Heating a metallic rod from one end by continuous heat flux

[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