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.

 

[itsmoked]

Hello Exhibitionist;

What I was getting at was if this place is already running and someone has figured out "we have a heat problem". Rather than going in circles over all these calculations just run some simple tests.

Mass flow in one door and temperature of mass.
Mass flow out another door and exit temperature.
Gives the actual, real heat load.

Measure the actual current (if you care) and then you can easily estimate if current is X the rooms heat load will Y.

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[Skogsgurra]

Hey, Exhibitionist. Make up your mind now! You were saying "what are the actual losses through the insulation" in your OP. Did you mean to ask what the thermal impedance of the insulation is? If that was your intent, it doesn't matter since all the heat produced in the cables will eventually need to be transferred out of the cable.

You cannot "keep the heat inside" using the insulation. The heat must get out - and it gets out through the insulation.

Your question has been answered over and over in the postings above but I answer again: Calculate I^2R losses and use that value for the cooling needed. No way around that.

Gunnar Englund

http://www.gke.org

 

[sugarshot]

Hi All,

Having overlooked the actual current (skim reader),
I re-visit the question;

Say there is some nominal length for each wire, let's call it 5 meters. If the voltage drop across 5 meters was
1/2 volt, each wire would be a '250 watt' radiator, loosely speaking. If such a stout wire drops so much voltage, I would have some difficulty imagining what kind of power delivery system could sustain that situation, unless the power transformer was next to the room.. as isn't often the case.

What comes to mind is;

A) I have some doubts that the wires get hot enough to need more than some air venting.. but I didn't attempt making estimates.

B) At such power levels, I think the right thing to do is to get someone who has done this before. If I was handed a problem which, if off by 25% the building might burn down - or more likely the expensive AC system would be poorly sized, I would request that an engineer with experience and necessary rubber stamps be paid to take care of it. If I were project manaager, I wouldn't want someone working on the problem if they had to ask questions - because in this instance mistakes sound expensive. Your questions are perfectly valid, of course, but this is my perspective.

Feel free to ask how to do the thermal analysis.

Geoff

 

[fflinders]

The majority of the loss will be carried by heat conduction through the insulation to the room or along the cables to the switchgear and to the room depending on how much dissipation there is in the switchgear itself. Generally there would be little if any power conducted away from the room along the copper conductor(unless it is connected to something very cold). These cables will dissipate more power and run hotter due to there 110oC rating (if selected for there full rating). If V-75 (75oC)cable was used the size would need to be 5x300mm2 or larger depending on other derating factors. The larger cable has lower resistance and hence less heat loss. The other problem with higher temperature rated cables is that at there rating they tend to also contribute significantly to the heating of the switchgear if it is not adequatly ventelated.

 

[VxA]

Even if you have a 100 feet of each conductor (15 carrying around 500A each at 110 degrees C) inside the room, I still can't get my I^2*R calculations to yield more than approx. 15kW of heating from the cables. Just out of curiousity, what kW value are you thinking you have based on your calculations from the I^2*R components?