# Heat exchanger consisting of Coaxial Cables

1. Feb 18, 2009

### TFM

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

A heat exchanger consists of two straight co-axial tubes, each of square cross-section. Liquid 1 enters at a rate m1 kg.s-1 at one end (x = 0) of the exchanger at temperature T10, and flows through the inner tube. Liquid 2 emerges at a rate m2 kg.s-1 at x = 0 at temperature T20, having flowed through the space between the two tubes. The outer tube is heavily lagged. The tube separating the two liquid streams has wall thickness, s, square cross-section of side, a, and its material has thermal conductivity, . By considering an element of length x to x = x of the exchanger operating in the steady state, calculate the heat which must flow transversely through the metal between the liquids. Hence show that the temperature difference, T(x) = T1(x) - T2(x), between the two liquid streams at distance x along the exchanger is

T(x) = (T10 - T20).exp(-x)

where  = (4a/s)[(m1C1)-1 – (m2C2)-1] and C1 and C2 are the specific heats of the two liquids.

You may make the two simplifying assumptions: (i) that due to efficient turbulent mixing, there are no transverse temperature gradients within the two liquids and (ii) that longitudinal heat conduction within the liquid and the solid along the direction of flow may be neglected, due to the smallness of the temperature gradients in the x direction.

2. Relevant equations

N/A

3. The attempt at a solution

I am trying to do this problem, but I am having trouble visualising what is happening. I have attached the diagram given.

I have included what I think is happening, the T10 fluid is lowing through the inner tube, and T20 out the outer tube. The insulation is around the outside. Does this look right?

TFM

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2. Feb 18, 2009

### Delphi51

Yes, sounds good. Lots of symbols I can't read in the text and nothing happens when I click on the attached thumbnails. Looks like a tricky calculus problem - good luck!

3. Feb 19, 2009

### TFM

Diagrams are now working.

And also, the question should have said:

$$\Delta T(x) = (T_1_0 - T_2_0).exp(-\alpha x)$$

where \alpha = (4a/s)\kappa[(m_1C_1)-1 – (m_2C_2)-1] and $$C_1$$ and $$C_2$$ are the specific heats of the two liquids.

Does my diagram look right, compared to the one given?

4. Feb 20, 2009

### TFM

Okay, so is the heat flowing from the outer tube, which is at T20, into the inner tube, T10, so that the water steadily gets hotter asd you go along the tube, which is parallel to the x-axis?

5. Feb 21, 2009

### TFM

Can anyone here help to show me how to start this question, cause I have looked through it and I am unsure how to start.

I know the heat is basically traveling from the hotter outer tube into the cooler tube.

I also know $$\Delta Q = cm\delta T$$

I don't know if this is rught, but I have also got:

Since the Temperature differnce should be the same,

$$\frac{Q_1}{c_1m_1} = \frac{Q_2}{c_2m_2}$$

Many Thanks,

TFM

6. Feb 21, 2009

### Delphi51

I'm having trouble with this one, too - sorry I'm not much help.
Seems to me we must begin with a bit of area between the two temps, say dA.
Do you have a nice formula for the rate of flow of heat through dA? Something like
dQ/dt = k(T2-T1)dA I suppose?

Next, how does that change the temperature of the flowing liquids? It will certainly involve the flow rates and heat capacities. Say mass dm will be in contact with dA.
Maybe we could pretend it pauses there for dt while dQ is exchanged, then jumps to the next dA.

I can't see the fancy symbols in your first post.

7. Feb 21, 2009

### TFM

Well the fancy symbols were just:

Thermal Conductivity, $$\kappa$$

x to x = $$\delta x$$

show that the temperature difference,
$$\Delta T(x) = T_1(x) - T_2(x)$$
between the two liquid streams at distance x along the exchanger is
$$\Delta T(x) = (T_1_0 - T_2_0)e^{-\alpha x}$$

Where $$\alpha = (4a/s)\kappa[(m_1C_1)-1 – (m_2C_2)-1]$$

Okay, so we have:

dQ/dt = k(T2-T1)dA

The Heat will thus flow through the inner tube, going from the hot to cold.

Any good?

What has sprung to my mind is that in this course, sometimes we use ideas from other similar areas. could we use an analogue to electrical resistance?

8. Feb 21, 2009

### Delphi51

Okay, we have expressions for the dQ flowing across our dA in time dt.
We know how this affects the temperature on each side: dQ = mCdT
The m1 and m2 are the rate of flow of mass in kg/s, so the mass in contact with dA for time dt is m1*dt and m2*dt. It should now be possible to put all this together and get expressions for the dT taking place in the dm over time dt on each side.

Maybe just a matter of integrating this over the length of the tubes!

9. Feb 21, 2009

### TFM

Okay so we have

dQ/dt = k(T2-T1)dA

dA = m1 dt + m2dt

$$dQ/dt = k(T2-T1)m_1dt + m_2dt$$

dt on both sides cancel

$$dQ = k(T2-T1)m_1 + m_2$$

I need to get a length factor dx in here...

10. Feb 21, 2009

### Delphi51

I don't see "dA = m1 dt + m2dt".
Check the definition of "a" in the question - I'm thinking that dA = a*dx.

11. Feb 21, 2009

### TFM

Wouldn't da = a dx = a volume?

Edit, no a is a side, sorry

12. Feb 21, 2009

### TFM

Okay so:

dQ/dt = k(T2-T1)dA

dA = a dx

dQ/dt = k(T2-T1) a dx

so now, I need to get m_1dt and m_2dt in here...

Would we use:

Q = MCT

$$dQ_1 = M_1dtC_1$$

$$dQ_2 = M_2dtC_2$$

13. Feb 21, 2009

### Delphi51

The definition of "a" isn't very clear, but certainly dA = some constant times dx, so let us go with the a*dx for now. I'm getting a nasty differential equation - hope you know something about that!

Oh, it came out in the math that T1 is closely related to T2 - maybe not so bad.

14. Feb 21, 2009

### TFM

Yeah, it does say a is square cross-section of side, a,

So do I need to use:

$$dQ_1 = M_1dtC_1$$

$$Code: dQ_2 = M_2dtC_2$$

15. Feb 21, 2009

### Delphi51

That doesn't look right.
We started with dQ = mass*C*dT, then said mass = m1*dt so we have
dQ = m1*dt*C1*dT
It looks bad, but I think it works out okay.
I wrote it as dT = .... and got rid of the dQ using dQ/dt = k(T2-T1)dA

16. Feb 21, 2009

### TFM

Okay, so,

$$dQ = m_1 dt C dT$$

$$dT = \frac{dQ}{m_1 dt C}$$

$$dQ = k(T_2 - T_2) a dx dt$$

$$dT = \frac{k(T_2 - T_2) a dx dt }{m_1 dt C}$$

$$dT = \frac{k(T_2 - T_2) a dx }{m_1 C}$$

Does this look better?

17. Feb 21, 2009

### Delphi51

That's it, but make that dT1 and C1.
So dT1/dx = u1(T2-T1) and dT2/dx = u2(T1-T2)
where u1 = ka/(m1*C1).

This is where I'm not sure how to handle the differential equations.
I solved the dT2 one for (T2 - T1) and substituted into the dT1 equation to get
dT1/dx + u1/u2*dT2/dx = 0
Since u1 and u2 are constant, this is
d/dx(T1 + u1/u2*T2) = 0
This must mean T1 + u1/u2*T2 = constant.
IF the constant is zero, progress can be made. If not, I'm in trouble.

18. Feb 21, 2009

### TFM

So:

$$dT_1 = \frac{k(T_2 - T_2) a dx }{m_1 C_1}$$

How did you get from this to:

$$\frac{dT_1}{dx} = u_1(T_2 - T_1)$$

?

Where has a, M, C gone?

19. Feb 21, 2009

### Delphi51

It doesn't make sense for the constant to be zero - then one temp would have to be negative! Call the constant d.
So T1 = d - u1/u2*T2. Put that in the earlier equation dT2/dx = u2(T1 - T2) and we get
dT2/dx = -(u1+u2)T2 + d*u2.
With d = 0, this is a classic exponential.
Is there a function whose derivative is a constant times itself plus another constant?

20. Feb 21, 2009

### Delphi51

I just lumped the constants together.
u1 = ka/(m1*C1).