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

[Physics_Kid]

i need some help understanding this not-so-simple heatsink model.

lets say i have a length of copper wire, say 8ft long, and i put tubular insulation around the wire for just a section of it, say a 4ft length and it is co-centered on the wire (2ft on each side is in ambient air), for giggles the insulation R value is 100.

now, i pass amps through the wire, from the wire properties i can calculate the heat being generated by the wire, hence the exothermic density (flux) is the same everywhere. as example, i know its watts/mm^2.

i know copper has low R value, yet i also know that the wire is exothermic everywhere, so help me understand the answer to my Q.

my question is, does the exposed copper outside of the insulation have significant impact of the temp of center of wire?

 

[berkeman]

i need some help understanding this not-so-simple heatsink model.

lets say i have a length of copper wire, say 8ft long, and i put tubular insulation around the wire for just a section of it, say a 4ft length and it is co-centered on the wire (2ft on each side is in ambient air), for giggles the insulation R value is 100.

now, i pass amps through the wire, from the wire properties i can calculate the heat being generated by the wire, hence the exothermic density (flux) is the same everywhere. as example, i know its watts/mm^2.

i know copper has low R value, yet i also know that the wire is exothermic everywhere, so help me understand the answer to my Q.

my question is, does the exposed copper outside of the insulation have significant impact of the temp of center of wire?

I'm no expert in heat flow, but I can offer a couple thoughts and a question for you.

You could run an experiment with an insulated section of wire and an uninsulated section of wire separately, to determine the Thermal Resistance from the wire to the air (in Degrees C per Watt). Obviously the insulated wire will have a much higher Thermal Resistance, and hence will get hotter for the same current than the uninsulated wire.

So, in your setup with the middle section of wire insulated, you know that the center section will get hotter. Your question asks how much effective heat sinking will the uninsulated wire sections provide, and how much will that help to lower the temperature of that center section (and raise the temperature of the uninsulated sections). What is it about this setup would mitigate the heat sinking effect of the uninsulated sections of wire and keep them from being effective as heat sinks?

 

[Physics_Kid]

i think you understand my Q, but let me clarify. in my setup the ends are clamped to the power source, using heavy metal clamps. the wire sticks out of the insulation some so that it can be connected. so there is some convection mode there outside of the insulation. i am trying to understand a couple of things. does the non-insulated sections act as a heatsink in a significant manner that it affects the temp in the middle? and if so, how long would the insulated section need to be so that the heatsink affects would be insignificant on the temp in the middle of the wire? my goal is to test the temp of the wire vs amps, but need to do so without heatsink affects changing the temp in the center. its not trivial to me because the wire itself is exothermic everywhere on the wire.

 

[berkeman]

does the non-insulated sections act as a heatsink in a significant manner that it affects the temp in the middle?

No. I already asked you why, and you have not responded.

 

[Chestermiller]

The temperature in the insulated section would not be affected significantly at distances more than a couple of insulation diameters in from the ends. But if a more precise answer is required, the system in the region near the ends of the insulated section can be modeled pretty easily.

Chet

 

[Physics_Kid]

hi Chet,
what is "insulation diameter" if the wire is laying in the middle of a 2ft x2ft x 4"thick insulation sandwich (sorry, i changed physicals of insulation)? from center the closest "out" for heat would be two inches on each side (lowest R value section of the sandwich), so is "insulation dia" = 4" and the heatsink mode would be insignificant beyond about 8" into the insulation from the end, but up to ~8" from end the heatsink mode will impact wire temp?

also noted, the R value of copper is magnitudes less than R value of the insulation.

 

[Chestermiller]

hi Chet,
what is "insulation diameter" if the wire is laying in the middle of a 2ft x2ft x 4"thick insulation sandwich (sorry, i changed physicals of insulation)? from center the closest "out" for heat would be two inches on each side (lowest R value section of the sandwich), so is "insulation dia" = 4" and the heatsink mode would be insignificant beyond about 8" into the insulation from the end, but up to ~8" from end the heatsink mode will impact wire temp?

also noted, the R value of copper is magnitudes less than R value of the insulation.

Sorry. I'm having trouble picturing the geometry. I thought it was a single wire with annular insulation around it, stripped for part of the wire length.

Chet

 

[Physics_Kid]

Hi Chet,
the physicals are simple.

12awg bare copper that is 4ft long, a 2x2ft x 2" foam board (two of them) sandwiches the wire co-centered, thus 12" of wire sticks out each side of the foam board sandwich. taking temp measurements of wire at center as amps through the wire varies. the ends of the wire are clamped to a variable power source, etc.

 

[berkeman]

Hi Chet,
the physicals are simple.

12awg bare copper that is 4ft long, a 2x2ft x 2" foam board (two of them) sandwiches the wire co-centered, thus 12" of wire sticks out each side of the foam board sandwich. taking temp measurements of wire at center as amps through the wire varies. the ends of the wire are clamped to a variable power source, etc.

What is the context of this question? That is an unusual setup. What is the reason for this question? It sounds like a specific situation, not a general heat flow question...

 

[Chestermiller]

Hi Chet,
the physicals are simple.

12awg bare copper that is 4ft long, a 2x2ft x 2" foam board (two of them) sandwiches the wire co-centered, thus 12" of wire sticks out each side of the foam board sandwich. taking temp measurements of wire at center as amps through the wire varies. the ends of the wire are clamped to a variable power source, etc.

Are you also worried about the heat loss through the two 2' x 4" rectangular areas at the ends? Do you need to know the answer in advance, or, can you first run the system and then, based on the temperature measurement at the center, use that to estimate the heat sink effect of the wires at the ends? If it is the latter, I can tell you how to make the estimate. If the former, there is a way of getting an upper bound to the temperature effect at the center, but it might be over-conservative.

Chet

 

[Physics_Kid]

i can answer the last two posts here.

post #9, i already stated the purpose, testing wire temps vs amps for a wire that passes through insulation, in this case a 2x2 x4" foam board insulation sandwich, that is my experiment. simple.

post #10, i am not worried about heat loss via the insulation sandwich, just trying to understand how if the ends of the wire are acting as a heatsink (which they are) does this impact the temp of wire in center in a significant way? i could run the system with the ends insulated and non-insulated to see if there's a diff, but i wanted to get something in form of basic equation that would suggest your notion of not more than X # of dia, etc.

 

[Chestermiller]

i can answer the last two posts here.

post #9, i already stated the purpose, testing wire temps vs amps for a wire that passes through insulation, in this case a 2x2 x4" foam board insulation sandwich, that is my experiment. simple.

post #10, i am not worried about heat loss via the insulation sandwich, just trying to understand how if the ends of the wire are acting as a heatsink (which they are) does this impact the temp of wire in center in a significant way? i could run the system with the ends insulated and non-insulated to see if there's a diff, but i wanted to get something in form of basic equation that would suggest your notion of not more than X # of dia, etc.

The answer to your question about the heat sink is intimately related to the heat transfer through the insulation sandwich. The temperature disturbance caused by the bare wire at the edge is dissipated through the insulation sandwich. Note, for example that, if the insulation were perfect, the temperature at the center would rise to a huge value, since the only way that heat could be removed would be by conduction along the wire (to the edges).

Chet

 

[Physics_Kid]

so, i already know a few things.

1) the R value of the insulation is magnitudes bigger than the R value of the copper, so up front we know some heat from inside the insulation will want to escape using conductive mode out down the wire to the outside where there is a "heatsink" clamped at the very end. some heat will also escape through the insulation.
2) equilibrium temp is governed +Q(total) + -Q(insulation) + -Q(copper) = 0
simply put, the exothermic wire is generating heat, and that heat can escape via several paths via two modes, doing so until the temp in center reached equilibrium.

now, here's where it is not trivial for me. there is a "heatsink" at the very ends of the wire, copper has a very low R value, the wire is exothermic (constant flux) everywhere, how does the heatsink at the very ends impact the temp at the center.

as a hypo, let's change the dimensions of the insulation sandwich, let's say its 20ft x 20ft x 10"thick, and the wire is 30ft long co-centered in the sandwich (5ft sticking out each end). same amps, same physical "heatsink" on the ends. we still agree that the copper will conduct heat out to the ends, but is this conductive mode significant such that it affects the temp of the wire in the center? the monkey wrench for me is, the wire is exothermic everywhere, so its not as simple as looking at this as if just the center of the wire is exothermic.

i can partially answer my own Q. we know it affects temp because some heat moves down and out the copper, but to what magnitude relevant to -Q(insulation)? a 50/50 split would certainly mean big significance, but a 92/8 split means much smaller significance.

 

[Chestermiller]

Let's try to get a handle on this by using a heat transfer model. Instead of a 2' x 2' x 4" sheet, suppose the wire is inside a 4" diameter annulus of insulation, and that the combined wire and insulation is 2' long. To start with, let's assume that the heat sinks at the ends of the wire do not exist, so that the two ends of the wire are considered insulated. In this case, the flow of heat out of the wire and through the insulation will be radial. We assume that the thermal conductivity of the insulation is k, and the heat transfer coefficient from the outer surface of the insulation to the room is h. We call r₀ the 2" outer radius of the insulation, and r_i the radius of the wire (roughly 0.04" for 12 gauge). The rate of heat generation in the wire per unit length of wire is Q. The radial heat conduction equation within the insulation for this situation is given by:

$$2\pi rk\frac{dT}{dr}=-Q$$
where r is the distance from the cylindrical centerline of the wire.

At the surface of the wire, the boundary condition is:$$2\pi r_0h(T_s-T_0)=Q$$
where T_s is the temperature at the outer surface of the insulation and T₀ is the ambient room temperature.

OK so far?

 

[Physics_Kid]

sure, makes sense, we can swap out Q for I^2R, so for 12awg(1inch) its 0.1325x10^-3 * I^2

now need to add the "heatsink" mode that is connected at the very end of the wire. let's assume that the conductive path of the copper that is outside of the insulation has zero heat loss along that path until it reaches the heatsink, so like water flow in a pipe, except in this model the main pipe has a bunch of smaller feeder radial pipes feeding the main pipe along the path (because the wire is exothermic everywhere).

 

[Chestermiller]

sure, makes sense, we can swap out Q for I^2R, so for 12awg(1inch) its 0.1325x10^-3 * I^2

now need to add the "heatsink" mode that is connected at the very end of the wire. let's assume that the conductive path of the copper that is outside of the insulation has zero heat loss along that path until it reaches the heatsink, so like water flow in a pipe.

Whoa pardner. Hold yer horses. Before we start looking at the heat sink case, I want to first get the solution for the no-heat-sink case. Why? If we can't solve that, we certainly won't be able to do the heat sink case. Plus, I want to get some results under our belts that we can look over and apply to the heat sink case. Please be patient.

If T_w is the temperature of the wire at its surface r = r_i, are you able to solve the equations I wrote to obtain T_w and T_s in terms of Q, r_i, r_o, h, and k?

Chet

 

[Physics_Kid]

2πr₀h(T_s−T₀)=Q

so you said "at the surface of the wire" for this eq, did you mean to say at the surface of the insulation, the boundary between insulation and ambient air?

not sure what i am solving for, are we solving for T_w to start with?

and isn't T_s considered to be temp of ambient air when calculating the heat gradient across an insulator of value R ? and from that we know the gradient is a function of Δt

 

[Chestermiller]

so you said "at the surface of the wire" for this eq, did you mean to say at the surface of the insulation, the boundary between insulation and ambient air?

That equation applies to the outer surface of the insulation, as I said in my post.

not sure what i am solving for, are we solving for T_w to start with?

Were are trying to solve the differential equation and boundary conditions for (a) the temperature as a function of radial position (b) T_w and (c) T_s.

and isn't T_s considered to be temp of ambient air when calculating the heat gradient across an insulator of value R ?

No. There is typically a convective heat transfer resistance between the insulation surface and the ambient room air.

I wanted to give you a chance to solve the equations yourself, but if you are uncomfortable and uncertain about that, I can do it. What's your preferrance?

Chet

 

[Physics_Kid]

sure, ok, please show me an example.

we can use Q_wire = 0.159watt = 0.159J/s = 572.4J/hr = 0.5724kJ/hr = 0.5425btu/hr
this is 10amps on 1ft of 12awg copper wire

but, i think you are driving at a answer to dT/dr so that we can find the temp of r_wire, am i correct?

 

[Physics_Kid]

i found this from a UK site

where d₁ is conductor diameter and d₂ is insulation diameter

 

[Chestermiller]

sure, ok, please show me an example.

we can use Q_wire = 0.159watt = 0.159J/s = 572.4J/hr = 0.5724kJ/hr = 0.5425btu/hr
this is 10amps on 1ft of 12awg copper wire

but, i think you are driving at a answer to dT/dr so that we can find the temp of r_wire, am i correct?

Yes.

Here's the solution to the differential equation:

$$T=T_w-\frac{Q}{2\pi k}\ln{\frac{r}{r_i}}\tag{1}$$

At the boundary r = r_o, T = T_s, so
$$T_s=T_w-\frac{Q}{2\pi k}\ln{\frac{r_o}{r_i}}\tag{2}$$
But, from the other boundary condition,
$$T_s=T_0+\frac{Q}{2\pi r_o h}\tag{3}$$
If we combine Eqns. 2 and 3, we get

$$T_w=T_0+\frac{Q}{2\pi k}\ln{\frac{r_o}{r_i}}+\frac{Q}{2\pi r_o h}\tag{4}$$
This equation can be used to estimate the temperature of the wire. You found that Q = $0.54\, \frac{BTU}{ft-hr}$ . I recommend a value of about $1\, \frac{BTU}{hr-ft^2-F}$ for h. You can look up the value of the thermal conductivity of the insulation k, and the wire radius, which is about 0.04". The insulation outer radius is 2". So, what do you calculate for $T_w$?

I'll continue with the analysis of the sink problem after you provide the result for the wire temperature.

Chet

 

[Physics_Kid]

i am on it, will compare this the equation i just posted. thank you.

 

[Physics_Kid]

ok, so some confusion on Q. why do you say its 1btu/hr-ft²-^oF
are you saying that Q is the the energy per unit area of the wire? why ^oF in there?

thus far i have
r_i = 0.0033ft
r_o = 0.167ft
h =1btu/ft²-hr-^oF
T₀ = 80^oF
Q=0.54btu/hr
k_insulation = 1/R = 1/(10ft-hr-^oF/btu) = 0.1btu/ft-hr-^oF

 

[Chestermiller]

ok, so some confusion on Q. why do you say its 1btu/hr-ft²-^oF
are you saying that Q is the the energy per unit area of the wire? why ^oF in there?

Ooops. That's the value I recommend for the heat transfer coefficient h, not Q. I accidentally omitted the words "for h" in my previous post. I just went back and added these words. I hope this clarifies things. Sorry.

Chet

 

[Physics_Kid]

i am editing my last post, but h_c(air) has conductivity value of 0.024W/mK
1 W/(m K) = 0.5779 Btu/(ft h ^oF)

and is Q a per unit area # ? i aks because the wire itself has radius r_wire thus has surface area for the 1ft length (0.021ft²), so is my Q/hr-ft need to be represented as btu-ft/hr ? or does Q simply represent the total energy as r_wire=0 over the 1ft length, btu/ft ?

 

[Chestermiller]

i am editing my last post, but h_c(air) has conductivity value of 0.024W/mK

h is not the conductivity. The heat transfer coefficient h is the conductivity divided by the thermal boundary layer thickness. The value I gave for h is a typical value based on experience.

Chet

 

[Physics_Kid]

h is not the conductivity. The heat transfer coefficient h is the conductivity divided by the thermal boundary layer thickness. The value I gave for h is a typical value based on experience.

Chet

ok, will use 1

 

[Chestermiller]

i am editing my last post, but h_c(air) has conductivity value of 0.024W/mK
1 W/(m K) = 0.5779 Btu/(ft h ^oF)

and is Q a per unit area # ? i aks because the wire itself has radius r_wire thus has surface area for the 1ft length (0.021ft²), so is my Q/hr-ft need to be represented as btu-ft/hr ? or does Q simply represent the total energy as r_wire=0 over the 1ft length, btu/ft ?

r of the wire is not equal to zero. It is equal to 0.04 inches. Q is expressed in W/m or BTU/hr-ft. It is the rate at which heat is released from the wire per unit length of the wire.

 

[Physics_Kid]

r_i = 0.0033ft
r_o = 0.167ft
h =1btu/ft²-hr-^oF
T₀ = 80^oF
Q=0.54btu/hr
k_insulation = 1/R = 1/(10ft-hr-^oF/btu) = 0.1btu/ft-hr-^oF

Tw = ~88.482^oF
this is about 3^o higher than my std calculation for just Q=AΔT/R doing it as total Q_wire being dissipated through the area of the insulation (in this case 1.0466ft²)

 

[Chestermiller]

Tw = ~88.482^oF
this is about 3^o higher than my std calculation for just Q=AΔT/R doing it as total Q_wire being dissipated through the area of the insulation (in this case 1.0466ft²)

Again, in the equation, Q has units of BTU/hr-ft. This will be needed when we do the sink analysis.

So, let me understand correctly. The predicted temperature rise of the wire is 8.4 F, compared to your simplified method which gives about 5 F. Are these temperature rises about what you were expecting when you actually carry out the experiment? Do you have any indication that these are in the right ballpark? How much of a sink effect are you willing to tolerate as far as the center temperature is concerned?

Once we're satisfied with this simple 1D heat flow problem, I will show you how to analyze the contribution of the sink.

 

[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