Calculating the Power Dissipation of a Wire

[Benjamin Sorensen]

I'm having trouble verifying an experiment I ran to determine the power dissipation of a heating element. 13.15W of power was applied to 3ft nichrome wire. Temperature readings were collected until they stabilized at 128.5F (room temp was 70F). I want to create a mathematical model of the system to confirm the results of the experiment. I am doing this by using the temperature readings and calculating the theoretical power dissipation. Maybe this isn't the best way to confirm the values.

What I have thus far:

3 modes of heat transfer: free convection to the air, radiation to the surroundings, and conduction to the blue painters tape used to attach the thermistor.

Convection:
I used the Rayleigh Number and Nusselt Number for cylinders and laminar flow. Followed by Newton's Law of Cooling to calculated power: 3.585W.

Radiation:
I went straight to Newton's Law of Cooling for radiation. Using the emmisivity constant for nichrome. 0.501W

Conduction:
I wrapped standard blue tape around the wire 3 times. So, using Fourier's Law I calculated the power dissipation to be: 0.734W

In total: 4.82W. Pretty far away from 13.15W. I feel like I am missing something major. I can post the equations I used if you think that will help.

I appreciate any help anyone can provide. Unfortunately, faculty at my university won't give me the time of day!

 

[jim hardy]

I'm having trouble verifying an experiment I ran to determine the power dissipation of a heating element. 13.15W of power was applied to 3ft nichrome wire. Temperature readings were collected until they stabilized at 128.5F (room temp was 70F).

It's difficult to troubleshoot from somebody else's observations.
Where's your raw data ?

13.15W of power

measured how ?

stabilized at 128.5F

that's the temperature of a thermistor inside three layers of tape wrapped around a nichrome wire .
Seems to me you're measuring at the hot spot created by taping on the thermistor..

When measured power in ≠ calculated power out,
the first question is "Do i trust either one ?"

old jim

 

[Baluncore]

I'm having trouble verifying an experiment I ran to determine the power dissipation of a heating element. 13.15W of power was applied to 3ft nichrome wire.

Welcome to PF.

How did you measure the reference 13.15 watt ? Did you use AC or DC current?
Did the power or the temperature of the wire change while making those different measurements?

 

[osilmag]

Have you thought of resistivity? And then using the current through that to get power dissipated?

$R=\rho \frac{l} {A}$

 

[jim hardy]

I would
Measure its resistance at two different currents
one so small it the heating is negligible
and the other enough to raise its temperature considerably

calculate the temperature by Nichrome's temperature coefficient of resistivity.

 

[Benjamin Sorensen]

Thank you for the helpful response! You bring up valid questions. Which raw data would you like to see? I have graphical data outputted by the arduino showing the predicted exponential decay to the steady state temperature value.

The power was calculated by first measuring the resistance of the wire to be 1.9Ω with volt meter. Then delivering 5V of DC power from a variable power supply. So, V²/R = 13.15W.

I had thought it might be a hot spot. Would I need consider the heat flow between concentric cylinders? I imagined insulating would be accounted for when calculating the heat transfer through conduction into the blue tape.

 

[Benjamin Sorensen]

Welcome to PF.

How did you measure the reference 13.15 watt ? Did you use AC or DC current?
Did the power or the temperature of the wire change while making those different measurements?

The power was calculated by first measuring the resistance of the wire to be 1.9Ω with volt meter. Then delivering 5V of DC power from a variable power supply. So, V2/R = 13.15W. The power delivered was constant for the 5 minutes or so it took to reach equilibrium temperature.

 

[anorlunda]

The power delivered was constant for the 5 minutes or so it took to reach equilibrium temperature.

I doubt that. Do you mean the voltage was constant?

As @jim hardy already pointed out, the resistance changes substantially with temperature.

 

[Benjamin Sorensen]

Have you thought of resistivity? And then using the current through that to get power dissipated?

$R=\rho \frac{l} {A}$

I have considered that. Is that a good way to validate an experiment though? My goal is to confirm that my wire is indeed dissipating 13.15W. The resisitivity would just be a second way to calculate theoretical power dissipation. I would still need a way to validate that second calculation.

 

[osilmag]

You might have to dig into a heat and mass transfer textbook. I know that you can model the flow of heat with an equivalent circuit.

 

[Benjamin Sorensen]

I doubt that. Do you mean the voltage was constant?

As @jim hardy already pointed out, the resistance changes substantially with temperature.

Sorry, I misspoke. The voltage delivered was constant. And the current drawn displayed on the power supply was also constant. I couldn't figure a good way to validate the current value displayed. Since when I hooked up a voltmeter, the voltmeter draw affected the readings. For example, the power supply stated a 2.52A draw. When I attached the voltmeter, that value changed to 3.1A and the voltmeter displayed a third value that I neglected to note down (lessen learned!). None of the readings lined up which is why I chose not to use them when calculating the power draw.

Do you have a reliable way to measure power draw as the temperature of the wire increases?

 

[Benjamin Sorensen]

I would
Measure its resistance at two different currents
one so small it the heating is negligible
and the other enough to raise its temperature considerably

calculate the temperature by Nichrome's temperature coefficient of resistivity.

That is a great point that resistance changes with temperature. Would attaching a voltmeter while a power supply is attached to the wire be a good way to measure resistance? I had trouble measuring current that way.

 

[anorlunda]

Do you have a reliable way to measure power draw as the temperature of the wire increases?

See #5. Get the change in resistance first.

 

[Benjamin Sorensen]

Thank you for the awesome welcome! Current plan is to measure the change in resistance to calculate the actual power draw at high temperature. As science goes, I am predicting the resistance will be higher and with a constant voltage in, the power draw will be lower. I will report back on my findings after the weekend. I will also try to measure the temperature of the wire without the use of insulation to try to remove one mode of heat transfer.

 

[jim hardy]

Would attaching a voltmeter while a power supply is attached to the wire be a good way to measure resistance? I had trouble measuring current that way.

The best way is to use two meters
an ammeter in series and a voltmeter in parallel.

Be sure your voltage reading does NOT include drop across the ammeter.

Ohms is of course voltmeter reading divided by ammeter reading.

These days i assume you have access to multi digit DMM's.
You can't rely on a 5% analog panel meter to give much precision.

 

[davenn]

o

Since when I hooked up a voltmeter, the voltmeter draw affected the readings. For example, the power supply stated a 2.52A draw. When I attached the voltmeter, that value changed to 3.1A and the voltmeter displayed a third value that I neglected to note down (lessen learned!).

unless you hooked it up incorrectly, that shouldn't happen

really is time for you to show a circuit and even better a photo or two showing how you hooked this up
what is your voltmeter ?
what is your ammeter ?Dave

 

[Babadag]

Since we have not sufficient data to calculate I have to improvise some.
I think the wire diameter will be 0.097" and ρ≈ 1 ohm*mm^2/m at 21oC and temperature factor 0.123. The resistance at 21.1oC[70oF] will be 0.214 ohm and 1.9 at 85.11 oC.[185.2oF instead of 128.5!].
The 3 rows of insulating tape will add another 6*0.005=0.030" and the final overall diameter will be 0.127".
According to Neher&McGrath [these formulae were revised in IEEE-835/1994] convection evacuation power will be:
Wc=0.072*Ds ^0.75*∆T^1.25*length[ft] (eq. 56)
Let's say Ts=82oC [outside surface temperature] then ∆T=61oC and then Wc=7.83 W.
Radiation:
Wr=0.10256*Ds*ε*∆T*[1+0.0167*Tm] [eq.55a] Wr=4.08 [W]
Total evacuated power =Wc+Wr=11.91 W
Conduction it could be from this wire to the battery, ampermeter and voltmeter connection 13.15-11.91=1.24 W.
This it could be a 1.43 ft 0.5 mm^2 copper conductor 0.5 mm insulation thickness at 60oC[average].[Wc=0.887 W;Wr=0.356 W]

 

[sophiecentaur]

@Benjamin Sorensen . Can I revisit your original question? If the temperature has reached equilibrium then there is no doubt that the dissipated power will be the same as the electrical power input. If you are trying to identify how the dissipation is shared by the various paths then you need to reduce / eliminate them one at a time and see how the equilibrium temperature of the wire is affected. This is not dissimilar to a simple model of climate.

You need to be able to measure the temperature of your wire without disturbing it so you need it to be its own thermometer. You can heat it to a range of temperatures inside an all-encompassing and highly insulating jacket and measure its resistance for a range of temperatures (using your thermistor, perhaps). The calibrated wire can do its work without being hooked up to the thermistor.

This is great as a thought experiment but it would involve a fair bit of construction effort. But the two transfer mechanism are difficult to separate

If you have access to a large vacuum chamber, you could eliminate convection but we are talking big money, I think. A cheaper way to study the effect of convection would be to operate in a large transparent box and change the temperature of the air in the box (very slow air flow). The radiated heat would pass straight through the air to the room. You can vary the air temperature so that it goes above and below the ambient temperature and see how the wire temperature is affected. That will give an indication of the effect convection.

To study the effect of radiation, you can surround the wire in a temperature controlled black jacket with free air circulation around it. The jacket could be made of black pipes with water circulating to control the temperature. Again, you can vary the temperature of the black jacket and see how the wire temperature is affected. The jacket would need to be in a transparent layer to eliminate convection from and to the black surface.

If you use the same temperature variations for the convection and radiation experiments then you should get perhaps two different graphs - one showing the effect of convection and one showing the effect of radiation.

Actually, you could do both convection and radiation study with the 'radiation ' set up.

 

[Babadag]

Sorry,
I forgot to show how I got 82oC the outside surface temperature.
The insulating tape thermal resistance –in T.O.F-thermal ohm.feet it is –as per Neher&McGrath eq.38:
Ri=0.012*ρi*log(Di/Dc). If ρi=500 W.cm/oC then Ri=0.702 T.O.F.
Ts=Twire-P/3*Ri=85.11-13.15/3*0.702=82oC

 

[Babadag]

On the other hand I found in the internet Ni-Ch wire ρ=1.1 Ω.mm^2/m and the resistance thermal factor it is 85/10^6.That means it is no change in resistance when it is hot. In this case the 128.5oF it could be 128.5oC. In this case the calculation fits the measurements.

 

[Benjamin Sorensen]

Since we have not sufficient data to calculate I have to improvise some.
I think the wire diameter will be 0.097" and ρ≈ 1 ohm*mm^2/m at 21oC and temperature factor 0.123. The resistance at 21.1oC[70oF] will be 0.214 ohm and 1.9 at 85.11 oC.[185.2oF instead of 128.5!].
The 3 rows of insulating tape will add another 6*0.005=0.030" and the final overall diameter will be 0.127".
According to Neher&McGrath [these formulae were revised in IEEE-835/1994] convection evacuation power will be:
Wc=0.072*Ds ^0.75*∆T^1.25*length[ft] (eq. 56)
Let's say Ts=82oC [outside surface temperature] then ∆T=61oC and then Wc=7.83 W.
Radiation:
Wr=0.10256*Ds*ε*∆T*[1+0.0167*Tm] [eq.55a] Wr=4.08 [W]
Total evacuated power =Wc+Wr=11.91 W
Conduction it could be from this wire to the battery, ampermeter and voltmeter connection 13.15-11.91=1.24 W.
This it could be a 1.43 ft 0.5 mm^2 copper conductor 0.5 mm insulation thickness at 60oC[average].[Wc=0.887 W;Wr=0.356 W]

Thank you for the awesome reply! Is the IEEE-835/1994 document available for free? I would like to read about the equations you listed in more detail. Here is more information about my setup:

Wire diameter: 20AWG (0.032in)
Resistance per manufacture's spec sheet: 0.6348 Ω/ft

3ft of wire is dangling in the air (to reduce conduction to objects). A DC power supply is connect on both ends applying 5V. A thermistor, connect to an arduino, is taped to the center of the wire. I'm not sure how to attach pictures, so I hope this description will do.

Following your equations (without 100% understanding because I haven't read the literature):

Ri = 1.723
Ts = 77.56C
Wc = 4.163W
Wr = 0.316
Wt = 4.479

I'm not sure what Tm represents in the radiation equation. I solved for it backwards and got 413C. In my calculations, I assumed it was the ambient temperature in the room.

Overall, these values generally line up with my previously calculated values. Since the wire diameter is much smaller than you assumed, the resistance is much higher.

I did run the experiment again and registered 4.5Ω at the increased temperature. This means the revised power input is 5.56W, which is better. What do you think about this new information?

 

[anorlunda]

Thank you for the awesome reply! Is the IEEE-835/1994 document available for free?

Unfortunately, no. IEEE standards are very expensive. That one is $535 for a PDF copy.

Expensive fees for standards is very controversial. But years and years of pressure have failed to get them repealed.

 

[Babadag]

IEEE 835/1994 it is expensive indeed. However Neher&McGrath theory it is not so expensive [only 33 $].
Neher,McGrath-Temperature and Load Capability of Cable Systems
https://ieeexplore.ieee.org/document/4499653

 

[Babadag]

Sorry. Tm=(Ts+Ta)/2. Actually Wr=0.10256*Ds*ε*∆T*[1+0.0167*Tm]*length[ft]-It has to multiply by length[ft].