《汽车理论》课程教学课件(英文讲稿)第2章 汽车燃油经济性 Fuel Economy 2.2 Calculation of Fuel Consumption of Vehicle

2.2 Calculation of Fuel Consumption of Vehicle1.TheEquation ofFuel ConsumptionT..ntqunit: Pe (kW);Tta(N-m); b(g/kW.h);P(kW)e9550p(kg/L) ;g(m/s2) ;pg(N/L);P.b(g / h)Qs(L/100km)。(g / kW.h)hfor gasoline: pg=6.96~7.15 (N/L)vehicle: Q,(L/100km); u.(km / h)for diesel:Pg = 7.94~8.13 (N/L)+Pb100PbPbh.g(kg /100km)g / km)1000ua10uakmhuadensity of fuel (kg / L)P,bLLkg10u.p100km100kmkgP.bQ(L/100km)1.02u.pg
2.2 Calculation of Fuel Consumption of Vehicle 1. The Equation of Fuel Consumption b g kW h ( / ) unit: Pe (kW);Ttq(N·m);b(g/kW·h); ρ(kg/L) ; g(m/s2 ) ; ρg(N/L); Qs (L/100km)。 for gasoline:ρg= 6.96~7.15 (N/L) for diesel: ρg = 7.94~8.13 (N/L) ( / ) e a P b g h g km u h km : ( /100 ); ( / ) s a vehicle Q L km u km h( / ) P b g h e ( ) 9550 tq e T n P kW 100 ( /100 ) 1000 10 e e a a Pb Pb kg km u u ( ) 10 100 100 e s a P b kg L L Q u km kg km —density of f el ( u kg / L) ( /100 ) 1.02 e s a P b Q L km u g

2.Fuel Consumption Performance of Engine1)Fuelconsumptionperformance-needed output power characteristicsofengine(负荷特性)b(g/kw.h)To describehow theperformanceofn1<n2<n3nlengine develops with the change of1n2n3load of the vehicle that is the neededoutput power of engine.u2Definition:b(min)uloutput-power ratio(负荷率)pe(kw)uoneededoutputpowerofenginemaximumpoweravailableofengineU2uo
2. Fuel Consumption Performance of Engine 1)Fuel consumption performanceneeded output power characteristics of engine(负荷特性) • To describe how the performance of engine develops with the change of load of the vehicle that is the needed output power of engine. Definition: 2 0 ( ) needed output power of engine = maximum power availableof engine output power ratio u u 负荷率

NoteForQ.(Q,=Pb/1.02u.pgP.is the correspondingY-coordinate value of (P+Pw)/nicurve in the Graph of Power Equilibrium;b could be available from the needed output powercharacteristics withgivennConclusion : (under given n)· P.increases with the decrease of b under the given n(in mostinstance);output-power ratio always increases with the decrease ofb(in mostinstance);b meets its minimum value when the output-power ratiocomecloseto80%
Note For Qs ( ) • Pe is the corresponding Y-coordinate value of (Pf +Pw )/ηT curve in the Graph of Power Equilibrium; • b could be available from the needed output power characteristics with given n. /1.02 Q Pb u g s e a Conclusion : (under given n) • Pe increases with the decrease of b under the given n (in most instance); • output-power ratio always increases with the decrease of b (in most instance); • b meets its minimum value when the output-power ratio come close to 80%

2) UniversalCharacteristics of Engine(万有特性)T+g<N.m)Pe(kw)(g/kw.h)Q:b3bKbe<b3b<2<b3n(r/min)n(r/min)
2)Universal Characteristics of Engine (万有特性)

Universal70Characteristicsof发动Engine60F机万有特性(等高线)50F40MX/db = const.h)I(kw.3050826022010F04008001200200024002800320036004000440048001600n/(r.min-l)50Jo100150ua/(km-h-l)
b const Universal Characteristics of Engine——发动 机万有特性 (等高线)

3)PowerEquilibriumWaP(F,+F,+F,+F)3600ntWhen level road & constant speed.uaP+F)H36001CpAu?u21.153600ntNote:P,is the power needed by the vehicle to maintain the constantspeed drive on the level road; For calculating the fuelconsumptionP,that exist in the formulaof Q,(Q,=P.b/1.02uapg)must be the needed power to maintain the constant speed drive onthe levelroad
Note: Pe is the power needed by the vehicle to maintain the constant speed drive on the level road; For calculating the fuel consumption Pe that exist in the formula of Qs (Qs =Peb/1.02ua ρg) must be the needed power to maintain the constant speed drive on the level road. 2 ( ) 3600 ( ) 3600 21.15 a e f w T a D a T u P F F u C Au Gf ( ) 3600 f i w j T a e F F F F u P 3)Power Equilibrium When level road & constant speed:

4)Fule consumption calculationofvehiclewith constant speeddrive onlevelroad(StableMode)(稳定行驶)(Q~U.)A. Method One (given ig)Q-UaVehiclePowerEquilibriumGraphFuel consumptionperformance-neededoutputpowercharacteristicsofengine(负荷特性图)(功率平衡图)Step:19987654320b(g/kw.h)Pe(kw)(P+P.)/n一n3(yoo))/0o:司UaiPeua(km/h)pe(kw)60120208010040CAu?u.ua/(km·h-1)P=Gf3600m21.15>PualelLinealInsertMethod>b,ua=0.377rniglonual二2 =b .Per /1.02-uar:pgOual....二(uar,Q,)u
Q-Ua Fuel consumption performance-needed output power characteristics of engine(负荷特性图) Vehicle Power Equilibrium Graph (功率平衡图) η 4)Fule consumption calculation of vehicle with constant speed drive on level road (Stable Mode)(稳定行驶)—— (Q~Ua) A. Method One (given ig ) 2 ( ) 3600 21.15 1 1 a D a e T u C Au P Gf a e u P Step: 1 . 1 1 ( , ) ( , ) a an u u a an n u Q u Q 0 0.377 1 1 a g r n u i i a u n 1 Lineal Insert Method b 1 1 1 1 /1.02 Q b p u g e a

b(g/kw.h) Pe(kw)(P+Pw)/n Tn1<n2<n30201e0Ua(km/h)PelPe<kw)Uai
η 2 < <

B. Method Two: (given ig)Q-UaStep:Universal Characteristic Graph of EnginePower Eguilibrium GraphPePe99816547(wy001)71/0二320ni12020608010040ua/(km·h-l)CoAuuaG21.153600nPuLinealalelInsertMethodua=0.377-r-nig-iou.>nalual-..waQ =b :Pei /1.02.ua:pg =(ual,Q)uan
Q-Ua Universal Characteristic Graph of Engine Power Equilibrium Graph B. Method Two: (given ig ) Step: 2 0 1 ( ) 3600 21.15 1 1 1 0.377 1 1 . 1 1 1 1 1 1 /1.02 ( , ) ( , ) a D a e T a g a an u C Au P Gf a e Lineal Insert Method r n u i i a u u e a a an n u P b u n Q b p u g u Q u Q

5) Fuel consumption calculation of the vehicle under the drive cycle(循环行驶工况的油耗计算)Basic drive cycle comprises of four parts:acceleration→constant speed→decelerationparkingUaacceleration constant speeddecelerationparking十basic drive cycle
5)Fuel consumption calculation of the vehicle under the drive cycle (循环行驶工况的油耗计算) Basic drive cycle comprises of four parts: acceleration →constant speed →deceleration→parking
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