Thermodynamic Cycles Air-standard analysis is a simplification of the real cycle that includes the following assumptions: 1) Working fluid consists of fixed amount of air (ideal gas) 2) Combustion process represented by heat transfer into and out of the cylinder from an external source 3) Differences between intake and exhaust processes not considered (i.e. no pumping work) 4) Engine friction and heat losses not considered SI Engine Cycle vs Air Standard Otto Cycle FUEL A I Ignition R Fuel/Air Mixture Combustion Actual Products Cycle Intake Compression Power Exhaust Stroke Stroke Stroke Stroke Qin Qout Air Otto TC Cycle BC Compression Const volume Expansion Const volume Process heat addition Process heat rejection Process Process Air-Standard Otto cycle Process 1→ 2 Isentropic compression Process 2 → 3 Constant volume heat addition Process 3 → 4 Isentropic expansion Process 4 → 1 Constant volume heat rejection Compression ratio: v v r = 1 = 4 v v Qin 2 3 Qout v v 2 1 TC TC BC BC First Law Analysis of Otto Cycle 1→2 Isentropic Compression AIR Q W (u − u ) = − (− in ) 2 1 m m W in = (u − u ) m 2 1 vr v 1 P v Pv P T v 2 = 2 = R = 2 2 = 1 1 → 2 = 2 1 v v r T T P T v r1 1 2 1 1 1 2 First Law Analysis of Otto Cycle 2→3 Constant Volume Heat Addition Qin W (u − u ) = (+ ) − Qin 3 2 m m AIR TC Q in = (u − u ) m 3 2 P P P T v = 2 = 3 → 3 = 3 RT2 RT3 P2 T2 First Law Analysis of Otto Cycle 3 → 4 Isentropic Expansion Q Wout (u − u ) = − (+ ) AIR 4 3 m m W out = (u − u ) m 3 4 vr v 4 4 P v P v P T v = = r 4 4 = 3 3 → 4 = 4 3 vr v3 3 T4 T3 P3 T3 v4 First Law Analysis of Otto Cycle 4 → 1 Constant Volume Heat Removal Qout W (u1 − u4 ) = (− ) − Q m m AIR out Qout = (u − u ) BC m 4 1 P P P P v = 4 = 1 → 4 = 1 RT4 RT1 T4 T1 First Law Analysis Net cycle work: Wcycle =Wout −Win = m(u3 −u4 )− m(u2 −u1) Cycle thermal efficiency: Wcycle Wout −Win (u3 − u4 )− (u2 − u1 ) th = = = Qin Qin (u3 − u2 ) (u3 − u2 )− (u4 − u1 ) u4 − u1 th = =1− u3 − u2 u3 − u2 Cold Air-Standard Analysis • For a cold air-standard analysis the specific heats are assumed to be constant evaluated at ambient temperature values (k = cp/cv = 1.4). • For the two isentropic processes in the cycle, assuming ideal gas with constant specific heat using Pv k = const. Pv = RT yields: k−1 k−1 k 1 → 2 : T2 v1 k−1 T2 P2 = = r = T1 v2 T1 P1 k−1 k−1 k−1 k 3 → 4 : T4 v3 1 T4 P4 = = = T3 v4 r T3 P3 cv (T4 −T1) T1 1 th =1− =1− =1− k−1 const cV cv (T3 −T2 ) T2 r Effect of Specific Heat Ratio 1 th =1− k−1 const cV r Specific heat ratio (k) Cylinder temperatures vary between 20K and 2000K where 1.2 < k < 1.4.