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If I pass a current through a copper conductor, how can I calculate how hot the conductor will get?

For example, if I have a 7.2kW load powered by 240VAC, the current will be 30A. If I transmit this power to the load via a \$2.5mm^2\$ copper conductor, how do I calculate how hot this conductor will get?

UPDATE:

From the comments and answer from Olin and Jason, I've created the following graph showing Watts per foot of \$2.5mm^2\$ copper wire:

Watts per foot

But how do I translate this into the the actual temperature rise. I understand that the missing variable is the rate of cooling, but I just need to get an idea of what the maximum safe current is that can be passed through copper cable of a given thickness.

Assuming a constant current, and that there is no cooling at all, how do I calculate the degrees of temperature rise per hour per Watt for the foot length of copper cable in question?

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    \$\begingroup\$ You will need additional parameters, like the thermal resistance between the copper conductor and the surrounding air. Then you can make a rough estimate just like with heatsinks. Or for better results, make some experiments and get a result with included convection. \$\endgroup\$ – 0x6d64 Nov 16 '11 at 17:59
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    \$\begingroup\$ As @ox6d64 said, you can't know temperature without thermal resistance. But you can start with power dissipation per length to get a feel for whether it is a issue or not. Look up the resistivity of copper and determine what the resistance of 2.5 mm^2 for one foot is. Then compute the power this foot of wire must dissipate by Watts = Amps^2 * Ohms. If you've only got a Watt or two per foot, clearly it's not going to get that hot. If it's 10s of Watts you need to sharpen the pencil and look carefully at cooling. \$\endgroup\$ – Olin Lathrop Nov 16 '11 at 18:24
  • \$\begingroup\$ The IEC 60287 series of standards (equivalent to BS 60287 in your country) is for Electric Cables - Calculation of the Current Rating. IEC 60287 Part 2-1 Thermal resistance - Calculation of thermal resistance gives the required formulas and figures to calculate the thermal resistance of a cable in various conditions. \$\endgroup\$ – Li-aung Yip Nov 18 '15 at 9:23
  • \$\begingroup\$ Do you really need to do all that math? Referring to the 2017 National Electric Code, Table 310.15(B)(16) says that, with 60C rated insulation, 10 AWG can safely carry 30 Amps, provided that the ambient temperature is no more than 30C and there are no more than 3 conductors in your cable or raceway. (BTW - 10 AWG is 2.59 mm) \$\endgroup\$ – Bill Wentz Sep 18 '17 at 17:23
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In your edit, what's missing is that the rate of cooling will depend on the temperature. In general, the cooling rate will increase as the temperature increases. When the temperature rises enough that the cooling rate matches the heating rate, the temperature will stabilize.

But the actual cooling rate is very difficult to calculate. It depends on what other materials the copper is in contact with (conductive cooling), the airflow around the conductor, etc.

As an added complication, the heating rate will also depend on temperature, because the resistance of the copper will increase at higher temperatures.

So without much more detailed information about your conductor and its environment, its not really possible to give a precise answer to your initial question, how hot will it get?.

As for the second question, how fast will it heat up if there's no cooling, you can calculate that from the heat capacity of copper, which Wikipedia gives as 0.385 J / (g K), or 3.45 J / (cm^3 K).

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Purely theoretically with no cooling at all:
\$ P=I^2*R(T) \$
\$ E(t)=\int{P dt}\$
\$ T=T0+dT \$
\$ dT=\frac{E(t)}{m*C} \$
\$ m=V*density \$
\$ V=l*A \$
\$ R(T)=l/A*r(T) \$

The above can be condensed into a linear approximation:
\$ R(T)~=l/A*(r+T*\alpha) -> R(dT)~=l/A*(r0+dT*\alpha) \$

combining all this: \$dT ~= \int{I^2*l/A*(r0+dT*\alpha) dt}/(l*A*density*C) = I^2/(A^2*density*C)*\int{r0+dT*\alpha dt} \$

if \$ dT*\alpha << r0 \$ then \$ dT ~= I^2*r0*dt/(A^2*density*C) \$

unless I messed up something :) and it would melt eventually

I: current, R:resistance, P: power, T: temperature, t:time, E:energy, m:mass, V:volume, l:length, A: cross section area of wire, C:heat capacity of copper

Of course some kind of heat transfer always exists: conduction, convection, radiation. A good rule of thumb is to allow 2.5A/mm^2 on a copper wire in a coil with multiple layers, 4..5 A/mm^2 for single layer (without heat insulation) and 8..9 A/mm^2 will require active cooling.

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    \$\begingroup\$ Welcome to Electrical Engineering! You've got quite a few equations in this answer, which is great. You may have noticed that it's a bit difficult to read - For this reason, we have support for LaTeX equations on this site: See the editing help and MathJaX Documentation for help. Give it a moment, and it will render in the preview. I've done the first block for you. \$\endgroup\$ – Kevin Vermeer Nov 18 '11 at 20:14
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Olin's comment has a good start on the quantitative analysis, but keep in mind that the effect of a watt or two per foot in an 18ga AWG wire (approx 1mm diam) is quite different from a 38ga wire (approx 0.1mm diam). 2.5mm^2 = approx 0.89mm radius 1.78mm diam = approx 13ga AWG wire which is pretty large and a watt per foot is probably fine, but let's see:

The wikipedia page for AWG = American wire gauge shows the National Electric Code copper wire "ampacity" (current capacity) at several temperatures for insulated wire, and 13AWG (not a standard product) is midway between the 12AWG rating of 25A at 60C-rated insulation, and the 14AWG rating of 20A at 60C-rated insulation, so my guess is that at 30A it would get pretty hot (probably >= 100C at 25C ambient) without convective cooling.

The wikipedia page also lists copper resistance of 13AWG as 2 milliohms per foot, so P = 2milliohms * 30A^2 = 1.8W/foot; the 22.5A "rating" at 60C rated insulation (average of neighboring ratings) has dissipation of very nearly 1W/foot.

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Moving away from the pure calculus, just look at the manufacturers rating. Most cables are limited by the insulation material as this melts long before the cable causing catastrophic failure.

Think of a fuse wire. A 30 A fuse wire is very thin and a lot thinner than the in property cabling. The difference? the fuse wire can run hot as there is no insulation and you want it to rupture accordingly. Distribution wires are rated taking into account a myriad of operating conditions (type of mounting, insulation material, number of cores, etc). All manufacturers will provide guidance on the rating and de-rating (depending on installation method and other factors) of their cables. Unless using open exposed copper bus bars any calculations are not really worth their salt, copper capacity is way above the cable capacity. e.g. 30 A fuse wire is only 0.4 mm^2 but you wouldn't wire the boiler with that. (incidentally 30A fuse wire needs approx 170 A to rupture in 1 Second, even if you increase permissible rupture time to 5 seconds this is still 125a for the wire to burn through (all from 0.4 mm^2 wire!))

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Approximation of temperature rise in wire.
AWG-- Fuse Current-- Temp rise °C/A
10- 333- 3.258258258
12- 235- 4.617021277
14- 166- 6.536144578
16- 117- 9.273504274
18- 82- 13.23170732
20- 58.6- 18.51535836
22- 41.5- 26.14457831
24- 29.2- 37.15753425
26- 20.5- 52.92682927
28- 14.5- 74.82758621
30- 10.2- 106.372549
32- 7.3- 148.630137
34- 5.1- 212.745098
36- 3.62- 299.7237569
38- 2.59- 418.9189189
40- 1.77- 612.9943503
Bare wire in free air.
Based on melting temperature of copper = 1085C
1085 / Fusing Temp = °C/A Note: PVC insulation commonly rated at 60° to 105°

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  • \$\begingroup\$ Is this degrees C rise in the first second, ms, hour..? \$\endgroup\$ – N-ate Jan 25 '18 at 21:51
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I understand that the missing variable is the rate of cooling, but I just need to get an idea of what the maximum safe current is that can be passed through copper cable of a given thickness.

without knowing the rate of cooling, there is no answer to your question.

Two things are at work here:

1) heating: the temperature rise is proportional to the power dissipated, thus proportional to the I^2, and secondarily the resistance, which itself is a function of the temperature. within a certain range, you may be able to ignore the 2nd term;

2) cooling: this is proportional to the temperature over ambient, assuming a static environment.

in equilibrium the two balances.

So I^2 = k (T-Tambient)

k would be determined by the factors mentioned above.

To show you how important cooling is, this approach is exactly what many MAF meters use to measure air flow in cars, where T - Tambient is sensed via resistance.

for your purpose, however, there are lots of tables for you to check out instead of going through all of this pain.

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How do I calculate the temperature rise in a copper conductor?

You don't. Make a test setup and measure.

Why not? Read this paper.

If you have a strong desire to calculate, the following is from a 1930 Hokkaido Imperial University paper
titled: Temperature rise of a conductor due to the electric current
Authors: Ikeda, Yoshiro; Yoneta, katsuhiko
Abstract:

Heat generated by electric current is partially dissipated in the surrounding medium through conduction, convection and radiation, and partially produces a temperature rise of the conductor. It is, however, destructive for most electric apparatus or machines to be at too high a temperature. Therefore it is importance to know the relation between the intensity of current and the amount of the temperature rise. Now we are going to treat the phenomena in the wider range of application in order to have an exact and simple form of solution.

For the unknown values you will need to download the paper because there is 35 pages of formulas preceding this final formula.

exact and simple form of solution

enter image description here enter image description here



For an approximation
enter image description here
enter image description here

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Although this is a 7 year old question, I thought I may contribute the approach I found inspired by some points mentioned in an application note from SIEMENS.

Steady state temperature approximation of a conductor

$$\Theta_{op}=\Theta_{amb}+\Delta\Theta_{max}\left(\frac{I_{op}}{I_{max}}\right)^2$$

$$I_{max} :\text{maximum continuous current, } I_{op} :\text{operating current}$$ $$\Theta_{x} :\text{x temperature, }\Theta_{amb}:\text{ambient, }\Delta\Theta_{max}:\Theta\text{ rise @ }I_{max}$$

Maximum continuous operational current

Cables have specified current carrying capabilities for continuous operation. Different cable insulations allow for different maximum operational temperatures. These can be calculated following an IEC norm, but we can use either our specific cable datasheet or general ones to get a ball-park value.

  • Specified here, 2 Single Core 2.5mm^2 PVC insulated cables have a current carrying capacity of 24 Amps (AC/DC) with the conductor operational temperature at 70ºC and an ambient temperature of 30ºC.

  • Specified at a Nexans application note, 2 Single Core 2.5mm^2 XLPE insulated cables have a current carrying capacity of 24 Amps with the conductor operational temperature at 90ºC and an ambient temperature of 45ºC

From this data we can extract the following: $$\text{PVC 2.5mm}^2@I_{max}=24A,\Delta\Theta_{max}=40^o\text{C, }\Theta_{op_{max}}\leq 70^oC$$ $$\text{XLPE 2.5mm}^2@I_{max}=24A,\Delta\Theta_{max}=35^o\text{C, }\Theta_{op_{max}}\leq 90^oC$$

If we assume that your cable is XLPE and in the air with a maximum ambient temperature of 25ºC: $$\Theta_{op}=25+35\cdot\left(\frac{30}{24}\right)^2\approx 80^oC$$ This is dangerously close to the maximum operational of the XLPE insulated cable. If it is the PVC insulated one, the calculation results in >87ºC, where the insulation will probably melt. PVC at temperatures above 60ºC becomes unstable.

Time required to reach steady state temperature

How long it will take to reach this temperature can be estimated by considering the short-circuit current rating of the cable. Looking it up in the tables, 2.5mm^2 @ 1second short = 358 Amps.

The heating transition of the cable follows approximately the following equation:

$$\Theta_{op}=\Theta_{amb}+\Delta\Theta_{max}\left(1-e^{\frac{-t}{\tau}}\right)$$

$$\tau\text{(min)}=\frac{1}{60}\cdot\left|\frac{I_{1s-short}}{I_{max}}\right|^2=\frac{1}{60}\cdot\left|\frac{358}{24}\right|^2\approx 3.7\text{min}$$

\tau defines the time it requires to reach 63% of the final temperature. Normally we estimate that at 5*\tau we are at around 99% of the final temperature. 5*3.7 min = 18.5 minutes.

$$\tau \text{ is valid for reaching any calculated steady state conditions}$$

$$\text{Time to reach any steady state temperature} \approx 5\cdot\tau \approx 18.5\text{min}$$


ballpark/estimated demonstration

Our calculated \tau was with values: Ambient temperature 45ºC, operating temperature = 90ºC. \Delta T = 35ºC. I_max = 24 Amps

Power dissipation follows a square rule, P=I^2*R , we could extrapolate that to say that rate of temperature rise follows a similar square rule. $$K_{\tau}\approx\left(\frac{I_{ref}}{I_{op}}\right)^2 = \left(\frac{24}{30}\right)^2 = 0.64$$

but our calculated \Delta T (temperature rise) is of 55ºC versus 35ºC. $$K_{\Delta\Theta}\approx\frac{\Delta\Theta_{op}}{\Delta\Theta_{ref}} = \frac{55}{35} \approx 1.5714$$

applying these to our \tau as follows would give us $$\tau_{op}=\tau_{ref}\cdot K_{\tau} \cdot K_{\Delta\Theta}=3.7\cdot 0.64\cdot 1.5714=3.72 \leadsto 5\tau = 18.6\text{ min}$$

Note that these formulas for the demo of a modified \tau was invented out of "thin air", by "feeling", by some "logical" considerations. This may be completely wrong, and if I have made an assumption that is "crazy" please do let me know so I can learn my mistake.


Resources

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