Efficiency of a cycle in TS diagram

[dingo_d]

Homework Statement

A working substance goes through a cycle within which the absolute temperature varies n-fold, and the shape of the cycle is

,
where T is the absolute temperature, and S the entropy. Find the efficiency cycle.

Homework Equations

\frac{\delta Q_R}{T}=dS, \eta=1-\frac{|Q_{out}|}{Q_{in}}

The Attempt at a Solution

The total heat is Q=\int T dS, that is the area of the surface in the picture. I could just say: it's a triangle so I'll use the formula for the triangle surface:
P=\frac{1}{2}ab.

The Q_{in} is easy to calculate:Q_{in}=T_0\int_{S_0}^{S_1}dS=T_0(S_1-S_0).

But how do I get the Q_{out}? The temperature changes. I have found in solution (without explanation) that the answer is:

Q_{out}=\frac{1}{2}(T_0+T_1)(S_1-S_0), but why T_0+T_1? and where does that 1/2 comes from? The triangle area formula? :\

 

[hikaru1221]

No, it's trapezium's area.
But why don't you look at it another way? The "total heat" you calculated is actually work done A! And for a cycle: A = Q_{in} - |Q_{out}|. So instead of doing an integration or wondering it's triangle or something else, just calculate a simple sum.

 

[dingo_d]

No, it's trapezium's area.
But why don't you look at it another way? The "total heat" you calculated is actually work done A! And for a cycle: A = Q_{in} - |Q_{out}|. So instead of doing an integration or wondering it's triangle or something else, just calculate a simple sum.

Ummm it's triangle :D

I don't get it how to calculate the sum when I don't have the work :\ A=W=pdV right? 1-2 is isotherm, 2-3 is isenthalp (no Q) and 3-1 is sth I don't know :D I could say that its a line with slope k, which is not that hard to calculate, but I don't have the y-intercept (T in my graph)...

 

[hikaru1221]

Eek, if you rotate your diagram 90 degrees, you will see a trapezium.
And again, the "total heat" you got is the work A. It's a complete cycle, so \Delta U = 0.

 

[dingo_d]

Eek, if you rotate your diagram 90 degrees, you will see a trapezium.
And again, the "total heat" you got is the work A. It's a complete cycle, so \Delta U = 0.

Maybe you cannot see it from my drawing, but it is a triangle, it's such in the original form. There's just no way to make it into trapezium. But that aside, I still don't see how to calculate work XD

 

[hikaru1221]

The trapezium corresponds to Q_{out}. The formula you got for Q_{out} is the trapezium's area. See the area you need to calculate? Rotate it 90 degrees.

For a complete cycle: \Delta U = Q_{total} - A = 0. You see?

 

[dingo_d]

I get it now, the area UNDER 3-1 line is Q_{out}! And that is trapezium. I didn't understand that XD Thnx anyhow :D