamma-1}{\gamma}=(\frac{1.2}{4.5})^\frac{1.4 -1}{1.4}=0.267^{0.286}=0.685$
Therefore $T_5=T_4\times0.685=318\times0.685=217.8$K
Isentropic efficiency of the cooling turbine,

$\eta_T=\frac{Actual temperature rise}{Isentropic temperature rise}=\frac{T_4-T_5’}{T_4-T_5}$

$0.75=\frac{318-T_5’}{318-217.8}=\frac{318-T_5’}{100.2}$

Therefore $T_5’=242.85$ K(Ans) 2. Power delivered to the ram air which is passed through the heat exchanger
We know that work delivered to the ram air which is passed through the heat exchanger,
$=m_ac_p(T_4-T_5’)=20\times1(318-242.85)=1506$kJ/min

Power delivered = 1503/60=25.05 kW (Ans)

$=m_ac_p(T_6-T_5’)=20\times1(298-242.85)=1103$kJ/min
$=\frac{1103}{210}=5.25$TR (Ans)