Temperature effect on resistance of Electrical Connector Conductor

[patrickewen]

Hi

I am currently working at a company for a work placement who make underwater mateable electrical connectors. The company recently have been asked to make a connector which can withstand high temperatures. Basically, the conductor parts of the connector will be in hot oil which will have a temperature of 140c. (Ambient temp 35c).

I was asked to find out how the resistance will change with this increase of temperature. I know of an equation which i think would help (R = r0 [1 + a(T-t0)] ) but I am not sure this is the right way to go about it. The american comapny who wanted the connector gave me a vague equation similar to R = I^2 * T which i have never come across. I think its because resistance will increase exponentially, as increasing the temp will increase resistance which will inturn make the conductors hotter...(i hope i am talking sense here)

Can anyone offer me guidence in this matter?

Any help would be greatly appriciated.

 

[NoTime]

I think they may be referring to P = I^2 * R, where R in this case is the difference R_140c - R_20c (assuming a room temp spec) of the connector. How much of a problem this is depends on the required currents, the thermal resistance of the connector shell and how well the oil bath will work as a heat sink.

 

[patrickewen]

Ok, i managed to get more information on this.

Basically, the company want to know how long the connector can last before the temperature inside the connector reaches critical temp of 180c. As said before the connector will be in oil at 140c and a current of 50A will be flowing through the conductors. With the high temperature of the oil and the heat coming off the conductors due to resistance means that the temperature will increase. I am also required to find out how long it will take for temp to reach the critical temperature at x2 and x4 current overload.

My main plan was to work out the resistance of the conductors at 140c, find out the heat quantity (Q = I^2 * R) then use the thermal conductance equation [ K = Q*L / (t * A * dT) ] and rearrange to get the time.

I have done this and got a very short time of 0.17s, which i don't think is correct.

Any ideas on what i could do?

 

[jim mcnamara]

This is not my area - but if the ambient temp is 35C, then how did the oil get to 140C? There must already have been current flowing. You don't get 140C temps in oil at ambient 35C without adding heat energy to the oil.

 

[NoTime]

My guess from your statements is that the connector can not handle 50A in any environment.
Not my area either, but if you post your entire derivation with all the factors you considered someone might check your work.
You certainly did not provide enough information here to even attempt a solution.

 

[patrickewen]

Yeah, sorry guys. Its just that the company I work for cannot get the information required straight from the customer. There is a middle-man which doesn't seem to know much about what is required.

I think the customer requires proof that connectors can withstand a temperature =>140c for a certain amount of time. They requested that the connector can have a current of 50A flowing through the conductors while submerged in hot oil. However, my company have told me that the critical temperature is 180c as some parts of the connector may start to warp. With the hot oil, and heat from the resisting conductors, how long will it be until the connector reaches the critical 180c.

The conductors are brass, insulated with PEEK and the connector shell is stainless steel

Im sorry I cannot be much clearer.

 

[NoTime]

I think some of the info you would need.
Conductor size, material, insulation, connector contact resistance, Contact resistance thermal coefficient, if the oil is flowing or stagnant, volume, viscosity.
This is the short list.

 

[TFH]

ANSI C119 provides some guidance in this area. Although the calculations or models may be fun and very helpful for guidance and perhaps design, in this case it seem to me that proof would be from a certifed test not a calculation.

 

[patrickewen]

ok, i have good news! My manager and I talked it over, and it is bloody difficult to calculate this all.

So, all I have to do now is this (for the time being):

Assume at time zero brass conductor is @ 140c with 50A flowing through it. How long will it take for the conductor to reach critical temp of 180c. (Im going to do two calculations, one for the connector in air, and one in oil) Ignore all other materials at the moment (insulator, connector shell, etc).

I have the cross sectional area and length of conductor and found the electrical resistance of brass. I then assume to use this thermal conductivity equation (to find time) here http://en.wikipedia.org/wiki/Thermal_conductivity, but I am not sure how to calculate the Q (quantity of heat). Any suggestions?

Again, thanks for helping me out! :)

 

[AlephZero]

Grabbing formulas off the web when you don't understand the basics is dangerous. IMO you really need to read up a bit about thermal conduction, specific heat, Newton's law of convective cooling, etc (probably in high school physics text) before you jump into this.

Here are some "ballpark" numbers (ignoring details like the operating temperature, what particular brass alloy you are using, etc!)

Assume a brass cable, diameter 4mm, that is thermally completely insulated.

Properties of "generic" brass:
density 8500 kg/m^3
electrical resistivity = 7e-8 ohm/m (approx 4x the resistivity of copper)
specific heat (Cp) = 380 J/kg.K

Cross section area of wire = 2e-3 x 2e-3 x pi = 1.26e-5 m^2 (= A)
Resistance of wire, per unit length = resistivity/A
= 7e-8 /1.26e-5 = 0.00556 ohms/m (= R)

Heat generated for 50A current = I^2R
= 50 x 50 x 0.00556 = 13.9W/m (= Q)

Mass of wire = density x A
= 8500 x 1.26e-5 = 0.107 kg/m (= M)

Temp rise = Q / M.Cp = 13.9/(0.107 x 380) = 0.342 degK/sec.

Temp rise from 140C to 180C = 40K

Time to rise 40K = 40/.342 = 117 sec.

That's the MINIMUM time, assuming perfect thermal isolation.

If the wire is in direct thermal contact with the oil (i.e. no insulation layer between them) the "thermal time constant" for the oil and wire temperatures to equalize would be a lot less than 117 sec, so it would be reasonable to assume that it happened instantaneously.

With that assumption, to include the effect of the oil heating up as well as the wire, just change the M.Cp term to M_wire.Cp_wire + M_oil.Cp_oil (that's the mass of oil surrounding 1m length of wire).

To include any more effects, including heat loss to the environment, would be a diffferent and bigger ball game. But the above assumptions are conservative for a failure analysis, since they give mimumum estimates of the time to heat up.

 

[patrickewen]

Thank you so much. Really helped me out!