Mechanical Engineering - Thermodynamics - Discussion

The efficiency will be lower if cutoff ratio is higher because the piston has already moved far down while the heat is still being added.

The Efficiency of diesel I think much greater then petrol so here we can say choose the value of cut off high as it decreases the efficiency of diesel which approaches to petrol. Am I right?

Correct answer is b.

Becomes its clearly mentioned APPROACHES not When EQUAL. We can't say when cutt of ratio is zero its totally useles the question is telling you when n of diesel Engine APPROACHES. This means you have to decrease cutt of i.e. Rc !

Then it vill approach towards otto, it does not tell you when efficency of diesel eng is equal to that of otto. Even if it had mebtioned we still can't put RC (CUTT OFF) equal to zero.

Hence answer is B.

If cut off ratio tend to 1, then find the limiting value of diesel cycle efficiency as cut off ratio tend to 1, it will be equal to otto cycle efficiency. Also for those who are considering the cut off ratio zero, right answer, if cutt off ratio is zero means the volume of flue gases after combustion is zero, how can it be possible that flue gases will have zero volume.

As an diesel cycle the cut off ration is the ratio between V3/V2.
V3 - Fuel Injection Into the Cylinder Stop While Power Stroke.
V2 - Fuel Injection Into the Cylinder Start While Power Stroke.

Those V3 & V2 are Smaller than The V1(Cylinder Volume).

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

V2/V1.

No,

Here the term used "approached". Not used equal to. Understand the key.

The Answer must be cut off ratio is 0.

Many guys can think, it can be 1 then also it can be right. But, when cut off ratio is 1, the cycle will be comprised of 3 processes, which is impossible. Along with that, it will reach in the limit of 0/0 form.

Here, the Correct answer must be: When cut-off ratio is 0, the diesel cycle approached to Otto cycle.

Answer is B.

When cutoff is decreased, diesel cycle efficiency approaches (not equal) to the otto cycle efficiency.

It is ratios of two constant volume of decrease and increases the cp and CF or acrobatics process.

In the options, they said cutoff(vol after combustion-vol before) not cutoff ratio (vol after combustion/vol before combustion).

Therefore as cutoff reaches zero it becomes constant volume process.