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V V R Ratio, N Compressio = Internal Combustion Engines

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V V R Ratio, N Compressio = Internal Combustion Engines Internal Combustion Engines V2 V1 V1 Compression Ratio, rc = V2 Bottom Top Dead Dead Center Center BDC TDC 1 Lenoir Cycle 1860 Intake Combustion Power TDC Exhaust TDC 0-1 1-2 2-3 3-0 p=constant v=constant Q=0 p=constant one revolution 2 Q Air Standard p v=c Q=0 Lenoir Cycle 0 3 1 Q p=c 2 v Air Standard Lenoir Cycle 2 Closed Thermodynamic System - quantity of mass Q q = Δu + w Q=0 p v=c 2 - 3 Combustion Process 0 v = constant Þ w = 0 3 q = u 2 - u1 1 Q p=c 2 - 3 Expansion Process, Power Stroke q = 0,adiabticprocess k k -1 k-1 T3 æ p3 ö æ v2 ö = ç ÷ = ç ÷ T2 è p2 ø è v1 ø w = -(u 2 - u3 ) 3-1 Exhaust Process p = c q = Δu + w q = (u1 - u3 ) + p3 (v1 - v3 ) = h1 - h3 w = q31 - (u3 - u1) 3 OTTO CYCLE , 2 Revolutions/cycle Compression Combustion Power Exhaust Exhaust Intake 1-2 2-3 3-4 4-1 1-0 0-1 TDC BDC TDC BDC revolution 1 revolution 2 3 p OTTO CYCLE spark ignition, SI fuel injection + SI 2 4 0 1 v 4 4 CYCLE, 2 REVOLUTIONS Revolution 1 Revolution 2 5 Air Standard Otto Cycle Closed System- a quantity of mass p p adiabatic and isentropic W Q v ignition exhaust opens W atmospheric pressure intake opens intake closes exhaust closes W v 6 AIR STANDARD OTTO CYCLE, 2 Revolutions/cycle Q in Qout V=c V=c W Win out Compression Combustion Expansion Exhaust 1-2 2-3 Power 4-1 3-4 p p Q adiabatic in W v=const isentropic 4 adiabatic 4 2 2 isentropic v=const Qout W 1 1 7 v v “COLD” Compression 1 ® 2, Δs = 0, pvk = c, q = 0, w = Δu k k Air Standard Otto Cycle p1v1 = p2v2 k-1 æ v ö ç 1 ÷ 3 T2 = T1ç ÷ è v2 ø Δs = 0 w 2-1 = Δu = u 2-u1 = cv (T2 - T1 ) T qin pvk = c Combustion 2 ® 3, v = c, w = pdv = 0, q =ΔU Q = 0, ò q2-3 = Δu = u3 - u 2 = cv (T3 - T2 ) 2 k 4 Expansion 3 ® 4, Δs = 0, pv = c, q = 0, w = ΔU k-1 q æ v ö out T = T ç 3 ÷ 4 3ç v ÷ 1 è 4 ø v w3-4 = Δu = u3 - u 4 = cv (T3 - T4 ) CLOSED SYSTEM Exhaust 4 ®1, v = c, w = ò pdv = 0, q =ΔU q = Δu + w q4-1 = Δu = u 4 - u1 = cv (T4 - T1 ) "COLD"Þ Cycle constant c c at T . qout q4-1 cv (T4 - T1 ) p, v 1 ηcycle =1- = 1- = 1- 8 qin q3-2 cv (T3 - T2 ) 3 “COLD”Air Standard Otto Cycle CLOSED SYSTEM, Q = ΔU + W "COLD"Þ T Compression 1 ® 2, Δs = 0, Q = 0, W = ΔU constant cp, cv at T1 Combustion 2 ® 3, v = c, W = ò pdv = 0, Q =ΔU Expansion 3 ® 4, Δs = 0, Q = 0, W = ΔU 2 4 Exhause 2 ® 3, v = c, W = ò pdv = 0, Q =ΔU 1.0 Ds2®3 = Δs4®1 1.0 1 æ T ö æ v ö æ T ö æ v ö ç 3 ÷ ç 3 ÷ ç 4 ÷ ç 4 ÷ cvlnç ÷ - R lnç ÷ = cvlnç ÷ - R lnç ÷ è T2 ø è v2 ø è T1 ø è v1 ø V2 V1 T T V v 3 = 4 Compression Ratio = r = 1 = 1 T T c V v 2 1 2 2 k-1 T æ v ö Q - Q Q 1 2 η =1- in out =1- out - = ç ÷ T2 è v1 ø Qin Qin 1 1 1 u 4 - u1 cv (T4 - T1 ) η =1- k-1 =1- k-1 =1- k-1 η =1- =1- æ v ö æcompressionö (r) u - u c T - T 1 3 2 v ( 3 2 ) ç ÷ ç ÷ è v2 ø è ratio ø æ T ö ç 4 ÷ w = p ´ V - V ç -1÷ net mean ( 2 1 ) T4 - T1 T1 è T1 ø η =1- =1- wnet T - T T æ ö Mean Effective Pressure (mep) = 3 2 2 T3 (V - V ) 9 ç -1÷ 1 2 è T2 ø What is the thermal efficiency k-1 æ v ö of a COLD air standard Otto cycle T = T ç 1 ÷ = 559.7´6.4 =1146o R operating between 2800 F and 2 1ç ÷ è v2 ø 100 F with a compression ratio of 6? w 2-1 = u 2-u1 = cv (T2 - T1 ) w = .171´(1146- 559.7) =100.3BTU/lb Closed System 2-1 q = u - u Q = ΔU + W 2-3 3 1 Compression 1® 2, Δs = 0, q = 0, w = Δu q2-3 = .171´(3259.7 -1146) = 361.4BTU/lb k-1 .4 Combustion 2 ® 3,v = c, w = pdv = 0, q =Δu æ v ö æ 1 ö ò ç 3 ÷ o T4 = T3ç ÷ = 3259.7ç ÷ =1591.9 R Expansion 3 ® 4, Δs = 0, q = 0, w = Δu è v4 ø è 6 ø Exhaust 2 ® 3, v = c, w = pdv = 0, q =ΔU ò w3-4 = u3 - u 4 w = .171(3259.7 -1591.9) = 285.2BTU/lb 3 3-4 2800 F q4-3 = u 4 - u3 q4-3 = .171(1591.9- 559.7) =176.5BTU/lb p W Q æ w ö æ 285.2-100.3ö ç net ÷ ηcycle = ç ÷ = ç ÷ = 51.1% è qin ø è 361.4 ø 2 4 Q 1 1 W η =1- =1- = 51.1% cycle k-1 61.4-1 1 æ v1 ö 100F ç ÷ 10 è v2 ø v AIR TABLE - Tables A-22, A-22E specific heats are integrated as variables of T- page 113, 230 1) Table base 0o F, 0o C 5) Entropy - ideal gas æ T ö æ v ö ç 2 ÷ ç 2 ÷ s2 - s1 = cvln ç ÷ + R lnç ÷ 2) Isentropic Process, è T1 ø è v1 ø pvk = constant æ T ö æ p ö s - s = c ln ç 2 ÷ - R lnç 2 ÷ p (p ) 2 1 p ç ÷ ç ÷ 1 = r 1 è T1 ø è p1 ø p (p ) 2 r 2 Entropy - Using Table A - 22 and Table A 22E, v (v ) 1 = r 1 dT v (v ) s = c (T) + R ln (volume ratio) 2 r 2 ò v T dT s = c (T) - R ln (pressure ratio) 3) Enthalpy ò p T h = cp (T) dT æ v ö ò 0 0 ç 2 ÷ s2 - s1 = s (T2 ) - s (T1) + R lnç ÷ è v1 ø 4) Internal Energy æ p ö s - s = s0 (T ) - s0 (T ) - R lnç 2 ÷ u = c (T) dT 2 1 2 1 ç ÷ 11 ò v è p1 ø @ 300o K, v = 621.2, u = 214.07 r1 1 9.1 RT .286´300 3 v1 = = = .858 m3/kg p p 100 1400 r=8.5 v v æ v ö kJ/kg r2 = 2 , v = v ç 2 ÷ = 621.2/8.5 = 73.08 v v r2 r1ç v ÷ 4 r1 1 è 1 ø 2 T vr u 75.5 496.62 1 73.08 503.06 kJ/g = u 3 100 kPa 72.56 504.45 v 300 K 73.08 - 72.56 .52 interpolation ratio = = = .1769 75.5 - 72,56 2.94 a) W = Qin - Qout =1400 - 674.25 W = 725.75kJ/kg Qin = u3 - u 2 = 1400 kJ/kg u3 =1400 - 503.6 =1903.06 kJ/kg æ Qout ö 725.75 b) η =1- ç ÷ =1- = 51.8 % InterpolateTableA - 22 near u3 =1903.6 kJ/kg è Qin ø 1400 vr3 =1.9176 W W c) mep = = æ v ö V - V æ v ö ç 4 ÷ 1 2 ç 2 ÷ vr4 = vr3ç ÷ =1.9176´8.5 =16.3 v1ç1- ÷ è v3 ø è v1 ø Interpolate TableA - 22 near v =16.3 725.75 r4 mep == = 958.7 psi u 4= 888.32 kJ/kg æ 1 ö .858´ç1- ÷ 12 Qout = (u3-u 4 ) = (888.32 - 214.07) = 674.25kJ/kg è 8.5 ø The initial temperature and pressure of Process 1-2 air inan ideal Cold Otto cycle having a compression ratio of10, is 60 F and 14.7 psia isentropic, adiabatic with pvk = constant respectively. Heat isadded in the constant k-1 æ V ö volume process at a rateof 800 BTU/lbm. ç 1 ÷ .4 o T2 = T1 ´ç ÷ = 520´10 =1305 Consider air to be an idea gas. è V2 ø a) Calculate: a) The change in internal energy per pound of air ΔU =cv (T2 -T1 ) = .17´(1305 -520) =134.2BTU/lb during the compression process. b) b) The maximum temperature and pressure of the cycle Combustion Heat,Q = cv (T3 -T2 ) k =1.4 800 = .17´(T3 -1305) cv =.171BTU/lbm-R T = 5893o R cp =.24BTU/lbm-R 3 Since V = V , from the ideal gas law Pv = mRT, 3 2 3 Expansion æ T ö ç 3 ÷ p3 = p2ç ÷ è T2 ø p Since 1- 2 is isentropic and adiabatic, Combustion k 2 Q=800 BTU/lb 4 æ V ö p = p ç 1 ÷ =14.7´101.4 = 367psia Compression Exhaust 2 1ç ÷ è V2 ø 1 14.7 psia 520 R æ 5983 ö 13 .1´V1 V1 p3 = 367´ç ÷ =1685psia V è1305 ø a) A 265 cubic in V8 gasoline engine (1500 -1200) BTU/power stroke/cylinder ´8 cylinders ´ 4600 rpm runs at 4600 rpm. Compression work HP = is 1200 ft-lb, expansion work is 550 ft - lb/HP ´ 2 rev/ power stroke ´ 60 sec/min 1500 ft-lb and heat input is 1.27 BTU HP = 167.3 HP per piston per cycle.
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