Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

To be clear, this is not a homework question, but it is something I am studying for an exam. The question is about a hilsch tube which separates a stream of high pressure air to a high temperature and cold temperature stream. After this, a carnot engine is placed between the streams to convert some of the heat from the hot stream to work and the rest of the heat is given to the cold stream such that the two streams are exiting at the same temperature.

I have given the full problem diagram at the end of the question.

The main question I have is this:

Most definitions of Carnot engine work have isothermal temperature reservoirs, cold and hot, at temperatures $T_{hot}$ and $T_{cold}$ such that

$\eta=W/Q_{hot}=1-T_{hot}/T_{cold}$

However since the temperatures here are constant, how does this relationship change if they continually change as energy is removed or added as in this heat exchanger between two streams?


The manner in which I have approached this problem is:

a) Apply an Energy Balance in which the Hilsch vortex tube is isenthalpic:

$H_{in} = H_{out}$

$\implies$

$n_AC_p(T_A-T_{ref})=n_BC_p(T_B-T_{ref})+n_CC_p(T_C-T_{ref})$

Since $C_p$ drops out, $n_A$ is given, and $T_{ref}$ is arbitrary (can be chosen to be 0), $n_A$ and $n_B$ can be found along with using $n_A=n_B+n_C$. Which for this specific example ends up being

$n_A=1mol/s$, $n_B=0.833mol/s$, $n_C=0.166mol/s$.

b) This is where the problem originates. If we use the equation for efficiency above, we can end up using another energy balance where we take advantage of the equation $Q=mC_p\Delta T$, so we have:

$Q_{hot}=n_CC_p(T_C-T_D)$

where $Q_{hot}$ is the heat being removed from the hot source, or stream $C$. So, similarly,

$Q_{cold}=n_BC_p(T_B-T_D)$

and

$W=Q_{hot}(1-T_{cold}/T_{hot})$

This would (if the temperatures were constant as reservoirs) be:

$W=Q_{hot}(1-T_B/T_C)=n_CC_p(T_C-T_D)(1-T_B/T_C)$

So, The energy balance would be:

$W = Q_{cold} + Q_{hot}$

$\implies$

$n_CC_p(T_C-T_D)(1-T_B/T_C)=n_BC_p(T_B-T_D)+n_CC_p(T_C-T_D)$

$\implies$

$n_C(T_C-T_D)(1-T_B/T_C)=n_B(T_B-T_D)+n_C(T_C-T_D)$

All variables in this equation are known except $T_D$, which allows us to calculate it, but I don't know if the assumptions are right to get here.

problem

share|improve this question
1  
A diagram would help. First, the temperature of the fluid is not constant. It changes from a hot (Th) and cold (not Tc) stream to a stream of intermediate (unknown) temperature. Th and Tc refer to the inlet temperature and the outlet temperature of the Carnot engine. What happens to the temperature of the fluid afterwards has no significance to the efficiently of the Carnot engine. –  LDC3 Mar 30 at 4:08
    
I added the image of the problem. The Wikipedia articles on Carnot engines seem to give a description in which heat is taken from a source at one constant hot temperature and expelled to a cold sink at another constant temperature (isothermal reservoirs). –  chase Mar 30 at 4:15
    
It seems like there is a difference here since along the length of the two streams there is a temperature gradient. When both streams enter, they are at a maximum temperature gradient, so more work is probably allowed to be extracted, but when they exit, they are both at $T_D$, so no work is being extracted any longer. So I would think there would be less work created from this process than a process with two resevoirs at the two inlet temperatures $T_C$ and $T_B$ –  chase Mar 30 at 4:19
1  
The diagram is different from what I thought you had described. The Carnot engine changes the temperature of the cold side to equal that of the high side (I thought that they were mixed). It is important that the first question is answered since it is not likely that Stream B and Stream C are equal. –  LDC3 Mar 30 at 4:29
    
I added the numerical result for the specific numbers given in the example here. You were right in that they will not be the same.You end up getting more moles on the hot side, which I think? is due to satisfying entropy requirements. –  chase Mar 30 at 4:57
add comment

1 Answer 1

up vote 1 down vote accepted

Well I don't quite know the details of Hilsch tube, but I can see in which direction the question is heading towards.

Here is what I think would solve the problem

The amount of heat extracted from the source $C$ is $$ Q_{out} = n_C C_p(T_C - T_D) $$

Also the amount of heat rejected to sink $B$ is

$$ Q_{in} = n_B C_p(T_D - T_B) $$

where $n_B$ and $n_C$ are $\frac{n_A(T_C - T_A)}{T_C-T_B}$ and $\frac{n_A(T_A - T_B)}{T_C - T_B}$ respectively.

Now for source $C$ change in entropy is given by

$$ \Delta S_C = \int_{T_C}^{T_D}n_CC_p \frac{dT}{T} = n_C C_p \ln{\frac{T_D}{T_C}} $$ Similarly the entropy change in sink $B$ is

$$ \Delta S_B = \int_{T_B}^{T_D}n_B C_p \frac{dT}{T} = n_B C_p \ln{\frac{T_D}{T_B}} $$

Since its a carnot engine, the change in entropy of working fluid is zero.

Hence $$ \Delta S_{total} = 0 $$ $$ n_C C_p \ln{\frac{T_D}{T_C}} + n_B C_p \ln{\frac{T_D}{T_B}} = 0 $$

Using this you can find out $T_D$, if it would have been same flow in source and sink i.e $n_B = n_C$ then this equation reduces to

$$ T_D = \sqrt{T_B T_C} $$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.