A large V8 SI fourstroke cycle engine with a displacement of

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: (1) Engine speed needed as a V4 to produce same brake power output. [RPM] Ans: (1)2245 RPM

Solution

Solution:

Volumetric efficiency = Volume displaced/volume swept

mechanical efficiency is given by = BP/IP

BP is equal to IPx efficieny

= P Vs N (eff) where P is mean effective pressure ,Vs is swept volume)

(BP)₈ = P 4.6/0.51 (1750) (0.75)

(BP)_⁴= P 2.3/0.86 (N) (0.87)

As given BP should be equal.

P 4.6/0.51 (1750) (0.75) = P 2.3/0.86 (N) (0.87)

N = 2543 rpm