Can Simulation determine this Power Generation problem?

[allyi]

Hi all,

Before I begin, I am new to this forum so if I am conveying my question incorrectly, please bear with me. I have been trying to acquire Solidworks Simulation for the past 3 months (I’m still awaiting my purchasing department), and I only have another 3 months left in my rotation assignment, so any knowledge I can obtain to help me get started the moment I get the program would help tremendously.

I am trying determining the change in resistivity due to a bad electrical connection [the interface of the connector is improperly crimped] on a cable joint.

======[]----------------[]======
^ ^ ^ ^ ^
1 2 3 4 5
1= Cable 1
2= crimped connector 1
3= yoke
4= crimped connector 2
5= cable 2

Given:
What I know is the power generated [I^2*R] by the conductors inside of both cable 1 and cable 2, as well as the surface temperatures of everything. Knowing the material of the cable, I’ve been able to determine the temperature of the cable conductor [by setting the total Q from heat transfer (conduction convection and radiation) to the environment equal to the power generated from I^2*R losses] I also know that the bad crimp had caused a severe increase in temperature in both connectors and yoke.

What I don’t know is the power generation, resistance, or temperature of the connector conductor [its 13kV, everything is insulated]. My question is: knowing the current, can Simulation be able to determine the power generation despite not knowing the resistance? Normally it would all be the same as generated in the cable, but it is altered in this case due to the increased resistance associated with the connection. From what I know, (sorry, I’m Mech E, not EE) I’d need one to determine the other. If I put them back into the heat transfer eq from above, I’ll have three unknowns, the third being the temp of the connector, so that still won’t help. Do any of you know if simulation would be able to find a way around this? If so, do you know how?

 

[GregLocock]

You needn't wait for Simulation to fail to turn up in time. Time to dig out your thermo books, if you know the temperature profile of the surface. For instance Rogers and Mayhew or Kay and Nedderman have pages discussing radiative and convective heat transfer from pipes to air (which is probably the case here).

Even if the software package does turn up you would still need to check your model using the hand calcs.

However I would have thought that a voltmeter would be the obvious solution.



Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[allyi]

My thermo book has helped me tremendously so far, thats how I have been able to determine the conductor temperature [I've double checked my numbers with the cable manufacturer to ensure that they are right]I've been able to correctly determine the heat transfer to the environment, but what I am not sure of is the power generation occuring inside because I am not able to open up the splice. [at the moment, we are only running tests that do not involve taking off the outer layering, I will be long gone from the department by the time they will be allowed to open it up, so using a voltmeter is out of the question for me]

My question really is what does simulation need to determine a power generation? If it can be solved with the given I have, I can work it backwards to get the resistance and use that as a reference for the department when they can finally put a voltmeter on it.

 

[GregLocock]

Why isn't the power generated at the splice = to the power radiated and convected to the environment?

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[allyi]

They are equal, but you cant determine power radiated or convected to the environment off surface Temperature alone. The amount of heat held within the material itself increases [delta T increases as Q-dot increases - ie the material is able to aborb more heat as generation increases] and since I dont know what the inner T is, I cant predict that.


This is what my plan is going to be:
I know power generated for the cable so that will be constant, but I will vary power generated for the connectors until I get the right surface T, and that will have the correct associated delta T and resistance will be found from the number I used for power generation.

 

[kellnerp]

Greg,

At 13kv I don't think a voltmeter is a good idea, at least not the kind you are probably thinking of.

Zorched

If a bad crimp leaves an open connection at this kind of voltage it doesn't matter. The current will jump the gap and plasma will form which tends to be really hot.

Electricians use non-contact thermal probes to find bad connections all the time. That is what he is after. The rubber insulation is good at insulating electricity and heat.

Considering the cable assembly the energy released by each item's contribution is I^2 * (R1 + R2 + R3 + R4 + R5) using the nomeclature above. However, R1 and R5 dissipate this energy over a large area whereas R2 and R3 dissipate their contribution over a small area. You can use an analogy to electrical conduction for removing the heat from the conductor and connectors where Qheat = I^2R, R is the thermal resistance of a section of insulation and Temp = Voltage.

I would assume the width of insulation carrying heat away from a bad connection to be fixed in size. Because the thermal resistance of the jacket over the wire is considerably smaller than that over the connections you should be able to get a solution fairly easily.

You can guess the width of insulation over a bad connection by perhaps looking at failed or burned connections.

TOP
CSWP, BSSE

http://www.engtran.com

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[GregLocock]

Are you seriously telling me the entire HV industry operates without being able to measure voltage?

"They are equal, but you cant determine power radiated or convected to the environment off surface Temperature alone."

Yes you can if it natural convection and black body radiation.

Cheers

Greg Locock

I rarely exceed 1.79 x 10^12 furlongs per fortnight

 

[kellnerp]

Greg,

Sorry if I assumed you were thinking of an OTS Radio Shack DVM. Yes, it involves special equipment. Just removing the insulation at 13kv can change things like allow for corona to form.

Allyl is thinking in terms of transient heat transfer. I would think this is steady state so it is a matter of conduction, not absorbtion of energy.

TOP
CSWP, BSSE

http://www.engtran.com

http://www.niswug.org

"Node news is good news."