Question

A 4-stroke, 6-cylinder gas engine with a stroke volume of 1.75 litres develops 26.25 kW at 506 r.p.m and the MEP is 600 kN/m². The number of misfires per minute per cylinder will be
1. 3
2. 4
3. 5
4. 6

Answer 1

Correct Answer - Option 1 : 3

Given:

 A 4-stroke, 6 – cylinder gas engine

Volume of stroke = 1.75 ℓ

Power developed = 26.25 kW, ω = 506 rpm.

Mean effective pressure (MEP) = 600 kN/m²

To find no. of Misfires per minute

Calculation:

Total volume swept (V_s) = 1.75 ℓ = 1.75 × 10^-3 m³

Let N be the no. of fires in which engine produces power per minute

\({\rm{Power}} = \frac{{\left( N \right)}}{{60}}\;{P_{mean}} \times {V_s}\)

\({\rm{Power}} = \frac{N}{{60}}\;\left( {600} \right) \times {10^3}\left( {1.75} \right) \times {10^{ - 3}}\)     -------- (1)

Also power = 26.25 × 10³ W      ------ (2)  (Given)

Now, equating equation 1 & 2

\(\therefore \left( {\frac{N}{{60}}} \right)\left( {600} \right) \times {10^3}\left( {1.75} \right) \times {10^{ - 3}} = 26.25 \times {10^3}\)

∴ N = 1500

Hence there should be 1500 fires.

Now, No. of fires expected per minute for on engine speed of 506 rpm is:

\({N_{actual}} = \left( {\frac{{506}}{2}} \right) \times No.\;of\;cylinders = \left( {253} \right)6\)

N_actual= 1518.

But by calculation we can see that there are 1518 fires.

∴ Number of misfires for engine per minute = N – N_actual = 18.

∴ Misfires for engine per minute per cylinder is \(\frac{{18}}{6} = 3\) (since there are 6 cylinders)

∴ Number of misfires per minute per cylinder will be 3.