Heating a metallic rod from one end by continuous heat flux

[Adel Makram]

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]

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

 

[Adel Makram]

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 continuous contact with a reservoir at T1.

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

 

[Chestermiller]

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 continuous 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

 

[Adel Makram]

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?

 

[Chestermiller]

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

 

[Adel Makram]

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, b_n of sin(π nx/L) of v(x,0), which is suppose to be odd function, be calculated?

 

[Chestermiller]

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, b_{n of}of 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

 

[Adel Makram]

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]

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

 

[Adel Makram]

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]

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

 

[Adel Makram]

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 canceled 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 b_n=-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 canceled as well as n.

 

[Chestermiller]

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 canceled 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 b_n=-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 canceled as well as n.

Your Fourier coefficients can't be correct. Show us your derivation of the Fourier coefficients please.

Chet

 

[Chestermiller]

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

 

[Adel Makram]

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!

 

[Chestermiller]

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?

 

[Adel Makram]

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

 

[Chestermiller]

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

Chet

 

[Adel Makram]

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]

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]

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

 

[Adel Makram]

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. Let's say I can choose it arbitrarily to be even function of x. So how we can retrieve b_n?

 

[Chestermiller]

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. Let's say I can choose it arbitrarily to be even function of x. So how we can retrieve b_n?

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

 

[Adel Makram]

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?

 

[Adel Makram]

Another technical thing seems wrong in the initial solution.

v(x,t)= ∑b_n 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).

 

[Adel Makram]

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?

 

[Chestermiller]

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]

Another technical thing seems wrong in the initial solution.

v(x,t)= ∑b_n 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]

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

 

[Adel Makram]

Hello Chet and thanks for your help,

I went through 2 versions of solution and non of them were satisfactory,

The first one I used BCs of a constant heat flux at x=0 and x=L and IC of u(x,0)=0. This time I used your trick of introducing a new function v(x,t) which could be easily solved by letting v_x(0,t) and v_x(L,t)=0 and v(x,0)=0.

However, the solution went in some complicated things especially in 2 points;
1) deriving a₀ of Fourier coefficient.
2) deriving a_n where it should be only odd number of n.
Also I would have been happy if λ=nπ/2L but it comes to be nπ/L which means cos(nπ/L)x can not be zero if x=L ( so as to be canceled and yields u(L,0)=0 :( )
Would you please see the attached file for my detailed calculation.

The second solution, I used BCs, v_x(0,t) and v(L,t)=0 and v(x,0)=0. I didn't upload it because I still see that my problem is of rod with a length L not reaching infinity.

(n.B: some typing mistakes in the file r(x/L -1) instead of r(1 - x/L)

 

[Chestermiller]

Hello Chet and thanks for your help,

I went through 2 versions of solution and non of them were satisfactory,

The first one I used BCs of a constant heat flux at x=0 and x=L and IC of u(x,0)=0. This time I used your trick of introducing a new function v(x,t) which could be easily solved by letting v_x(0,t) and v_x(L,t)=0 and v(x,0)=0.

However, the solution went in some complicated things especially in 2 points;
1) deriving a₀ of Fourier coefficient.
2) deriving a_n where it should be only odd number of n.
Also I would have been happy if λ=nπ/2L but it comes to be nπ/L which means cos(nπ/L)x can not be zero if x=L ( so as to be canceled and yields u(L,0)=0 :( )
Would you please see the attached file for my detailed calculation.

The second solution, I used BCs, v_x(0,t) and v(L,t)=0 and v(x,0)=0. I didn't upload it because I still see that my problem is of rod with a length L not reaching infinity.

v is not equal to zero at time = 0. What is it equal to?

Chet

 

[Adel Makram]

yes sorry for that, it is again typing error v(x,0)=(x-L)P/KA

 

[Chestermiller]

yes sorry for that, it is again typing error v(x,0)=(x-L)P/KA

Yes. That's correct. So now, let's see your solution for v.

Chet

 

[Adel Makram]

But before proceeding with the solution, I think v(x,0)=(x-L)P/KA is derived when the right end of the rode is kept at a constant temperature u(L,t)=0 which is not the case in hand now. As we have a constant flux to the left end (x=0) and from the right end (x=L). This means that the temperature is not kept at a constant u(L,t)=0.

But for a while I attached my solution with v(x,0)=(x-L)P/KA

 

[Chestermiller]

Is that really the boundary condition you want for the rod... Constant flux of P/A at both ends? We can do that, but it's the same thing as having a constant temperature of T2 at x = L/2. So we might as well solve it for constant T2 at L.

Incidentally, for the case you defined, your v(x,0) equation is not correct. It should be $(x-\frac{L}{2})\frac{P}{KA}$

Chet

 

[Adel Makram]

For simplicity, we assume that at the steady state u(x,t)= a-bx where a is the temperature at x=0 and b is the constant flux.
u(x,t)= a-bx +v(x,t)
v(x,0)= bx-a given that u(x,0)=0

Calculating Fourier coefficients a₀ and a_n;
a₀=1/L ∫(bx-a) dx = 1/L (½ bL²-aL)= ½bL-a
a_n= 2/L ∫(bx-a) cos(nπ/L) dx = -2b/n²π² for n=1,3,5,...
v(x,t)= ½bL-a -2b/π² Σ1/n² cos(nπ/L) Exp (-n²π²/L²)t

u(x,t)=a-bx + ½bL-a -2b/π² Σ1/n² cos(nπ/L) Exp (-n²π²/L²)

Now putting t=∞ to simulate the steady state leads to u(x,t)=(a-bx + ½bL-a) not (a-bx) as desired ! This might indicate that if a₀=0, this would solve the problem.

 

[Adel Makram]

your v(x,0) equation is not correct. It should be $(x-\frac{L}{2})\frac{P}{KA}$

Chet

Not sure why? please explain.

 

[Chestermiller]

Not sure why? please explain.

If you are always adding P/KA at one end of the rod and always removing P/KA at the other end of the rod, the amount of heat in the rod doesn't change, and the average temperature remains T2. So the steady state solution is $T_2+(\frac{L}{2}-x)\frac{P}{KA}$.

Chet

 

[Adel Makram]

The equation of the steady state can be written in many correct but different ways depending on which constant we prefer to appear in.
For example, u=T₁ -bx where T₁ is the final temperature at x=0------------ (1)
Where b=P/KA.

substitute x=L, u= T₁ -bL = T₂ where T₂ is the final temperature at x=L
so T₁= T₂ +bL and then plug it in (1) leads to;
u= T₂ +bL-bx =T₂ +b(L-x) ------------(2)

Similarly, u= T_½ +½bL-bx =T_½ +b(½L-x) -------------(3)
where T_½ is the temperature at the middle of the rod.

But in any case, the expansion of v(x,t) will show up an a₀ term that will cancel one of constant terms in the steady state equation and will not result to its desired form once t->∝. This what I don`t understand.

 

[Chestermiller]

In the equation I gave in post #39, b = P/KA, and a = LP/2KA. What does that tell you about a₀ and T_1/2?

Chet

 

[Adel Makram]

That means T2=0, but this will not be the average because the average =(T1+T2)/2= T½. That is why I found that your equation in post#39 is more suited when defining T2 as T½ ( the temperature at the middle not the end of the rod)!

 

[Adel Makram]

Also, I still think my problem is finding the Fourier series of v(x,0)

Say u(x,t)= something +v(x,t) where something is the steady state.
u(x,0)= something +v(x,0)
v(x,0)= - something because u(x,0)=0
But I can not find v(x,0) to cancel this "something" because a₀ is problematic as far as I can see.

 

[Adel Makram]

In other words,

at t=0 v(x,0)= a₀+∑a_n cos(nπx/L)
at t=∝ v(x,0)= a₀ which will not let u(x,t)= steady state unless a₀=0 or the (a₀+∑a_n cos(nπx/L)) is multiplied by e(^-n2π2/L2)t

 

[Chestermiller]

That means T2=0, but this will not be the average because the average =(T1+T2)/2= T½. That is why I found that your equation in post#39 is more suited when defining T2 as T½ ( the temperature at the middle not the end of the rod)!

You can't have both T2 being equal to the temperature at the end of the rod and the fluxes at both ends of the rod being P/KA for all times. Something has got to give. If the fluxes at both ends of the rod are P/KA for all times (equal amounts of heat entering and leaving for all time), then the average temperature of the rod must be T2 for all times. But T2 can't be both the average temperature of the rod and the temperature at the end of the rod at any time except t = 0. Once you come to grips with this, you will understand why a₀ has to be equal to zero.

I am unwilling to discuss how you solved the equations (which is probably correct) until you realize this.

Chet

 

[Adel Makram]

If the fluxes at both ends of the rod are P/KA for all times (equal amounts of heat entering and leaving for all time), then the average temperature of the rod must be T2 for all times.

Chet

Why, it is suppose to be (T1+T2)/2

What I understood is, the flux is kept fixed at both ends while the temperature rises as time goes by till reaches the values of the steady state.

I don't want to keep the temperature at x=L fixed that is why I didn't put it as a boundary condition, instead I kept the flux fixed.

However, this could not be physically real as well, because the steady state is reached when we substitute t=∝ so how can we use a fixed flux at both ends as our boundary conditions at all times to derive the solution of the equation that gives that fixed flux only at time =∝?

Is that a paradox? if we kept the temperature at x=L fixed, the flux could not be fixed and the steady state could not be reached. And if we keep the flux fixed at all times, the solution gives that only at t=∝!

 

[Chestermiller]

Why, it is suppose to be (T1+T2)/2

Who said anything about the average temperature being (T1+T2)/2. For the problem as you defined it, the average temperature is T2 (for all time). Did you not read my post #39. This is the steady state profile for the problem as you defined it.

What I understood is, the flux is kept fixed at both ends while the temperature rises as time goes by till reaches the values of the steady state.

No. For the way you defined the problem, the temperature rises in the half of the rod from 0 to L/2, but drops for the half of the rod between L/2 and L. The positive flux at x = L causes cooling of that end.

I don't want to keep the temperature at x=L fixed that is why I didn't put it as a boundary condition, instead I kept the flux fixed.

No problem. The formulation as you defined it does not have the temperature at x = L fixed. In your formulation, it is decreasing with time.

However, this could not be physically real as well, because the steady state is reached when we substitute t=∝ so how can we use a fixed flux at both ends as our boundary conditions at all times to derive the solution of the equation that gives that fixed flux only at time =∝?

I don't understand this question.

Is that a paradox? if we kept the temperature at x=L fixed, the flux could not be fixed and the steady state could not be reached.

Sure it could. But, the flux at x = L could not be fixed. It could only be fixed at x = 0. So, at steady state, the temperature profile would be $T2+\frac{P}{AK}(L-x)$. Tell me why that can't be the final steady state profile, with the flux you want at both ends (and everywhere else for that matter).

Chet[/quote][/quote][/QUOTE]

 

[Adel Makram]

No. For the way you defined the problem, the temperature rises in the half of the rod from 0 to L/2, but drops for the half of the rod between L/2 and L. The positive flux at x = L causes cooling of that end.

This is not my problem as I don't expect any cooling anywhere.

My problem is to describe heating a metallic rode from one side only by a continuous heat source deriving a constant flux. As if you are holding a needle and approaching it from one side to a heat source. Then what are the boundary conditions ( especially regarding the end which is not exposed to the source)? I am very much interested to describe the evolution of temperature along the needle. Will the needle reach a steady state? and When?

 

[Chestermiller]

This is not my problem as I don't expect any cooling anywhere.

You don't want it to cool anywhere, but the equations you formulated are consistent only with cooling at x = L. So your boundary condition at x = L is not consistent with what you want. Sorry about that.

My problem is to describe heating a metallic rode from one side only by a continuous heat source deriving a constant flux. As if you are holding a needle and approaching it from one side to a heat source. Then what are the boundary conditions ( especially regarding the end which is not exposed to the source)? I am very much interested to describe the evolution of temperature along the needle. Will the needle reach a steady state? and When?

Here are a selection of possible boundary conditions at x = L, and the characteristics of the steady state solution (if it exists):

A. Insulation (zero flux) at x = L : This will not reach steady state, but, nonetheless, can be solved analytically.

B. Flux = P/A at x = L: This is the boundary condition you specified. The final steady state solution for this case is $T_2+\frac{P}{AK}(\frac{L}{2}-x)$. The final steady state value of temperature at x = L for your specified boundary condition is $T_2-\frac{P}{AK}\frac{L}{2}$. Sorry about that. (I can prove to you mathematically that this is the steady state solution to the set of equations you specified).

C. Temperature = T₂ at x = L. The final steady state solution for this case is $T_2+\frac{P}{AK}(L-x)$. The final steady state value of temperature at x = L for this specified boundary condition is $T_2$. The initial flux at x = L is zero.

Choose which boundary condition you want to use to solve the problem. Also, you seem to feel that, if you specify T = T2 at x = L, this will cause a discontinuity at that boundary. That will not be the case.

Chet

 

[Adel Makram]

C. Temperature = T₂ at x = L. The final steady state solution for this case is $T_2+\frac{P}{AK}(L-x)$. The final steady state value of temperature at x = L for this specified boundary condition is $T_2$. The initial flux at x = L is zero.

Choose which boundary condition you want to use to solve the problem. Also, you seem to feel that, if you specify T = T2 at x = L, this will cause a discontinuity at that boundary. That will not be the case.

Chet

This is near to my problem except for one thing. u(L,t)=T₂ at all times, right?, so how could be that consistent with my needle example?
Physically wise, the temperature of the needle end at x=L should rise as time goes because of the heat energy reaches it from the source at x=0 after some time. So, it will not stay at its initial temperature except if the length of the needle is infinite. So my question, how u(L,t) is a boundary condition at all time but the temperature is expected, intuitively, to depend on the time? And what would be the ideal boundary condition in the case where u(L,t) rises from 0 at t=0 to T2 finally?