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 kl. (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

Firstly pressure and temperature is calculated for each states. Then heat input is calculated. And work done in each process is obtained. Then net work is calculated. Using this, thermal efficiency is obtained.

o 3 2 Given, Cold-air standard Diesel cycle. Mass of alr, m = Ikg. Cv = 0.7 kJ/kg-Ki K 31.4 v T=300K Given states. State: p=100K PL State 2: P = 2000K Pa State 3: Ta1100K For Process 1-2, it is adiabatic compression process, [f=301 147 so T, TI IPL PL 1200 100 > T2 = 300 x 120) & T = 706.06K Process 2-3 is constant pressure process, So, PoEP, = 2000KPA Doroby ca) Heat addition = m Gp AT Qin = m (KG) AT Ft (1,43 017) Xp100-706) Hot Qin = 386.06 kJ 3-4 is, adiabatic expansion proces, T4 T3 (V4) 113 km

VE Tz TE Tz f T24 Vg vi EV - 7. In = T2 & 13, V2 Te VE + 444 T. Pi K Fire 11:44 KY E1100/1100 V2 _706.08) =1100*11100 100 206-04 42000) FT4: 558.07 K Now, During proce ocek 1-2. [ Adiabatic ] Qin co from 1it law of thermodynamics Qin =AU & W mcv (T2-T) 1x 0.7*1706.06-300 Wizz- 284.242 kJ 7 W= [ Work done during T2 During process 2-3 { constant pressure}, 3,86.06 m Gv (T3-T) + W 2-3. W2g = 386 06- 38646-1X0.7-(1100 - 706.06) 7 W2-3 = 110.3 kJ During process 3-4 [Adiabatic] Qin = AUTW mG (T4 -T3) + W 34 1 x 0.784558.07–1100) 0 * W34 G W34 = 379.35kJ

During process 44. [Constant Vetilme]. W# =0 Work done in each processes are, W, 2=-284.24 Wzz = 110.3 kJ W34= 379.85kJ W4y=0 (6) The net work ocutput is Whet = Wo+W23 & W34*W41 -284+24 +110.3 + 379.35 +0 >Wnets 205. HKS (C) Thermell efficiency . Whet 205c4l nh *1907 386-06 7th = 53,21% Qin