Question

Problem 9.011 SI Consider a modification of the air-standard Otto cycle in which the isentropic compression and expansion processes are each replaced with polytropic processes having n = 1.25. The compression ratio is 10 for the modified cycle. At the beginning of compression, P1 = 1 bar and T1 = 310 K. The maximum temperature during the cycle is 2200 K. Determine: (a) the heat transfer and work in kj per kg of air for each process in the modified cycle. (b) the percent thermal efficiency. (c) the mean effective pressure, in bar. Part A Your answer is partially correct. Try again. Determine the heat transfer and work in kj per kg of air for each process in the modified cycle. Note the air is the closed system. Process 1-2: Q12 X = 0 W12 kJ/kg m -276.9814 m KJ/kg Process 2-3: Q23 x W23 = 0 0 m k]/kg m kJ/kg Process 3-4: Q34 W34 = 0 1105.35 m kJ/kg m kJ/kg Process 4-1: 941 X 665.6937 W41 kJ/kg m 0 kJ/kg m SHOW HINT SHOW SOLUTION Attempts: 2 of 15 used SAVE FOR LATER SUBMIT ANSWER

Answer 1

p/12820 4 Vilve 210 2 putas ac Pialbar Ti=3lok. V For Ti= 310k, "1=221.25 kJ/kg (Table A-22) ni T2 = 21018) (10):25-1 T2=551.27k. For T2=551.27k, U2 = 397.83 kJ/kg At T3 = 22ook, U3 = 1872.4 kJ lkg . T4= Co Tz - Go) (2200) 1.25-1 = 1237.15k V4 U4 = 966:40 kJ/kg wiata pdv = RCT2-TI) 0.287 kJ/kgk) C551.27-3100K 1-1.25 WIZ - 276.98 kJ/kg Qiz (12-21) + Wiz (397.83 –221.25) + (-276.98) 100.39 kJ lkg QiZ -

For process procers 2-3 Was anstant volume heat addition. Com - (Us -Ue) = 1872.4 -397.83 Q23 - 1474.57 kJ/kg Process 34 W94 W34 = RC 74_Ts) Spalu = 0.287 (1237.15-22007 l-1.25 W34 = 1105.35 Kolkg Qa4 (14-us) + wish to (966.4 766.40 – 1872.4) + 110 5.35 834 = 1999.35 kJ/kg m. WKI =0 Faso poocers 4 Qula 2,-44 = 221-25 - 966.4 Q4) -745.15 kJ/kg

by the wycielm Qilm wa cycle Way 9-276.98 + 1109.34 Wacle 928.36 kJ/kg 1673.92 kyllg Que or on boys 1474.57+199.35 tha 828-36 kolky = 6.4948 1673.92Kulkg 7th = 49.48% (c) mep= .0.89m3/kg Wasce lm vi(I-Valvi) RUI= RTI =Vis 0.287 x 310 Ixloo mep= 828-36 kJ/kg (0.89 melkg) Cl- to) 1034.15 kPa.x Ibar lookla imep = 10:34 bar