Summation of kVA and kVAR with Different Power Factors

  • Thread starter traderza
  • Start date
  • #1
traderza
5
0
TL;DR Summary
I need the formula for each of these summations please
I am measuring the incoming power from 2 transformers using smart meters.

The client wants to know the total kVA and total kVAR of his facility

I know that the total kW is a simple arithmetic sum but I don't know how to summate kVA and kVAR because of the different power factors.

Can anyone help please?
 
Engineering news on Phys.org
  • #2
Factor (P.F.) is the ratio of Working Power to Apparent Power.
KVAR is reactive power. The power of your transformer uses to produce magnetizing flux.
KVA is apparent power and that is Vectorial Summation of KVAR and KW.
KW is working power. The power that performs useful work.
I like the beer and analogy.
photo_1693218226698.png
 

Attachments

  • PowerFactorBasics.pdf
    109.3 KB · Views: 49
  • Like
Likes CalcNerd, nsaspook and russ_watters
  • #3
Delta Prime said:
Factor (P.F.) is the ratio of Working Power to Apparent Power.
KVAR is reactive power. The power of your transformer uses to produce magnetizing flux.
KVA is apparent power and that is Vectorial Summation of KVAR and KW.
KW is working power. The power that performs useful work.
I like the beer and analogy.
View attachment 331125
Thanks for the reply and the thirsty analogy! My question is; how do I summate the two kVA readings to provide a composite total kVA for the facility in question?
 
  • #4
traderza said:
I know that the total kW is a simple arithmetic sum but I don't know how to summate kVA and kVAR because of the different power factors.
kVA is the product of average voltage and in-phase current.
kVAR is the product of average voltage and quadrature current.

If supply 1 and 2 are at the same voltage ...
kW1 + kW2 = kW total.
kVA1 + kVA2 = kVA total.
kVAR1 + kVAR2 = kVAR total.
But that assumes the phase of the current through the two meters is the same.
 
  • #5
Many thanks for that. I was under the impression that the kVA summation was not a simple arithmetic addition but I was wrong clearly.
 
  • #6
traderza said:
Many thanks for that. I was under the impression that the kVA summation was not a simple arithmetic addition but I was wrong clearly.
You were not wrong. To add the VA or VAR you must assume that:
1. The voltages are the same, (or you must compute and add the currents).
2. The phase of the current through the two meters is identical.

Imagine two independent meters, both reading 1 kVA. If meter #2, had 180 degree phase current with reference to meter #1, the currents would cancel to zero kVA.
 
  • Like
Likes CalcNerd
  • #7
Baluncore said:
You were not wrong. To add the VA or VAR you must assume that:
1. The voltages are the same, (or you must compute and add the currents).
2. The phase of the current through the two meters is identical.

Imagine two independent meters, both reading 1 kVA. If meter #2, had 180 degree phase current with reference to meter #1, the currents would cancel to zero kVA.
Thanks for the reply.

The two feeders will be supplying different combinations of inductive and resistive loads but i have no way of knowing the phase difference between the two.

The parameters I can measure for each feeder are:

Volts per phase
Amps per phase
kVA
kVAR
kW
Power Factor

Given these are what I have to work with, how do I calculate the total kVA and kVAR for the facility
 
  • #8
traderza said:
Given these are what I have to work with, how do I calculate the total kVA and kVAR for the facility
You have a right angle triangle. kW2 + kVAR2 = kVA2;
You can check your data for either metered feeder is consistent.

traderza said:
The two feeders will be supplying different combinations of inductive and resistive loads but i have no way of knowing the phase difference between the two.
That precludes capacitive loads, so it tells you kVAR is positive.

You can independently sum kW and kVAR since they are orthogonal.
kWt = kW1 + kW2;
kVARt = kVAR1 + kVAR2;
From that compute; kVAt = √( kWt2 + kVARt2 );
P.F. = cos(θ) = kWt / kVAt;
 
  • #9
Baluncore said:
You have a right angle triangle. kW2 + kVAR2 = kVA2;
You can check your data for either metered feeder is consistent.That precludes capacitive loads, so it tells you kVAR is positive.

You can independently sum kW and kVAR since they are orthogonal.
kWt = kW1 + kW2;
kVARt = kVAR1 + kVAR2;
From that compute; kVAt = √( kWt2 + kVARt2 );
P.F. = cos(θ) = kWt / kVAt;
That's what I'm looking for! Many thanks!!
 

Similar threads

Replies
2
Views
989
  • Electrical Engineering
Replies
17
Views
5K
Replies
35
Views
7K
  • Electrical Engineering
Replies
5
Views
2K
Replies
4
Views
2K
Replies
8
Views
995
  • Electrical Engineering
Replies
22
Views
8K
Replies
1
Views
1K
  • Engineering and Comp Sci Homework Help
Replies
7
Views
2K
  • Electrical Engineering
Replies
22
Views
2K
Back
Top