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

Diesel Cycle : Pro kpa 4 (P-v diagram) 100kpa r = 1.4 V2 V Vi= Vy P=0 V=C. Si=52 -Sy s stale : P = 100 kpa, T, = 300k stale 2: P2 = P₂ = 2000 kPa. state 3 i P2 = 2000 kpa. T3 = 11000k. for Process 1-2: ( Isentipoic Proces). Compression. 14. 1.4 Pivi P₂ V₂ MRT. 1.4 P. (mesos 3*+ Pas Cher V= (MR)T P. POCO X2

1.4 1.4 1.4 Ti (MR) 1.4 T2 (MR) 0.4 P P20.4. 1.4 0.4 ( 77 (6) 0.4 T2 2000 Toy 100 300 T2= 300 X 2.353 T2 = 706.06K work done (1-2) Piv, - P₂ V2 Y-1. Wi-2 MR (T. -T2) 7-1. - 1x 0.287* (300 - 706.06) 1.4-1 Wi-2 = -291.35 kJ C-ve indicates won the Process).

heat transfer in 1-2 91-2 (asentropic Compression)

for Process 2-3: Constant Pressure heat addition, P₂ = P2 = 2000 kpa. Workdone W2-3 - T3 = 110ok. Се 3 P dv (p= 1.4x07 [(p= 0.9865/66 P (vz-V₂) MR (Tg - Tz) 1 x0.287 *( 1100 - 706.06) W2-3 = 113.06 KJ! 92-3 mcp (T3-T2) (P=c) 1x 0.98 (1100 - 706.06) Q2-3 .= 386.06 KJ

for 3-4 Procen (Isentropic Expansion ansion Process). 1.4 P₂ V₃ 1.4 Pu Vu 1.4. P₂. MR. Tz P3 MR. Ty Pu L (me. Pu 1.4 1.4 T3 Tu 1) 2014 Pu 0.4 1.4. 0.4 IT: (P3 Py こ Tч V2 V3 Vu Vy 2-3 (P=0) ¥ =]. Vy=vi. TB2 VI vo Жч M100 100 You. Jul 706.06 2000 0.1833 V3 Vy 1.4 Py = 2000 (0-1433] Py 186.02 kPa

from eq 0.4 P3 1.4 T3 Py Ty 0.4 2000 Ty 186.02 1100 1.9710 Tu Ty = 558.07k Workdone (3-4) P3 V₃ - Puvy W3-4 = MR (T3-T4) 11 V-1 1 x 0.287 X (1100 - 558.07) 1.4-1 W3-4 388.83 KT

heat transfer in in nu 93-4 = 0 (asentropic Expansion).

for 4-1 (Procm) Const volume heat rejection Wargao (vec) spava Pu-l- mc, CT, -Tu) 1x0-7 0.7 ( 300 - 558.07) Pu-l - 180.649 C-ve indicates heat rejection Process) 6 net work output: + W3-4 + Wurl - + W 2-3 1-2 -291.35 + 113.06 + 388.83 +0 Whet 210.54 kJ therneal efficiency n thermal Whet P supplied - Whet 92-3 210.54 386.06 0.5453 1+ th 54-53%
I hope you will understand the answer and i gave it in a detailed manner and please thumbs up...