# Something > Carnot efficiency ? WTF?

1. Dec 19, 2008

### ironizer

Something > Carnot efficiency ????? WTF?

So I've made a very simple, ideal model of a heat engine, and my calculations just don't match up to the carnot thing: efficiency = 1-(Tc/Th)

It looks like this: I supposed a rectangular block, base of 1cm by 1cm, and a piston free to move up and down this block without friction. Also suppose that the walls are completely insulated thermally (no heat can transfer through, as to not screw up our calculations). Now we know that 1 atm = 1.03323 kg/sq cm, and the piston is 1x1cm so it must weigh 1.03323kg in order to apply a downward pressure of 1 atmosphere constantly. Assume a vacuum outside our piston, only gravity pushes the piston down.

So I've done 2 calculations, one with helium and one with chlorine. PV=nRT, P is 1, n is 1 because we calculate using one mole each, and R is constant. I know that the specific heat capacity for chlorine is 33.95 J/mol for each degree Kelvin. So I did this: suppose we apply 100 joules of heat INSIDE the rectangular cylinder, and the new temperature would be 100/33.95=2.945 degrees Kelvin increase. The 33.95 capacity says its for room temperature, so I calculated the volume of the gas at room temperature (R*(273+25)) and subtracted that from the new volume (after we applied our 100J heat) so it's R*(273+25+2.945) and I get a .24170839 Liters increase in volume, after applying just 100Joules of heat. If our rectangle is 1x1cm, it means that the height (distance the piston moves) is increased by 2.41708 meters, and we use work=F*D to find our increase in potential energy. Remember our piston weighs 1.03323 kg, so we do 1.03323*9.8*2.41708 = 24.47308 JOULES mechanical energy (potential). This is 24.47308% efficiency (we started with 100J)

WTF?! Did I screw up somewhere? I checked it over about 18 times and nothing looks wrong. But when we use the equation efficiency = 1-(Tc/Th) we get 1 - (273+25)/(273+25+2.945) = 0.009785 or 0.978% efficiency theoretically possible. 0.978% is a bit different than my 24.47308%.

I did the same thing for 1 mole of helium, and I got 39.97339%, because helium has lower specific heat capacity (20.786 J/mol K).

So carnot's theorem says that the efficiency is based solely on the difference in operating temperatures, I'm finding that my efficiency depends on the specific heat capacity and that's about it.

I know carnot's engine is a bit different, he does some weird crap, but my simple system can be made to cycle as well. I could have a "valve" that would let the temperature return to room temperature (273+25) as it was before we applied 100J, and then we could repeat the process.

I can't draw any lines. Is my stuff better than carnot's engine, or did I just fail at math?

Help! Thanks.

2. Dec 19, 2008

### vanesch

Staff Emeritus
Re: Something > Carnot efficiency ????? WTF?

I think the problem is that your system is not cyclic. After the fact, your piston is high, and the gas is hot. So now how are you going to bring that piston down again ? If you cool the gas, then the piston lowers again, and delivers exactly the amount of work that it did when it rose. So you have then 0% efficiency (you put back all the work that you got out of it).

3. Dec 19, 2008

### ironizer

Re: Something > Carnot efficiency ????? WTF?

Oh, you are right. Dam why am I so stupid?

I'll calculate it with a removable mass. See what I get. I still think its weird how I get different efficiencies just by using different working gasses (chlorine and helium).

4. Dec 20, 2008

### ironizer

Re: Something > Carnot efficiency ????? WTF?

Okay I rethought this thing and here's what I got:

Suppose we want to make an "elevator" that lifts 1.0332kg balls a distance of 5 meters.
We have 1 mole of gas in the same square shaft 1cm x 1cm. To do this, we have these balls at a height of the cylinder when it's at room temperature and we "roll" them on to the piston, THEN we add heat to lift it 5 meters above its original height. When we "roll" this ball onto the piston, it will drop to half its height (because we double the pressure). We know the volume of the gas around room temperature (298K) is ~24.5 Liters. So we want to raise the volume to 25Liters to get the 0.5 Liter increase therefore 5 meters distance upwards. PV=nRT, so now we have pressure of 2 atm (we added an extra 1.0322kg on top of the 1cm^2 piston) so 2(25)=1*R*T => 50/R=T= 609.31K. Remember we started with about 298K so this tells us we must increase the temperature 609.31-298 = 311.31K.
Now we want to see how many Joules of energy we need to raise the temperature of this 1 mole 311.31 degrees Kelvin. If it is helium, it would take 20.786(311.31) = 6470.9 Joules. If it is chlorine, it would take 33.95(311.31) = 9635.04 Joules.

First question, why does it take more energy to do the same work (lifting our 1.033Kg ball 5 meters) if we use Chlorine as opposed to Helium? Doesn't Carnot's theorem disregard the working substance?

Second question is about efficiency. If we lift the 1.0332Kg ball 5 meters, that is 1.0332*9.8*5 = 50.6268 Joules we get in potential. 50.6268/6470.9 = 0.007823 or 0.7823% efficiency (suppose we use Helium). Now we do 1-(298/609.31)= 0.51 or 51%.
Now this is really screwed up. What is the reason for this difference?

If someone could explain this and/or point where I screwed up, I would much appreciate it. Thanks in advance.

5. Dec 20, 2008

### Mapes

Re: Something > Carnot efficiency ????? WTF?

One way to look at is that exciting the rotational mode of the chlorine molecule (which doesn't happen with helium) gets you nowhere when using a machine that exploits gas pressure, which is related to translational energy.

6. Dec 20, 2008

### ironizer

Re: Something > Carnot efficiency ????? WTF?

Well, nitrogen has a capacity of around 29 J/mol K and oxygen around the same. The point is, if there was a gas with a heat capacity of close to 0 then we could get 100% efficiency.

I'm just trying to make a connection between this and carnot's theorem.

7. Dec 20, 2008

### Mapes

Re: Something > Carnot efficiency ????? WTF?

But you must see the contradiction here. In effect, you're asking for an increase in atomic/molecular speed (to provide sufficient pressure to make the machine work) without having to contribute the necessary energy.

Why not model an actual Carnot cycle? The (half) cycle you describe above isn't a Carnot cycle; it's not surprising that you're finding that it's working-fluid dependent.

8. Dec 21, 2008

### ironizer

Re: Something > Carnot efficiency ????? WTF?

So why isn't carnot's engine dependent on the working fluid? That's what I don't get.

9. Dec 21, 2008

### Mapes

Re: Something > Carnot efficiency ????? WTF?

I've always figured it's because of the symmetric, reversible heating/cooling and expansion/compression processes. If the working fluid has a higher heat capacity, then more energy is transferred in and transferred out. The magnitudes change but the efficiency doesn't.

10. Dec 21, 2008

### ironizer

Re: Something > Carnot efficiency ????? WTF?

Okay I can see that. But I don't quite get why the Carnot engine does the isothermal compression stage. I don't see why that is necessary.

11. Dec 21, 2008

### Staff: Mentor

Re: Something > Carnot efficiency ????? WTF?

It is what makes the working fluid move.

12. Dec 21, 2008

### Mapes

Re: Something > Carnot efficiency ????? WTF?

As vanesch mentioned above, you need to compress again for the whole process to be cyclic. It is during compression that you unload entropy to the cold reservoir.

The whole point of the Carnot cycle is that you unload as much entropy as you took on. However, you unload less energy to the cold reservoir than you took on from the hot reservoir, and the difference ends up as the work that the engine performs. The books end up balanced because you import entropy at a relatively high temperature and export the same amount at a relatively low temperature.