Question

A 4-stroke diesel engine has length of 20 cm and diameter of 16 cm. The engine is producing power of 25 kW when it is running at 2500 r.p.m. The mean effective pressure of the engine will be nearly
1. 5.32 bar
2. 4.54 bar
3. 3.76 bar
4. 2.98 bar

Answer 1

Correct Answer - Option 4 : 2.98 bar

Concept:

\({P_{mean}} = \frac{{work\;output}}{{volume\;swept}}\)

Calculation:

Given:

4-stroke diesel engine

Length (L) = 20 cm, Engine speed (ω) = 2500 rpm ⇒ ω = 161.8 rad/s

Diameter (D) = 16 cm, Power = 25 kW

To find mean effective pressure

\({P_{mean}} = \frac{{work\;output}}{{volume\;swept}}\)      ----- (1)

Now,

volume swept per minute \(= \frac{{N}}{2} \times \frac{\pi }{4}{{\rm{D}}^2}L\)

∴ Volume swept per second \( = \left( {\frac{N}{{2 \times 60}}\frac{\pi }{4}{D^2}L} \right)\)     ------ (2)

Substituting equation 2 in equation 1

\({P_{mean}} = \frac{{25 \times {{10}^3}}}{{\frac{{\left( {2500} \right)}}{2}\left( {\frac{\pi }{4}} \right)\frac{{{{\left( {0 \cdot 16} \right)}^2}\left( {0.2} \right)}}{{60}}\;}}\)

\({P_{mean}} = 2.98 \times {10^5}\)

∴ P_mean = 2.98 bar