# Temprature-Entropy cyclic process

1. May 6, 2010

### astrozilla

1. The problem statement, all variables and given/known data
There is a temperature-Entropy graph (T-S) (attachment),which illustrates a hypothetical cyclic process.
a) Calculate the heat input or output along each of the paths.

b) Find an expression for the efficiency η of the complete cycle in terms of T1 and T2 only.

2. Relevant equations

for b ,η=(Q1/Q2)-1=(T1/T2 )-1,HEAT INPUT: DH=DQ (ONLY in case of isobaric change)

3. The attempt at a solution

File size:
5 KB
Views:
105
2. May 6, 2010

### physicsworks

[PLAIN]http://img202.imageshack.us/img202/3227/temperatureentropygragh.jpg [Broken]
I suppose the most complicated path is the linear process II-III (see attached image) in which both T and S change:
$$T(S)= \alpha S + T_0$$
where $$\alpha, T_0 >0$$ are some constants.
So, in this process we have
$$\frac{dS}{dT} = \frac{1}{\alpha}$$
and hence
$$\delta Q_{II-III} = TdS = \frac{1}{\alpha} T dT$$

When T changes from $$T_1$$ to $$T_2$$ you have
$$Q_{II-III} = \frac{1}{\alpha}\int_{T_1}^{T_2} TdT = -\frac{1}{2\alpha} \left(T_1^2 - T_2^2 \right)$$
That is, the heat is extracted in this process.
Now for the process III-I where the entropy is constant it is trivial that $$Q_{III-I}=0$$. It is in fact an adiabatic process.

Try to find the heat for isothermal process I-II. Note that the change of the entropy
$$S_{II} - S_{I}$$
you can express in terms of $$T_2$$, $$T_1$$ and $$\alpha$$ just from the following equations for III and II points on the graph:
$$T_2 = \alpha S_{I} +T_0$$

$$T_1 = \alpha S_{II} + T_0$$
Note also that $$S_{I}=S_{III}$$

Last edited by a moderator: May 4, 2017
3. May 6, 2010

### astrozilla

That's amazing.Thank you very much for both answers !