Mechanical Engineering - Thermodynamics - Discussion

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 ratio is nothing but ratio of volume after heat addition to vol before heat addition. V3/v2 (in diesel cycle).

Why cut-off ratio of an diesel cycle is always greater than 1.0?

Because, generally the volume will obviously increase after heating. And so it is greater than 1.0.

What is meant by combustion ratio?

If cut off ratio tends to unity then the efficiency of diesel cycle approaches to otto cycle.

Diesel cycle efficiency = 1 - ((1/Y*r^ (y-1) * lim ((Ρ^y -1) / (Ρ-1))).

Ρ-> 1.

As a limit term become 0/0, so applying L hospital's law. We get any term which cancels out why in the first term. So diesel cycle efficiency exactly becomes equal to otto cycle.

Practically I don't whether it is possible or not.

I am satisfied with your explanation, Thanks @Anjani Kumar.

When cut off ratio is 1 then the efficiency of diesel cycle is maximum which is equal to the efficiency of the otto cycle. So right answer should be when cut off ratio is 1.

How the cut-off will become zero?

If compression ratio increases then diesel cycle efficiency is?