It's written that at constant pressure & temperature for Otto & Diesel cycles the efficiency of diesel is more than efficiency of Otto.
Efficiency = Work / Heat added = 1- (Heat rejected / Heat added )

from fig :
-(4->1) Heat rejected is same
-(2'->3) Heat added for diesel is greater than (2->3) )heat added for Otto

so for diesel the term (Heat rejected / Heat added ) is less

so efficiency of diesel is more.

What make it not clear for me is that
work ( Cv(T4-T3) ) (3->4) is same for both & heat added for diesel is more than that of Otto.
So if we use eff= Work/(heat added)
Eff of otto will be more since less heat will be added and same work is done,
What makes it wrong?

P-V For same pressure and temperature

source : http://slideplayer.com/slide/10748942/ slide 30

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    $\begingroup$ I think you overlooked that a substantial chunk of work is done between 2' and 3. Admittedly some work is absorbed from 2 to 2' but the net work (area of the loop) is greater for the diesel cycle. $\endgroup$ Dec 21 '16 at 19:23
  • $\begingroup$ Thanks , got it now while studying , i didn't know that 2'->3 also produce useful work (p*dv) $\endgroup$ Dec 29 '16 at 16:31
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    $\begingroup$ @MohamedRefaat Please make the question status as solved if you have got the answer. Because the questions is still in the unanswered section. $\endgroup$ Jan 29 '17 at 12:17

As @Brian Drummond said ,i overlooked the work done between 2' and 3


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