Question

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

ASSUMPTIONS:

1.) All process are internally reversible.

2.) The workout will be partially from constant pressure heat addition and adiabatic expansion process.

3.) The working fluid air behaves as an ideal gas.

4.) Combustion process is replaces by constant pressure process.

Sof: Given, Gold - Air Standard Diesel cycle. air 1 kg к = Mass of (m) Gv of air 0.7 KJ kg-k 8 of air 1.4 Temperature at Inlet (Ti) 300K Temperature at State 3 (T3 (01) Tmon) = 1100k Pressure at Inlet (Pi) 100 kPa Prenure at outlet of Compressor (P2) 2000 kPa. T-S Diagram T3 A S=C 3 + P=0 C 2 S=C Ti 1 Process 1-2 → Internally reversible Adiabatic Comprenion Procen 2-3 → Internally reversible Isobaric heat addition Process 3-4 → Internally reversible Adiabatic Expansion Process 4-1 Internally reversible Isochoric heat Rejection (a) heat addition and work done in each Process : Process 1-2:- Heat addition (Q) = 0 (Adiabatic)

Cv (T2-Ti) Work Input (WI) that, ТР we know 1-) =c 3,72 Topi = Tp Pa -104 100 1.4 2000 300 x ey Ti - 127,46 K Work I put (nog) 2:0.+ (127146-100) 2 19.22 RJ ка Wp mwa 1x 19.22 2/19:22 KG Process 2-3 :- addition (95) 1 x 0.98 (1100-127.46) a. Heat m Gp (T3 - To Qg = Qs - 953.08 kJ We know that, P2 pi 2000 Too polt 8 ..10 8.49 from Ideal Gas Equation, PV = mrti

100 x 10° x V, 1 X 287 X 30 0 : Vi 0.861 m 3 » 8- V 8. 0.861 V2 0.861 À Va |V2 = 0.001 m3 8.49 Now, Cutt- off ratio (3) V3 Tz Te 1100 124.46 8.63 8063 There fore, V3 Oitol 3 = 8163 V₂ 0.8716 m ... Work during this Process w 2-3 P2 (Vz-V2) 2000 (0.8716 - 0.101) W 2-3 w 2-3 154.12 KJ Process 3-4:- Heat Transfer (@3-4) -0 Work done (hy) me, (T3-T4) 는

Since V=V4 = 0.861 m3 9-1 To Th TTOO lik- T4 0.861 0.8716 . Tx=1105.39k W 1 x 0.7 (1100 - 1105:39) 3-4 W 3-4 31773 kJ Process 4-1- Heat Rejection (QR) may CT4-7) 1x014 ( 1105,39 – 300) QR * :- QR 563,77 kJ Work Interaction is. Zero in Iso choric Process. b. Net work output :-(EW) فيع t W W 2-3 3-4 Wo I EW 154:12 + 3,773 - 19:22 Ew 2 138.67 KG C.) Thermal efficiency :- non 1- qR qo * - You 563,77 1 : Non = 40.84% O 953.08