Question

The following are the results of a Morse test conducted on a four-cylinder, four-stroke petrol engine at a common constant speed in all cases:

The brake power of the engine when all the cylinders are firing is 80 kW. The brake power of the engine when each cylinder is cut off in turn is 55 kW, 55.5 kW. 54.5 kW and 55 kW, respectively.

The mechanical efficiency of the engine when all the cylinders are firing will be
1. 90%
2. 85%
3. 80%
4. 75%

Answer 1

Correct Answer - Option 3 : 80%

Concept:

Indicated power = Brake power + Frictional power

Let Brake power per cycle = B,

Indicated Power per cycle = I,

Friction Loss per cycle = F

Then,

When all the cylinders are working then

4I = 4B + 4F

When one cylinder is cut one by one, 4 different reading is taken for 3 working cylinders then the average value will be

\(3{\rm{B\;}} = {\rm{\;}}\frac{{3{{\rm{B}}_1} + {\rm{\;}}3{{\rm{B}}_2}{\rm{\;}} + {\rm{\;}}3{{\rm{B}}_3}{\rm{\;}} + {\rm{\;}}3{{\rm{B}}_4}{\rm{\;}}}}{4}\)

For three working cylinders

3I = 3B + 4F

For four working cylinders

4I = 4B + 4F

From two equations of I

I = 4B – 3B

4I = 4 × (4B – 3B)

Calculation:

Given that, 4B = 80 kW, B₁ = 55 kW, B₂ = 55.5 kW, B₃ = 54.5 kW and B₄ = 55 kW

Then \(3{\rm{B\;}} = {\rm{\;}}\frac{{55 + 55.5{\rm{\;}} + {\rm{\;}}54.5{\rm{\;}} + {\rm{\;}}55{\rm{\;}}}}{4} = 55\;kW\)

I = 4B – 3B = 80 – 55 = 25 kW

4I = 25 × 4 = 100 kW

\({\eta _m} = \frac{{Net\;B.P}}{{Net\;I.P}} = \frac{{80}}{{100}} = 80\% \)