How to Calculate Connected Loads

[freakydeakie]

I am currently training as an electrician and have been set a question and was hoping someone please explain to me how to calculate the connected load if I have a mixture of single phase and three phase loads, and I have a three phase service connection from the utility.

For instance, is the following correct:
1. Single phase load (between line-neutral) 100% on each phase, trying to distribute as uniformly as possible between the 3 phases.
2. 3-phase load (irrespective if delta or Y) 33.33% of each on each phase.
3. (1-phase load (2-pole-load) line-line): 50% of each on the 2 phases involved.
3. Add up loads for each phase
Example:
5 Loads (Line-Neutral) (kVA): 1,5,4,2
3 Loads (3-phase)(kVA): 3,6,2
2 Loads (Line-Line) (kVA): 3,4
An example of load distribution between phases would be:
Phase A: 1 + 2 + 3/3 + 6/3 + 2/3 + 4/2 = 8.66 kVA
Phase B: 5 + 3/3 + 6/3 + 2/3 + 3/2 = 10.16 kVA
Phase C: 4 + 3/3 + 6/3 + 2/3 + 3/2 + 4/2= 11.16 kVA
The total connected load (for the entire 3-phase system) is 29.98 kVA. The connected load for each phase is as indicated above.
Dividing the total load by 3 (number of phases) (29.98/3 = 9.99) you get the average load.
1) If someone asks me what are the supply requirements that they would have to provide for the system, would it be 3 phase 400v + E at 29.98 KVA or 9.99KVA??
2) What size would the protective device for the main incoming supply be??
Thanks in advance

 

[malcolmsanford]

We do need a bit more information as a large part of your calculation will involve power factor correction so find the KW rating or consumption of the 3-phase circuit if the power factor of the device is 0,8 as identified on the nameplate say of a motor

Use the formula, (KW = KVA * power factor) to find the result. KW is equal to say 6.9 KVA times 0.80, = 5.52KW

Also rather than break down your calc into 9.99KVA per phase you can keep it simply as a 3 phase calc as shown

9.99* 0.8 (PF) = 7.992KW which equates to 7992/230 = 34.74

29.98 * 0.8(PF) = 23.984KW which equates to 23984/ 400*1.732 = 34.61

So you looking at perhaps 40amp per phase supply.

So you will need to know this type of information, before you can size your protection device. Quite often though protection devices will be sized solely on KVA.

Your balancing looks quite good, though you need to pay attention of the 2 phase supplies, these can have dramatic effect on balancing and effect a neutral on that circuit, I assume these maybe for spot welders.

 

[freakydeakie]

Hi Malcolm and thanks very much for the quick reply. The equipment on the 2 phase are transformers, 400vac primary, 120vac secondary. I would assume a PF of 0.8 is a good guess
for these without manufactures data.
As for the other equipment, these are mainly VFD drives. They provide a input rating in KVA and Amps, but no PF or consumption in watts. I have been told never to use
amps in load calculation, therefore I am assuming I convert the KVA to KW with a PF of 1.
Other 3 phase equipment are items such as power supplies, Air Conditioning units which do
have the power consumption in watts. From what you have explained, the best way to calculate the total consumption is to get the manufactures data in KW's, and if this is not
given convert the KVA to KW. If the PF is not given, assume 0.8. Is this correct??

 

[malcolmsanford]

As we know there is never been, as of yet a PF of 1, and the normal rule of thumb is indeed 0.8.

I personally design to PF 1 if I don't have sufficient information to do otherwise. Though with a lot of installations now utilizing PF correction units, you may easily get a PF of 0.9-0.95 especially with VFDs.

Not knowing you site or it's demand you have not even gone down the road of diversity etc etc, but as you said this seems to be a simple question on balancing and finding a protection device to suit, and it maybe that i'm going to deep.

Personally I would show the calc with a PF of 1 for this, unless your told differently

 

[freakydeakie]

Malcolm, thanks very much for your help, its much appreciated.

 

[freakydeakie]

Hi Malcolm. Could you please have a look at the attached PDF and your comments would be much appreciated.
I think I may have finally got this correct.

Regards

Russell

 

[malcolmsanford]

The balancing looks a dream.

Your calculation should be though I = kVA x 1000 ÷ V x √3 so 106.65 x 1000 / 692.8 = 153.94 amps

I gather also that this assignment only asked for loading specs and not wanting diversity taken into it