A 1.8 L SI engine running at 3.000 rpm produces 60 kW of brake power. It is oper
ID: 1841843 • Letter: A
Question
A 1.8 L SI engine running at 3.000 rpm produces 60 kW of brake power. It is operating with an air fuel ratio of 14. The fuel has a heating value of 43,000 kJ/kg. The volumetric efficiency at this speed is 80%. The inlet temperature is 60 C and the inlet pressure 1 atm. A supercharger is being considered at this operating condition. It is proposed that the supercharger with the map below be used and will be driven off the engine at 2x the engine speed. A cooler will be placed between the compressor and engine to drop the temperature of the air from the compressor back down to 60 C before entering the engine. Assume the same AF and bsfc are to be maintained in supercharged operation. You may assume that C_p = 1 kJ/kgK and k = 1.4 for air what is the best pressure ratio to run the supercharger at what is the thermal efficiency of the naturally aspirated engine based on the brake power; that is use the standard definition of thermal efficiency equals the ratio of useful work out to heat in - use the brake work which is the useful work what is the power required from the engine to drive the supercharger how much cooling is required to bring the air temperature down to 60 C what is the efficiency of the supercharged engine (based on useful work out)Explanation / Answer
Engine Capacity = 1800cc = 0.0018 cu.m
N= 3000 rpm
Pb= 30 kW
Air fuel ratio= 14
Heating value = 43000 kJ/kg
nvol= 0.8
ti= 60Oc
P1c= 1 = 101.325 Kpa
C.P = 1
k= 1.4
a)Pressure ratio c = P2c/P1c
P2c = Absolute discharge pressure
= Psea level + P1c
Psea level = 1418.5 Kpa
P2c = 1519.825 Kpa
c = 1.07
b) Thermal efficiency,
nth = 3600 x Pb / (FC x C.V )
Air flow = Engine Disp x (rpm) x (nvol) / engine stroke
= 2.16 cu.m
AFR = 14
Fuel Consumption (FC) = 2.16x14= 30.24 cu.m = 30240 kg/hr
C.V = c.p /K
= 0.714
nth = 5
c) Desired Power for supercharger, Ps
= Pb x Air flow x Nvol
= 30 x 2.16 x .8
= 52 kW
d) Efficiency ,
n = 1 – Pb/ Ps
= 0.421
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