Answer to Question #110695 in Thermal Power Engineering for Mohd Amir Hussain

Question #110695
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
1
Expert's answer
2020-04-24T13:38:41-0400

Solution.

p2 = 1 bar;

p1 = 5 bar;

T3 = 283 K;

T1 = 298K;

pv1.3=constant;

L/d = 1.5;

cp = 1.005 kJ/kgK;

cv = 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\n= cp (T3 \u2013 T2) = 1.005 (283 \u2013 196) = 87.44 kJ\/kg."

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

"R = (c_p \u2013 c_v) = 1.005 \u2013 0.71 = 0.295 kJ\/kg K;"

"Refrigerating effect produced by the refrigerating machine\n= 6 \u00d7 14000 = 84000 kJ\/h;"

"Hence mass of air in circulation\n= 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.





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