Question

7. The principal states in a cold air-standard Diesel cycle are given below. The mau of artist 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 kJ. (c) the thermal efficiency State 1: p = 100 kPa; T = 300 K State 2: p = 2000 kPa State 3: T = 1100 K

Answer 1

Solutions m= 1 kg 3 const po 1 cerst volume G= 0.7 KJkg K=1.4 = 8 State 0 P = lookia, T=300k State o h=2000kPa State Prouss 1-2, I = TIN pectR (cp=1.ooks T = look KI for aur rol al Pe B trobic 300 RTI RT2 2000 Joo 6000 0.4 T=300l Z = 300 x 6000 r = 8.498 T. TOM= 9736.036 To. rc = 3 = V3 T 14 = 9736.036- The 70G.OGL Perfor-p re= 1100 1.56 process 3-4 E 1 h 706.06 Heat addition Ty = T () Qs = m Cp (55-52) = 1*1*(1100-706-06) = To 393.94 KJ QR = mcv ( Ty Ti) = 1x07/55636) = 1100 l is not a 558:36 187.852 kJ

Workdene Net work out but = 393.94- 87.852 = 206.088 KJ The smal efficiency = 1 L (8.498,004 88_1 Lrac (1.5614 "27.4(1.561) -|-0-424B8(1.10.164) = 0.53193 = 53.193% 1-2 (Isentropic) ( Heat addition will bezeso a zo W1-2= Rivi - PzV₂ = MRITE = 14 0.3 (200-706d) - 1210818 kJ (const Pressure) Wez= m R (T3-IZ) = 1+0.3 (1100-706) Qg=22-3 = 118.2 kJ 3-4 (Isentropic prags Q3450 W3-4 = MR (T3T4) = 100.3(1100-558.36 162.49 KJ ५- २५-1 - 20 const volume Whi= Pdv = 0 Y-1 ун