Compare otto cycle with Diesel cycle:

Sol.: These two cycles can be compared on the basis of either the same compression ratio or the same maximum pressure and temperature.

        1 - 2 - 3 - 5 = Otto Cycle,

          for the identical heat rejection Q2  the higher the

1 - 2 -4 - 5 = Diesel Cycles, heat given Q1,

the higher is cycle efficiency.

So from T-S figure for cycle 1 - 2 - 3 - 5, Q1 is more than that for 1 - 2 - 7 - 5 (area under curve represents Q,).

Hence  η_Otto > η_Diesel

For same heat rejection by otto and diesel cycles both.

Both can be compared on the basis of same maximum pressure and temperature.

1 - 2 - 3 - 4  = Otto Cycle; Here area under curve

1 - 2  - 3 - 4 = Diesel Cycle

1 - 2  - 3 - 4 is more than 1 - 2 - 3 - 4

So diese l > otto; for the same T_max  and P_max