Mechanical Engineering - Thermodynamics - Discussion

I think the right answer is B.

The cut off ratio is defined on basis of heat addition process. So, say if its isochoric as in petrol cut off ratio is one. And for diesel it is Isobaric process giving ratio greater than one, in my opinion when it reaches unity the efficiency reach to that of petrol Otto cycle.

Mind it I say reaches, not 'is equal to', as it would be like Lenoir cycle with 3 processes,not exactly like Lenoir.

It is safer to say the diesel cycle efficiency will reach that of petrol cycle (or a combined cycle to be precise) as cutoff ratio approaches unity

Cut off is the ratio of the volume before heating and volume after heating.

Cut off ratio is volume after combustion process to volume before combustion v2/v1.

The efficiency of Otto cycle is 1-(1/r^(Y-1)) Y = Gamma.

And that of Diesel cycle is 1-[(1/r^(Y-1))[(ρ^y-1)/y(ρ-1)] ρ=rho= cut-off. Here if ρ=0, then Efficiency of Diesel cycle becomes 1-(1/Y(r^(Y-1))).

So we can say the efficiency of Diesel cycle approaches to that of Otto Cycle.

It is ratio of volume after heat addition to volume before heat addition in diesel cycle at constant pressure.

Efficiency of diesel cycle is:
1- [(p^(Y)-1)/r^(y-1)Y(p-1)].

When we apply limit with L-hospital rule we get,

Efficiency of otto:
1-[1/r^(y-1)]

Where: r = compression ratio = Volume before compression/Volume after compression.
p = cut-off ratio = volume after injection/volume before injection.
y = Ratio of specific heats or adiabatic index.

Actually right answer is 1, when cutoff is 1, efficiency of diesel cycle approaches efficiency of Otto cycle at same compression ratio.

If cut off ratio becomes 1 that cycle would be consist of only 3 processes. What about that?

If cut off ratio = 1 then efficiency is undefined. So make sure your answer.