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

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In summary: The insulation diameter is the length of the wire divided by the thickness of the insulation. In this scenario, the insulation diameter is 2ft.
  • #36
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
 
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  • #37
Physics_Kid said:
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
 
  • #38
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


ri = 2.7x10-3ft
ro = 4.4x10-3ft
h =1btu/ft2-hr-oF
T0 = 55oF
Qwire = 262*0.00259=1.75W=1.75J/s=6303J/hr=6.303kJ/hr , 1kJ=0.9478btu, =5.97btu/hr-ft
kinsulation = 0.0982btu/ft-hr-oF, approx R=10.1 = 1/k

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

not even close to what my test has shown. with some error i think my wire is around 90oF[/SUP]
 
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  • #39
Physics_Kid said:
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


ri = 2.7x10-3ft
ro = 4.4x10-3ft
h =1btu/ft2-hr-oF
T0 = 55oF
Qwire = 262*0.00259=1.75W=1.75J/s=6303J/hr=6.303kJ/hr , 1kJ=0.9478btu, =5.97btu/hr-ft
kinsulation = 0.0982btu/ft-hr-oF, approx R=10.1 = 1/k

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

not even close to what my test has shown. with some error i think my wire is around 90oF[/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?
 
  • #40
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.
 
  • #41
Physics_Kid said:
before was 12awg @10amps

the later is 14awg @26amps
OK. What about the 4.4 E-3, and the 2.7 E-3?

Chet
 
  • #42
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.
 
  • #43
Physics_Kid said:
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.
 
  • #44
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
 
  • #45
Physics_Kid said:
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
 
  • #46
i have not yet ran any sandwich testing. the 14awg test is wire in open air, simply some 14awg copper surrounded by tubular insulator.
 
  • #47
we can use the std equation of Qwire=AΔT/R at equilibrium

where Qwire = I2⋅rohms
A = area of the insulation exposed to air = ro⋅2π⋅length
ΔT = Twire - Tair
R(ro-ri) = thermal resistance of the insulation of thickness ro-ri
 
  • #48
Physics_Kid said:
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
 
  • #49
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.
 
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  • #50
Physics_Kid said:
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
 
  • #51
hi Chet,
yes, ready to see heat transfer term for the heat sinks.
Cu conductivity is 401W/m⋅K
the contact pad between the Cuwire 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 Aclamp

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.
 
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  • #52
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 ro-min and a long side ro-max
to use this annular equation do you think it is ok to approximate the rectangular sandwich using 2* ro-min ?

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

cant use 2x as a general rule, would just solve for r of a circle having equivalent area of the rectangle.
your thoughts?
 
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  • #53
Physics_Kid said:
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 ro-min and a long side ro-max
to use this annular equation do you think it is ok to approximate the rectangular sandwich using 2* ro-min ?

for a 24"x4" sandwich = 962in
Acirc=6.28 r2
r=3.9
which is ~ 2*ro-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
 
  • #54
Physics_Kid said:
hi Chet,
yes, ready to see heat transfer term for the heat sinks.
Cu conductivity is 401W/m⋅K
the contact pad between the Cuwire 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 Aclamp

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.
 
  • #55
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.
 
  • #56
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 kw 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?
 
  • #57
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?
 
<h2>1. How does exposed copper affect the temperature of the center of a heatsink?</h2><p>The exposed copper on a heatsink can have a significant impact on the temperature of the center. Copper is a highly conductive material, meaning it can transfer heat quickly. If the exposed copper is in direct contact with the heat source, it can absorb and dissipate heat more efficiently, resulting in a lower temperature at the center of the heatsink.</p><h2>2. Is exposed copper necessary for a heatsink to function effectively?</h2><p>No, exposed copper is not necessary for a heatsink to function effectively. While copper is a good conductor of heat, other materials such as aluminum can also be used in heatsinks. The design and construction of the heatsink, as well as the airflow in the system, are more important factors in determining its effectiveness.</p><h2>3. Can the amount of exposed copper on a heatsink affect its cooling performance?</h2><p>Yes, the amount of exposed copper on a heatsink can affect its cooling performance. A larger surface area of exposed copper means more heat can be absorbed and dissipated, resulting in better cooling. However, the design and overall construction of the heatsink also play a significant role in its cooling performance.</p><h2>4. Does the size of the exposed copper area on a heatsink matter?</h2><p>Yes, the size of the exposed copper area on a heatsink can impact its cooling performance. A larger exposed copper area means a larger surface area for heat to be dissipated, resulting in better cooling. However, the overall design and construction of the heatsink also play a role in its effectiveness.</p><h2>5. Are there any potential drawbacks to using exposed copper in a heatsink?</h2><p>While exposed copper can improve cooling performance, there are some potential drawbacks to consider. Exposed copper can be more expensive and may add weight to the heatsink. It can also be more prone to corrosion, which can affect its longevity. Additionally, if not properly designed and constructed, exposed copper may not have a significant impact on cooling performance.</p>

1. How does exposed copper affect the temperature of the center of a heatsink?

The exposed copper on a heatsink can have a significant impact on the temperature of the center. Copper is a highly conductive material, meaning it can transfer heat quickly. If the exposed copper is in direct contact with the heat source, it can absorb and dissipate heat more efficiently, resulting in a lower temperature at the center of the heatsink.

2. Is exposed copper necessary for a heatsink to function effectively?

No, exposed copper is not necessary for a heatsink to function effectively. While copper is a good conductor of heat, other materials such as aluminum can also be used in heatsinks. The design and construction of the heatsink, as well as the airflow in the system, are more important factors in determining its effectiveness.

3. Can the amount of exposed copper on a heatsink affect its cooling performance?

Yes, the amount of exposed copper on a heatsink can affect its cooling performance. A larger surface area of exposed copper means more heat can be absorbed and dissipated, resulting in better cooling. However, the design and overall construction of the heatsink also play a significant role in its cooling performance.

4. Does the size of the exposed copper area on a heatsink matter?

Yes, the size of the exposed copper area on a heatsink can impact its cooling performance. A larger exposed copper area means a larger surface area for heat to be dissipated, resulting in better cooling. However, the overall design and construction of the heatsink also play a role in its effectiveness.

5. Are there any potential drawbacks to using exposed copper in a heatsink?

While exposed copper can improve cooling performance, there are some potential drawbacks to consider. Exposed copper can be more expensive and may add weight to the heatsink. It can also be more prone to corrosion, which can affect its longevity. Additionally, if not properly designed and constructed, exposed copper may not have a significant impact on cooling performance.

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