Maximizing Efficiency in Tandem Carnot Engines

[mopar969]

Please help with the following:
Prove that the efficiency of two carnot engines operating in tandem, one going from a cold reservoir of temperature Tc to a hot reservoir of Ti, then the second going from a cold reservoir of temperature Ti, to a hot reservoir of temperature Th must be less than that of a single carnot engine going from Tc to Th.

 

[rock.freak667]

How would you find the efficiency of a single carnot engine?

 

[Andrew Mason]

Please help with the following:
Prove that the efficiency of two carnot engines operating in tandem, one going from a cold reservoir of temperature Tc to a hot reservoir of Ti, then the second going from a cold reservoir of temperature Ti, to a hot reservoir of temperature Th must be less than that of a single carnot engine going from Tc to Th.

The proof cannot be given - because it is not true. Are you sure you are not being asked to prove that the overall efficiency of the two Carnot engines operating in tandem must be equal to the efficiency of the single Carnot engine?

AM

 

[mopar969]

That is the problem I was given in class so maybe he ment to prove what you said. Either way please assists. Thanks.

 

[OmCheeto]

That is the problem I was given in class so maybe he ment to prove what you said. Either way please assists. Thanks.

Since you said "tandem", it is two Carnot engines in series.
It's interesting how many times this question has been asked here at the forum. There was even one person that asked about an infinite number of Carnot engines in series.

I just spent 3 hours trying to figure out the answer to your question. Then blink! I love those lightbulb moments.

But anyways, you have to show that you've done some work on the problem. To paraphrase rock.freak667, what is the equation for Carnot engine efficiency?

 

[mopar969]

I have gotten some help from a friend and this is are work for the problem please let me know if we did something wrong or if this is the correct answer? My question is why did we multiply the two engines though how would you know to do that?

 

[mopar969]

Is my work correct?

 

[OmCheeto]

Is my work correct?

You've found the equation for Carnot efficiency. That is correct. But your image is such that I cannot read the subscripts, so I cannot comment on whether or not the conclusion of the image is correct.

I'm a bit senile, so this is just hearsay now, but I believe that I came to the same conclusion as Andrew Mason last weekend. You cannot prove your given OP, because it is not true.

Prove that the efficiency of two carnot engines operating in tandem... must be less than that of a single carnot engine...

I ran some theoretical numbers in order to get a grasp of the problem, and I think that's why it took me so long to solve. It struck me as peculiar that lowering the operating temperature of the engine, with the same delta T, increased it's efficiency.

one engine
Th = 400K Tc = 300K efficiency = 25%

two engines in tandem
Th = 400K Tc = 350K efficiency = 12.5%
Th = 350K Tc = 300K efficiency = 14.3%

It made it look like an infinite series of tandem Carnot engines might improve efficiency to unheard of levels!

But it's a matter of understanding what heat is, and it's relationship to temperature, that solved the problem. Not yours mind you, but mine.

And it didn't hurt that I discovered a cool google feature last week.

google: site:physicsforums.com carnot efficiency

It's an oft asked, and many times unanswered question.

 

[mopar969]

Sorry the pic was fuzzy. Using substitution for n1 and n2 into n3 I found n3 = Th/Ti (n1)+ n2. The last line says Tk>Ti therefore n3 > n1+n2. My friend came to this conclusion but I am confused with how he got that?

 

[OmCheeto]

Sorry the pic was fuzzy. Using substitution for n1 and n2 into n3 I found n3 = Th/Ti (n1)+ n2. The last line says Tk>Ti therefore n3 > n1+n2. My friend came to this conclusion but I am confused with how he got that?

I'm sorry too. Without seeing all of the subscripts, I can't explain why n3 > n1+n2.

In your image, it looks as though you've written the efficiency of engine two = 1 - Tk/Ti

But then you stated that Tk>Ti

The equation for carnot efficiency is 1-T_cold/T_hot

You've implied that engine 2 has an efficiency of 1-T_hot/T_cold, which is wrong.

 

[mopar969]

I though that the equation for the second carnot engine is 1- Th/Ti because the second engine is going from a cold reservoir to a hot reservoir?

 

[rock.freak667]

I though that the equation for the second carnot engine is 1- Th/Ti because the second engine is going from a cold reservoir to a hot reservoir?

The cold reservoir from the 1st engine is going as the hot reservoir to the second engine. The second engine then rejects heat to some other cold reservoir.

 

[OmCheeto]

I though that the equation for the second carnot engine is 1- Th/Ti because the second engine is going from a cold reservoir to a hot reservoir?

After looking at your blurry subscripts a second time, I've decided that all of your equations are backwards.

 

[mopar969]

I thought the where the engine is working from goes on the bottom of the fraction and where it is going to goes on top of the fraction for a Carnot engine equation. If this is not correct then what equations should I have to finish this problem and how do I finish it?

 

[OmCheeto]

I thought the where the engine is working from goes on the bottom of the fraction and where it is going to goes on top of the fraction for a Carnot engine equation. If this is not correct then what equations should I have to finish this problem and how do I finish it?

Rewrite them all:

The equation for carnot efficiency is 1-T_cold/T_hot

 

[mopar969]

Okay so for engine one the cold reservoir is tc and the hot is Ti so n1 = 1- Tc/Ti
for n2 the cold is ti and the hot is th so n2 = 1 - ti/th
for n3 the cold is tc and the hot is th so n3 = 1 - tc/th

through subbing n1 and n2 into n3 I got n3 = n1 + tc/ti n2

Now what do I do? thanks for all the help much appreciated.

 

[Andrew Mason]

Start with an arbitrary amount of heat flow, Qh, from the hot reservoir and follow it through: eg. The heat flow to the intermediate reservoir is Qi = Qh(Ti/Th) and the work done is Qh-Qi = Qh(1-Ti/Th).

Qi then is the input for the second engine so the heat flow to the cold reservoir is Qc = Qi(Tc/Ti) = Qh(Ti/Th)(Tc/Ti) = Qh(Tc/Th). The work done in the second engine is Qi-Qc = Qi(1-Tc/Ti).

The total work is the sum of the work for each engine. The total heat flow in is Qh so determine the overall efficiency of the two engine system. It is the same as for the single Carnot engine.

AM

 

[mopar969]

I was wondering why my old method for solving the problem won't work?

 

[mopar969]

Here is the answers I have so far:
Okay so for engine one the cold reservoir is tc and the hot is Ti so n1 = 1- Tc/Ti
for n2 the cold is ti and the hot is th so n2 = 1 - ti/th
for n3 the cold is tc and the hot is th so n3 = 1 - tc/th

through subbing n1 and n2 into n3 I got n3 = n1 + tc/ti n2.
Then, tc > ti therefore n3 > n1 + n2.

Is this correct?

 

[mopar969]

Is my reasoning correct?

 

[OmCheeto]

Is my reasoning correct?

Apparently not, since neither AM nor I agree with your conclusion.

I wouldn't put all that much weight in my opinion, but I've seen AM's explanations in the other 15 Carnot engine threads. He groks Carnot. Trust him.

 

[mopar969]

Okay but AM says it is the same as for a single engine I need to show that the single engine is greater than the two according the the problem.

 

[mopar969]

So is this problem not solvable the engines are the same?

 

[OmCheeto]

So is this problem not solvable the engines are the same?

Why not put some actual numbers into the equations you came up with and find out.

n3 = 1 - tc/th
n3 = n1 + tc/ti n2.

Here, use mine:

one engine
Th = 400K Tc = 300K efficiency = 25%

two engines in tandem
Th = 400K Ti = 350K efficiency = 12.5%
Ti = 350K Tc = 300K efficiency = 14.3%

 

[Andrew Mason]

Why not put some actual numbers into the equations you came up with and find out.

Here, use mine:

Ok. I am not sure what is wrong with algebra. But let's use numbers.

W = Q_h\eta_1 + Q_h(1-\eta_1)\eta_2 = Q_h(\eta_1 + \eta_2 - \eta_1\eta_2)

Using your numbers:

W = Q_h(.125 +.143 - .143 \times .125) = Q_h(.125 + .125) = Q_h(.25)

\eta = W/Qh = .25

AM

 

[mopar969]

Also the carnot problem I posted the professor told me to do (1 - (tc/ti))(1-(ti/th)). So I am factoring that now but what do I do after I factor it.