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

[Physics_Kid]

i have some test data, 1sec while i run the equation using my test #'s

 

[Physics_Kid]

ok, so here's a test i ran earlier using some electrical wire with THWN insulation. it was 14awg. the PVC insulation has a very thin layer of nylon over the top.

the nylon is ~10x thinner than the PVC
PVC k=0.19
nylon k=0.25
so net conductivity k = ~0.17W/mK = 0.0982btu/ft hr ^oF

wait while i finish here

 

[Chestermiller]

the k value i gave earlier in post # 23 is wrong, the R=10 is per 2" so i need to multiply by 12 to get it into 1ft terms, so R=120

ok, so here's a test i ran earlier using some electrical wire with THWN insulation. it was 14awg. the PVC insulation has a very thin layer of nylon over the top.

the nylon is ~10x thinner than the PVC
PVC k=0.19
nylon k=0.25
so net conductivity k = ~0.17W/mK = 0.0982btu/ft hr ^oF

wait while i finish here

You're trying to be too precise.

Do you have any actual experimental measurements of the temperature rise for the wire?

Chet

 

[Physics_Kid]

yes, 14awg @26amps had a ΔT of 28^oF, ambient air was 56^oF, but the TC probe i am using is a metal tube (i think its stainless) and 0.1"dia x 2.5" long, the probe was wrapped with some insulation but the metal was heatsinking it, etc. this test was free air test of 14awg with THWN insulation on it.

the equations you gave, should k be in terms of feet, or the actual k value for the thickness of the insulation? notice i used R=10 for 2", that's R=60 for 1ft.

 

[Chestermiller]

yes, 14awg @26amps had a ΔT of 28^oF, ambient air was 56^oF, but the TC probe i am using is a metal tube (i think its stainless) and 0.1"dia x 2.5" long, the probe was wrapped with some insulation but the metal was heatsinking it, etc. this test was free air test of 14awg with THWN insulation on it.

the equations you gave, should k be in terms of feet, or the actual k value for the thickness of the insulation? notice i used R=10 for 2", that's R=60 for 1ft.

The units in each term of the equations have to be consistent with each other, so the final units are temperature. What is the equation for k in terms of the R value?

Two questions:

1. In the experiment, if you don't feel that 28 F was the actual temperature rise, what is your best guesstimate of what the actual temperature rise was?
2. For the parameters of this experiment, what does our equation predict for the temperature rise?

Chet

 

[Physics_Kid]

R=1/k
but it was not clear to me if we needed actual R of the insulation thickness, or R per ft. it sounds like we need 1/R per ft of insulating material. this is quite different than the std Q=AΔT/R equation because this basic equation uses actual R value of the dimensions used, not a normalized R value, etc.

i was getting to the calculation of pvc coated wire, 1sec

 

[Chestermiller]

R=1/k
but it was not clear to me if we needed actual R of the insulation thickness, or R per ft. it sounds like we need 1/R per ft of insulating material. this is quite different than the std Q=AΔT/R equation because this basic equation uses actual R value of the dimensions used, not a normalized R value, etc.

i was getting to the calculation of pvc coated wire, 1sec

The thermal conductivity is related to R by $k=\frac{δ}{R}$, where δ is the thickness of the insulation (in our case 2"). When working with R values, you need to know whether R is given in US units or SI units. The two are not equivalent. If R is in US units, its units are $\frac{hr-ft^2-F}{BTU}$. Which is it for the insulation you are using, US or SI units?

Chet

 

[Physics_Kid]

14awg with THHN insulation @26amps (1ft)
i am using k value for the insulation
1 W/(m K) = 1 W/(m oC) = 0.85984 kcal/(h m oC) = 0.5779 Btu/(ft h oF) = 0.048 Btu/(in h oF)

and http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
does not show k using area

r_i = 2.7x^10-3ft
r_o = 4.4x10^-3ft
h =1btu/ft²-hr-^oF
T₀ = 55^oF
Q_wire = 26²*0.00259=1.75W=1.75J/s=6303J/hr=6.303kJ/hr , 1kJ=0.9478btu, =5.97btu/hr-ft
k_insulation = 0.0982btu/ft-hr-^oF, approx R=10.1 = 1/k

T_w=55 + 5.97/6.28*0.0982 * ln(4.4/2.7) + 5.97/6.28*4.4x10^-3*1
T_w=55 + 9.68 * 0.488 + 216.1
T_w=55 + 4.7 + 216.1
T_w= 275.8^oF

not even close to what my test has shown. with some error i think my wire is around 90^oF[/SUP]

 

[Chestermiller]

14awg with THHN insulation @26amps (1ft)
i am using k value for the insulation
1 W/(m K) = 1 W/(m oC) = 0.85984 kcal/(h m oC) = 0.5779 Btu/(ft h oF) = 0.048 Btu/(in h oF)

and http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
does not show k using area

r_i = 2.7x^10-3ft
r_o = 4.4x10^-3ft
h =1btu/ft²-hr-^oF
T₀ = 55^oF
Q_wire = 26²*0.00259=1.75W=1.75J/s=6303J/hr=6.303kJ/hr , 1kJ=0.9478btu, =5.97btu/hr-ft
k_insulation = 0.0982btu/ft-hr-^oF, approx R=10.1 = 1/k

T_w=55 + 5.97/6.28*0.0982 * ln(4.4/2.7) + 5.97/6.28*4.4x10^-3*1
T_w=55 + 9.68 * 0.488 + 216.1
T_w=55 + 4.7 + 216.1
T_w= 275.8^oF

not even close to what my test has shown. with some error i think my wire is around 90^oF[/SUP]

Something looks wrong with this calculation.

How did you get the 5.97, when it was 0.54 in the previous case?

Where did you get the 4.4x10^-3 from?
Where did you get the 2.7x10^-3 from?

 

[Physics_Kid]

before was 12awg @10amps with hypo 2" radius insulation that had R=5/inch, the non-SI R value

the later is 14awg @26amps with real thickness of THWN insulation, measured with my digital caliper, the nylon coating is about 10x thinner than the pvc under it.

 

[Chestermiller]

before was 12awg @10amps

the later is 14awg @26amps

OK. What about the 4.4 E-3, and the 2.7 E-3?

Chet

 

[Physics_Kid]

those are the dimensions of 14awg with THWN insulation on it.

before was 12awg @10amps with hypo 2" radius insulation that had R=5/inch, the non-SI R value

the later is 14awg @26amps with real thickness of THWN insulation, measured with my digital caliper, the nylon coating is about 10x thinner than the pvc under it.

 

[Chestermiller]

those are the dimensions of 14awg with THWN insulation on it.

before was 12awg @10amps with hypo 2" radius insulation that had R=5/inch, the non-SI R value

the later is 14awg @26amps with real thickness of THWN insulation, measured with my digital caliper, the nylon coating is about 10x thinner than the pvc under it.

I have no idea what you're talking about, but shouldn't that 4.7E-3 be 2" = 0.167 ft.

 

[Physics_Kid]

so, in the 1st scenario we said 12awg with a radial insulator of 2" radius, length = 1ft, this to simplify the equations for a exothermic wire in a tubular insulator.

in my test data i used real 14awg wire from romex cable, the insulation on the wire is rather thin, no where near 2" radius. the insulation is made up of a layer of pvc and then coated in nylon. the nylon layer is approx 10x thinner than the pvc layer. this scenario is really no different than the 1st except for amps, dimensions of the wire, and insulator thermal conductivity.

the last term in your equation is very strong compared to the others. a little Q and the term is rather large. ~2watts and the term is in the 200's

 

[Chestermiller]

so, in the 1st scenario we said 12awg with a radial insulator of 2" radius, length = 1ft, this to simplify the equations for a exothermic wire in a tubular insulator.

in my test data i used real 14awg wire from romex cable, the insulation on the wire is rather thin, no where near 2" radius. the insulation is made up of a layer of pvc and then coated in nylon. the nylon layer is approx 10x thinner than the pvc layer. this scenario is really no different than the 1st except for amps, dimensions of the wire, and insulator thermal conductivity.

the last term in your equation is very strong compared to the others. a little Q and the term is rather large. ~2watts and the term is in the 200's

Are you saying that the wire is not embedded between two sheets of 2" thick insulation in the second example?

Chet

 

[Physics_Kid]

i have not yet ran any sandwich testing. the 14awg test is wire in open air, simply some 14awg copper surrounded by tubular insulator.

 

[Physics_Kid]

we can use the std equation of Q_wire=AΔT/R at equilibrium

where Q_wire = I²⋅r_ohms
A = area of the insulation exposed to air = r_o⋅2π⋅length
ΔT = T_wire - T_air
R(r_o-r_i) = thermal resistance of the insulation of thickness r_o-r_i

 

[Chestermiller]

i have not yet ran any sandwich testing. the 14awg test is wire in open air, simply some 14awg copper surrounded by tubular insulator.

A value of h equal to 8 - 10 is not atypical either. It looks like this would be a more accurate choice, given your data. Try a value of 8 and see what you get. But, if this is correct, the outside of the insulated wire should be warm to the touch. Was it?

Chet

 

[Physics_Kid]

at ~90F measured, yes, it was a light warm to the touch

h=8 does then get real close to measured, including some error for the metal TC probe

i measured 83F, equation with h=8 says 86.7F, which is in alignment with some heatsink affect from the metal TC probe.why a change in h ? is it something that swings that much depending on the environment of air to convect heat away from the wire?

in this specific calculation, the #'s i used, the wire was 84" long, ran this test with a "long" wire thinking the heatsink affect at the ends would be very insignificant. but now i am interested to see how you factor in heatsinks on the ends of the wire. i suspect the factor will be a minus to the existing equation.

 

[Chestermiller]

at ~90F measured, yes, it was a light warm to the touch

h=8 does then get real close to measured, including some error for the metal TC probe

i measured 83F, equation with h=8 says 86.7F, which is in alignment with some heatsink affect from the metal TC probe.why a change in h ? is it something that swings that much depending on the environment of air to convect heat away from the wire?

Yes. Exactly. There are air currents in the room from the ventilation (so-called forced convection) and also natural convection resulting from the buoyancy effect of the insulation surface being at a different temperature than the room air. What we've done here is use your experimental data to calibrate the model (i.e., get an estimate of the value of h). This result tells us that, with the actual 2" of insulation in place, this term will have a much smaller effect than the heat conduction through the insulation in terms of influencing the wire temperature.

in this specific calculation, the #'s i used, the wire was 84" long, ran this test with a "long" wire thinking the heatsink affect at the ends would be very insignificant. but now i am interested to see how you factor in heatsinks on the ends of the wire. i suspect the factor will be a minus to the existing equation.

Exactly right again. You're on a roll.

What we are going to do is solve two separate heat transfer problems to get the combined effect of current flowing through the wire, plus with a heat sink at the ends. We can do this because the heat transfer equations are linear in the temperature. When we add the solutions to the two problems together, we get the answer to the problem you want. We have already solved the first heat transfer problem, namely, current flowing through the wire, but with no heat sink. Next we will solve the heat transfer problem of no current through the wire, but with cooling at its ends.

How does that grab you so far?

Chet

 

[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?