Question

Use engineering methodology and show units calculations.

An air standard Otto cycle at the initial condition of P,=1 bar, T,=290 °K, Vi= 400 cm?. The maximum temperature of the engine is 2200 °K and compression ratio is 8. The mass of air is assumed to be 4.8x104 kg. Assume isentropic process exits in both compression and expansion strokes. (a) Sketch the P-V and T-S diagram of the cycle (5%), (b) determine the total heat addition to the gas in kj (20%). (c) Net cycle work output of the engine (5%) and (d) the thermal efficiency of the cycle (5%).

Answer 1

Solution Air standard otto given initial condition Pi = 1 bar, Ti=290k vi= 400 cm3 400x106m3 = 4 x 10-4 m3 compression ratio=&= V1 = va 3 maximum temperature ty= 22ook mall of our = 4.8x104 Kg = 48x 10 x 10 que molecular 4 -0.48 gn weight of Arm=29 gr/mol mod of ouy, 0.01655 mol n= 148 gu 29 gr/mol comprelion ratio, ✓ 8 4x104m3 N2 = sxio-sm? Pov diagram (Pa) PP3 3 T3=2200k Wout frain 4 P _2 > Rout Pi - win Veus vema u = V4 = 4x100 m3 T3=22ook T-S diagra diagram 3 Dit wout 14 tt E Win TT=290k Yout si

B) 1-2 procul in itsentropic proud to heat added 8,90 Amount of heat addition in proculi 2-3 823=mc CT3-T2) = 4.8x10 + Mg x 0.718 kJ x (2200 - T₂) y = for air Y = 1.4 Prous 1-2 is an it entropic proum, so we can Pina = P2 VY P = P1 . (A) = + barx (8)"4 apply = 18.38 bar According To to ideal gas law P2 U2 = nRT2 P2 kg 18.38x10sx sxlos pax m3 MR 0.0lbssmolx 8.314 J molek [T2 = 667.89k so, 023 10-4 kg x = 418 810-4 X0.78 T (2200 – 667.89) kgik = 0.528 KJ = 528J or din = 528 Joule total amount of heat added to the system will be 528 Joule 823 work done in proces 1-2 ciffentropic procum) in isentropic procell > 0,2 = 0 According to first law of thermodynamics

DU = 92+ k 12 DUR - ot W12 W 12 = mx (T2-T) 1,2 = 4.8x10-4, 4.8 x 104 kg xo. R0.H8 kJ x (667.89-290) kgik W 12 = 0.130 kJ = 130 joule, Cor! [Win=13051 1412=1305 work done in prouu 2-3 ccontant volume proum) W2z=0 work done in procum 3-4 (Ittentropic compression) for issentropic procum Azq=0 so = DU 034 W34 Wout Wout = mcu CT4 - Tg) According to ideal gas domes he equation at point 3 P3 V3 = nRT3 P3 MRT3 0.016 ss molx 8.3 14 J X 22ook molik V3 sx165 m3 P3 = 60.54 bar for proul 3-4 P3 V3 Pq V4 P3 x 14) 3.2abar PA N3 Cha) = 60.54 barx (+)"4 194 -

According to ideal gas equation, at pointa Pq Vq = nRT4 3.29x105 T4 = Pax 4x10m MR 0.016ssmol x 8.314 I molok ITA = 956.4K Pq V4 Wout - ma (Tq-TB) = 4.8x104 kg x 0.718 kJ (956.4 - 22000k kgik = -0.4206 kJ Wout - 428.6ks! 4- in prouM A (constant volume procup At work done W34 0 cycle Met work done in the otto Whet - W124 W23 +34 + Wal AN het = 130to-42816 to Whet - 298.6 Joule sign shows that took done by Air on System d) thermal efficiency of cycle Mau 298.6 What din s 28 = 0.56ss or s6.55% Thermal efficieney of otto cycle will be soiss% Ang