Question

Q. 2 An air cooling system for a jet plane cockpit operates on the simple cycle. The cockpit is to be maintained at 25 oC. The ambient air pressure and temperature are 0.35 bar and – 15 oC respectively. The pressure ratio of the jet compressor is 3 . The plane speed is 1000 km/hr. The pressure drop through the cooler coil is 0.1 bar. The pressure of the air leaving the cooling turbine is 1.06 bar and that in the cockpit is 1.01325 bar. The cockpit cooling load is 58.05 kW. Calculate- (i) Stagnation temperature and pressure of the air entering the compressor
(ii) Mass flow rate of the air circulated
(iii) Volume handled by the compressor and expander
(iv) Net power delivered by the engine to the refrigeration unit
(v) COP of the system

Answer 1

Answer:

The air is passed through a heat exchanger after compression and cooled to its original condition entering into the air jet. The pressure loss in heat exchanger is 0.1 bar. the pressure of air leaving the cooling turbine is 1.013 bar and is also the pressure in the cockpit. the cooling load in the cockpit is 70kW. Determine i)Mass flow rate of air circulated to the cabin. ii)Net power delivered to the refrigeration system. iii)The COP of the system.

Answer 2

Mass Flow Rate 7200 kg/hr

Net Power = 11100 cu.m./hr

Volume handled = 5220 cu.m. / s

COP = 0.306

Explanation:

• Mass Flow Rate of Air:

We have, (1000*1000/3600) == 278m/s (speed)

T' = T + V.V/2Cp, therefore T' = 296.5 K (temperature)

p' = 0.57 bar, p(3) = 1.71 Bar, T(3) = 406K

therefore, referigeration effect = 35.002 = 35 kJ/kg

Therefore Flow Rate = 70 * 3600/35 =  7200 kg/hr

• Net power

We have  $V_c = mRT_2/p_2, V_e = mRT_5/p_5 \ etc.$

Therefore Net Power = 11100 cu.m./hr

• Volume handled = 5220 cu.m. / s

• COP = Refr. eff. / Net Work = 70/229 = 0.306