Example 9
The reduced ambient air refrigeration system used for an aircraft consists of two cooling turbines, one heat exchanger and one air cooling fan. The speed of aircraft is 1500 km/h. The ambient air conditions are 0.8 bar and 10° C. The ram efficiency may be taken as 90%. The rammed air used for cooling is expanded in the first cooling turbine and leaves it at a pressure of 0.8 bar.
The air bled from the main compressor at 6 bar is cooled in the heat exchanger and leaves it at 100° C: The cabin is to be maintained at 20° C and 1 bar. The pressure loss between the second cooling turbine and cabin is 0.1 bar. If the isentropic efficiency for the main compressor and both of the cooling turbines are 85% and 80% respectively, find:
1. mass flow rate of air supplied to cabin to take a cabin load of 10 tonnes of refrigeration ;
2. quantity of air passing through the heat exchanger if the temperature rise of ram air is limited to 80 K ;
3. power used to drive the cooling fan; and
4. C.O.P. of the system.
Solution
Given : V=1500 km/h=417 m/s ;$p_2=0.6 bar ;T_1=10° C=10+273=283 K;$
$\eta_R=90$%$=0.9 ;p_3=p_4=6 bar ;T_4=100° C=100+273=373 K;$
$T_6=20° C=20+273=293 K ;p_6= 1 bar ;\eta_C=85$%$=0.85 ;\eta_{T1}= 80$%$=0.8 ; Q=10 TR$
The T-s diagram for the reduced ambient air refrigeration system with the given conditions is 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 air entering the main compressor.

We know that $T_2’=T_1+\frac{v^2}{2000c_p}=283+\frac{417^2}{2000\times 1}=370 K$

For the isentropic ramming process 1-2,
$\frac{p_2}{p_1} =(\frac{T_2}{T_1})^\frac{\gamma}{\gamma-1} =(\frac{370 }{283})^\frac{1.4}{1.4-1}=(1.31)^{3.5} =2.57$
Therefore $p_2=p_1\times 2.57=2.06 bar$
We know that ram efficiency,
$\eta_R=\frac{Actual\space rise \space in\space pressure}{Isentropic \space rise\space in \space pressure}=\frac{p_2’-p_1}{p_2-p_1}$

$0.9=\frac{p_2’-0.8}{2.06-0.8}=\frac{p_2’-0.8}{1.26}$
Therefore $p_2’=1.93 bar$
The T-s diagram for the expansion of ram air in the first cooling turbine is shown in Fig. 8.20.the vertical line 2'-1' represents the isentropic cooling process and the curve 2'-2'' represents the actual cooling process.
now for isentropic process 2'-1',

$\frac{T_2’}{T_1’} =(\frac{p_2’}{p_1’})^\frac{\gamma-1}{\gamma} =(\frac{1.93 }{0.8})^\frac{1.4-1}{1.4}=(2.4)^{0.286} =1.284$ ….$(p_1’=p_1)$
$T_1’=T_2’/1.284=370/1.284=288 K$

Isentropic efficiency of the first cooling turbine,
$\eta_{T1} =\frac{Actual\space rise \space in\space temp}{Isentropic \space rise\space in \space temp}=\frac{T _2’-T _2”}{T_2’-T_2”}$
$0.8=\frac{370-T_2”}{370-288}=\frac{370-T_2”}{82}$

therefore $T_2”=304 K$
For the isentropic compression process 2'-3,

$\frac{T_3}{T_2’} =(\frac{p_3}{p_2’})^\frac{\gamma-1}{\gamma} =(\frac{6 }{1.93})^\frac{1.4-1}{1.4}=(3.11)^{0.286} =1.38$
Therefore $T_3=T_2’\times 1.38=370\times 1.38=511 K$

T-s diagram of first cooling tube
We know that isentropic efficiency of the compressor, $\eta_{C } =\frac{Isentropic \space rise\space in \space temp}{Actual\space rise \space in\space temp}=\frac{T _3-T _2’}{T_3’-T_2’}$
$0.85=\frac{511-370}{T_3’-370}=\frac{141}{T_2’-370}$

Therefore $T_2’=536 K$

Since there is a pressure drop of 0.1 bar between the second cooling turbine and the cabin, $p_5=p_5’=p_6+0.1=1+0.1=1.1 bar$

Now for the isentropic expansion of air in the second cooling turbine (process 4-5), $\frac{T_4}{T_5} =(\frac{p_4}{p_5})^\frac{\gamma-1}{\gamma} =(\frac{6 }{1.1})^\frac{1.4-1}{1.4}=(5.45)^{0.286} =1.62$
Therefore $T_5=T_4/1.62=373/1.62=230 K$

We know that the isentropic efficiency of the second cooling turbine, $\eta_{T2} =\frac{Actual\space rise \space in\space temp}{Isentropic \space rise\space in \space temp}=\frac{T _4-T _5’}{T_4-T_5}$

$0.8=\frac{373-T_5’}{373-230}=\frac{373-T_5’}{143}$

Therefore $T_5’=258.6 K$

1. Mass flow rate of air supplied to cabin
We know that mass flow rate of air supplied to cabin,
$m_a=\frac{210Q}{c_p(T_6-T_5’)}=\frac{210\times 10}{1(293-258.6)}=61 kg/min$ (Ans)

2. Quantity of ram air passing through the heat exchanger
Let $m_R$= Quantity of ram air passing through the heat exchanger.

The compressed air bled off at temperature $T_3’=$ 536 K is cooled in the heat exchanger to a temperature $T_4=$ 373 K by the ram air from the first cooling turbine at a temperature $T_2”=$ 304 K. The temperature rise of ram air in the heat exchanger is limited to 80 K. Considering perfect heat transfer in the heat exchanger,
$m_R\times c_p\times 80=m_a\times c_p(T_3’-T_4)$

$m_R\times 1\times 80=61\times 1(536-373)=9943$
Therefore $m_R=9943/80=124.3 kg/min$ (Ans)

3. Power used to drive the cooling fan
Since the work output of both the cooling turbines is used to drive the cooling fan, therefore work output from the first cooling turbine,

$W_{T1}=m_R\times c_p(T_2’-T_2”)$
$=124.3\times 1(370-304)=8204 kJ/min$

and work output from the second cooling turbine,
$W_{T2}=c_p \times c_p(T_4-T_5’)$
$=61\times 1(373-258.6)=6978 kJ/min$

Therefore Combined work output from both the cooling turbines, $W_T=W_{T1}+W_{T2}=8204+6978=15182 kJ/min$

and power used to drive the cooling fan
=15182/60=253 kW(Ans)

4. 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\times 10}{61\times 1(536-370)}=0.21$ (Ans)

Example 10
The reduced ambient system of air refrigeration for cooling an aircraft cabin consists of two cooling turbines, one heat exchanger and one fan. The first cooling turbine is supplied with the ram air at 1.1 bar and 15 ° C and delivers after expansion to the heat exchanger at 0.9 bar for cooling the air bled off from the main compressor at 3.5 bar. The cooling air from the heat exchanger is sucked by a fan and discharged to the atmosphere. The cooled air from the heat exchanger is expanded upto I bar in the second cooling turbine and discharged into air cabin to be cooled. The air from the cabin is exhausted at 22° C. The refrigerating capacity required is 10 tonnes. If the compression index for the main compressor is 1.5 and the expansion index for both the cooling turbines is 1.35, determine
1. Mass flow rate of the cabin air ;
2. Cooling capacity of the heat exchanger and flow rate of the ram air when compressed air is to be cooled to 60° C in the heat exchanger and temperatue rise in the heat exchanger for the ram air is not to exceed 30 K ;
3. Combined output of both cooling turbines driving the airfan with transmission efficiency of 60%; and
4. C.O.P. of the refrigerating system considering only power input to the compressor.
Solution
Given :
$p_2=1.1 bar ;T_2=15° C=15+273=288 K ;p_2’'=0.9 bar ;p_3=p_4=3.5 bar ;$
$p_5=p_6=1 bar ;T_6=22° C=22+273=295 K ; Q=10 TR ; \gamma_{1}=1.5 ;$ $\gamma_{2}=1.35 ;\eta_T=60$%$=0.6 ;T_4=60° C=60+237=333 K$
The T-s diagram for the reduced ambient system of air refrigeration for cooling an aircraft cabin with the given conditions is shown below. The process of cooling the ram air in the first cooling turbine is shown by the curve 2-2'. Let
$T_2’$= temperature of ram air after expansion in the first cooling turbine, and
$T_3$= temperature of air bled off from the main compressor

We know that for the expansion in the first cooling turbine ( process 2-2'),
$\frac{T_2}{T_2’} =(\frac{p_2}{p_2’})^\frac{\gamma-1}{\gamma} =(\frac{1.1 }{0.9})^\frac{1.4-1}{1.4}=(1.22)^{0.286} =1.053$
Therefore$T_2’=T_2/1.053=288/1.053=273.5 K$

For the process 2-3,
$\frac{T_3}{T_2} =(\frac{p_3}{p_2})^\frac{\gamma-1}{\gamma} =(\frac{3.5 }{1.1})^\frac{1.4-1}{1.4}=(3.18)^{0.286} =1.47$
Therefore $T _3=T_2\times 1.47=288\times 1.47=423.4 K$

and for the process 4-5,
$\frac{T_4}{T_5} =(\frac{p_4}{p_5})^\frac{\gamma_{2}-1}{\gamma_{2}} =(\frac{3.5 }{1})^\frac{1.35-1}{1.35}=(3.5)^{0.26} =1.385$
Therefore $T_5=t_4/1.385=333/1.385=240.4 K$

1. Mass flow rate of cabin air We know that mass flow rate of cabin air,
$m_a=\frac{210Q}{c_p)T_6-T_5’)}=\frac{210\text10}{1(295-240.4)}=38.5 kg/min$ ….($c_p$ for air=1 kJ/kg K)

2.Cooling capacity of the heat exchanger and flow rate of ram air
We know that cooling capacity of the heat exchanger
$=m_ac_p(T_3-T_4)=38.5\times 1(413.4-333)=3480 kJ/min $ (Ans)

In order to find the flow rate of ram air (MR), equate the enthalpy lost by compressed air to the enthalpy gained by ram air.
We know that enthalpy lost by compressed air
$=m_ac_p(T_3-T_4)=38.5\times 1(413.4-333)=3480 kJ/min $ ..........(i)
Since the temperature rise for the ram air is not to exceed 30 K,
therefore enthalpy gained by ram air
$=m_R\times 1\times 30=30m_R kJ/min$ ........(ii)
Equating equations (i) and (ii),
$m_R=116 kg/min$ (Ans)

3. Combined output of both cooling turbines
We know that output of first cooling turbine
$=m_ac_p(T_2-T_2’)=116\times 1(288-273.5)=1682 kJ/min$
$=1682/60=28 kW$

and output of second cooling turbine
$=m_ac_p(T_4-T_5)=38.5\times 1(333-240.4)=3565 kJ/min$
$=3565/60=59.4 kW$

Since the transmission efficiency=60%,
therefore combined output of both the cooling turbines
=(28+59.4)0.6=52.44 kW (Ans)

4. C.O. P. of the refrigerating system
We know that C.O.P. of the refrigerating system
$\frac{210Q}{m_ac_p(T_3-T-2)}=\frac{210\text10}{38.5\times 1(423.4-288)}=0.4$ (Ans)