analytical calclation of connector temperature rise

[electricpete]

To sharpen up my heat transfer skills calculation skills, I tried to solve the problem of determining temperature rise for a specific example of high-resistance electrical connector as posted in the following link:

http://www.brazosport.edu/~pschimpf/forums/heat_xfer_wire_connector.PDF

It turned out to be longer than I hoped - 11 pages, 300k pdf file. It should be easy reading if you are familiar with Maple or other symbolic math languages. Otherwise it may be difficult to get past the syntax to see what I'm doing.

I would be interested in hearing your comments - are there any big errors or big assumptions not identified?
Thx.

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[hacksaw]

without going through the step by step details, it is not a bad nor a particularly good approach to the problem.

all of the solutions appear to be linked to the NEC tables as a basis for the boundary conditions for a heat flow problem. that is a bit bogus, but at least if gives you an estimate based on the criteria used to develop those tables.

the best and least misleading approach is to calculate the heat rise based on the effective surface area of the joint, but that is my own view.

 

[sailoday28]

Without reading the many pages of the problem/solution can you give in simple terms with boundary conditions what the problem is?

 

[electricpete]

That's a good idea to summarize it since it is lengthy. I am glad someone is interested to read it and provide feedback. He is my summary:

physical problem: Assume an infinitely long THHW-insulated 10 AWG copper wire carrying 10A in free air located along the x-axis with hot spot at position x=0 generating 0.375 watts.

Differential equation:
Qin-(1/Rt)*T(x)+Ac*k*dT^2 /dx^2=0

Solution of differential equation
T(x) = Qin*Rt+C1*exp(x/sqrt(Ac*k*Rt))+C2*exp(-x/sqrt(Ac*k*Rt))

Method 1: Assume thermal resistance is uniform everywhere along the wire and all of the heat is generated precisely at x=0.

Method 1 Boundary Conditions are denoted BC1.X, for X from 1 to 2 as follows:
BC1.1 - Since the solution cannot grow without bound as x->Infinity, we know C1=0.
BC1.2 - Since the heat is generated at an infinitessimally small point at x=0, all of the heat is disippated radially at x=0. Half goes in the-x direction and half goes in th +x direction. The boundary condition is therefore:
Qhotspot/2 = -k_wire*Ac_wire*subs(x=0,diff(Tsolution1,x));


Method 2 - Divide the solution into two regions: wire region and connector region, with differing thermal resistances. Also the hotspot heat 0.375 watts is distributed uniformly along the length of the connector region. So we have different values of Qin for the two regions as well as differerent Rt in the two regions.

For the two regions we have 4 constants: C1wire, C2wire, C1connector, C2connector, and therefore we need 4 boundary conditions.

Method 2 Boundary Conditions are denoted BC2.X for X from 1 to 4 as follows:.
BC2.1 - Similar to BC1.1 above we conclude C1wire must be 0 to prevent solution from increasing without bound.
BC2.2 - due to symmetry around x=0 we don't expect any heat transfer along x-axis at x=0 and the derivate of T with respect to x at x=0 must be 0:
dT/dt = 0 at x=0.
BC2.3 - Temperature is continuous at the point x=L/2 which forms the boundary between the connector region and the wire region so we have
Tconnector(x=L/2) = Twire(x=L/2)
BC2.4 - Axial heat conduction is continuous at the point x=L/2 which forms the boundary between the connector region and the wire region so we have:
[-k*Ac*dT/dx]wire = [-k*Ac*dT/dx]connector @ x=L/2

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[Tunalover]

electricpete-
Why not turn the problem over to a mechanical engineer who has formal education and substantial experience in the thermal sciences?


Tunalover

 

[electricpete]

hacksaw - I did use the surface area of the joint times an estimated convective film coefficient in the connector region in method 2. However I am uncertain of the thermal conductivity of electrical insulating materials so I relied on the NEC tables for portions of the problem which included electrical insulation over copper. I have an article which suggests that cables in free air operated at their NEC free-air ampacity reach approx 90% of their rated temperature rise.

tunalover - It would be helpful to me if you identify one or more areas that you believe are incorrect.

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[chicopee]

Looking at the problem, it seems to me that you have a condition the wire is similar to a cylindrical fin, therefore, shouldn't the temperature solution be a hyperbolic function along the length of the wire?

 

[sailoday28]

Differential equation:
Qin-(1/Rt)*T(x)+Ac*k*dT^2 /dx^2=0

Is T temp of conductor minus T ambient?
Is Rt, thermal resistance in radial direction?
Is Qin, heat generation based on I^2*electrical resistance?

 

[electricpete]

chicopee - I don't know enough about the cooling fin solution to compare it. This solution does have exp(+cx) and exp(-cx) components which are the building blocks of hyperbolic functions.

sailoday -
"Is T temp of conductor minus T ambient?"
Yes.

"Is Qin, heat generation based on I^2*electrical resistance?"
Yes. Qin represents distributed (lengthwise) I^2*R heat generation (watts/meter). In method 1 and in method 2 wire region it applies to the normal conductor I^2*R heating due to normal conductor resistance. In method 2 wire region it applies to abnormal connector I^2^R heating due to abnormally high connector resistance assumed in method 2 to be distributed over length..

Is Rt, thermal resistance in radial direction?
Yes.... in a length-related manner. Rt in such a way that it would satisfy dT = Qin*Rt where again Qin is a per-length heat generation.

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