Heat exchanger consisting of Coaxial Cables

In summary: I was thinking that the rate of heat flow as a function of x would be the same on both sides of that difference equation, but I just remembered that T1 is the temperature at the point of interest, and T2 is the temperature at the other end of the pipe. That's not going to be the same, is it?Okay, so we have:dQ/dt = k(T2-T1)dAdA = 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 dQ/dx = k
  • #1
TFM
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Homework Statement



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.


Homework Equations



N/A

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
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?
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
Diagrams are now working.

And also, the question should have said:


[tex] \Delta T(x) = (T_1_0 - T_2_0).exp(-\alpha x) [/tex]

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

Does my diagram look right, compared to the one given?
 
  • #4
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
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 [tex] \Delta Q = cm\delta T [/tex]

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

Since the Temperature differnce should be the same,

[tex] \frac{Q_1}{c_1m_1} = \frac{Q_2}{c_2m_2} [/tex]Many Thanks,

TFM
 
  • #6
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
Well the fancy symbols were just:

Thermal Conductivity, [tex] \kappa [/tex]

x to x = [tex] \delta x [/tex]

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

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


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
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
Okay so we have

dQ/dt = k(T2-T1)dA

dA = m1 dt + m2dt

[tex] dQ/dt = k(T2-T1)m_1dt + m_2dt [/tex]

dt on both sides cancel

[tex] dQ = k(T2-T1)m_1 + m_2 [/tex]

I need to get a length factor dx in here...
 
  • #10
I don't see "dA = m1 dt + m2dt".
Check the definition of "a" in the question - I'm thinking that dA = a*dx.
 
  • #11
Wouldn't da = a dx = a volume?

Edit, no a is a side, sorry
 
  • #12
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

[tex] dQ_1 = M_1dtC_1 [/tex]

[tex] dQ_2 = M_2dtC_2 [/tex]
 
  • #13
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
Yeah, it does say a is square cross-section of side, a,

So do I need to use:

[tex] dQ_1 = M_1dtC_1 [/tex]

[tex] Code: dQ_2 = M_2dtC_2 [/tex]
 
  • #15
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
Okay, so,

[tex] dQ = m_1 dt C dT [/tex]

[tex] dT = \frac{dQ}{m_1 dt C} [/tex]

[tex] dQ = k(T_2 - T_2) a dx dt [/tex]

[tex] dT = \frac{k(T_2 - T_2) a dx dt }{m_1 dt C} [/tex]

[tex] dT = \frac{k(T_2 - T_2) a dx }{m_1 C} [/tex]

Does this look better?
 
  • #17
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
So:

[tex] dT_1 = \frac{k(T_2 - T_2) a dx }{m_1 C_1} [/tex]

How did you get from this to:

[tex] \frac{dT_1}{dx} = u_1(T_2 - T_1) [/tex]

?

Where has a, M, C gone?
 
  • #19
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
I just lumped the constants together.
u1 = ka/(m1*C1).
 
  • #21
So u in this case isn't the internal energy...?
 
  • #22
No. Would you prefer a different letter?
 
  • #23
Breakthrough! I noticed the question doesn't ask for the formula for T1 and T2, only for T2 - T1. So back to dT1/dx = u1(T2 - T1) and dT2/dx = u2(T1 - T2).
Now the objective is to get a d/dx(T2 - T1) on one side and a (T2 - T1) on the other. Not too bad, and we have our exponential! I haven't checked to see if it is exactly the same as the objective given.
 
  • #24
Okay, fair enough.

[tex] dT_1 = \frac{k(T_2 - T_2) a dx }{m_1 C_1} [/tex]

[tex] U_1 = \frac{ka}{m_1 C_1} [/tex]

Thus:

[tex] dT_1 = U_1(T_2 - T_1) dx [/tex]

When did we ho through dT_2? I have read through all the posts, and can't find any mention of it?
 
  • #25
Just the same except for reversing T2-T1.
dT2 = U2(T1 - T2)dx

Watch out for a sign problem. I'm getting a difference in sign on one of the u's in the final answer as compared to the given answer. Am I right in saying (T1 - T2) for the dT2? Or should it be (T2 - T1) again? I don't think so.
 
  • #26
So,

[tex] dQ = m_2 dt C_2 dT_2 [/tex]

[tex] dT_2 = \frac{dQ}{m_2 dt C_2} [/tex]

[tex] dQ = k(T_2 - T_1) a dx dt [/tex]

[tex] dT_2 = \frac{k(T_2 - T_1) a dx dt }{m_2 dt C_2} [/tex]

[tex] dT_2 = \frac{k(T_2 - T_1) a dx }{m_2 C_2} [/tex]

Insert V = ka/m2C2

[tex] dT_2 = V(T_2 - T_1) dx [/tex]

Look okay?
 
  • #27
Seems to me it should be T1 - T2, but you'll agree with the given answer if you keep going with that.
 
  • #28
So:

[tex] dT_1 = U_1(T_2 - T_1) dx [/tex]

and

[tex] dT_2 = V(T_2 - T_1) dx [/tex]

Okay, so now should I subtract one from the other:

[tex] dT_2 - dT_1 = V(T_2 - T_1) dx - U_1(T_2 - T_1) dx [/tex]

[tex] d(T_2 - T_1) = V(T_2 - T_1) dx - U_1(T_2 - T_1) dx [/tex]


?
 
  • #29
Yup, and factor the right side to (V-U1)(T2-T1)dx
Why do you have V and U1? V1, V2 or U1, U2 would make more sense.

Remember, the idea is to get it in the form dy/dx = ky
so the solution is y = Ae^-kx
The A can be found by letting k = 0.

I had a missing factor of 4 in the constants. Something was said about the "a" being for a side, so likely the "a" we were using should be "4a" to include the 4 sides. We didn't use the s in our dQ/dt = k(T2-T1)dA. I think our k should be kappa/s or something like that - certainly the dQ/dt should be inversely proportional to the wall thickness s.

Good luck finishing it!
 
Last edited:
  • #30
Okay, so:

[tex] d(T_2 - T_1) = V(T_2 - T_1) dx - U_1(T_2 - T_1) dx [/tex]

[tex] d(T_2 - T_1) = (V(T_2 - T_1) - U_1(T_2 - T_1)) dx [/tex]

factorise out the U and V

[tex] d(T_2 - T_1) = (V - U_1)(T_2 - T_1) dx [/tex]

Insert modified V and U1


[tex] d(T_2 - T_1) = (\frac{\kappa 4a}{sm_2C_2} - \frac{\kappa 4a}{s m_1 C_1})(T_2 - T_1) dx [/tex]

Would the solution to this be:

[tex] (T_2 - T_1) = e^{(\frac{\kappa 4a}{sm_2C_2} - \frac{\kappa 4a}{s m_1 C_1}(T_2 - T_1))dx} [/tex]
 
  • #31
The solution to dy/dx = ky is y = Ae^kx
so your solution should by T2 - T1 = Ae^kx, k = (constant expression with k4a's on top)
Let x = 0 to see what A should be.

Missing the closing bracket on the constants (after the C1). Instead of (T2 - T1)dx, you should have just x. Missing the constant before e.

Compare with the given answer in your first post. Looks like the answer has T1 - T2 instead of T2 - T1; probably doesn't matter.
 
  • #32
okay, so differnce is that they should be other way around...

[tex] (T_1 - T_2) = Ae^{(\frac{\kappa 4a}{s m_1 C_1} - \frac{\kappa 4a}{sm_2C_2}) (T_2 - T_1))dx} [/tex]

Now this is looking similar to the required equation, except T1 - T2 is on the other side, where A is, and instead on the left is delta T. Also, there exponential is:

[tex] (4a/s)\kappa [(m_1C_1) -1 – (m_2C_2) - 1]x [/tex]
 

Related to Heat exchanger consisting of Coaxial Cables

1. What is a heat exchanger consisting of coaxial cables?

A heat exchanger consisting of coaxial cables is a device that uses two concentric tubes, one inside the other, to transfer heat between two fluids. The inner tube carries the hot fluid, while the outer tube carries the cold fluid. The two fluids are separated by the walls of the tubes, but heat is transferred through the walls from the hot fluid to the cold fluid.

2. How does a heat exchanger consisting of coaxial cables work?

A heat exchanger consisting of coaxial cables works by utilizing the principle of conduction. The hot fluid inside the inner tube heats up the walls of the tube, which then transfer the heat to the cold fluid in the outer tube. This process continues until the two fluids reach an equilibrium temperature.

3. What are the advantages of using a heat exchanger consisting of coaxial cables?

There are several advantages to using a heat exchanger consisting of coaxial cables. Firstly, it has a compact design, making it suitable for use in small spaces. It also has a high heat transfer efficiency, as the two fluids are in close contact with each other. Additionally, it is easy to install and maintain, and can be used for a wide range of applications.

4. What are the common applications of a heat exchanger consisting of coaxial cables?

A heat exchanger consisting of coaxial cables has various applications in industries such as HVAC, refrigeration, chemical processing, and power generation. It is commonly used for heating and cooling fluids, as well as for recovering waste heat from industrial processes.

5. How do I choose the right heat exchanger consisting of coaxial cables for my application?

Choosing the right heat exchanger consisting of coaxial cables depends on several factors, including the type of fluids being used, the required heat transfer rate, and the available space for installation. It is important to consult with a heat exchanger expert to determine the best design and size for your specific application.

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