Question

A four-stroke single-cylinder SI engine of 6 cm diameter and 10 cm stroke running at 4000 r.p.m. develops power at a mean effective pressure of 10 bar. The power developed by the engine is
1. 9.45 kW
2. 5.54 kW
3. 4.92 kW
4. 2.94 kW

Answer 1

Correct Answer - Option 1 : 9.45 kW

Concept:

Mean Effective Pressure: It is the imaginary pressure which when remains constant will give us the same work done like that of the actual cycle for the same volume as that of the actual cycle.

Mean effective pressure \( = \frac{{work\;done}}{{swept\;volume}}\)

Swept Volume of the cylinder is given by V_s = \(\frac{\pi }{4}{D^2} \times L \times N \times K\)

Where D is the diameter of the cylinder and L is the stroke of the cylinder N is r.p.m. and K is the number of cylinders.

For a 4-stroke engine since the two revolutions of crank complete one cycle, therefore, there will be only one suction stroke, then effective r.p.m. \(N = \frac{N}{2}\)

Calculation:

Given, D = 6 cm = 6/100 m, L = 10 cm = 10/100 m, N = 4000 r.p.m. and Mean effective pressure = 10 bar = 10 × 100 kPa

\({P_m} = \frac{W}{{\frac{\pi }{4}{D^2} \times L \times N \times K}}\)

\( \Rightarrow 10 \times 100 = \frac{W}{{\frac{\pi }{4}{{\left( {\frac{6}{{100}}} \right)}^2} \times \frac{{10}}{{100}} \times \frac{{4000}}{{2 \times 60}} \times 1}}\)

W = 9.42 kW