Watts Losses from insulated cable

[Exhibitionist]

Can someone please offer advice on how to calculate watts losses from insulated power cables. It is easy to calculate the watts loss of the actual conductor based on current and resistance (I squared R), but when the cable is insulated, what are the actual losses through the insulation.

We are trying to calculate the amount of cooling required in a large switchroom and using the straight I2R calculation the amount of cooling require seems too high, therefore expensive.

The type of cable we are using is a flexible fire rated power cable (110 deg rating). Each circuit is aroung 2500A and we are using 5 x 240mm sq cable per phase & neutral (415V 3ph 50hz system).

We have sought help from the cable manufacturer but they had never been asked the question before and could not give us the answer. The engineer there wanted to put a length of cable in an oven to see how hot it got (basically had no idea).

Look forward to input on this issue.

 

[waross]

The greatest part of your losses will be the I^2R that you have already calculated. Your insulation resistance losses will be minute. Total insulation resistance losses based on a very conservative figure of resistance to ground is less than 100 Watts.
The capacitive effect of the insulation will add some heat but probably even less.
yours

 

[jraef]

It seems to me that you are implying (or hoping?) that having insulation will reduce the thermal loses in your cable? That answer would be no. Thermoplastic insulation will slightly resist the conduction of the I[sup]2[/sup]R heat losses through the cable, but that only equates to a delay in raising the surrounding air temperature since ultimately the cable is not ventilated elsewhere. Now if you put the cable in an enclosed duct and had airflow through the duct that was not common to your switch room, then you have something.

But how much actual cable is in the switch room that you are so worried about it? Don't forget that when calculating your I[sup]2[/sup]R losses for heat gain in the room, you must base it only on the length of cable actually in that room, not the entire cable length. Unless of course you are using enclosed duct and the switch room is the uppermost terminus!

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[electricpete]

EPRI 5036 volume 4 offers a pretty comprehensive guide to heating effects to be considered in cable sizing:

- skin effect
- proximity effect (change in current distribution within one cable due to another cable close by)
- shield eddy curents
- shield circulating currents (when shorted at both ends)
- losses in rigid steel conduits

You can see this includes some fairly minor effects.. but no mention of insulation losses. Safe to say IMHO that it is negligible as mentioned by waross.

You could begin to get an order of magnitude by assuming 1 Megaohm insulation resistance in which case P = sqrt(3)*V^2/R = sqrt(3)*480^2/1E6 < 0.5 watts. This is not an exact calculation since "absorption" losses are not included (associated with realigning of dipoles). Once again I still agree with waross it is negligible.


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[electricpete]

actually with 1 megaohm system insulation resistance, you would have 3 megaohms to ground on each phase individually (parallel combination gives 1). So P= sqrt(3)*480^2/3E6

Not that the calculation is that exact... just wanted to correct my error.

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[Exhibitionist]

Thanks for the advice every one.

 

[sitl]

One way of lowering the heat dissipation is maybe to use insulated busbars instead of cables. In that way you should be able to transfer more power with significantly less copper (no reduction coefficients for cables). Since copper is today very expensive, by ordering even brand name busbars (like Schneider Electric Canalis), it should not be more expensive than cables.

 

[dpc]

Just in case someone tries to over-generalize: Dielectric losses are somewhat more significant at medium voltage, especially for very long cable runs.

EPR losses are quite a bit higher than XLPE - one of the few characteristics where EPR comes in second.

 

[electricpete]

I guess I’m under the impression they are still fairly insignificant (compared to the other losses mentioned above) for medium voltage cables.

Can you quantify it at all?


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[sugarshot]

Initially your inquiry is one of thermal analysis. Heat in a wire has two options.. radiate out through it's insulation surface, or crawl along the length of the metal towards a cooler wire.
To simplify matters, we think of the thermal losses of the power lines as a continuous thing. To compute the heat contribution to the room, you need to ascertain the temperature of the wire within the conductor, and subtract the ambient temperature from the wire core temp. Using R value of such and such wire insulation, and
come up with an R-value for thin air films on tubular surfaces (0.65 or higher in still air environments).
The temperature differential and r-values with some constants are used in simple calculations to ascertain the heat contribution in MBTU (thousands of BTU) which can be
interpreted as added baseline cooling load. Just measuring the current and voltage through a power line carrying AC currents to a highly inductive load.
The primary thing I wonder is why is this a concern.
It seems that if a wire gets warm enough to make a significant contribution, the gauge is too small.

There is an vast abundance of information available on the internet which can guide you to some kind of analysis. The method I roughly outlined is e.a.s.y. (some acronym?) and will produce a reasonable result. For a test, if you wrap 30 feet of wire in 6 inches of fiberglass batt, if you take the surface temperature at the midpoint, it will be fairly close to the temperature of the wire. A simply wattage calculation for the wires would also work pretty well, providing a good estimate, and it's e.a.s.i.e.r. Don't use 11.? MBTU/kwh (you'll find this number in the field), use 3.4-something (something very close to that), in figuring the heat contributed. Think of the wire as a resistor in a box. If the wire is said to dump energy across it's length (heat dissipation), then you can be sure that that energy amount is the amount contributed to the room. If the wire had anything interferring with the heat flow out of it, it would get hot.. same as if it's ability to self heat exceeded it's ability to radiate.

As an aside, I find it interesting that different metals have different heat holding capacities, and since it's not dependant only on mass and density, one could think molecular structure was itself a storage medium... adding something to the 'vibrating spheres' model that first comes to mind.

Some physical properties of materials are searchable at matweb.com

Geoff

 

[waross]

Hi sugarshot;
You do understand that there are 20 cables in question, each close to 3/4" in diameter, and of these, 15 are carrying 500 amps each? You noticed that the cable is rated to run safely at 110C, hotter than boiling water?
yours

 

[electricpete]

"It is easy to calculate the watts loss of the actual conductor based on current and resistance (I squared R), but when the cable is insulated, what are the actual losses through the insulation. "

I guess we all spent the first part of this thread assuming you meant losses generated in the insulation. Now seeing sugarshot's response I see there is a different possible/likely interpretation.

Are you asking how much heat is generated within the insulation itself or are yo askig how much of I^2*R heat generated in the copper leaves the cable within the room (vs how much is conducted along copper path to outside of room).

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[Skogsgurra]

In the latter case, put all cables in a large diameter tube and blow air though it so that the heat is concected out of the tube to somewhere you can dissipate it. Open air, for instance. Make sure you have a second cooling unit in case the main unit fails. And have an air flow or temperature alarm. Also, test the second unit once a month.

Still cannot understand why the cooling of the cable losses is such a big thing. It usually is the equipment in an electric room that produce the most heat.

Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

concected -> convected

Gunnar Englund

http://www.gke.org

 

[waross]

Original post;

We are trying to calculate the amount of cooling required in a large switchroom

Hi skogsgurra;
That would work. It is an ingenious adaptation of the method used to cool control cabinets that have a substantial heat source such as a constant voltage transformer.
Pros; Much cheaper than air conditioning.
Cons; Dirt accumulation, even with filters.
Reduced cooling when the filters get dirty.
The filters are often discarded instead of being cleaned or replaced and the dirt gets worse.
What are your thoughts re; Relative Humidity? Should RH be monitored and/or alarmed? Can the heat of the cables be depended upon to prevent condensation when the ambient RH is high?
respectfully

 

[Skogsgurra]

Bill,

The problem with filters is always there. As soon as you got filters, you got maintenance. I see no reason to have filters in such an application. Dust amasses anywhere, especially on cables in cable trays. Blowing air along the cables may or may not increase or even decrease dust collection. RH problems should be out of the question since the cables are a lot warmer than surrounding air, otherwise you wouldn't need to cool them.

The biggest problem, and one that I haven't got answered, is why the cables are supposed to be the worst heat source. Transformers, VFDs and other apparatus usually give off more heat than the cables.


Gunnar Englund

http://www.gke.org

 

[Skogsgurra]

OOPS! Jraef already mentioned this solution in his 24 May 06 14:53 posting. So, now we are at least three that think that it is a good solution

But why? Why?

Gunnar Englund

http://www.gke.org

 

[waross]

Hi Gunnar and all;
I understand that the original poster is calculating the heating load in the switch room for the purpose of sizing the air conditioning unit for the electrical room. I assume he is a professional who has already determined the heat rejection of the switch gear by some means and was surprised by the amount of heat rejected by the cables. As a professional who is not sure of something he has asked other professionals for advice.
The answer that has been given to the main question is that the main heating component is the I^2R losses, and other losses are in addition to this, but are small enough to be neglected.
The possible exception to this is skin effect but if the OP has used voltage drop tables to find the effective resistance, this will have been factored in.
From there we are into redesigning the system, which we are prone to do on this forum.
I think the options are as follows;
1> Bite the bullet and instal the calculated amount of cooling.
2> Reduce the heat rejection of the cables by adding more parallel cables. This will also reduce the cost of ownership.
3> Enclose the cables in a duct and ventilate. Be aware that if the inspection department considers your duct a raceway, your cable ampacity may fall from the free air rating to the conductor in raceway rating.
In some installations this may be a good solution. In other installations it may be ridiculous. "It depends".
4> Ventilate the entire room instead of mechanical cooling.
5> A combination, use an HVAC type air handling mixer with fresh air make-up and use ventilation for the entire room on all but the hottest days.
Respectfully

 

[itsmoked]

Exhibitionist; Does this facility exist or is it just in development?

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Exhibitionist]

itssmoked, this is a real facility / real project.

Electricpete, to anwser your question, "Are you asking how much heat is generated within the insulation itself or are yo askig how much of I^2*R heat generated in the copper leaves the cable within the room (vs how much is conducted along copper path to outside of room)". I was never concerned with heat generated within the insulation itself, only the I^2*R losses.