1. Mar 25, 2008

### LandOfStandar

I just typed a whole question and as I typed it I solved it, lol!

1. The problem statement, all variables and given/known data

In the 1st stage of a 2-stage Carnot engine, energy is absorbed as heat Q1 at temp T1, work W1 is done, and energy is expelled as heat Q2 at lower temp T2. The 2nd stage absorbs that energy as heat Q2, does work W2 and expels energy as heat Q3 at a still lower temp T3. Prove that the efficiency of the engine is (T1 - T3)/T1

Stage 1 Qin=Q1
Qex=Q2
T2 less then T1
W1

Stage 2 Qin=Q2
Qex=Q3
T3 less then T2
W2

2. Relevant equations

e =Qex/W = (Qin-Qex)/Qin = (Tin-Tex)/Tin

3. The attempt at a solution

e(stage1) = (T1-T2)/T1
e(stage1) = (T2-T3)/T2

How do you put them together?

or

is this all wrong? how do I approach this?

2. Mar 25, 2008

### Andrew Mason

Start with the definition of efficiency for each of the cycles and the combined cycle:

E1 = W1/Qh1
E2 = W2/Qh2

What is Ec in terms of W1, W2, Qh1 and Qh2?

What are Qh1 and Qh2 in terms of T1, T2, T3?

AM

3. Mar 25, 2008

### LandOfStandar

E1 = W1/Qh1 = (Qh1-Ql1)/Qh1 = (T1 - T2)/T1

what do you mean?

E2 = W2/Qh2 = (Qh2-Ql2)/Qh2 = (T2 - T3)/T2

4. Mar 26, 2008

### Andrew Mason

So what is the overall efficiency in terms of the total work done (W1+W2) and the total heat flow into the system (Qh1+Qh2)?

AM

5. Mar 26, 2008

### LandOfStandar

(T1 -T2 + T2 - T3) / (T1 + T2) = (T1 - T3) / (T1 + T2)

the problem is the T2 on the bottom

6. Mar 26, 2008

### Andrew Mason

Since the heat flow into the system is just Q1 (the heat flow into the second engine is the output heat of the first):

$$\eta_{total} = (W_1 + W_2) / Q_1$$

But W1 = Q1-Q2 and W2 = Q2-Q3, so

$$\eta_{total} = (Q_1-Q_3) / Q_1 = (T_1-T_3)/T_1$$

AM

7. Mar 26, 2008

### LandOfStandar

thank you that makes since

I now understand the question