# Heating a metallic rod from one end by continuous heat flux

What is the solution of heat transfer equation when there is continuous heating ( flux of energy) to one end of a metal rod? what will be the temperature profile before reaching the steady state? when I try the boundary condition where one end (x=0) is kept at T1, I got the general solution of the space dependent component X(x)= C1 sin(ux)+C2cos(ux). substituting x=o leads to C2=T1, then I stopped !! How to retrieve the other constant and the general solution too? will the rod reach steady state given the other end is exposed to room temperature.

Chestermiller
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What is the solution of heat transfer equation when there is continuous heating ( flux of energy) to one end of a metal rod? what will be the temperature profile before reaching the steady state? when I try the boundary condition where one end (x=0) is kept at T1, I got the general solution of the space dependent component X(x)= C1 sin(ux)+C2cos(ux). substituting x=o leads to C2=T1, then I stopped !! How to retrieve the other constant and the general solution too? will the rod reach steady state given the other end is exposed to room temperature.
Please specify where the heat flux is applied. What is the boundary condition at the other end of the rod? What is the initial condition? What is the steady state solution to the heat equation for the prescribed boundary conditions?

Chet

The rod of length L, is put at the x-axis with its left end at the origin and its right end is at x=L. The boundary conditions (BCs): u(0,t)=T1, u(L,t)=T2. where T1>T2. Inital condition u(x,0)=T2. The steady state is expected at t approach infinity: U(x,t)= T1 - (T1-T2/L)x.

One point which I could not figure out and it might affect BCs, is that if the left end is in contact with hot reservoir at temperature T1, then that end should not approach T1 because if it does there would not be continuous flux to it which is not the case when steady state is reached! so what is the maximum temperature the end at x=0 could reach if it is in continous contact with a reservoir at T1.

Also, very important, what equation describes u(x,t) before reaching the steady state?
Thanks.

Chestermiller
Mentor
The rod of length L, is put at the x-axis with its left end at the origin and its right end is at x=L. The boundary conditions (BCs): u(0,t)=T1, u(L,t)=T2. where T1>T2. Inital condition u(x,0)=T2. The steady state is expected at t approach infinity: U(x,t)= T1 - (T1-T2/L)x.

One point which I could not figure out and it might affect BCs, is that if the left end is in contact with hot reservoir at temperature T1, then that end should not approach T1 because if it does there would not be continuous flux to it which is not the case when steady state is reached! so what is the maximum temperature the end at x=0 could reach if it is in continous contact with a reservoir at T1.
The temperature at the boundary jumps from T2 to T1 at time zero, and it stays at that temperature for all time. The heat flux at the boundary jumps from zero to an infinite value initially, and then decreases monotonically to the steady state value at long times.

The way to solve a problem like this is to represent the temperature in the rod at some arbitrary time as:

u(x,t) = T1 - (T1-T2/L)x + v(x,t)

You then solve for v(x,t), which is much easier to obtain. Boundary conditions on v are v = 0 at x = 0 and v = 0 at x = L. The initial condition on v is

v(x,0) = -[T1 - (T1-T2/L)x]

Show that, as with u(x,t), v(x,t) satisfies the transient heat conduction equation.

For v, you now have a transient heat conduction problem in which there is an initial temperature profile in the rod, but the ends are set to v = 0 at t = 0.

Chet

For v, you now have a transient heat conduction problem in which there is an initial temperature profile in the rod, but the ends are set to v = 0 at t = 0.
But v(x,0) at ends (x=0) and (x=L) does not satisfy u(x,0) because u(x,0) at all location is T2 not 0.
The temperature at the boundary jumps from T2 to T1 at time zero, and it stays at that temperature for all time. The heat flux at the boundary jumps from zero to an infinite value initially, and then decreases monotonically to the steady state value at long times.
the heat flux q= -k du/dx, so if the end at x=0 rise to T1 which is equal to the reservoir temperature, du/dx=0 and so does q? Also how about the flux at x=L at different times?

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Chestermiller
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But v(x,0) at ends (x=0) and (x=L) does not satisfy u(x,0) because u(x,0) at all location is T2 not 0.
Ooops. Correction: v(x,0) = T2 - [T1 - (T1-T2/L)x]
the heat flux q= -k du/dx, so if the end at x=0 rise to T1 which is equal to the reservoir temperature, du/dx=0 and so does q? Also how about the flux at x=L at different times?
No. The reservoir is assumed to have infinite conductivity, so it can supply whatever flux is necessary. The temperature gradient within the rod in the vicinity of x = 0 is very high at short times, and this corresponds to a very high heat flux.

In the transient behavior of the rod, at short times, the temperature will decrease very rapidly with distance from x = 0. Beyond a certain distance, both the temperature change and the heat flux will be very close to zero. So the boundary layer region near x = 0 will grow with time until it penetrates the entire rod. At short times, the heat flux at x = L will be very close to zero. As time progresses, it will rise to the final steady state value.

Chet

Thank you, that was very helpful.

Technically speaking, what if v(x,0) is an even function, for example what if we choose u(x,0) so as v(x,0)=cos x. Then how the coefficient, bn of sin(π nx/L) of v(x,0), which is suppose to be odd function, be calculated?

Chestermiller
Mentor
Thank you, that was very helpful.

Technically speaking, what if v(x,0) is an even function, for example what if we choose u(x,0) so as v(x,0)=cos x. Then how the coefficient, bn ofof sin(π nx/L) of v(x,0) be calculated?
You have to take v as an odd function in order to satisfy the boundary condition on v at x = 0. A half-wave sine series is involved in satisfying the initial condition on v.

Chet

The heat flux at the boundary jumps from zero to an infinite value initially, and then decreases monotonically to the steady state value at long times.

In the transient behavior of the rod, at short times, the temperature will decrease very rapidly with distance from x = 0. Beyond a certain distance, both the temperature change and the heat flux will be very close to zero. So the boundary layer region near x = 0 will grow with time until it penetrates the entire rod. At short times, the heat flux at x = L will be very close to zero. As time progresses, it will rise to the final steady state value.
Chet

Hereby I show as you said that the flux is infinite at x=o at t=0 and then reduces to a finite value of steady state as time goes by to infinity. However, at x=L and t=0, the heat flux is T1-T2/L ???

the solution to that problem is u(x,t)= T1-(T1-T2/L)x + v(x,t)
= T1-(T1-T2/L)x + -2(T1-T2)/π Σ(1/n) sin (nπx/L) Exp (-n2π2t/L2

and from Fourier law, the heat flux, φ= -K∂u/∂x

so φ(x,t)= T1-(T1-T2/L) + 2(T1-T2)/L (Exp (-n2π2t/L2) Σ cos (nπx/L)

so φ((0,0) = ∞ exactly as you kindly said.
φ((L,0) = (T1-T2)/L which is the finite final value which is not zero
φ((0,∞) = (T1-T2)/L
and finally, φ((L,∞) = (T1-T2)/L

Chestermiller
Mentor
Hereby I show as you said that the flux is infinite at x=o at t=0 and then reduces to a finite value of steady state as time goes by to infinity. However, at x=L and t=0, the heat flux is T1-T2/L ???
No, at x = L and t = 0, the heat flux is zero.
the solution to that problem is u(x,t)= T1-(T1-T2/L)x + v(x,t)
= T1-(T1-T2/L)x + -2(T1-T2)/π Σ(1/n) sin (nπx/L) Exp (-n2π2t/L2
and the heat flux, φ= -K∂u/∂x
so φ(x,t)= T1-(T1-T2/L) + 2(T1-T2)/L (Exp (-n2π2t/L2) Σ cos (nπx/L)

so φ((0,0) = ∞ exactly as you kindly said.
φ((L,0) = (T1-T2)/L which is the finite final value which is not zero
φ((0,∞) = (T1-T2)/L
and finally, φ((L,∞) = (T1-T2)/L
No. We know that at t = 0 , φ = 0 at x = L because T = T2 for all x < L. So, you must have made a mistake somewhere. For one thing, the first term in your equation for φ, T1, should not be there. Secondly, what happened to the summation term in your equation for the flux at t = 0?

Chet

yes sorry, the true flux is φ(x,t)= (T1-T2/L) + 2(T1-T2)/L (Exp (-n2π2t/L2) Σ cos (nπx/L).

At t=0, the exponential term =1 but all cos (nπx/L) at x=L goes to zero, because cos(nπ)=(-1)n. So even-n will cancel out odd-n which leads Σ cos (nπx/L) to be zero at x=L. However, still (T1-T2/L) is not zero which mean φ(L,0) is not zero too.

Chestermiller
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yes sorry, the true flux is φ(x,t)= (T1-T2/L) + 2(T1-T2)/L (Exp (-n2π2t/L2) Σ cos (nπx/L).

At t=0, the exponential term =1 but all cos (nπx/L) at x=L goes to zero, because cos(nπ)=(-1)n. So even-n will cancel out odd-n which leads Σ cos (nπx/L) to be zero at x=L. However, still (T1-T2/L) is not zero which mean φ(L,0) is not zero too.
Your Fourier coefficients are incorrect. There should be π's and n's in them.

Chet

Your Fourier coefficients are incorrect. There should be π's and n's in them.
n&π appear in the equation of v(x,t) as well as u(x,t), but as we differentiate u(x,t) to x to get the flux from the equation of φ= - K ∂u(x,t)/∂x, n&π will be cancelled out.

u(x,t)= T1-[x.(T1-T2)/L] + (-2)(T1-T2)/π Σ (1/n) sin nπx/L. Exp (n2π2/L2. t

so true, the Fourier coefficient bn=-2(T1-T2)/nπ

But, taking the first differential to x, ∂u(x,t)/∂x= -(T1-T2)/L + (-2)(T1-T2)/π Σ nπ/L (1/n) cos nπx/L. Exp (n2π2/L2. t

so π will be cancelled as well as n.

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Chestermiller
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n&π appear in the equation of v(x,t) as well as u(x,t), but as we differentiate u(x,t) to x to get the flux from the equation of φ= - K ∂u(x,t)/∂x, n&π will be cancelled out.

u(x,t)= T1-[x.(T1-T2)/L] + (-2)(T1-T2)/π Σ (1/n) sin nπx/L. Exp (n2π2/L2. t

so true, the Fourier coefficient bn=-2(T1-T2)/nπ

But, taking the first differential to x, ∂u(x,t)/∂x= -(T1-T2)/L + (-2)(T1-T2)/π Σ nπ/L (1/n) cos nπx/L. Exp (n2π2/L2. t

so π will be cancelled as well as n.

Chet

Chestermiller
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I checked your Fourier coefficients and they are correct. I can see what's happening at x = L. Even though the solution for the temperature converges correctly at x = L, the solution for the flux does not converge at this location at time zero. The slope fluctuates between -(T1-T2)/L and +(T1-T2)/L as n increments by 1. The average as n varies is zero.

Chet

• The slope fluctuates between -(T1-T2)/L and +(T1-T2)/L as n increments by 1. The average as n varies is zero.
For me, I can see what fluctuating is the second term of -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L Σ cos nπx/L. Exp-(n2π2/L2. t)
so at t=0 and x=L, the average of the second term is 0 leaving ∂u(L,0)/∂x = (T1-T2)/L which is not zero!

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Chestermiller
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For me, I can see what fluctuating is the second term of -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L Σ cos nπx/L. Exp-(n2π2/L2. t)
so at t=0 and x=L, the average of the second term is 0 leaving ∂u(L,0)/∂x = (T1-T2)/L which is not zero!
What do you get if you only include the term n=1 in the summation? What do you get if you include only the terms up to n=2? n=3?

What do you get if you only include the term n=1 in the summation? What do you get if you include only the terms up to n=2? n=3?

for n=1, -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L (cos π) = (T1-T2)/L + 2(T1-T2)/L(-1) = -(T1-T2)/L
for n=1&2, -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L (cos π+cos 2π) = (T1-T2)/L +2(T1-T2)/L(-1+1) = (T1-T2)/L
for n=1->3 -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L (cos π+cos 2π+cos 3π) = (T1-T2)/L + 2(T1-T2)/L(-1+1-1) = -(T1-T2)/L

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Chestermiller
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So now you see what I'm saying, right?

Chet

• You have to take v as an odd function in order to satisfy the boundary condition on v at x = 0. A half-wave sine series is involved in satisfying the initial condition on v.
Yes thank you for your help. The issue of flux is over now.

Now, if the initial condition of v(x,t) is an even, then no Fourier coefficients could be calculated right? This means the initial condition could not be chosen arbitrarily? This also mean that the initial condition is dependent on the boundary conditions which I don't understand.

Chestermiller
Mentor
For me, I can see what fluctuating is the second term of -∂u/∂x= (T1-T2)/L + 2(T1-T2)/L Σ cos nπx/L. Exp-(n2π2/L2. t)
so at t=0 and x=L, the average of the second term is 0 leaving ∂u(L,0)/∂x = (T1-T2)/L which is not zero!
That's only if you truncate the summation after an even number of terms. If you include an odd number of terms, you get minus that. Who says an even number of terms is more correct than an odd number of terms. For example, suppose I say that the summation is -2(T1-T2)/L. What makes you more correct to say that the summation is 0.

chet

Chestermiller
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Your saying that the summation term is zero is tantamount to saying that infinity is an even number. The truncated summation term fluctuates between -2(T1-T2)/L and zero as the upper counting limit on the summation varies; the average of these two values is -(T1-T2)/L.

Chet

Yes I agree with you about the issue of summation. So the average heat flux at x=L and t=0 is zero. This matter is cleared for me now.

Now my point is something different, I am wondering what if the initial condition v(x,0) is an even function. Lets say I can choose it arbitrarily to be even function of x. So how we can retrieve bn?

Chestermiller
Mentor
Yes I agree with you about the issue of summation. So the average heat flux at x=L and t=0 is zero. This matter is cleared for me now.

Now my point is something different, I am wondering what if the initial condition v(x,0) is an even function. Lets say I can choose it arbitrarily to be even function of x. So how we can retrieve bn?
The only way you can know if the initial condition on v(x,t) is an even function is if you know how v(x,0) looks at x < 0. Besides, an odd function on v(x,0) is consistent with the boundary condition that v(0,t) is zero, and an even function is not, unless it is a very unusual even function that is zero only at the isolated point x = 0. I don't think a Fourier series can be fitted to that.

Chet

Hello again, I though about my problem once again and I found that this solution u(x,t)=T1-[(T1-T2)/L]x + v(x,t) can not be practical, though it satisfies my BCs and IC. The temperature at x=0 can not jump from T2 ( initial temperature) to T1 instantaneously once the metallic rod comes to contact with the source and the heat flux at t=0 can not be infinite at x=0 then no matter how big is the heat conductivity of the source.
I expect a gradual increase in the temperature at x=0 from T2 to T1 until reaching the steady state after some time. This might change my BCs to a constant heat flux from the source rather than constant temperature at x=0. So how we can propose a solution for this?

Another technical thing seems wrong in the initial solution.

v(x,t)= ∑bn sin (nπx/L) e(-n2π2/L2)t
substituting x=o, t=o=> v(0,0)=0 not T2-T1 as given from the IC of v(x,t).

in this post, I found a similar solution, https://www.physicsforums.com/threads/heat-equation-with-const-heat-source.267812/
The heat flux is fixed at x=0, u(x,0)=0 and not sure what value for u(L,t) in his solution.

I followed Fourier solution as usual and I got,
u(x)= C1sinλx + C2cosλx
u(x=0)x= λC1cosλx - λC2sinλx
But u(x=0)x=-P/KA where P is the fixed power used to derive the heat flux, K is the thermal conductivity and A is the cross area of the rod. (not sure why it is considered u(x=0)x=P/KA in that post although the power should always positive because the gradient of temperature along x is negative)
now -P/KA= λC1cosλx so how to solve for C1?

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Chestermiller
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Hello again, I though about my problem once again and I found that this solution u(x,t)=T1-[(T1-T2)/L]x + v(x,t) can not be practical, though it satisfies my BCs and IC. The temperature at x=0 can not jump from T2 ( initial temperature) to T1 instantaneously once the metallic rod comes to contact with the source and the heat flux at t=0 can not be infinite at x=0 then no matter how big is the heat conductivity of the source.
Yes it can, at least ideally. All you would need to do would be to bring your rod into contact at x = 0 with an identical rod initially at the temperature 2T1-T2 at time t = 0. The interface between the two rods would come to the average temperature ((2T1-T2)+T2)/2 = T1, and it would stay at that temperature for all time. (The other end of the identical rod at -L would be held at 2T1-T2). At the instant that the two rods were brought together, the heat flux at their boundary would be infinite, but only at that very instant.

So, I'm sorry that you are not happy with the problem that you originally posed, but be aware that it is not as unrealistic as you think.

I expect a gradual increase in the temperature at x=0 from T2 to T1 until reaching the steady state after some time. This might change my BCs to a constant heat flux from the source rather than constant temperature at x=0. So how we can propose a solution for this?
Well this problem can be done also. The easiest version would be to impose the steady state heat flux at x = 0 for all time and hold the temperature T2 at x = L.

Chet

Chestermiller
Mentor
Another technical thing seems wrong in the initial solution.

v(x,t)= ∑bn sin (nπx/L) e(-n2π2/L2)t
substituting x=o, t=o=> v(0,0)=0 not T2-T1 as given from the IC of v(x,t).
Of course we know that the initial temperature profile is discontinuous at t = 0, x = 0. So I don't see why this should bother you.

Chet

Chestermiller
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in this post, I found a similar solution, https://www.physicsforums.com/threads/heat-equation-with-const-heat-source.267812/
The heat flux is fixed at x=0, u(x,0)=0 and not sure what value for u(L,t) in his solution.

This method only applies to a rod that is infinitely long, so that the temperature profile never reaches steady state. It also provides an approximate solution to your problem at short times.
I followed Fourier solution as usual and I got,
u(x)= C1sinλx + C2cosλx
u(x=0)x= λC1cosλx - λC2sinλx
But u(x=0)x=-P/KA where P is the fixed power used to derive the heat flux, K is the thermal conductivity and A is the cross area of the rod. (not sure why it is considered u(x=0)x=P/KA in that post although the power should always positive because the gradient of temperature along x is negative)
now -P/KA= λC1cosλx so how to solve for C1?
If you are using Fourier, then you should do this problem the same way that we did the other problem. That is use the steady state solution plus the deviation v from the steady state solution. So let's see what you come up with for the boundary and initial conditions on v.

Chet