AZINGLY, for the Joule cycle there is no heat flow at all, so it is 1-0 = 1.

[Abigale]

Diesel, Otto , Joule TRICK?

Hi Guys,
I try to callculate the efficiency for a cycle Process.
For Diesel, Otto and Joule Processes you can write the efficiency as a really small equation.

I have seen that for example sometimes the adiabatic equation is used,
but i don't understand in which case, or better, how I have to connect the procceses?

Is there a trick or do I have to learn it by memorization?

THX

 

[Andrew Mason]

Hi Guys,
I try to callculate the efficiency for a cycle Process.
For Diesel, Otto and Joule Processes you can write the efficiency as a really small equation.

I have seen that for example sometimes the adiabatic equation is used,
but i don't understand in which case, or better, how I have to connect the procceses?

Is there a trick or do I have to learn it by memorization?

THX

Efficiency = W/Qh and, since there is no change in U in a complete cycle, W = net heat flow = Qh+Qc = |Qh|-|Qc|.

So all you have to worry about is the heat flow.

Since both the Diesel and Otto cycles involve adiabatic compression and expansion, heat flow only occurs in two of the four parts:

For Diesel the constant pressure expansion part (1 - heat flow in) and constant volume part (3 - heat flow out). So it is just: (Cp|ΔT1|-Cv|ΔT3|)/Cp|ΔT1| = 1 -(1/γ)|ΔT3|/|ΔT1|

For Otto heat flow is in the two constant volume parts, so efficency is (|ΔT1|-|ΔT3|)/|ΔT1| = 1 - |ΔT3|/|ΔT1|

AM