# Homework Help: Thermodynamics - engine efficiency

1. Sep 12, 2008

### portofino

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

A reversible engine contains 0.20 mol of ideal monatomic gas, initially at 600 K and confined to 2.0 L. The gas undergoes the following cycle:

Isothermal expansion to 4.0 L.
Isovolumic cooling to 300 K.
Isothermal compression to 2.0 L.
Isovolumic heating to 600 K.

Determine the engine's efficiency, defined as the ratio of the work done to only the heat absorbed during the cycle.

2. Relevant equations

efficiency, e = W/Q_h = (Q_h - Q_c)/Q_h = 1 - (Q_c/Q_h) where W is work, Q_h is heat energy (high), and Q_c is cool heat energy (low)

3. The attempt at a solution

in the problem leading to the question, i was asked to determine net work and net heat added during the cycle, there values are as follows:

net heat added during cycle: Q = 350 Joules

net work done during cycle: W = 350 Joules

now when i try and determine the efficiency i did this:

e = W/Q_h = (Q_h - Q_c)/Q_h = 1 - (Q_c/Q_h)

e = 350/600 = (600 - 300)/600 = 1 - (300/600), i used the 350/600 = 57.6% because the problem asked for the ratio of work and heat energy

i realized that 350/600 is not equivalent to the other ratios. since (600 - 300)/600 = 50% = 1 - (300/600) =/= 57.6%

which ratio am i supposed to use, the W/Q_h = 350/600 = 57.6% is incorrect? are my values correct for each variable?

thanks

2. Sep 13, 2008

### Ygggdrasil

$$\frac{Q_h - Q_c}{Q_h}$$ is the maximum efficiency possible. The engine cycle you are using is not a Carnot cycle, so its efficiency will be lower than the maximum possible efficiency.

3. Sep 13, 2008

### portofino

the question wants me to use a ratio of work and heat energy though, why should i use (Q_h - Q_c)/Q_h? so should i be getting 50% then is use the efficiency equation you suggest?

4. Sep 13, 2008

### portofino

using (Q_h - Q_c)/Q_h = (600 - 300)/600 = 50% was not correct? what am i doing wrong?

5. Sep 13, 2008

### scrplyr

the equation for efficiency is e=W/Qh
where W is the net work you found, and Qh is the sum of the heat added (only add the positive Q values you found for the process - which should be two values)
*just wondering do you go to umbc?

6. Sep 13, 2008

### portofino

nah i go to old dominion, i already tried W/Q_h where W = 350 joules, my original Q_h was wrong i used temp not Q, as for Net = 691.5-748.3+748.3-345.8 = 345.8, so should e = 350/345.8 = 101.2 % or 350/350 = 100%? or should it be Q_h = 691.5 + 748.3 = 1439.8 so e = 350/1439.8 = 24.3%?

7. Sep 13, 2008

### scrplyr

right, you jsut use the positive Q values, so that last part is right, i got 24.3% and it showed up as right on MP.
hmm, its weird tho, cuz i saw some of your other posts, and u have a lot of the exact same homework problems as i do - including engineering, lol.

8. Sep 13, 2008

### portofino

oh i should've been more specific, i do know people who go to umbc, i am actually posting on behalf of one my old school mates who goes to umbc, he works nights and cant get stuff done

btw 24.3 worked

9. Sep 13, 2008

### gnarkil

whoah, scryplr, i go to umbc too! had one look at the blackboard discussion board and realized how crap it was, i came here cos this place works

you should post there and get more people to join here, everyone could collab so much better

do you know how do to 19.54????

Last edited: Sep 13, 2008