 Science
volumetric efficiency

### Question Description

how is volumetric efficiency within a 4 stroke engine calculated Volumetric efficiency in the internal combustion engine design refers to the efficiency with which the engine can move the charge into and out of the cylinders. More specifically, volumetric efficiency is a ratio (or percentage) of the quantity of air that is trapped by the cylinder during induction over the swept volume of the cylinder under static conditions. Volumetric Efficiency can be improved in a number of ways, most effectively this can be achieved by compressing the induction charge (forced induction) or by aggressive cam phasing in naturally aspirated engines as seen in racing applications. In the case of forced induction volumetric efficiency can exceed 100%

Example:  A single cylinder, 4-stroke diesel engine running at 18000 rpm has a bore of 85 mm and a stroke of 110 mm. It takes 0-56 kg of air per minute and develops a brake power output of 6 kW while the air fuel ratio is 20 : 1. The calorific value of the fuel used is 42550 kJ/kg and the ambient air density is 118 kg/m3. Calculate:

(a) The volumetric efficiency and

(b) Brake specific fuel consumption

Solution:  Given that

Engine speed, N = 1800 rpm

Bore, d = 85 mm

Stroke, L = 110 mm

Air flow rate, m°= 0-56 kg/min

Brake Power Output, BP= 6kW

Air fuel ratio (A/F) = 20: 1

Calorific value of fuel used, C = 42550 kJ/kg

Ambient air density,  p = 118 kg/m3

(a)   Volume of cylinder,   V1 = π/4 (0.085)2 X 0.110 = 6.242 X 10-4 m3

volume of air taken per stroke = mo/p .1/(N/2) = 0.56 / 1.18 (1 / 1800/2)

= 5-273 X 10-4

volumetric = volume of air taken in per stroke / volume of cylinder

5.273 X 10–4 / 6.242 X 10–4 = 0.845 (or 84.5%)

Brake specific fuel consumption = BHP / kg of fuel

Fuel consumption/Hour = 0-56 X 1/20 X 60 = 1.68 kg

∏ Brake specific fuel consumption = 1.68/6 = 0.28 / BHP- Hr UIUC Anonymous
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