Heating a metallic rod from one end by continuous heat flux

[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?

 

[Chestermiller]

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?

This could be accomplished if the needle were immersed in a constant temperature bath (T2) at x = L.

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?

The temperature is not expected to depend on time at x = L if it is fixed at that location by means of a constant temperature bath at whatever temperature you wish.

And what would be the ideal boundary condition in the case where u(L,t) rises from 0 at t=0 to T2 finally?

Let me understand this correctly. You want the initial temperature to be u(x,0) to be zero everywhere along the rod, but you want the temperature at x = L to rise from 0 to T2 over time in some natural way. Correct? If this is what you want, I can provide a boundary condition that do this.

Chet

 

[Adel Makram]

Let me understand this correctly. You want the initial temperature to be u(x,0) to be zero everywhere along the rod, but you want the temperature at x = L to rise from 0 to T2 over time in some natural way. Correct? If this is what you want, I can provide a boundary condition that do this.

Chet

Yes this is what I want. Thanks.

 

[Chestermiller]

Yes this is what I want. Thanks.

$$k\left(\frac{\partial u}{\partial x}\right)_{x=L}=-\frac{P}{A}\frac{u(L,t)}{T_2}$$

Chet

 

[Adel Makram]

$$k\left(\frac{\partial u}{\partial x}\right)_{x=L}=-\frac{P}{A}\frac{u(L,t)}{T_2}$$

Chet

I follow your formula to derive the boundary condition for v(x,t)

u(x,t)= T₂+(L-x)(P/KA) +v(x,t) ------(1)this is the general formula
u_x(x,t)= - (P/KA) +v_x(x,t) --------(2)
Divide (1) on (2);
u(x,t)/u_x(x,t)= T2+(L-x)(P/KA) +v(x,t) / -(P/KA) +v_x(x,t)
at x=L
u(L,t)/u_x(L,t)= T2+v(L,t) / -(P/KA) +v_x(L,t) -----(3)

But according to you, u(L,t)/u_x(L,t)= T₂ / (-P/KA)
which mandates v(L,t) & v_x(L,t) =0 in equation (3), correct?
But if v_x(L,t) =0 then u_x(L,t)= -(P/KA) which again is not fitting our needle model as you described before!

Where did I go wrong? and how did you derive your formula?

 

[Chestermiller]

I think you might have made a mistake in algebra. I get:
$$kv_x=-\frac{P}{A}\frac{v}{T_2}$$
at x = L

Chet

 

[Adel Makram]

I think you might have made a mistake in algebra. I get:
$$kv_x=-\frac{P}{A}\frac{v}{T_2}$$
at x = L

Chet

here is my calculation attached.

 

[Chestermiller]

I have no idea what your motivation was for doing what you did, but here's my (much more straightforward) analysis:
$$u_x=-\frac{P}{AK}+v_x$$
At x = L,
$$u=T_2+v$$

So, substituting into the boundary condition for u at x = L:

$$K(-\frac{P}{AK}+v_x)=-\frac{P}{A}\frac{(T_2+v)}{T_2}$$
So,
$$-\frac{P}{A}+Kv_x=-\frac{P}{A}-\frac{P}{A}\frac{v}{T_2}$$
So, at x = L,
$$Kv_x=-\frac{P}{A}\frac{v}{T_2}$$

Any questions?

Chet

 

[Adel Makram]

So, at x = L,
$$Kv_x=-\frac{P}{A}\frac{v}{T_2}$$

Any questions?

Chet

Ok, This is fine.

I then attached a solution. But I didn't get values for λ in a closed form.

 

[Chestermiller]

Ok, This is fine.

I then attached a solution. But I didn't get values for λ in a closed form.

$$\cot(λL)=\frac{T_2KA}{PL}(λL)$$

See Table 4.20 in Abramowitz and Stegun.

Chet

 

[Chestermiller]

I have a much easier way of achieving what you wish to achieve in your rod problem (i.e., constant flux at x = 0 and temperature rising gradually from 0 to T2 at x = L in a natural way. The solution to this problem is much simpler. I will get back to you a little later with the details.

Chet

 

[Chestermiller]

1. Add a fictitious section of length Δ to the end of the existing rod, featuring the same thermal properties as the existing rod.

2. Hold the temperature at x = L' = L + Δ constant at u = 0 for all times.

3. At final steady state, the temperature profile in the combined rod will be $u = \frac{P}{KA}(L+Δ-x)$. Choose Δ such that, at final steady state, u = T2 at x = L:

$$Δ=T_2\frac{KA}{P}$$

4. Solve the problem for the combined rod of length L' = L + Δ, but only consider the solution for the region between x = 0 and x = L.

Chet

 

[Adel Makram]

1. Add a fictitious section of length Δ to the end of the existing rod, featuring the same thermal properties as the existing rod.

2. Hold the temperature at x = L' = L + Δ constant at u = 0 for all times.

3. At final steady state, the temperature profile in the combined rod will be $u = \frac{P}{KA}(L+Δ-x)$. Choose Δ such that, at final steady state, u = T2 at x = L:

$$Δ=T_2\frac{KA}{P}$$

4. Solve the problem for the combined rod of length L' = L + Δ, but only consider the solution for the region between x = 0 and x = L.

Chet

This seems to be a brilliant simplification. In fact I thought of it but I discarded it because I felt that the thermal proprietaries of the room air around my rod of length L, is different from the thermal properties of the rod material of that additional fictitious length Δ.
I will follow it once again and I will post the calculation.

 

[Adel Makram]

I would be so glad if you have time to check my solution.

 

[Chestermiller]

I didn't have it to go through all the details of your analysis, but it certainly looks like you had the right idea. One thing I would do is, instead of saying that the sum is over the odd values of n, replace n by (2n-1) and say that the sum is over all values of n.

Chet

 

[Adel Makram]

But this solution will not satisfy the mixed boundary condition that was proposed in an earlier post (#53). As the ratio between u(L,t) and u_x(L,t) is not satisfied at all times but only at t is large enough.

 

[Chestermiller]

But this solution will not satisfy the mixed boundary condition that was proposed in an earlier post (#53). As the ratio between u(L,t) and u_x(L,t) is not satisfied at all times but only at t is large enough.

So... Is that a problem? The two problem specifications are slightly different, and give two slightly different solutions. Who's to say that one is more realistic than the other. The new formulation assumes that you add a small additional piece of rod to the end of the existing rod. This small additional piece of rod has thermal inertia. The previous formulation essentially assumes the same thing, except that the thermal inertia of the small additional piece is negligible. So what?

Chet

 

[Adel Makram]

So... Is that a problem? The two problem specifications are slightly different, and give two slightly different solutions. Who's to say that one is more realistic than the other. The new formulation assumes that you add a small additional piece of rod to the end of the existing rod. This small additional piece of rod has thermal inertia. The previous formulation essentially assumes the same thing, except that the thermal inertia of the small additional piece is negligible. So what?

Chet

So what is the model that best describe the reality? Which model the needle will follow in nature, the one with mixed BCs or the one with fictitious additional length with the same thermal properties?

 

[Chestermiller]

So what is the model that best describe the reality? Which model the needle will follow in nature, the one with mixed BCs or the one with fictitious additional length with the same thermal properties?

It doesn't pay to compare them, because they both give just about the same results. It would be splitting hairs. The actual boundary condition at x = L depends on what you deliberately decide to impose there.

Chet

 

[Adel Makram]

So, Thank you for your help in solving this problem.
Adel