A large V8 SI four-stroke cycle engine with a displacement of 4.6 liters is equi

Question

A large V8 SI four-stroke cycle engine with a displacement of 4.6 liters is equipped with cylinder cutout, which converts the engine to a 2.3 liter V4 when less power is needed. At a speed of 1750 RPM the engine, as a V8, has a volumetric efficiency of 51%, a mechanical efficiency of 75%, an air-fuel ratio of 14.5, and produces 32.4 kW of brake power using gasoline. With cylinder cutout and operating at higher speed as a V4, the engine has a volumetric efficiency of 86%, a mechanical efficiency of 87%, and uses an air-fuel ratio of 18.2. Indicated thermal efficiency can be considered the same at all speeds, and combustion efficiency is 100%. Calculate: Engine speed needed as a V4 to produce same brake power output. [RPM]

Explanation / Answer

Lets Say,

Mass of air at NTP for volume equal to volume of one cylinder        = m Unit

Calorific value of gasoline                                                                  = C Unit

Indicated Thermal Efficiency                                                             = n

RPM for 4-V operation                                                                       = R

Now for 8 cylinders in operation

Mass of fuel intake / burnt per cycle                                       = (mass of air intake / Air-fuel ratio) x no. of cylinders

                                                                                                = (m x 0.51 / 14.5) x 8

Power out put

= mass of fuel x calorific value x Combustion efficiency x Indicated thermal efficiency x mechanical efficiency x no. of cycles per sec

= (m x 0.51 / 14.5) x 8 x C x 1 x n x 0.75 x (1750/60)/2                                ……. (i)

For 4 cylinders in operation

Similarly, Power out put

= mass of fuel x calorific value x Combustion efficiency x Indicated thermal efficiency x mechanical efficiency x no. of cycles per sec

= (m x 0.86 / 18.2) x 4 x C x 1 x n x 0.87 x (R/60)/2                         ……. (ii)

Equating (i) and (ii)

(m x 0.51 / 14.5) x 8 x C x 1 x n x 0.75 x (1750/60)/2 = (m x 0.86 / 18.2) x 4 x C x 1 x n x 0.87 x (R/60)/2

Or, (0.51 x 8 x 0.75 x 1750) / (14.5 x 60 x 2) = (0.86 x 4 x 0.87 x R) / (18.2 x 60 x 2)

Or, R = 2245

Therefore, engine speed for V-4 = 2245 RPM

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