Summation of kVA and kVAR with Different Power Factors

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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?
 
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  • #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
 

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  • #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.
 
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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.
 
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  • #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!!
 

1. What is the difference between kVA and kVAR?

Both kVA (kilovolt-ampere) and kVAR (kilovolt-ampere reactive) are units used to measure electrical power. The main difference between them is that kVA measures the total power, including both real (kW) and reactive (kVAR) power, while kVAR measures only the reactive power. In other words, kVA represents the total power supplied by the source, while kVAR represents the power that is not converted into useful work.

2. How do power factor and power triangle relate to kVA and kVAR?

Power factor is a measure of how efficiently electrical power is being used. It is the ratio of real power (kW) to apparent power (kVA). The power triangle is a visual representation of this relationship, where the real power (kW) is the horizontal side, the reactive power (kVAR) is the vertical side, and the apparent power (kVA) is the hypotenuse. In other words, the power triangle shows how much of the total power (kVA) is being used for useful work (kW) and how much is being lost as reactive power (kVAR).

3. How does power factor affect the summation of kVA and kVAR?

The power factor affects the summation of kVA and kVAR because it determines the ratio of real power (kW) to apparent power (kVA). When the power factor is 1 (or 100%), the real power (kW) and apparent power (kVA) are equal, and there is no reactive power (kVAR). However, as the power factor decreases, the amount of reactive power (kVAR) increases, and the total power (kVA) increases. This means that the summation of kVA and kVAR will be greater for a lower power factor compared to a higher power factor.

4. How do you calculate the total power (kVA) when given the real power (kW) and power factor?

To calculate the total power (kVA), you can use the formula kVA = kW / power factor. For example, if the real power is 100 kW and the power factor is 0.8, the total power would be 100 kW / 0.8 = 125 kVA. This means that the reactive power (kVAR) would be 25 kVAR, as 125 kVA is the hypotenuse of a power triangle with 100 kW as the horizontal side.

5. How does a low power factor affect an electrical system?

A low power factor can have several negative effects on an electrical system. First, it results in higher currents, which can overload the system and cause damage to equipment. Second, it increases the amount of reactive power, which is not used for useful work and can lead to higher energy costs. Third, it reduces the efficiency of the system, as more power is needed to supply the same amount of real power. To avoid these issues, it is important to maintain a high power factor through proper power management and the use of power factor correction devices.

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