Carnot heat engine - calculate work

[skrat]

Using Carnot heat engine we connect two objects with the same mass m, specific heat capacities c. At the beginning the body 1 has temperature T_{1} and body 2 T_{2}. Calculate:
a) How much mechanical work can this engine give?


I tried like this, but i believe this is very wrong :(
dW=-\eta_{c}Q=(1-\frac{T}{T_{1}})mcdT
\int_{0}^{W}dW=-\int_{T_{2}}^{T}(1-\frac{T}{T_{1}})mcdT
W=(T_{2}+\frac{T_{1}}{2}(T^{2}-\frac{T^{2}_{2}}{2})-T)

Can anybody please help me to calculate this the right way?

Thanks!

 

[TSny]

A very similar question was asked yesterday here: https://www.physicsforums.com/showthread.php?t=680648

You might see if anything there is helpful.

 

[Andrew Mason]

Using Carnot heat engine we connect two objects with the same mass m, specific heat capacities c. At the beginning the body 1 has temperature T_{1} and body 2 T_{2}. Calculate:
a) How much mechanical work can this engine give?


I tried like this, but i believe this is very wrong :(
dW=-\eta_{c}Q=(1-\frac{T}{T_{1}})mcdT
\int_{0}^{W}dW=-\int_{T_{2}}^{T}(1-\frac{T}{T_{1}})mcdT
W=(T_{2}+\frac{T_{1}}{2}(T^{2}-\frac{T^{2}_{2}}{2})-T)

Can anybody please help me to calculate this the right way?

Thanks!

I am not sure you can do it using calculus. But there is an easy way. You just have to recognize that the total change in entropy of the two reservoirs is: ____.

You can determine the final temperature of both reservoirs using this principle. From that you can determine the change in internal energy of the reservoirs. From the first law, you can then determine the amount of work that has been produced.

AM

 

[skrat]

I am not sure you can do it using calculus. But there is an easy way. You just have to recognize that the total change in entropy of the two reservoirs is: ____.

That is actually part b) of my homework: What is the total change of entropy?

I haven' really thought about that part yet, but my idea was to start like this:
\Delta S_{1}=\int \frac{dQ}{T}=mc\int \frac{dT}{T}
Where this is change of entropy for one body only, so:
\Delta S=\Delta S_{1}+\Delta S_{2}=mc(ln\frac{T}{T_{1}}+ln\frac{T}{T_{2}}) where the first logarithm is negative (assuming that T_{1}>T_{2} and T somewhere in the middle).

You can determine the final temperature of both reservoirs using this principle. From that you can determine the change in internal energy of the reservoirs. From the first law, you can then determine the amount of work that has been produced.

And the final temperature is part c)

So if I find out how to calculate the final temperature I can than calculate the change in internal energy and there by from first law also the amount of work produced.
Well I am assuming that the total change of entropy is equal to 0, but I don't really understand why would this be true and even if it is, how can I than determine the final temperature from the equation above?

Thanks!

 

[Andrew Mason]

That is actually part b) of my homework: What is the total change of entropy?

I haven' really thought about that part yet, but my idea was to start like this:
\Delta S_{1}=\int \frac{dQ}{T}=mc\int \frac{dT}{T}
Where this is change of entropy for one body only, so:
\Delta S=\Delta S_{1}+\Delta S_{2}=mc(ln\frac{T}{T_{1}}+ln\frac{T}{T_{2}}) where the first logarithm is negative (assuming that T_{1}>T_{2} and T somewhere in the middle).

Correct. So \Delta S/mc = ln\frac{T}{T_{1}}+ln\frac{T}{T_{2}} = 0 which reduces to:

\ln{\frac{T}{T_{1}}} = \ln{\frac{T_{2}}{T}}

So what is T in terms of T1 and T2 (hint: apply the inverse of ln to both sides)?

And the final temperature is part c)

So if I find out how to calculate the final temperature I can than calculate the change in internal energy and there by from first law also the amount of work produced.
Well I am assuming that the total change of entropy is equal to 0, but I don't really understand why would this be true and even if it is, how can I than determine the final temperature from the equation above?

ΔS = 0 because all cycles are reversible Carnot cycles. The system and reservoirs are in equilibrium at all times.

AM

 

[skrat]

ΔS = 0 because all cycles are reversible Carnot cycles. The system and reservoirs are in equilibrium at all times.

So every reversible cycle has ΔS=0. But Carnot cycle is reversible, thereby the change of total entropy is equal to zero. Right?

Ok, I hope this is right:
mcln(\frac{T}{T_{1}})+mcln(\frac{T}{T_{2}})=0
ln(\frac{T^{2}}{T_{2}T_{1}})=0
\frac{T^{2}}{T_{2}T_{1}}=1
T^{2}=T_{2}T_{1}
T=\sqrt{T_{2}T_{1}}(result for part c) final temperature)
Where T is always positive.

We already said that ΔS=0, but let's make sure (mainly because it's part b) of my homework):
\Delta S=mcln(\frac{T^{2}}{T_{1}T_{2}})=mcln(\frac{T_{1}T_{2}}{T_{1}T_{2}})=0

Let's move to part a) so total amount of work given... hmm weird, but my result is 0?
If I understand correctly, you use letter U for internal energy? Funny. =)
\Delta U=\Delta Q+\Delta W
\Delta W=\Delta U-\Delta Q
\Delta W=(mc(T-T_{1})+mc(T-T_{2}))-(mc(T-T_{1})+mc(T-T_{2}))
\Delta W=mc(2T-T_{1}-T_{2})-mc(2T-T_{1}-T_{2})
\Delta W=0

Now I will be happy if anybody corrects me again.=)

 

[TSny]

Think about what ΔU should be for the engine as it goes through one cycle. U is a state variable, and when something goes through a cycle it returns to its initial state.

 

[skrat]

Think about what ΔU should be for the engine as it goes through one cycle. U is a state variable, and when something goes through a cycle it returns to its initial state.

Of course! Hah, completely forgot about that.

So \Delta W=\Delta Q=mc(2T-T_{1}-T_{2})

 

[TSny]

That's close. Check the overall sign.

 

[Andrew Mason]

Of course! Hah, completely forgot about that.

So \Delta W=\Delta Q=mc(2T-T_{1}-T_{2})

That is one way to look at it.
Or you could apply the first law to the whole system of engine + reservoirs as an isolated system, in which ΔQ = 0 (no heat flow into or out of the system). In that case, ΔU = W. The change in internal energy of the reservoirs is just:

ΔU = mC(ΔT_h + ΔT_c)

AM

 

[skrat]

What if the heat engine works n times worse than carnot engine?

Is in that case total change of entropy still 0 and final temperature the same as before, so: T=\sqrt{T_{1}T_{2}} ?

and \Delta W=-\eta \Delta Q where \eta =\frac{\eta _{c}}{n} ?

 

[Andrew Mason]

What if the heat engine works n times worse than carnot engine?

Is in that case total change of entropy still 0 and final temperature the same as before, so: T=\sqrt{T_{1}T_{2}} ?

and \Delta W=-\eta \Delta Q where \eta =\frac{\eta _{c}}{n} ?

I am not sure what n times worse than the Carnot means. The entropy would definitely not be 0 if it is an irreversible process.

AM

 

[skrat]

I am not sure what n times worse than the Carnot means. The entropy would definitely not be 0 if it is an irreversible process.

It means that we have an carnot heat engine that has effciency defined as \eta _{c}=1-\frac{T_{C}}{T_{H}}. A similar heat engine that works n times worse that carnot heat engine is an engine whose efficiency \eta _{x} is defined as \eta _{x}=\frac{1-\frac{T_{C}}{T_{H}}}{n}.
So Carnot heat engine has maximum efficiency, now I have to calculate the amount of work, final temperature and total change of entropy if the heat engine has lower efficiency.

At least, that't how I understand it...
I hope my english is not too bad to understand it.

 

[Andrew Mason]

It means that we have an carnot heat engine that has effciency defined as \eta _{c}=1-\frac{T_{C}}{T_{H}}. A similar heat engine that works n times worse that carnot heat engine is an engine whose efficiency \eta _{x} is defined as \eta _{x}=\frac{1-\frac{T_{C}}{T_{H}}}{n}.
So Carnot heat engine has maximum efficiency, now I have to calculate the amount of work, final temperature and total change of entropy if the heat engine has lower efficiency.

At least, that't how I understand it...
I hope my english is not too bad to understand it.

So efficiency = W/|Qh| = (|Qh|-|Qc|)/|Qh| = (1/n)(Th-Tc)/Th. This means that:

That appears to reduce to:

|Qh|/Th = |Qc|/Tc - (n-1)W/Tc which, since Qc > 0 and Qh < 0, means:

Qh/Th + Qc/Tc = (n-1)W/Tc

If the the heat reservoirs are such that there is negligible change in temperature of each reservoir in each cycle, the change in entropy in one cycle is:

ΔS = ΔS_h + ΔS_c = Qh/Th + Qc/Tc = (n-1)W/Tc

AM