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