Question

1. A single cylinder 1.6 lt, spark ignition engine is operating on the ideal Otto cycle at a compression ratio of 9. The ambient air around the engine is at 1 bar and 20 °C and the fuel consumption of the engine is 13 L/h gasoline at 5600 rpm. Evaluate a. The highest pressure and temperature in the cylinder throughout the cycle b. Net work output of each revolution and power output of the engine in horsepower C. Thermal efficiency by using net work and heat added to the system d. Thermal efficiency of Otto cycle in literature is given as 1-a where CR is the compression ratio. Compare the efficiency calculated with this formula and your result in part c. If there is a difference comment on the reasons and which one is more accurate Crk-1

The solution belongs to the 1st question, i need the solution for the 2nd question.

Answer 1

The solution is shown below if any doubts please post in comment section.

02). Ty = 108 l. 413k., Pilbar Ti - 293k. 15 = 2bar A 5 5' S $ Torbo Charger turbine РЧ T4 11 V P. Pup = P1 1081.43 (1) - 3.7bar 293 ? (P5 РЧ, T5 T4 solut (31) To loslo 43 30, T5 = 907.115. 7 0085 = Т - TsІ T4 - T5 त Th – 085 (Tu-TG) = 1081143 - 0985 [1081343 - 907.10) Terbine So Tol - 983,258k axit temperature he (mating up [Ty - Tat = [0.0896 + 278064103] xl005 (108)43 -933.75 TNT = 13,76kw.l w and (1) Compressor Exit temperature pressure for ist iteration, KG = We work by turbine 1 - work in for compresor na co [T, - To = 6:0896 X 1.005 [1,-293] 201 c = 2935 Ti => 445.775 V 13.76

exit pressure ne = Ti-To T - TO .85 = T1 - 293 445.77-293 So Ti - 492.8545k = Pi РО To 004 [isen tropic temperature] 1.4 apol 4.22 8545 (Po-ther) (2 293 TPI 3.61103 bar 3. Gbar (C) Converged turbocharger and compresor Erit tempo and pressure gince, B - constant and 8 = compression Tatio constant = re - Expansion raho. constant (There is no change in these values] Newengine. Bl. No so Ty is also remoins same 8- Since , T4 Ts vu I've rel and catso Raatso consturnt. So, PA T4 PI TI with New P, and To No3 13 1 (repeatig same procedure] - P4 'T4 PI 6 11 1081.43 2445.77 X 3.6 P4 Py = 8.734bar

Ry 8.734 bar = P4 1081.42 - T4 15 Ty N 14-1 4 80734 T5 1081.43 To= 709, 72K n=0.85 = T4-T51 T4-T5 T5l = Tu - 185 [14-15] 1081143 - 0085 [1081.43 – 709, 72 ] T Isla 759.5265kw combined turbine outlet temper atere, G gasoline. New flowrate Now TIF 445.77K P= 8.6barap, Vi 9 4 S Y2 So at To c V2 T2 - Fi 0.4 445.77 XG) 10 13:5145, na Cv Tz + ing (ev) = (ma+mf) Cv Tz 0:0896 X 0.714 X 1073,514 + + mif C45.881039 of J 0.118 (2604:32 To= constant 2604.32 for both engines [before rerbocharger & also of ter turbocharger] 2.242 x 103 kg (6.0896+ New in the

ulturbine - (oria tmp) C C Tu-15] (0.0896 + 2.2024103) (1.005) [1081.43 - 759.5265] Band iteration - 29.7 kW vane] 29.77 = Wc = trica Cp [T, - To 00896 X1.005 [T, - 293] T = 6 2 3. 6 6 : "Compressor Exit temperatur Converged" Ti-To ТІ- То T! 0.85 (623.6 - 293) +293 after the No = 0.85 = - 574 K. Р. Ро F TH To 0.4 a compressor TPI = Piw (34) 10.52 bar Exit te mossuthse power output of Engine with designet Kerboekanger ofter converged. We = HiS - HER! & Heat Supplied Heat Rejected mifrele cu) - (ma Mare Gov [ta-T) = 102.684 (0.0896 + 2. 242x103) (0.78 08lou3 6236 102.684 30.19 WE 11 72.5 kW