Mechanical Engineering - Thermodynamics - Discussion

What is cut off?

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.

The compression ratio is defined as the total cylinder volume when the piston is at the bottom dead center to the clearance volume.

Cutoff ratio-Say Fuel is injected into combustion chamber where only air compressed and is at high temperature. Fuel is injected for a duration of time, say T. The piston would not have reached the bottom dead centre in time T. The fuel is cut off when the volume is, say V2. The clearance volume is V1. The ratio V2/V1 is cut off ratio.

As far as the inefficiencies go the formula for for both diesel and Otto cycle have compression ratio in their formula. I don't understand how cut off ratio comes into play. If any one can find a suitable answer please post it and till that time even I am looking into this question.