Question

7. The principal states in a cold air-standard Diesel cycle are given below. The mass of air is 1 kg. The Cv of air is 0.7 kJ/kg.K and k = 1.4. Determine: (a) the heat addition and work done in each process, in kJ. (b) the net work output, in kl. (c) the thermal efficiency. State 1: p= 100 kPa; T = 300 K State 2: p= 2000 kPa State 3: T = 1100 K (35 points)

Answer 1

Qs OR. given Pressure P = lookpa T, = 300k f = 2000kpa T = look m=lky using ichal PV = MRT = 1008 V = 1X0.287X300 Volume gas equation , V = 0.86/m3 SII 200 g Tz T. P2 Pl = 300 100 22 = 7066 064k P₂V = MR Tz 2000 xV₂ = 140.28787060064 V2 = 0.10132 m3 Process 2-3 is constant pressure prouss.

12 TE 0.10132 V (100) Tz V3 * To V₂ T3 706.064 V₂ = 0.15784 m3 a) Heat addition = mcp ( 13 -T2) = 1Xky (13 (2) 1x 148 0.7 (1100 386-05728 kJ - 1x 7062 064 PA) Wi-z = malt, -12) MRIT,-T2) 1X0.287 (300—706.064) KY 1.4-1 1-2 -291. 3509 kJ W 2-3 P2 ( V z ve = 2000 (0.15784-1 4~0010132) W2-3 = 113.04 kJ

V = V4 = 0.86/m3 T = 3 Nynky 1:47 loo 10.861 0.15784 V3 Ty = 558.0566k, W3y =- MR (Tu-13 -180,287 (558.0566-1100) 2 서 104-1 Wzy = 388.8443 KJ Wy-1=0k] Net work = -291. 3509 +113.04+388.8443 +0 (W W = 210.5334 kJ 210.5334 = 0.54 534 y Efficiency QA 386.05728 ly 54.5342%