Consider a internal combustion engine modeled as follows an a P-v diagram. The c

Question

Consider a internal combustion engine modeled as follows an a P-v diagram. The compression ration is 10. 1 to 2 intake Isobaric P1=kPa, V1=4.8x10-5 Find V2=? Find W1to2=? (work in expansion) 2 to 3 Compression P2=100kPa n=1.4 Find P3 in kPa Find W2to3=? (Work in compression) 3 to 4 Isochoric heating P4=8000kPa Find W3to4 (isochoric work) 4 to 5 Power expansion n=1.4 Find P5 in kPa Find W4to5 (work in expansion 2 to 1 Exhaust compression isobaric Find W2to1 (work in compression) Find the net work, known as the work cycle, Wcycle, which is the shaded area. Wcycle=W1to2+W2to3+W3to4+W4to5+W5to2+W2to1

Explanation / Answer

r =10 = v2/v1 v2 = 4.8*10^-4 W1-2 = P(delV) = 10^3*(4.8*10^-4 - 4.8*10^-5) = 0.432 J P3V3^1.4 = P2V2^1.4 P3 = 10*(4.8*10^-4/4.8*10^-5)^1.4 = 10*10^1.4 = 251.2 kPa W2-3 = P2V2 ((V3/V2)^(1.4-1) - 1) = 10^3 *4.8*10^-4 ( (10^-5/10^-4)^0.4 - 1) = -2.89 *10^-3 KJ = -2.89 J W3-4 = V(del P) = 4.8*10^-5 (8000-1) = 383.952 J P4V4^1.4 = P5V5^1.4 P5 = P4 (V4/V5)^1.4 = 8000(4.8*10^-5/4.8*10^-4)^1.4 = 318.5 kPa W4-5 = P4V4( (V5/V4)^(1.4-1) - 1) = 8000*4.8*10^-5 ( (10^0.4) - 1 ) = 580. 56 J W5-2 = V (del P) = 4.8*10^-4 * (318.5 - 1) =152.4 J W2-1 = P (del V) = 1000*(4.8*10^-5 - 4.8*10^-4) = -0.432 W =0.432 + 2.89 + 383.952 - 580.56 - 152.4 - 0.432 = -346.118 J negative sign indicates work by the system

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