# Newton's law of cooling for a cylinder

• fayled
In summary: Yes, you need to solve the differential equation. However, you can also express the heat flow in terms of the gradient. This will allow you to relate the heat flow out of the wire with the temperature difference according to Newton's cooling law.

## Homework Statement

Suppose we have a cylinder of radius R, in an environment with temperature Tenv, and heat is generated in the wire at a rate P per unit volume (for example due to a current - the exact nature is irrelevant). The heat flux from the surface of the wire is A(T(R)-Tenv) for constant A (Newton's law of cooling). Find the temperature T(r).

## Homework Equations

Thermal diffusion equation,
∂T/∂t=(κ/C)∇2T+H/C
where κ is the thermal conductivity of the medium, C the heat capacity per unit volume of the medium, and H the rate of production of heat per unit volume in the medium.

κ∇2T+H=0

## The Attempt at a Solution

I'm assuming we want the steady state solution - however, if we had say a cup of tea and we had Newton cooling, the temperature would decay exponentially to Tenv, so the steady state would just be Tenv - is it the fact that we have some heat production P that allows a non trivial steady state?[/B]

Anyway, for some reason I believe I am supposed to convert the heat flux into a heat gained per unit volume, which is done by
-A(T(R)-Tenv)*(2πRl/πR2l)=-2A(T(R)-Tenv)/R
If correct, why am I allowed to do this - how do I know it is lost uniformly throughout the cylinder?
Then
H=P-2A(T(R)-Tenv)/R
Using the laplacian in cylindrical polars (no z or angular dependence)
(κ/r)d/dr(rdT/dr)=-H
T=-(Hr2/4κ)+Alnr+B
We need A=0 for T to be finite at r=0, and we need T(R) at r=R so B=T(R)+(HR2/4κ).
Then
T(r)=T(R)+(H/4κ)(R2-r2)
with
H=P-2A(T(R)-Tenv)/R.

However I don't really like leaving T(R) in the answer - how am I supposed to find out what this is?

Thanks for any help with the bold bits :)

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fayled said:
is it the fact that we have some heat production P that allows a non trivial steady state?

Yes. The heat production introduces an inhomogeneity into the equation. A linear PDE without any inhomogeneity (either in the PDE itself or in the boundary/initial conditions) always has the trivial solution as the only solution.

As you have been given no information about the wire length, I would say you can make the assumption that it is infinite in length (or rather, you can treat it as a 2D problem). You really do not need to solve the differential equation to know the temperature of the surface. It is simply going to be given by Newton's cooling law and the fact that in the steady state you need to lose the same amount of heat to the environment as is produced inside the wire ... (Of course T(R) should be part of your answer, it should be standing alone on one side of an equal sign as it is what you have been asked to find...)

Orodruin said:
Yes. The heat production introduces an inhomogeneity into the equation. A linear PDE without any inhomogeneity (either in the PDE itself or in the boundary/initial conditions) always has the trivial solution as the only solution.

As you have been given no information about the wire length, I would say you can make the assumption that it is infinite in length (or rather, you can treat it as a 2D problem). You really do not need to solve the differential equation to know the temperature of the surface. It is simply going to be given by Newton's cooling law and the fact that in the steady state you need to lose the same amount of heat to the environment as is produced inside the wire ... (Of course T(R) should be part of your answer, it should be standing alone on one side of an equal sign as it is what you have been asked to find...)

Oh no, really sorry - I should have typed 'find the temperature T(r)' not T(R).

Then, yes, you need to solve the differential equation. However, you can also express the heat flow in terms of the gradient. This will allow you to relate the heat flow out of the wire with the temperature difference according to Newton's cooling law.

Orodruin said:
Then, yes, you need to solve the differential equation. However, you can also express the heat flow in terms of the gradient. This will allow you to relate the heat flow out of the wire with the temperature difference according to Newton's cooling law.

Ok, so as in my original post, I solved the DE to obtain T(r), but this contains T(R) in it. I don't get whether the question wants me to find an expression for T(R) or just leave it like this.

fayled said:

## The Attempt at a Solution

I'm assuming we want the steady state solution - however, if we had say a cup of tea and we had Newton cooling, the temperature would decay exponentially to Tenv, so the steady state would just be Tenv - is it the fact that we have some heat production P that allows a non trivial steady state?[/B]

Yes.

Anyway, for some reason I believe I am supposed to convert the heat flux into a heat gained per unit volume, which is done by
-A(T(R)-Tenv)*(2πRl/πR2l)=-2A(T(R)-Tenv)/R
If correct, why am I allowed to do this - how do I know it is lost uniformly throughout the cylinder?

The problem statement says that the rate of heat generation per unit volume is uniform. It is not lost throughout the cylinder, it is uniformly generated throughout the cylinder.
Then
H=P-2A(T(R)-Tenv)/R

What is P, and how does it differ from H? For this equation, I get:

H=+2A(T(R)-Tenv)/R

Using the laplacian in cylindrical polars (no z or angular dependence)
(κ/r)d/dr(rdT/dr)=-H
T=-(Hr2/4κ)+Alnr+B
We need A=0 for T to be finite at r=0, and we need T(R) at r=R so B=T(R)+(HR2/4κ).
Then
T(r)=T(R)+(H/4κ)(R2-r2)
with
H=+2A(T(R)-Tenv)/R

However I don't really like leaving T(R) in the answer - how am I supposed to find out what this is?

Eliminate T(R) between the previous two equations.

Chet

Chestermiller said:
Yes.

The problem statement says that the rate of heat generation per unit volume is uniform. It is not lost throughout the cylinder, it is uniformly generated throughout the cylinder.What is P, and how does it differ from H? For this equation, I get:

H=+2A(T(R)-Tenv)/R

Eliminate T(R) between the previous two equations.

Chet

H is the heat produced per unit volume per unit time in the cylinder. The problem tells us that heat per unit volume per unit time P is produced, but there is also a loss of heat from the surface, given by the heat flux A(T(R)-Tenv).

This heat loss needs a negative because it is being lost, so H=P-2A(T(R)-Tenv)/R, but I don't quite understand the steps to get from the heat flux A(T(R)-Tenv) to the term 2A(T(R)-Tenv)/R. It appears to involve a multiplication by 2πRl/πR2l=2/R, which seems to be saying that we can treat this heat loss from the surface to be equivalent to some heat lost per unit volume, but how can we know this.

Also I can't just eliminate T(R) from the two equations, because H is an unknown. I suppose this problem isn't very nice - I will turn it into a more obvious example in the next post.

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fayled said:
H is an unknown.

There is no H in your original problem. The production of heat per unit volume is P.

Hopefully this is clearer.

Suppose we have a cylinder of radius R, in an environment with temperature Tenv, carrying a uniform current I. The heat flux from the surface of the wire is C(T(R)-Tenv) for constant C (Newton's law of cooling). Find the temperature T(r), given that the wire has resistivity ρ.

Thermal diffusion equation in steady state:
κ∇2T+H=0
(κ/r)d/dr(rdT/dr)=-H (cylindrical symmetry).

Ok so to proceed we need to find H (in case it varies with r). This is where I have issues. So there is clearly a heat produced per unit volume from the current, and the current is uniform, so this heat is uniform, and is simply I2ρ/(πR2)2, using the resistivity to get the resistance and then converting this to unit volume. But there is another term in the heat produced per unit volume H, due to the heat flux C(T(R)-Tenv). I think this is supposed to come out as -A(T(R)-Tenv)*(2πRl/πR2l)=-2A(T(R)-Tenv)/R but I don't get why (negative because it is lost, and what I've effectively done is fnd the heat flux for some length l, then divide by the volume over this length l to give the heat lost per unit volume). Then the total heat gain H=I2ρ/(πR2)2-2A(T(R)-Tenv)/R and so as this is constant with respect to r, I can solve the above to give
T=-(Hr2/4κ)+Alnr+B
and use the boundary conditions T finite at r=0, so A=0, T(R) at r=R, so B=T(R)+(HR2/4κ) so that
T(r)=T(R)+(H/4κ)(R2-r2)
with H=I2ρ/(πR2)2-2A(T(R)-Tenv)/R, i.e
T(r)=T(R)+[I2ρ/4κ(πR2)2-2A(T(R)-Tenv)/4κR](R2-r2)
but as before, this leaves me with a T(R) I don't want.

So the two issues: converting the Newton cooling heat flux into a term that gives me a heat loss per unit volume, and what to do with the T(R).

The Newton cooling law is not giving you any additional heat loss per volume, it is a boundary condition and not an inhomogenity in the PDE itself. The heat transfer PDE relies upon the infinitesimal version of the continuity equation, which essentially states that the change of heat inside the volume is equal to the production of heat minus the outflux. This also holds for finite volumes, including your wire.

Now, for the steady state that you are after there is no change in the heat contained (this is straight from the definition of the steady state). What does this tell you about the relation between the produced heat and the outflux? What does this in turn tell you about T(R)? How can you use this to reexpress the boundary condition and solve your PDE?

Here's what's happening:

Heat is being generated within the wire at a rate ##I^2\rho##, where ##\rho## is the resistance of the wire. The heat is being generated uniformly throughout the wire at a rate per unit volume of:
$$H=\frac{i^2\rho}{\pi R^2l}$$
If you prefer, you can call this P, but, in the differential equation you wrote, it is H. P and H are the same thing.

In order for the system to be at steady state, all this heat has to be removed from the wire. Outside the wire, there is an added thermal resistance to heat flow between the the surface of the wire (at temperature T(R)) and the bulk air outside (at temperature Tenv). This resistance is caused by a thin "boundary layer" of air surrounding the wire. The heat flow through this thermal resistance is driven by T(R)-Tenv. All the heat generated in the wire has to pass through this thermal resistance. The rate of heat flow out per unit area of wire surface is given by A(T(R)-Tenv). So the total heat flow out of the wire is given by

$$2\pi Rl\left[A(T(R)-T_{env})\right]=I^2\rho=\pi r^2l H$$

This gives you an equation for calculating the temperature at the surface of the wire T(R).

Chet

Orodruin said:
x

Chestermiller said:
x

Thanks a lot - this definitely clears up my confusion regarding this problem.

So for my initial general problem (P instead of current heating),
PπR2l=2πRl[A(T(R)-Tenv)]
T(R)=(PR/2A)+Tenv

Solve the DE to give
T=-(Pr2/4κ)+Alnr+B
using the boundary conditions T(r=0) finite, T(r=R)=(PR/2A)+Tenv so that
A=0,B=PR2/4κ+PR/2A+Tenv and then
T(r)=P(R2-r2)/4κ+PR/2A+Tenv
as the solution. Does this look good?

However this seems to raise an issue with a different problem. Suppose we're not necessarily in the steady state and we're told that the heat flux from a cylinder is given by Newton's law of cooling. Is there a way to use this information to determine the heat H produced per unit volume per unit time?

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fayled said:
Thanks a lot - this definitely clears up my confusion regarding this problem.

So for my initial general problem (P instead of current heating),
PπR2l=2πRl[A(T(R)-Tenv)]
T(R)=(PR/2A)+Tenv

Solve the DE to give
T=-(Pr2/4κ)+Alnr+B
using the boundary conditions T(r=0) finite, T(r=R)=(PR/2A)+Tenv so that
A=0,B=PR2/4κ+PR/2A+Tenv and then
T(r)=P(R2-r2)/4κ+PR/2A+Tenv
as the solution. Does this look good?
Yes.
However this seems to raise an issue with a different problem. Suppose we're not necessarily in the steady state and we're told that the heat flux from a cylinder is given by Newton's law of cooling. Is there a way to use this information to determine the heat H produced per unit volume per unit time?
Yes. It's still the same P. H and P are the same. However, the boundary temperature T(R) will no longer be given by the steady state equation, because some of the generated heat is going into heating up the cylinder. In all cases, both transient and steady state, the boundary condition at r = R is really given by:

$$-k\left(\frac{\partial T}{\partial r}\right)_{r=R}=A(T(R)-T_{env})$$

This will automatically reduce to your steady state result at long times. Substitute your temperature profile equation into this equation to confirm that what I am saying is correct.

Chet

Chestermiller said:
Yes.

Yes. It's still the same P. H and P are the same. However, the boundary temperature T(R) will no longer be given by the steady state equation, because some of the generated heat is going into heating up the cylinder. In all cases, both transient and steady state, the boundary condition at r = R is really given by:

$$-k\left(\frac{\partial T}{\partial r}\right)_{r=R}=A(T(R)-T_{env})$$

This will automatically reduce to your steady state result at long times. Substitute your temperature profile equation into this equation to confirm that what I am saying is correct.

Chet

That makes sense. How about the following though.

Suppose we now just have a cylindrical rod of radius R, with no energy dissipated per unit volume (for example no current). We will impose some boundary condition at one end, i.e a fixed temperature Tend at x=0, and so the temperature now only varies with x. We are given that the heat flux from the rod is still A(T-Tenv).

Generally,
∂T/∂t=D∇2T+H/C.

Now I would like to say H=0 here, and so
∂T/∂t=D∇2T
and we can solve this using the boundary condition and the fact that T remains finite as x→infinity.

However I won't have used the heat flux here in anyway - my book seems to suggest I should have obtained the diffusion equation as
∂T/∂t=D∇2T-2A(T-Tenv)/CR
and I have no idea of the logic behind this.

I believe they have made a one-dimensional approximation, i.e., approximating that the temperature is only dependent on the coordinate along the cylinder and not the radial coordinate. The heat loss at a given x-coordinate is then given by Newton's cooling law and thus you get a temperature sink as term in the diffusion PDE.

Orodruin said:
I believe they have made a one-dimensional approximation, i.e., approximating that the temperature is only dependent on the coordinate along the cylinder and not the radial coordinate. The heat loss at a given x-coordinate is then given by Newton's cooling law and thus you get a temperature sink as term in the diffusion PDE.

Ah, that is what I originally thought but then the use of Newton's cooling law differently in the above problem put me off.

However when converting the heat flux A(T-Tenv) into the heat gain H (i.e going from unit area to unit volume) we multiply by -2πR/πR2 (-ve for heat loss), giving a constant (for a given x) H=-2A(T-Tenv)/R - this means we are thinking of the heat being lost uniformly over the cross section for a given x Is this valid because T only varies with x, and so T is constant with radius r, and so this H has to be uniform for a given x otherwise T would have to start varying with r, which it doesn't?

fayled said:
Ah, that is what I originally thought but then the use of Newton's cooling law differently in the above problem put me off.

However when converting the heat flux A(T-Tenv) into the heat gain H (i.e going from unit area to unit volume) we multiply by -2πR/πR2 (-ve for heat loss), giving a constant (for a given x) H=-2A(T-Tenv)/R - this means we are thinking of the heat being lost uniformly over the cross section for a given x Is this valid because T only varies with x, and so T is constant with radius r, and so this H has to be uniform for a given x otherwise T would have to start varying with r, which it doesn't?
No. This isn't the way the math on the problem works out.
Before you assume that the radial variations of temperature are negligible, you start out with the same two equations:
$$\frac{\partial T}{\partial t}=D\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)+\frac{\partial ^2T}{\partial z^2}\right]$$
$$-k\left(\frac{\partial T}{\partial r}\right)_{r=R}=A(T(R)-T_{env})$$
Note that there is no heat generation in these equations.
Now, if you multiply the first equation by 2πrdr, integrate between 0 and R, and then divide by πR2, you obtain:
$$\frac{\partial \bar{T}}{\partial t}=D\left[\frac{2}{R}\left(\frac{\partial T}{\partial r}\right)_{r=R}+\frac{\partial ^2\bar{T}}{\partial z^2}\right]$$
where ##\bar{T}## is the temperature averaged over the cross section of the bar. If we now make the approximation that the temperature is nearly constant over the cross section at any value of z, we find from the boundary condition equation that:
$$-k\left(\frac{\partial T}{\partial r}\right)_{r=R}=A(\bar{T}-T_{env})$$
See what happens if you substitute this into the differential equation.

Chet

fayled
Chestermiller said:
No. This isn't the way the math on the problem works out.
Before you assume that the radial variations of temperature are negligible, you start out with the same two equations:
$$\frac{\partial T}{\partial t}=D\left[\frac{1}{r}\frac{\partial}{\partial r}\left(r\frac{\partial T}{\partial r}\right)+\frac{\partial ^2T}{\partial z^2}\right]$$
$$-k\left(\frac{\partial T}{\partial r}\right)_{r=R}=A(T(R)-T_{env})$$
Note that there is no heat generation in these equations.
Now, if you multiply the first equation by 2πrdr, integrate between 0 and R, and then divide by πR2, you obtain:
$$\frac{\partial \bar{T}}{\partial t}=D\left[\frac{2}{R}\left(\frac{\partial T}{\partial r}\right)_{r=R}+\frac{\partial ^2\bar{T}}{\partial z^2}\right]$$
where ##\bar{T}## is the temperature averaged over the cross section of the bar. If we now make the approximation that the temperature is nearly constant over the cross section at any value of z, we find from the boundary condition equation that:
$$-k\left(\frac{\partial T}{\partial r}\right)_{r=R}=A(\bar{T}-T_{env})$$
See what happens if you substitute this into the differential equation.

Chet

Oh I understand! Thank you very much for all of your help :)

## 1. What is Newton's law of cooling for a cylinder?

Newton's law of cooling for a cylinder states that the rate of heat loss from a cylinder is proportional to the difference between the temperature of the cylinder and the ambient temperature.

## 2. How does Newton's law of cooling apply to a cylinder?

Newton's law of cooling is applicable to a cylinder because it describes the rate of heat loss from an object with a specific shape, in this case a cylinder.

## 3. What factors affect the cooling rate of a cylinder according to Newton's law?

The cooling rate of a cylinder is affected by factors such as the surface area, material of the cylinder, and the temperature difference between the cylinder and the surrounding environment.

## 4. Can Newton's law of cooling be used to predict the temperature of a cylinder over time?

Yes, Newton's law of cooling can be used to predict the temperature of a cylinder over time as long as the initial temperature, ambient temperature, and other relevant factors are known.

## 5. How is Newton's law of cooling for a cylinder different from the law for other shapes?

Newton's law of cooling for a cylinder is different from the law for other shapes because the surface area and volume of a cylinder are different from other shapes, which affects the rate of heat loss and the overall cooling process.