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Edited: May 17, 2022

# Why thermal efficiency of the Diesel cycle decreases with an increase in the cut-off ratio?

Why thermal efficiency of the Diesel cycle decreases with an increase in the cut-off ratio?

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Let's draw a diesel cycle on P-v and T-s diagram.

The cut-off ratio is the ratio of cylinder volumes after and before the combustion process. It is given by the following formula.

Now suppose we increase the cut-off ratio up to point 3' as shown in the figure below.

(The values of clearance volume V2 and BDC value V1 are assumed to be fixed and can not be changed here).

The point after the expansion process(power stroke) also shifts to 4' in P-v and T-s diagrams.

We know that the area under the curve on the T-s diagram gives the heat transfer. Hence, the area under the curve 2-3 gives the heat added during the constant pressure and the area under the curve 4-1 gives the heat rejected in the original cycle.

The thermal efficiency of the cycle is given by,

Now, as we have increased the cut-off ratio, point 3 now shifts to 3' and point 4 shifts to 4'. So both the heat added into the cycle as well as heat rejected are increasing simultaneously. We can not decide if thermal efficiency increases or decreases.

But, we know that the slope of the constant volume process is more than the slope of the constant pressure process on the T-s diagram.Hence, the rate of heat addition is less than the rate of heat rejection. Which means