# Summation of kVA and kVAR with Different Power Factors

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.

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.

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• PowerFactorBasics.pdf
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CalcNerd, nsaspook and russ_watters
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?

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.

Many thanks for that. I was under the impression that the kVA summation was not a simple arithmetic addition but I was wrong clearly.

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.

CalcNerd
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.

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

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.

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;

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!!

## What is the difference between kVA and kVAR?

kVA (kilovolt-amperes) is a unit of apparent power, which represents the total power in an AC circuit, both usable (real power) and unusable (reactive power). kVAR (kilovolt-amperes reactive) is a unit of reactive power, which represents the power that oscillates between the source and the load, not doing any useful work but necessary to maintain the voltage levels in the system.

## How do you sum kVA and kVAR when they have different power factors?

To sum kVA and kVAR with different power factors, you need to convert them to a common base, typically by calculating the real power (kW) and then using vector addition. The relationship between kVA, kW, and kVAR is given by the power triangle, where kVA is the hypotenuse, kW is the adjacent side, and kVAR is the opposite side. You can use Pythagoras' theorem to find the resultant kVA from the summed kW and kVAR.

## Why is it important to consider power factor when summing kVA and kVAR?

Power factor is important because it affects the efficiency of the power system. A low power factor indicates a higher proportion of reactive power (kVAR) relative to real power (kW), leading to higher losses and reduced system efficiency. When summing kVA and kVAR, considering the power factor ensures accurate representation of the total power and helps in proper system design and analysis.

## Can you provide an example of summing kVA and kVAR with different power factors?

Suppose you have two loads: Load 1 with 100 kVA at 0.8 power factor (80 kW and 60 kVAR) and Load 2 with 50 kVA at 0.6 power factor (30 kW and 40 kVAR). First, sum the real power (kW): 80 kW + 30 kW = 110 kW. Then, sum the reactive power (kVAR): 60 kVAR + 40 kVAR = 100 kVAR. Finally, use the power triangle to find the total kVA: sqrt(110^2 + 100^2) = 148.66 kVA.

## How does varying power factor affect the total kVA required in a system?

Varying power factor affects the total kVA required because a lower power factor means more reactive power (kVAR) is present, increasing the total apparent power (kVA) needed. This requires larger capacity equipment and results in higher energy losses. Improving the power factor reduces the kVA demand, leading to more efficient system operation

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