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I am trying to create a program which estimates the final temperature of a magnet made of copper wire. I have different configurations in mind, but I will start off with a straight square wire with a circular cooling channel inside of it. Inside runs water which cools it off. The coil is wrapped in some type of electrical insulator. The wire is exposed to air. I am trying to model this using unsteady-state conditions and run it up to a steady state position to calculate the final temperature of the inside (middle if it makes it easy) of the wire, the temperature of the outside of the wire as well as the final temperature of the water coolant.

I have been at this, and coming up with some equations, but I have not taken heat transfer before and am struggling somewhat on this problem. I know that for unsteady state heat transfer

∂T/∂t = α ∇^2(T)

where the partial of temperature versus time is the laplacian of temperature times a factor alpha. There are some numerical methods to go about doing this and I am able to include a convective heat transfer coefficient, but I am not sure how to add in the rest of the geometries nor how to include a heat generation term. It just seems like there re so many things that have to be simultaneously solved and I am not sure where to go. I am not sure if the textbook I am reading is not sufficient or if I am not seeing where to go in general. Thanks in advance.

general sketch

The wire is square with an inner cooling channel. A layer of insulation wraps around the wire which may be neglected for now. I'm trying to reconcile both the convection from inside the wire, the convection outside the wire and the conduction inside the wire. No temperature is fixed which makes this very difficult for me to try to design around.

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  • $\begingroup$ Please post a representative 3-D picture (sketch) of the system geometry. In particular, how are the components oriented relative to each other (water channel inside a wire standing in air)? The steady state solution will be far easier to set up and solve, and it will likely be insightful enough to get you started. $\endgroup$ – Jeffrey J Weimer Jun 27 at 0:16
  • $\begingroup$ I uploaded a picture above. Like I mentioned the outer insulation layer can be neglected. I have wrapped my head around this a lot and know how to solve many steady state solutions and even many unsteady state solutions, but each of those has at least one temperature, whether that is a heat sink or a fluid temperature fixed while here, the actual wire heats up, the fluid heats up, I guess the air heats up and so on. Maybe I can assume that the air stays fixed at room temperature, but I don't know how to work around the fact that heat is dissipated in two directions. $\endgroup$ – alcopo63q Jun 27 at 18:31
  • $\begingroup$ Thank you. What is the distribution of power generated in the wire? Is it uniform in the radial and axial directions (or only one or neither)? Also, what is more important to you to determine first in the wire, the radial temperature variation or the axial temperature variation? Finally, can you accept an analysis with a cylindrical shaped wire as a starting point? $\endgroup$ – Jeffrey J Weimer Jun 27 at 22:41
  • $\begingroup$ The wire is just a normal current-carrying wire so power generated put me I^2R. The current state is uniform throughout the entire usage of the wire. And definitely radially is more important in this case. I am trying to simultaneously determine the average temperature which the coolant rises to as well as how hot the copper wire gets, weather I am able to tell the distribution radially or if I simplify and find a bulk/average temperature of it. I know using a cylindrical wire would be easier, so I think that would be a good place to start. $\endgroup$ – alcopo63q Jun 27 at 22:47
  • $\begingroup$ You can take two different approaches to solve this at steady state. The easier is conditions where radial temperature gradients can be ignored. The temperatures in the water and the wire only depend on position $z$ along the channel. The harder is conditions where radial gradients are also important. The solution for $T(z,r)$ requires separation of variables with $T(z,r) = Tz(z) Tr(r)$. Analysis of $Tr(r)$ is done in cylindrical coordinates. Can you accept a nearly full solution for the easier case and background + insights to develop an answer for the harder case? $\endgroup$ – Jeffrey J Weimer Jun 29 at 18:14
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Foundations

The heat transfer problem has three considerations. First is steady state versus unsteady state. The former case is far easier and is all that I will consider here. Second is the temperature profiles axial in $z$ and radial in $r$. The best approach is to presume that separation of variables will work. Finally, the last consideration is the influence of heat transfer internally and externally to the cylinder while generating uniform power in the shell $P = i^2R$.

Also, for simplicity, neglect the (electrical) insulation layer on the outside of the wire.

The starting point is to build each of the cases one step at a time.

Axial versus Radial Gradients - Biot Number Estimates

In the ideal case, we allow for separation of variables to make $T(r,z) = Tr(r)Tz(z)$. A starting point for this is to determine the effective Biot numbers for radial heat transfer.

$$Bi_{rw} = h_w \Delta R/k \hspace{1cm} Bi_{ra} = h_a \Delta R/k $$

In these, $h_w, h_a$ are the convection coefficients of the flowing water or surrounding air, $\Delta R$ is the outer minus inner radius $\Delta R = r_o - r_i$, and $k$ is the thermal conductivity of the wire. When $Bi < 0.1$, the wire functions in a lumped analysis. The radial gradients can be neglected.

Axial Gradients Only

Water Cooling Only

Consider the system for the case that $Bi_{rw} < 0.1$ with perfect thermal insulation around the wire. This will give the maximum temperature that can be expected in the wire. The energy balance along the flow says that total power input $P$ must be taken out by the flowing fluid (water) with mass flow $\dot{m}$ and specific heat $\tilde{C}_p$. The fluid temperature changes from input $T_{wi}$ to output $T_{wf}$.

$$ \dot{m}\tilde{C}_p(T_{wf,max} - T_{wi}) = P$$

The subscript "max" indicates the case of maximum output temperature on the fluid. You can also use this to establish that, in this case, the temperature profile in the fluid is linear along the axis.

$$ \dot{m}\tilde{C}_p(T_w(z) - T_{wi}) = P \frac{z}{L}$$

or

$$ T_w(z) - T_{wi} = P\left(\frac{z}{L}\right)\left(\frac{1}{\dot{m}\tilde{C}_p}\right)$$

The differential energy balance on the wire with no radial gradients and with the external wall as a perfect thermal insulator is as below with $r_i$ as the radius of the channel.

$$ h_w 2\pi r_i\left(T_z(z) - T_w(z)\right) dz = P \frac{dz}{L}$$

This gives the temperature in the wire as below with $A_i$ as the area of the water channel.

$$ T_w(z) - T_{wi} = P \left(\frac{z}{L}\right)\left(\frac{1}{\dot{m}\tilde{C}_p} + \frac{1}{h_w A_i}\right)$$

Including External Convection to Air

When we add convection to air, the energy balance to the internal fluid (water) written in differential form is as below with $r_o$ as the external radius of the wire and $T_a$ as the air temperature.

$$ \dot{m}\tilde{C}_p dT_w + h_a 2\pi r_o (T_z(z) - T_a) dz = P \frac{dz}{L}$$

The differential energy balance on the wire at any axial position $z$ is the equation below with $r_i$ as the channel radius.

$$ h_w 2\pi r_i(T_z(z) - T_w(z))dz + h_a 2\pi r_o(T_z(z) - T_a)dz = P \frac{dz}{L}$$

These are two coupled differential equations to solve for $T_w(z)$ and $T_z(z)$. In all cases, the water and wire temperatures will be lower than the case for water cooling only because a portion of the power generation is now removed by the air.

Radial Gradients

The radial gradient problem is easily solved for a hollow cylinder with fixed internal and external wall temperatures $T_{ci}$ and $T_{co}$. The steady state profiles are well-documented in textbooks on heat transfer (see Chapter 17 in Welty, et.al). At any position $z$, the radial temperature profile will have the form below.

$$ \frac{T_r - T_{co}}{T_{ci} - T_{co}} = \frac{\ln(r/r_o)}{\ln(r_i/r_o)} $$

Following the format of separation of variables, the radial gradient is superimposed on the axial profile from above.

The link between the wall temperatures $T_{ci}, T_{co}$ and the water or air temperatures occurs as convection. This is an entirely separate issue to tackle. It can only be done once a proof is established that separation of variables is valid.

Unsteady State

This level of analysis is best done step-wise. Include time on the axial profile with insulated boundary, then include time on the axial profile including air convection, and finally consider time only on the radial boundary.

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  • $\begingroup$ So this is saying that the two coupled differential equations, if solved through whatever means, would simultaneously produce the temperature of the water as well as the temperature of the wire at any radial position starting from the center of the wire, correct? $\endgroup$ – alcopo63q Jul 2 at 17:17
  • $\begingroup$ Also, power generated from current is I^2R, not IR^2 $\endgroup$ – alcopo63q Jul 2 at 17:43
  • $\begingroup$ I fixed the power term, thank you. I also modified the radial gradient equation. The proof you must first make is that separation of variables is valid. The radial gradient is then defined as per the statement. $\endgroup$ – Jeffrey J Weimer Jul 2 at 20:11
  • $\begingroup$ So with the two differential equations with just axial direction flow if I could solve these numerically somehow I could theoretically get the temperatures of the water and the wire at any length along the wire. Then, if I assumed that separation of variables was valid and the radial conditions followed the given formula, I could assume that say Tci is the temperature of the water at that position and Tco is the temperature of the wire at that length as that will most likely be the maximum and I could then find the actual radial gradient inside the wire. Is that what you would suggest? $\endgroup$ – alcopo63q Jul 3 at 11:30
  • $\begingroup$ Also you can theoretically simplify that into one differential equation which is not dependent on input power. That does not seem right to me. Or can we not do this as we resolved the equations into differentials? $\endgroup$ – alcopo63q Jul 3 at 12:53

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