Question

A reciprocating compressor having 0.20 m bore and 0.25 m stroke runs at 600 rpm. If the actual volume delivered by compressor is 4 m³/min, its volumetric efficiency will be about
1. 70%
2. 75%
3. 80%
4. 85%

Answer 1

Correct Answer - Option 4 : 85%

Explanation:

Volumetric efficiency:

The volumetric efficiency of an engine 

• It is defined as the ratio of the actual air capacity to the ideal air capacity.
• It is also the mass of air that enters in suction stroke to the mass of free air equivalent to the piston displacement at intake temperature and pressure conditions.

\({η _{vol}} = \frac{{{V_{actual}}}}{{{V_{swept}}}}\)

which can be further written as

\({η _{vol}} = \frac{{\dot m\ \times\ {V_1}}}{{\frac{\pi }{4}{D^2}L \ \times \ \frac{N}{{60}} \ \times \ K}}\)

where \(\frac{\pi }{4}{D^2}\) represents the area of the piston.

Calculation:

Given:

V_actual = 4 m3/min, D = 0.20 m, L = 0.25 m, N = 600 rpm

\({η _{vol}} = \frac{{{V_{actual}}}}{{{V_{swept}}}}= \frac{{V_{Actual}}}{{\frac{\pi }{4}{D^2}L \ \times \ N \ }}\)

\({η _{vol}} = \frac{{V_{Actual}}}{{\frac{\pi }{4}{D^2}L \ \times \ N \ }}=\frac{4}{\frac{\pi}{4} \times 0.2^2 \times0.25 \times600}\)

η_vol = 0.8488 = 85 %