# Heat engines and finding efficiency

1. Dec 7, 2013

### BOAS

Hello,

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

Engine A receives four times more input heat, produces six times more work, and
rejects three times more heat than engine B. Find the efficiency of (a) engine A and
(b) engine B.

2. Relevant equations

e = |W|/|QH|

e = 1 - |Qc|/|QH|

3. The attempt at a solution

I can't really see how i'm supposed to find any numerical answer to this question. It seems overly simple just to state the efficiency of A in terms of B and vice versa.

What relationship am I not seeing that will help me solve this problem?

2. Dec 7, 2013

### voko

Express the efficiency of A in terms of B via the two different equations for efficiency. That will give you two equations for two unknowns.

3. Dec 8, 2013

### ehild

What is the change of internal energy during the whole circle? Write up the First Law with QH, QC and W for both systems and use the relations given between the input heat, rejected heat and work.

ehild

4. Dec 8, 2013

### BOAS

Thanks for the response,

$e_{a}$ = $\frac{6|W_{b}|}{4|Q_{Hb}|}$ = 1 - $\frac{3|Q_{Cb}|}{4|Q_{Hb}|}$

$\frac{6|W_{b}|}{4|Q_{Hb}|}$ = 1 - $\frac{3|Q_{Cb}|}{4|Q_{Hb}|}$

$\frac{6|W_{b}|}{4|Q_{Hb}|}$ + $\frac{3|Q_{Cb}|}{4|Q_{Hb}|}$ = 1

$\frac{6|W_{b}| + 3|Q_{Cb}|}{4|Q_{Hb}|}$ = 1

$6|W_{b}| + 3|Q_{Cb}|$ = $4|Q_{Hb}|$

$6|W_{b}|$ = $4|Q_{Hb}| - 3|Q_{Cb}|$

The above is the same as a cons. energy statement, but i'm not sure if i'm getting any closer to an actual answer.

Last edited: Dec 8, 2013
5. Dec 8, 2013

### voko

You do not want to deal with the heats and work. You want to relate $e_A$ and $e_B$ in two different ways.

6. Dec 8, 2013

### BOAS

$e_{a}$ = $\frac{6|W_{b}|}{4|Q_{Hb}|}$

$e_{a}$ = $1 - \frac{3|Q_{Cb}|}{4|Q_{Hb}|}$

$e_{b}$ = $\frac{4|Q_{Ha}|}{6|W_{a}|}$

$e_{b}$ = $1 - \frac{4|Q_{Ha}|}{3|Q_{Ca}|}$

I don't understand what to do with this info.

7. Dec 8, 2013

### voko

Since $e_B = \dfrac {|W_B|} {|Q_{HB}|}$ by definition, you have $e_A = \dfrac 6 4 e_B$.

8. Dec 8, 2013

### BOAS

So I was unnecessarily confusing matters by talking about b in fractions of a(which was in turn multiples of b...)?

This has turned out a lot simpler than I was making it, but I am concerned that there are 5 marks available here.

9. Dec 8, 2013

### voko

Have you been able to form another equation for $e_A$ and $e_B$? Have you solved the system?

10. Dec 8, 2013

### BOAS

$e_{a}$ = 1 - $\frac{3|Q_{Cb}|}{4|Q_{Hb}|}$

$e_{b}$ = 1 - $\frac{|Q_{Cb}|}{|Q_{Hb}|}$

$e_{a}$ = $\frac{3}{4}$$e_{b}$ + $\frac{1}{4}$

I have a feeling that my equation for the efficiency of a is wrong...

Then doing simultaneous equations I obtain that ea = 1/2

so eb = 1/3

11. Dec 8, 2013

### voko

All correct.

12. Dec 8, 2013

### BOAS

Thank you Voko,

for being so patient and leading me through it.

BOAS.

13. Dec 8, 2013

### ehild

A bit different approach: During a cyclic process, the change of internal energy is zero: Qh(A)-Qc(A)-W(A)=0, Qh(B)-Qc(B)-W(B)=0. With the given relation between engines A and B: 4Qh(B)-3Qc(B)-6W(B)=0. Eliminate Qc(B): Qh(B)-3W(B)=0. The efficiency is defined as η=W/Qh, so it is 1/3 for engine B. You know already the ratio of the efficiencies...

ehild

14. Dec 8, 2013

### BOAS

Hi,

I don't know what you mean by cyclic process, or how I can know my question describes one. It is always a good idea to know different ways of solving a problem, so I would be interested in understanding your method.

BOAS.

15. Dec 8, 2013

### ehild

How does a thermal engine work? Is it not repeating itself, and going back to its initial state after performing the process?

How do you define QH, QC and W? They are not meaningless letters. Your two equations of efficiency were derived from the condition that the net change of the internal energy during the process is zero.

Eliminate e from the two equations for efficiency

e = |W|/|QH|, e = 1 - |Qc|/|QH| ,

You get that QH-QC-W=0. *

The definition of efficiency refers to a process where heat is taken in one stage, heat is rejected in other stage, (the net heat transferred to the system is Q= QH-QC) work W is done.

The change of the internal energy in a process is ΔU=Q-W, according to the First Law of Thermodynamics. Eguation * means that the internal energy does not change.

ehild