Question

A diesel engine has a bore of 0.1 m, a stroke of 0.14 m, a compression ratio of 19:1 and runs at 2000 RPM (revolutions per minute). Each cycle takes two revolutions and has a mean effective pressure of 1400 kPa. With a total 6 cylinders, the work done per cylinder per power stroke is

• A. 9.24 kJ/cycle
• B. 0.25 kJ/cycle
• C. 6.16kJ/cycle
• D. 1.54 kJ/cycle

Answer 1

The correct option is D 1.54 kJ/cycle

Work from mean effective pressure,

$M.E.P=\frac{W_{net}}{V_{max}-V_{min}}\Rightarrow W_{net}=M.E.P(V_{max}-V_{min})$

The displacement is

$\Delta V=\pi \times bor{e}^2\times 0.25\times stroke$

$=22/7\times {0.1}^2\times 0.25\times 0.14=0.0011m^3$

Work per cylinder per power stroke

$W=P_{m,eff}(V_{max}-V_{min})$

$=1400\times 0.0011kPa{m}^3=1.54kJ/cycle$