Heat transfer from a conductor 6

[Jgrovdal]

Hi
I am currently trying to calculate the temperature a conductor will reach when applying 50 amps to it. The conductor is an 16mm^2 copper stranded wire. The length is 2m. I am trying to make a model in Matlab to calculate it, is it possible to make it that way? Or is there a better way?

The model I am trying to use is the one from this paper:

https://jffhmt.avestia.com/2020/006...8h2ixV_BSjyU-UAkNXsbN6VYkFfs4mOLGBqWJ7T0AL2TY

But it has not been successful yet.

My problem seems to be that there still is some variables that is missing and therefore my calculations don't give me a result. For now I miss the Prandtl number and can't continue.

The variables I have found:
L = 2; %Length
I = 50; %Current
R_ref = 5.54830155*1^-4; %Reference resistance
alfa_T = 0.00393; %Resistance at temperature rise
T_ref = 21; %Reference temperature for Resistance
r_c = 1/2*12.1*10^-3; %Radius core
r_i = 1/2*15.45*10^-3 ; %Radius whole
g = 9.81; %Gravitation
l_alfa = 2*r_i; %Characteristics of the length of structure
sigma = 5.6704*10^-8; %Stefan Boltzmann constant
epsilon = 0.95; %Assumed after an paper
rho = 3.19*10^-8; %Ohm per meter
pi = 3.14; %Pie
lambda_c = 386; %Thermal conductivity conductor
c_c = 3.4*10^6; %Specific heat capacities conductor
c_i = 2.245*10^6; %Specific heat capacities isolation
lambda_i = 0.21; %Thermal conductivity conductor
M = 5; %Parameter - Gaver-Stehfest-algorithm (Dimensionless)
T_E = 25; %Temperature ambient
T_EK = T_E+273.1;
T_S = 25; %Temperature surface
T_SK = T_S+273.1;
T_0 = 21; %Reference temperature
beta = 1/T_EK; %Coefficient of thermal expansion
lambda_air = 25.90; %Thermal conductivity Air maybe 0.0246

Thanks in advance

 

[BrianE22]

I can at least help with this:

https://www.omnicalculator.com/physics/prandtl-number

 

[7anoter4]

See attached file also

 

[LittleInch]

Isn't there an easier way of doing this?

In steady state which is where you're going I think, the heat generated by the cable will equal heat loss to the air, mainly by convection if it's not a heating element. you can add losses from IR if you want or just add say 10%.

Convection to air is approx 50W/m2/K. the m2 is the outside area of the cable.

Now there is some element of iteration as the resistance varies with temp but pick a number and start working with it.
The actual temp of the conductor will be higher than the outside of the cable due to thermal resistance of the insulation, but this is easy to find once you've got your heat loss worked out.

The previous paper seemed to be interested in the transient temperature rise from a cold start, but if all you want is final temperature then you can simplify this quite a lot I feel.

Of course as soon as you start to move the air by any sort of wind or fan, then your heat losses from the cable increase and hence your temp comes down.
It also assumes the air doesn't heat up inside your cubicle / device.

Why do you need to know?


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Also: If you get a response it's polite to respond to it.

 

[7anoter4]

You are right, LittleInch. However, they are situations in which the loading time is less than “infinity”. For instance, in short-circuit case when a very high current is permissible. In IEEE 242 /2001 9.5.2.3 Intermediate and long-time zones
it is written:
Taking into account the intermediate and long-time ranges from 10 s out to infinity, the definition of temperature versus current versus time is related to the heat dissipation capability of the installation relative to its heat generation plus the thermal inertias of all parts. The tolerable temperatures are related to the thermal degradation characteristics of the insulation. The thermal degradation severity is, however, related inversely to time. Therefore, a temperature safely reached during a fault could cause severe life reduction if it were maintained for even a few minutes. Lower temperatures, above the rated continuous operating temperature, can be tolerated for intermediate times.
Here IEEE presents an approximate formula for IE/IN [IE-Emergency current for time more than 10 sec]
So, for short time -more than 10 sec-the current may be more than rated.

 

[LittleInch]

Ok, maybe I took the words "conductor will reach" to indicate that he or she is talking long term, but maybe not.

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[Jgrovdal]

Hi,
I were also thinking that there might be an easier way to do this. The reason why I would like to know are that i am writing a paper on a model made in COMSOL that researches the rise of temperature in a conductor, but i would like to validate the results with theoretical calculations.

As of now I am doing a physical test where the cable is inside an almost closed box, this is so that there should not be any disturbances while the test runs. So with that I do have a confined airspace that should make it more accurate and easier to calculate with.

The convection to free air is between 5-25W/m^2/K have i read at various sites as there are no fans.

There is some calculations on how much energy that would be needed to change the temperature with dQ = m*c_p*deltaT and then use the loss we have on a physical test to calculate how long it should take. But I would like to have a more precise way to do it.

Edit: I am currently trying to make a model that finds the maximum temperature that can be achieved when applying any amount of current. Taking the precaution that standard parameters is known.

 

[IRstuff]

Still air convection is closer to 3W/m^2-K; it's often bumped up to 5 W/m^2-K to account for radiative loss and fudge-factoring. If the cable gets sufficiently hot enough, something on the order of 7-10 W/m^2-K

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[LittleInch]

The question which hasn't been asked is how is the cable supported or suspended?

A cable which sits on the bottom of something like a pipe or duct is going to have a different air convection rate than one suspended in mid air inside a large volume.

And on its own or clustered with other cables?

I didn't think the convection rate per K changed with the actual temperature of the element concerned?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[electricpete]

I'll throw out some random thought that may or may not have relevance (I'm not sure what you're trying to do)

[li]Is there electrical insulation present? If so it has thermal insulating properties[/li]
[li]you may need to consider skin effect if the frequency is high enough.[/li]
[li]You may have an approxiatae 2D calculation option which would be simpler than 3D. It would be conservatively high assuming there is a component of heat transfer flowing axially outwards at the end.[/li]
[li]What is the actual boundary condition of the electrical circuit? Current flows in loops. Copper (if that's what this is) is a good heat conductor. The connected circuit may influence the temperature in this portion of the circuit[/li]


=====================================
(2B)+(2B)' ?

 

[7anoter4]

See IEEE 242 Transient loading

 

[7anoter4]

Note : in the above formula θK has to be θ/K and θ it is the duration in hours.

 

[IRstuff]

I didn't think the convection rate per K changed with the actual temperature of the element concerned?

There should be an increase in both convection and radiation. The convection because hotter air is even less dense, so higher buoyancy forces, although that might be mitigated by rising ambient temperature in a confined space; radiation goes as 4th power of absolute temperature, so that increases pretty fast with temperature. At around room temperature radiation is a small fraction of total, but at higher surface temperatures, it can dominate the heat transfer coefficient.

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[LittleInch]

Yes there is an increase in heat transfer, but that is dealt with by the r/ m2 K where K is the difference in temperature from conductor to air.

But just because the temperature of the conductor goes up, why would the heat output from convection increase faster and not in a linear fashion?

Remember - More details = better answers
Also: If you get a response it's polite to respond to it.

 

[FacEngrPE]

Prandtl number is a dimensionless number, Per this page for air use Air at room temperature has a Prandtl number of 0.71.
If you are working on a paper you should research the original source. There is some indication on the linked page that a good place to look is property tables, which is not an original source.

,

the units should cancel to dimensionless if you start with the correct units. The number may be temperature dependent.

For the experiment to match your equation you should figure out how to estimate / measure or calculate thermal conduction out the ends of the test cell.

 

[7anoter4]

According to IEC 60228 for copper conductor Table II Stranded conductors for single-core and multicore cables the reference resistance at 20oC for 16 mm^2[maximum] it is only 1.15 Ω/km [0.00115 Ω/m]. That means R_ref = 5.54830155*1^-4; %Reference resistance electric per unit length resistance at the temperature 𝑇ref (Ω/m) it is the resistance of 0.0055483/0.00115=4.824 m conductor it is not per unit [1m].
The conductor radius in m 12.1/2000=0.00605 m. This is a half of 12.1 mm [close to 95 mm^2 conductor and not for 16 mm^2-IEC 60228 for copper conductor Table I].
Then, at first put the R-ref to 0.00115*1.0393=0.0011952 Ω/m and rc=5.3/2=2.65 mm. State the final temperature or final loading time.
For instance, let's say what will be the temperature in still air at 25oC [air speed=0] of 16mm^2 1,2/0.6 kV rated PVC insulation loaded 50 A within an hour?
First let's see what IEC 60287-2-1 and 1-1 will say. Approximate 80 A for 70oC [may be 88A neglecting skin and proximity effect].
Using IEEE 242 formula the conductor temperature will be 40oC for 50 A[k=0.33 for less than #2 in free air].

 

[prc]

Let me raise a question not exactly connected with the theme- please pardon me.
I have a metal part in air heated by eddy currents from leakage flux ( bushing turrets on a large transformer) Inside of the turret is oil at 75C. But outside metal temperature( small area 10-20 Cm2) is 100C . Both temperatures at an ambient air temperature of 30C. Will the metal temperature vary when air steady-state temperature varies from 0-50C? If so how much? OIl temperature will vary almost linearly with air temperature.

 

[7anoter4]

Interesting indeed, prc. May be this is the answer:

https://www.heatsinkcalculator.com/blog/how-to-calculate-the-temperature-rise-in-a-sealed-enclosure/

 

[prc]

Thank you 7anoter4.

 

[IRstuff]

To the first order, thermal masses are like low-pass filters, i.e., they increase the elapsed time to a "steady state" temperature, and everything attempts to shift temperature proportionally with ambient temperature.

prc, a picture might help; you mention metal in a couple of places and it's not obvious whether it's all the same metal. Your temperature distribution seems odd, with interior being cooler than exterior, so is the oil circulated with a separate heat exchanger? Usually, heat flows from interior to exterior, so you'd expect interior oil to have the higher temperature.

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