Combined power factor

[mull982]

At my site I have a utility transmission line feeding two seperate utility transformers. The utility transmission line is tapped to feed the primary of both transformers. Both of these transformers individually feed two seperate plants with about a 14MVA load each.

On the secondary of each of these transformers we have power meters installed. Each of these power meters shows the power factor for each plant connected to each utility transformer.

My question is, what would the overall power factor be for these two plants looking from the incoming transmission line to the two transformers? In other words if each plant has a different power factor, how are these combined to represent an overall combined power factor transmitted onto the utility transmission service?

If I had one good power factor at say .90 and a poor one a say .50, what would the overall power factor be as seen by the utility?

 

[jghrist]

You would need to know the kVA and kW of each plant individually at the same time. The overall power factor would be the sum of kW divided by the sum of kVA. The overall power factor will change with time and will be different at the time of the first plant's peak, at the second plant's peak, and at the total peak.

 

[gobblerhuntr]

On our system (utility) we have several customer configured as you and we do not do combined metering, therefore the pf at each meter is what we bill by. 85% pf is the breaking point, so in you above example the .9 would be ok and you would get a reactive penalty for the .5. We do have some customers that are primary metered before the two transformers and the pf of this meter would represent the combined total of each bank. This also lets the customer pay for line and transformer losses. Currently it appears you would not be paying for these losses.

 

[Cold Fusion]

(cut) what would the overall power factor be for these two plants looking from the incoming transmission line to the two transformers? (cut)

With the information you have given, here is what I would figure:
For the .9pf, mva = 14/.9 = 15.6
For the .5pf, mva = 14/.5 = 28

mva(total) = 43.6
mw(total) = 28

pf (for combined load) = 28/43.6 = .64

certainly not dead accurate - but perhaps close enough for what you are looking for.

cf
(not to be confused with "pf" )