Question

An aircraft air conditioner works between the pressure limits of 1 bar and 5 bar. The
temperature of the air entering the compressor from the cabin and expansion
cylinder are 10o
c and 25oC respectively. The expansion and compression follow the
pv1.3=constant. Find the following:
1. The theoretical C.O.P of the refrigeration cycle:
2. If the load on the system assuming the actual C.O.P is 50% of the theoretical
C.O.P
3. The stroke length and the piston diameter of single acting compressor
running at 300 r.p.m and the volumetric efficiency is 85%.
(Take L/d = 1.5 : cp = 1.005 kJ/kgK ; cv = 0.71 kJ/kgK

Answer 1

Solution.

p₂ = 1 bar;

p₁ = 5 bar;

T₃ = 283 K;

T₁ = 298K;

pv1.3=constant;

L/d = 1.5;

c_p = 1.005 kJ/kgK;

c_v = 0.71 kJ/kgK;

$\dfrac{T_1}{T_2} = (\dfrac{p_1}{p_2})^ \dfrac{\gamma-1}{\gamma};$

$\dfrac{298}{T_2} = (\dfrac{5}{1})^ \dfrac{1.35-1}{1.35} =1.52 ;$

$T_2 =298/1.52 = 196K;$

$theoretical C.O.P =\dfrac{T_2}{T_1 - T_2} ;$

$theoretical C.O.P = \dfrac{196}{298 - 196} = 1.92$ ;

$actual C.O.P = 0.5 \sdot 1.92 = 0.96;$

$Refrigerating effect per kg of air = cp (T3 – T2) = 1.005 (283 – 196) = 87.44 kJ/kg.$

$V3 = \dfrac{mRT}{p_2};$

$R = (c_p – c_v) = 1.005 – 0.71 = 0.295 kJ/kg K;$

$Refrigerating effect produced by the refrigerating machine = 6 × 14000 = 84000 kJ/h;$

$Hence mass of air in circulation = 84000/(87.44\sdot60) = 16.01 kg/min.$

$V_3 =( 16.01\sdot0.295\sdot1000\sdot283)/1.0\sdot10^5 = 13.36 m^3/min;$

Swept volume per stroke =13.36/(2$\sdot$ 300) =0.02m³;

V = 0.02$\sdot$ 0.85 = 0.019 m³;

$\pi/4\sdot d^2\sdot l = 0.019; l = 1.5d;$

$d = 0.016m; l =0.024m;$

Answer: theoretical C.O.P =1.92;

actual C.O.P = 0.96;

d=0.016m;

l = 0.024 m.