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