Mechanical Engineering - Thermodynamics - Discussion

@All.

Option C is correct.

Refer PK Nag (Gas power cycles).

Condition is given the same compression ratio and heat input so the comparison would be done on Q2 i.e. Heat rejection. Lower Q2 means higher efficiency. Q2 Otto is less than that of diesel so efficiency of Otto is high. Option D is correct.

@Prakash.

I agree with your point and D is the Correct answer.

Of the options listed, the most direct reason for the Otto cycle's higher efficiency (compared to Diesel with the same compression ratio and heat input) is:

Combustion is at constant volume.

Here's why this is the key factor:

Rapid Temperature Rise: Constant volume combustion in the Otto cycle causes a very rapid and significant increase in temperature. This high peak temperature allows for a more efficient conversion of heat energy into work.

Higher Average: The higher peak temperature also leads to a higher average temperature during the expansion stroke. This results in less heat being wasted and rejected during the exhaust stroke.

While the other options are relevant to the Otto cycle, they don't directly explain its higher efficiency compared to the Diesel cycle:

Expansion and compression are isentropic: This is an idealization in both Otto and Diesel cycles. In reality, there are always some losses due to friction and heat transfer.

Maximum temperature is higher: This is a consequence of constant volume combustion, not the primary reason for the efficiency difference.

Heat rejection is lower: This is also a result of the higher peak and average temperatures caused by constant volume combustion.

In summary: The core reason for the Otto cycle's higher efficiency (under the given conditions) is the constant volume combustion, which leads to a rapid temperature rise and a higher average temperature during the cycle.