Lecture Note
University
Swansea UniversityCourse
EG-086 | Engineering SciencePages
4
Academic year
2023
Lloyd Reader
Views
0
Glory to you Q. Compression - Ratio of an otto cycle is 8. The temperature and pressure of air before compression are 288 K and 1 Bar respectively. If 1860 KJ/kg of heat is added to the gas per cycle. Find Pv and temp at each point of the cycle, heat rejection and thermal efficiency? P Given, Re=8. Qs P1. bar 2 Ti= 288K. world 4 Qs = 1860 EJ/kg. Qr -0381 In diagram u To find, 0821 P2, P3 Ph, T2, T3, T4, Qr and Netti of -012 Process 1-2, Reversible adiabatic compression, 5 Pizr = P2 P2 (R) (375-HS) P1 V2 P2 = T,(R) = 1 x(8)14 to P2= 18.379 bar in,9 5-1 ver = Rc) also, = TI 1.4-1 659.52K. T2 TI (R) (8) = Volume heat addition process. 2-3, constant Process ; bres Qas mer (Ts-72) 1x 0.718 (75- 659.25) wt (m=1vg) loves provision 1860 grea T3 3252.17K We have at constant volumi, P2VP = P3 V/3 T2 T3 P3 P2 x T3 18.379 x 3252.17 = - T2 659.52 P3 90.33 Bar
Now, Thermal efficiency N=1- 1- (Rc) 5-1 (8) n 0.564 or 56.4% oth To calculat amount of heat rejected: We have n Work done Qs - Qr - Heat supplied Q 0.564 = 1860-Qr composits 1560 Qv = 810.96 15/kg Consider process 4-1 constant volume heat rejection. Qr = m W (T4-II) 810.96 = 0.718 (T4-288) (i) J4= 1415.66K also at Constant volume 2/4 P, V/ V4=14 I - T1 T4 IX 1415.66 P1 is PH= x Tn - T1 288 P4 4.915 bar A 4- cylinder pethol Engine has swept Volume of 500 cc and learance volume 60 ll. If the Ps and temp at the Beginning t compression are 1.02 Bav and 297 K. man Cy cu temp is 673 K. Calculate air standard efficiency and mean effective vessure.
Given: Us=500cc To find: n=9 Vc 60 cc Pm-? P1= 1.02 Bar T1= 297 K T3= 1673 K. Compression Ratio: R==" = = V+Vs = [1+'500] 500 U2 ve Rc= 9.33 E.P.P 1-1 n-1-1 (R) (933)04 n= 59% mean effective Pressure (Pm) = W.D Qs-Q strope Volume (49-V2) very W.D = Qs- Prat hot Qs= me (T3-T2) T2 To = (v) T2=T,(R), T2 297 x (9-33)44 3/1/2020 at T2=725.61 K Qs- IX0-718(1673-72562) Mally Qs 680.21 15/kg T-1 r-1 Q.-mlv (T4-TI) , T4 T3 (S) 23 Vh - We 24 5-1 (Re) e TH= 1673 Hig (9.33) 04 T4= 684.76K us Qv= = 1 x 0 . 718 ( (684-76-297) H.1 Qr = 278.41 25/kg (Nixpf.di)
W.D = Q-Q - = 680.21 - 278.41 = 402.344 kj/kg P,U, = MRT, 2, = MRTI - 1x287 x297 P, 1.02x10*5 V1 0.835 m'/kg Rx V, U2 V2 = 0.835 = 0.0894 m3/kg 9-33 Pm= 4020344 = 539.62 KN/M2 (0.835-0-0594) Pm 5.39 bar
Problem of Thermal Efficiency
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