Does the Exposed Copper Affect the Temperature of the Center of a Heatsink?

[Physics_Kid]

hi Chet,
yes, ready to see heat transfer term for the heat sinks.
Cu conductivity is 401W/m⋅K
the contact pad between the Cu_wire and clamp is approx 2mm²
the clamp is steel (low carbon) and has conductivity of 43W/m⋅K
the clamp is in free air and has area A_clamp

there's actually a tad more than that going on because the clamp is holding the copper to the power supply lug ends, which are lugs connected to 2awg insulated copper stranded wire. but for this problem i don't think we need to worry about it.

i might conclude that the shiny nylon jacket with black pvc underneath has h between 8-10 as you said. h=8 seems to fit my test data rather well.

Surface coefficient
Approximate surface heat transfer coefficients that are reasonable accurate for low temperature surfaces in still air ( which are the conditions experienced for insulated systems are listed below)

h = 5.7 for low emissivity surfaces such as polished aluminium
h = 8.0 for surfaces of medium emissivity - galvanised steel , aluminium paint, stainless steel etc
h = 10 for surfaces of high emissivity - matt black surfaces, brick and plain insulated surfaces.

 

[Physics_Kid]

hi Chet,
so with a flat rectangular sandwich what we end up with (looking at the end where the wire exits) an infinite # of radii from centerline to the insulation surface. its a simple triangle with a short side r_o-min and a long side r_o-max
to use this annular equation do you think it is ok to approximate the rectangular sandwich using 2* r_o-min ?

for a 24"x4" sandwich = 96²in
A_circ=6.28 r²
r=3.9
which is ~ 2*r_o-min

cant use 2x as a general rule, would just solve for r of a circle having equivalent area of the rectangle.
your thoughts?

 

[Chestermiller]

hi Chet,
so with a flat rectangular sandwich what we end up with (looking at the end where the wire exits) an infinite # of radii from centerline to the insulation surface. its a simple triangle with a short side r_o-min and a long side r_o-max
to use this annular equation do you think it is ok to approximate the rectangular sandwich using 2* r_o-min ?

for a 24"x4" sandwich = 96²in
A_circ=6.28 r²
r=3.9
which is ~ 2*r_o-min

cant use 2x as a general rule, would just solve for r of a circle having equivalent area of the rectangle.
your thoughts?

What I think is that you are trying to get too exact at this stage. If, after we're done with the sink effect, you want a more accurate estimate of the relation between the heat flow and the temperature difference (which properly takes into account the effects of this geometry), we can solve the 2D heat transfer problem either analytically or numerically.

Chet

 

[Chestermiller]

hi Chet,
yes, ready to see heat transfer term for the heat sinks.
Cu conductivity is 401W/m⋅K
the contact pad between the Cu_wire and clamp is approx 2mm²
the clamp is steel (low carbon) and has conductivity of 43W/m⋅K
the clamp is in free air and has area A_clamp

there's actually a tad more than that going on because the clamp is holding the copper to the power supply lug ends, which are lugs connected to 2awg insulated copper stranded wire. but for this problem i don't think we need to worry about it.

i might conclude that the shiny nylon jacket with black pvc underneath has h between 8-10 as you said. h=8 seems to fit my test data rather well.

Surface coefficient
Approximate surface heat transfer coefficients that are reasonable accurate for low temperature surfaces in still air ( which are the conditions experienced for insulated systems are listed below)

h = 5.7 for low emissivity surfaces such as polished aluminium
h = 8.0 for surfaces of medium emissivity - galvanised steel , aluminium paint, stainless steel etc
h = 10 for surfaces of high emissivity - matt black surfaces, brick and plain insulated surfaces.

I don't know where you got these values (or their units), but radiative transfer is not going to be a significant factor for the system we are examining. We are going to be dealing with convective heat transfer. In any event, for our situation, in practice, it seems like your experimental data suggest that the dominant effect will be the conductive resistance of the insulation, and convective heat transfer will also be secondary.

 

[Physics_Kid]

i agree. but there is a conductive mode in the copper, from the center all the way to the end of the copper wire, heat has to conduct down the copper making its way to the heatsink, and there will be more convective mode from the wire being exposed directly to the air.

the copper outside the insulation is also exposed to open air, so there will be convective mode there too, and conductive mode to the "heatsink" clamps at the end, and the the clamps will convect to the air.

 

[Chestermiller]

Heat Sink Effect

To evaluate the heat sink effect, we are going to solve the following model:

1. No electrical heating
2. Ambient air temperature T₀
3. Temperature of wire at entry to insulation sandwich is equal to T₁ (≠ T₀)
4. Heat conduction along wire
5. Heat conduction to or from wire through insulation

The coordinate along the wire is z, and the coordinate perpendicular to the wire is r. z = 0 represents the entry location to the insulation sandwich. T(z) represents the temperature of the wire at distance z from the entry to the sandwich.

The rate of heat conduction along the wire is given by
$$q=-\pi r_i^2k_w\frac{dT}{dz}$$
where k_w is the thermal conductivity of the copper wire.
Heat is being conducted along the wire, but is also being lost by conductive cooling in the radial direction. In Part 1, we found that the rate of heat loss per unit length of wire in the radial direction is related to the temperature difference between the wire and the ambient air by:
$$rate\, of\, heat\, loss\, per\, unit\, length=2\pi r_iU(T(z)-T_0)$$
where $$\frac{1}{U}=\frac{r_i}{k}\ln{\frac{r_o}{r_i}}+\frac{r_i}{r_o h}$$
Based on this information, if we perform a differential heat balance on the section of wire between axial locations z and z + dz, we obtain:
$$\frac{d^2T}{dz^2}=\frac{2U}{k_wr_i}(T-T_0)$$
The solution to this equation subject to the boundary condition at z = 0 is given by:

$$T=T_0+(T_1-T_0)\exp{\left(-\sqrt{\frac{2Ur_i}{k_w}}\frac{z}{r_i}\right)}$$
Note from this equation that the temperature disturbance associated with the sink dies out exponentially with distance along the wire (measured from the entry to the sandwich). This equation gives us the information we need to estimate this quantitatively.

Questions?

 

[Physics_Kid]

ah, i have to absorb all this. when you say it dies out exponentially with distance along the the wire measured from the sandwich entry, which direction, towards the center or towards the heatsink?