Otto Cycle Calculation Problem statement: The compression ratio of an air-standard Otto cycle is 9.5. Prior to the isentropic compression process, the air is at 100kPa, 35oC, and 600 cm3. The temperature at the end of the isentropic expansion process is 800 K. Determine the highest temperature and pressure in the cycle; the amount of heat supplied per cycle (in kJ); the thermal efficiency; and the mean effecti ve pressure. Perform the calculations assuming 1) cold air standard conditions and 2) air standard conditions. p State 3 Given: State 4 Otto cycle, r = 9.5 State 2 State1: o p1 = 100 kPa; T1 = 35 C; V1 = 600 cc State 1 State 4: v T4 = 800 K TDC BDC Assumptions: Mhd1)ldiMethod 1) cold air stan dddard Mhd2)iMethod 2) air stan dddard To find: a) Tmax, Pmax b) QH (kJ) c) ηth d) MEP p State 3 Solution: State 4 State 2 System mass: State 1 v Method 1) Cold air standard Isentropic process(1 – 2), ideal gas: Isentropic process(3 – 4), ideal gas: p State 3 State 4 State 2 State 1 v Note that Tmax = T3 = 1970K Å ∴ pmax =←?kPa Heat transferred: p State 3 State 4 State 2 First law: U3 −U 2 = Q2−3 −W2−3 State 1 v but, and, QH = ? ← p State 3 By the same logic, State 4 State 2 State 1 v Qc = ? Conservation of energy, Efficiency of a heat engine, W η = = ? ← QH Mean effective pressure ( MEP) p State 3 Method 2) Air standard State 4 Recall system mass, m = 6.79 X 10-4 kg State 2 -4 3 Recall BDC volume, V1 = V4 = 6.00 X 10 m -5 3 Recall TDC volume, V1 = V4 = 6.32 X 10 m State 1 vT( ) v v Isentropic process(1 – 2), ideal gas: r 2 = 2 = ? vTr ( 11) v Interpolating from the table for ideal gas properties for air: TKuT=⇒=308 ?kJ and vT= ? 11( ) kg r ( 1 ) vTr ()2 = ? ⇒=TKuT?& = ?kJ 22( ) kg p State 3 Isentropic process(3 - 4), ideal gas: Sta te 4 vr (T3 ) v3 1 State 2 = = vr ()T4 v4 r State 1 v From the table for ideal gas properties for air: and vr (T4 ) = ? TKuT44=⇒=800( ) ? v T = ? ⇒ TK==?&u T ?kJ r ( 3 ) 33( ) kg p State 3 Tmax = T3 = ? K Å Sta te 4 State 2 mRT3 ppmax= 3 ==? V3 State 1 v ⇒=pmax ?kPa ← Heat added per cycle: Q = ?kJ H kg Cycle efficiency: QC ηth =1?−= QH Mean effective pressure: W MEP= =←? kPa VV12−.