0
$\begingroup$

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

$\endgroup$
3
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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
-1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.