Power factor three phase supply - single phase loads

[ruggedscot]

Im curious as in how to give information relating to this sort of complexity.....

Right simple explaination of system. Various single phase loads connected to a 415v three phase supply. supply being four wire as neutral is included.

So we can have say a total of 15A on phase one 34A on phase two and 24A on phase three. What would the power being consumed from the supply be ? Im looking at the actual watts figure as this is what we use to base the cooling requirements on. Id rough a guess that it would simply be VxA for each phase to give the power dissipated. Power factor would need to be factored and this would have to be done as per single phase wouldn't it? or would the powerfactor given for all three phases suffice.

so it would be (Ia + Ib + Ic) x 240 x pf /1000 = kW figure
for watts dissipated by loads.....

Rugged

 

[waross]

The utility measures KVA and kWatts.
Your KVA will be (Ia+Ib+Ic)x240/1000
This formula also works for 415 volt loads connected phase to phase, as 240V/415V = root three, so the root three factor for 415 loads is built in.
Your kWatts will be KVA x Power factor.
yours

 

[tinfoil]

Bear in mind that the different currents will create slightly different voltages due to I*R voltage drops, and that you also have to allow for voltage drops in your neutral when the vector sum is nonzero, which always occurs when the currents are not balanced. This means you must use something like Power = Ia*Va + Ib*Vb + IcVc which is not quite the same as (Ia+Ib+IC)*V,

For your numbers, this will not have much effect, but I mention it in case someone extrapolates from this thread to a situation where is DOES have a major impact.

 

[waross]

Good point tinfoil
I agree with you.
respectfully

 

[ruggedscot]

I know where you are coming from with the equations now - its just that we have been looking at a few ideas and skirting round a few ways at producing a loading display for a set area of process equipment. You know power consumed equals heat load to be dissipated. best way being I think to add all the powers being consumed and give this as a true power figure to allow the power in that area to be realised. By demarkation of the floor space area ratios Im sure that a w per meter square can be achieved.

rugged

 

[waross]

Your method will work for most equipment but not for motors. If you have lighting transformers you can use the %imp. figure on the nameplate times the transformer KVA rating to estimate the losses. The error may be 2:1 or more depending on the R/X ratio of the transformer but if you are estimating heating for airconditioning the error will put you on the safe side.
If this is as I suspect for an air conditioning load check, I would ignore power factor in the calculations. The error will again be on the safe side.
yours