A bootstrap cooling system of 35 T is required for an aircraft. Temperature and pressure of the atmosphere is 18°C and 0.75 bar. Pressure of air is increased 0.75 bar to 0.92 bar due to ramming. Pressure of air leaving the main compressor and auxiliary are 3.2 and 5.4 bar respectively. Isentropic efficiency of both compressors is 84% and of turbine is 81%. Heat removed from air leaving the compressor is 60% in the first heat exchanger and 34% in the second heat exchanger which is after auxiliary compressor. Assuming ramming to be isentropic and cabin pressure 1.03 bar, find power required to take cabin load and COP of the system. Temperature of the air leaving the cabin is 28°C.

Mumbai University > Mechanical Engineering > Sem8 > Refrigeration and Air Conditioning

Marks: 12M, 15M

Year: Dec2012, May2014

Capacity=35 TR

T1=291 K, P1=75 kPa

P2=92 kPa, P3=320 kPa, P5=540 kPa

Process 1-2= Isentropic ramming of air

Process 2-3’=Isentropic compression in main compressor

Process 2-3=Actual compression in main compressor

Process 3-4=Constant pressure heat rejection in heat exchanger 1

Process 4-5’=Isentropic compression in secondary compressor

Process 4-5= Actual compression in secondary compressor

Process 5-6=Constant pressure heat rejection in heat exchanger 2

Process 6-7’ =Isentropic expansion in cooling turbine

Process 6-7=Actual expansion in cooling turbine

Process 7-8=Constant pressure heat addition in cabin

Assuming isentropic ramming

$η_c=0.84 \text{ for both compressors and} η_t=0.81 \\ Q4=0.4 × Q3 \\ P8=103 kPa \ \ and \ \ T8=301 K$

For 1-2,

$\frac{T2}{T1}=(\frac{p2}{p1})^{(\frac{γ-1}γ)}$

Assuming $\gamma$=1.4 we get T2=308.49 K

For 2-3,

$$\frac{T3i}{T2}=(\frac{p3i}{p2})^{(\frac{γ-1}γ)} \\ T3i=440.47 K $$

Also, $η_c=\frac{(T3i-T2)}{(T3-T2)}$

We get, T3=465.61 K

Assuming no pressure loss in heat exchanger so P4=P3= 320 kPa

For process 3-4,

$$Q4=0.4×Q3 \\ ∴ Cp×t4=0.4×Cp×t3 \\ t4=0.4×192.61 ℃ \\ T4=77.043℃=350.043 K$$

For process 4-5,

$$\frac{T5i}{T4}=(\frac{p5i}{p4})^{(\frac{γ-1}γ)} \\ T5i=406.488 K $$

Also, $η_c=\frac{(T5i-T4)}{(T5-T4)}$

We get, T5=417.24 K

Assuming no pressure loss in heat exchanger so P6=P5= 540 kPa

For process 5-6,

$$Q6=0.66×Q5 \\ ∴Cp×t6=0.66×Cp×t5\\ t6=0.66×144.24 ℃ \\ T6=95.198℃=368.198 K$$

Assuming no pressure loss during process 7-8 so P8=P7= 103 kPa For 6-7,

Assuming Cp= 1.005 kJ/kgK

We get, $m ̇_a$=2.6925 kg/s

Substituting in equation (1),

$$\dot{W }_{net}=472.492 kW \\ Now, \ \ C.O.P=\frac{\dot{Q }_a}{\dot{W }_{net}} =\frac{(3.5×\text{capacity})}{472.492}=0.2593$$