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 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

Solution : - Given :- P = lookea Ti = 300k Gv = 0.7 kJ/kgk m = 1 kg P2= 2000 kPa K= 1,4 T3 = 1100k Diesel cycle ! P TA 2 P₂ = P3 3 2 4 4 1 s Vi = 14 V By Meyer's equation Cv 11 R K-1 R

R = (k-1) Cv = (1.4-1) X 0.7 R = 0.28 kJ/kgu For Process 1-2 kal k Tz P2 ( Ti Pi Tz 1.4-1 1.4 2006 = 300 100 Tz = 706.06 k PIVI -MRTI Vi= MRTI pi = I X 0.28 x 300 100 Vi = 0.84 m3 - V4

P₂ V₂ = MRTz V₂= т RT, P2 1x 0.28 x 706-06 = 2000 V₂=0.0988 m3 P3 V₃ = MRT3 V₂: = MRT3 P3 Ix0.28 X 1100 = 2000 Vz= = 0.154 m3 A] Heat addition and Workdone in each Process Heat added > in process 2-3 (Q) Q = m Cp AT

Q = mx Cp (T3-T2) Cp = R+C =0.28+ 0.7 = 0.98 kJ/kgk . Q = 1x 0.98 (1100 – 706.06) Q = 386.06 KJ Ans Work done in process 1-2 :- Wiz Wi-2 PV, -P₂ V2 = K-1 - (100x 0.84) -(2000X0.0988) 1.4-1 Wiz = -284 kJ - Ans

W2-3 Workdone in process 2-3 W2-3 = P2(03-1₂) 2000 (0.154 - 0.0988) = W2.3 2-3 = 110.4 KJ - Ans W3-4 = Work done in process 3-4 Wsu PzV₂ - = Pq V4 K-1 K k-l K T4 T3 P4 V3 P3 Ja Pų 1.4 0.154 = Pa 0.84 (Vi= U)

Pa = P3 x 0.093 - 2000X0.093 Ph=186.02 kPa W3-4 = (2000x 0.154) = (186.02 x 0.84) 1-4-1 W3-4 = 379.35 KJ Ans Warl > Workdone in process 4-1 Whol -0 KJ -ARS Because constant Volume Process

B] The Net work output Whet Whet = Wrzt Wz-z + W3-4 + Wars = (-284)+ (110.4) +(379.35) +o Whet = 205.75 KJ • Ans c] The Thermal efficiency (Mon): nih Whet - 205.75 - Heat added 386.06 り2 = 0.5329 or 53.29%