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 (35 points)

Answer 1

All subdivisions are solved below.

P Pac 3 P2 = P3 / 2. Given State : P = 100 kPa Ti = 300k state 2: P2 = 2000 kPa State 3: Tz = 1100 K. Note: C of air = 0.7 KJ/KG.K. Pre=c K= 1.4. solution: we know that p =k Cr 7 p = 1.4 V=V4 (P-v diagram for air Cp- 0.98 KJ/69.6. std. diesel cycle) pre 4. Y-C 0.7 we know that, Cp - G=R + R = 0.98 – 0.1 = 0.28 KJ/KOK Process 1-2 : Isentropic compression (P.V= PVK) so Tz = (P2 At state, TO Pii= m RTI T2 100 (V) = 1 (0.28) (300) K-1 K R 1-4-1 1.4. 12000 300 100 V, = 0.84 m? T2 = 706.064K = (222) ve - And V = 1P2 ) Yk V2 0.84 000 1.4 12= = 0.099 m m3 V2 100 Process 2-3 : constant pressure heart addition 50, P2 = Pz = 2000 kPa and T3 = V₂ T2 Vz 1100 V = V3 = 0.1542 m 706.064 0,099

so, TA K-1 1.4-1 1100 Process 3-4: [sentropic expansion (PVL = PAVK) - Cha) Note: Process 4-1 is Tz constant Volume heat Tq (0:1542) rejection process 0.94 Hence V4 = V =0.84 m3 T4 = 558.361 K aj Heat addition Qin): for process 2-3] Qin = m Cp (T3-T2) Qin = 1 (0.98) (1100 - 706.064) Qin = 386.0573 KJ Heat rejected, Qrey') : [for process 1-4] Dorej = m G (TA-T) Qrej = 1 6.7) (558.361 – 300) Qrej = 180.8527 KJ Work done for process 3-4 (Wout): For isentropic Wout = m R (Tz-Ta) process. KI ba= m RCT3-T) Wout= (0.28) (1100 - 558.361) K-1 1.4-1 Hout= 379.1473 kJ 3+ b] Net work output (Wnet); we know that, Net work done = Qin- Orej - 386.0573 - 180.8527 Whet Whet = 205, 2046 KJ

c] Thormal efficiency (": Whet 205. 2046 n = Qin 386.0573 0.53154 = 53.154 % diesel cycle

Note:

• There will be no heat transfer in process 1-2 and 3-4, as they are reversible adiabatic process (isentropic process). Hence, the heat transfer values for process 1-2 and 3-4 are 0 KJ.
• There will be no work done in process 2-3 and 4-1, as they are purely heat addition and heat rejection process respectively. Hence, the work done values for process 2-3 and 4-1 are 0 KJ