# Heat Transfer Between Two Rods in Contact (Equilibrium Temperature Distribution)

1. Problem statement
I've already posted this question in the advanced calculus help but it was suggested that I ask this in the physics forum. I can upload the pdf of my work again so I'll link to it https://www.physicsforums.com/attachment.php?attachmentid=31702&d=1296270924" from my other post. Two homogeneous rods have the same cross section, specific heat c, and density LaTeX Code: \\rho but different heat conductivities LaTeX Code: \\kappa 1 and LaTeX Code: \\kappa 2. Let kj=LaTeX Code: \\kappa j/(cLaTeX Code: \\rho ) be their diffusion constants. They are welded together so that the temperature u and the heat flux LaTeX Code: \\kappa ux at the weld are continuous. The left-hand rod has its left end maintained at temperature zero. The right-hand rod has its right end maintained at temperature T degrees.

Find the equilibrium temperature distribution in the composite rod.

## Homework Equations

qx=-kA$$\frac{dT}{dx}$$

## The Attempt at a Solution

Solution is in linked above as a pdf with picture because it was easier to put the equations together using an external tool.

Equilibrium temperature distribution just means find the governing equation for the temperature of the rod, right?If that's the case then I just equate both q's and solve for T2. Another question I had about this problem is the specific heat c. Where does this come into play for solving this problem? Thanks for any help with this topic. It's been a little while since I've done any heat transfer stuff.

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Sure, your approach sounds good. Sorry I am having trouble viewing your pdf for some reason (I need a new compy sometime soon), perhaps you could post the steps you've taken in tex. The format here ought to be the same as what you wrote up.

I've attached a jpeg version of the pdf. I really hate using the tex tools on this site because for long complicated equations I can't see the preview on the fly. This site http://eriskgroup.com/latex/" [Broken] is really good.

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Just about everything you have there looks good. However, I think you may have gone a bit weird with your definite integrals. At one point it looked good, then in the next step it looked strange.

$$\int_{x_1}^{x_2} q dx = \int_{T(x_1)}^{T(x_2)} -k A dT$$
gives
$$q(x_2-x_1) = -k A (T(x_2)-T(x_1))$$

It looked like you did this, then you didn't, then you did, I personally got confused following whether you kept T as a function or not. Sometimes it can be nice to just take the indefinite integral and use the boundary values explicitly to solve for the unknown constant.

Here's what I came up with based on what I believe you were given in the problem statement (T(-L)=0 T(L)=T_L and T(0)=continuous)
$$T(x) = \left\{ \begin{array}{lr} -q\frac{x+L}{k_1 A} & -L<x<0 \\ T_L + q\frac{(L-x)}{k_2 A} & 0<x<L \end{array} \right.$$

The continuity condition constrains the heat to be
$$q=-\frac{T_L L}{A}\frac{1}{1/k2+1/k1}$$

Did the problem statement want something like this?

Oh yeah, and I think it mentions that the same density, same specific heat, etc., so that you can assume that q is constant rather than a function of x. If I'm wrong about this, I think we'll have to use Green's functions or some other trick to solve this.

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The problem statement want me to find the equilibrium temperature distribution. I have tried to search and also look in my heat transfer book but I can't seem to find a clear answer on what this means. I suspect it has something to do with setting the two equations that I found equal to each other with T_2(as a function of x). I then solve this equivalency for T_2. After that I'm not sure. I won't be able to ask my professor for any help because he handed this out on Friday and it's due Monday.

The second part of this problem ask me to sketch it as a function of x for given values of k1, k2, L1, L2, and T. I'm just confused with that distribution part. To me it sounds like an equation for the equilibrium temperature.

This is the equilibrium temperature. The heat equation is

$$\frac{\partial T}{\partial t} = \frac{k}{c_p \rho} \nabla^2 T$$

Since we want steady state, take out the t, we only want one dimension, so nabla goes to second order ODE.

$$\frac{k}{c_p \rho} \frac{d^2 T}{dx^2} = 0$$

$$\frac{k}{c_p \rho} \frac{dT}{dx} =C_1$$

$$\frac{k}{c_p \rho}T(x) = C_1 x + C_2$$

Use BC (for left hand side)

$$0=-L*C_1+C_2$$

Now the whole equation is

$$\frac{k_1}{c_p \rho} T(x) = C_1(x-L)$$

simplify

$$T(x) = \frac{c_p m}{k_1 A} C_1(x-L)$$

that is heat up top

$$T(x) = \frac{q}{A k_1} (x-L)$$

oh yeah, and also x<0 for this part, so now finally

$$T(x) = -\frac{q}{A k_1} (x+L)$$

see, they are identical results, just that you didn't know the fourier equation was for steady state. Oh, and you'd do the same for the right side.

So my attempt was wrong? Question, you did come up with individual equations for the temperature dependent on x for each material. Why is that? The book for this class is terrible, no examples.

Your attempt was fine, you were just confused. I don't think you were wrong at all, you just had some trouble expressing your findings.

The equation has to be piecewise because they are two different halves. The left half may only vary from 0-30 while the right half may vary in temperature from 30-300. It depends on the constants, and the constants are different based on where you are in space, whether you are on the left half or the right half.

Ok, this makes sense. I'll attempt to do it again in a much more clearer fashion and attach a plot to see if it makes sense. Thanks for the help.

I think I just confused myself trying to work this again. Equilibrium temperature distribution means find an equation for the equilibrium temperature at any point x, where the equilibrium temperature is Tx2? So if I were to solve both equations for the q's and set them equal together under the condition that x=0 (that's the x where the temperature and heat flux are continuous), I can then solve this new equation in terms of T(x2).

Also, for the piecewise function for 0<x<L I got

$$T_L - q\frac{(L-x)}{k_2 A}$$.

Equilibrium temperature distribution is another way of saying steady-state, technically, after t->infinity.

Also, I think you are confusing yourself with your definite integrals. Work the problem using indefinite integrals, and use the boundary conditions to solve for the constant on each half.

I may have gotten the sign wrong, but I am pretty sure I did right, check yourself on that too when you quickly redo the problem with indefinite integrals.

Here's my indefinite integral approach.

$$\int{qdx}=\int-kAdT_x$$
Absorbing the constants into one constant I get:

(1) $$qx+c=-kAT_x$$

At $$x=-L_1, T_x=0$$

So,

$$-qL_1+c=0$$
Therefore $$c=qL_1$$

Substituting what I got for c back into (1) and solving for $$T_x$$:

$$T_x=-\frac{q(x+L_1)}{kA}$$

Now if I do this for the second half using the same integral where $$x=L_1+L_2, T_x=T$$

$$\int{qdx}=\int-kAdT_x$$

I get $$q(L_1+L_2)+c=-kAT$$

So then $$c=-k_2AT-q(L_1+L_2)$$

Plugging this back into (1) and solving for $$T_x$$:

$$T_x=\frac{q(x-L_1-L_2)}{k_2A}+T$$

I discovered that the back of this book actually has some answers. Here is what the back of the book says:

(a) $$\ddot{u_1}=0$$ in $$0\leq{x}\leq{L_1}$$. $$\ddot{u_2}=0$$ in $$L_1\leq{x}\leq{L_1+L_2}$$.
Note: They had the left side boundary condition be x=0 and the right hand boundary condition as $$L_1+L_2$$.

Together with four boundary and jump conditions, hence $$u_1(x)=ax+b$$ and $$u_2(x)=cx+d$$. Solve for a and b using the four conditions.
$$k_1=2, k_2=1, L_1=3, L_2=2, T=10$$

They of course don't graph it but they do give me the equations:

$$u_1=\frac{10x}{7}$$

$$u_2=\frac{10(2x-3)}{7}$$

This I do not understand...how do they get the value for q and I don't understand how they get a seven in the denominator?

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Yours is fine, given the new coordinate system, except that for the first equation you used an [email protected] boundary condition when it should just be [email protected] It looks like the book used the second approach that I went over (with a=q/k_1A and c=q/k_2A).

$$u_1(0)=0=b$$
$$u_1=ax$$

for u2

$$u_2(L_1+L_2)=T=c(L_1+L_2)+d$$
$$d=T-c(L_1+L_2)$$
$$u_2 = c(x-(L_1+L_2))+T$$

Now use the fact the the equations are continuous
$$u_1(L_1)=u_2(L_1)$$
$$L_1 a = c(L_2)+T$$

and that the flux is continuous
$$u'_1(L_1)=u'_2(L_1)$$
$$a=c$$

then the system of equations needs to be solved
$$a=c$$
$$L_1 a = cL_2+T$$

$$a=c=\frac{-T}{L_2-L_1}$$

Hmm, looks like I might have made an error too because I L2-L1≠7. Well, I did just now on the compy rather than paper, so I wouldn't be surprised if I messed up a factor somehow. Don't know how much time you have before you turn in your assignment, but you're definitely close to the right answer.

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Well for that denominator I was thinking that they might have made a mistake because I see no way in which a 7 could appear. This makes much more sense now. Thank you very much for making this clear.

Edit: I just got back from class and it's due Wednesday, haha. For some reason I was thinking today was the 2nd. I will be meeting with the professor today to ask him about some discrepancies with my answer and the books answer.