Reactive power (var) calculation

given the real power P and power factor, how would i calculate the reactive power?

The basic equations are:

Pf = W / VA = cos(a)
W is watts or real power, P

Where (a) is the angle between the VA and the Watts (various symbols are used)

VA^2 = W^2 + VAR^2

From those you can get:
W = VA X cos(a)
VAR = VA X sin(a)
VAR = W X tan(a)

To calculate VAR from W and Pf use:

VAR = ((W/Pf)^2 - W^2)^0.5

or

VAR = W X tan(acos(Pf))

Volume II - AC » POWER FACTOR »
True, Reactive, and Apparent power


We know that reactive loads such as inductors and capacitors dissipate zero power, yet the fact that they drop voltage and draw current gives the deceptive impression that they actually do dissipate power. This “phantom power” is called reactive power, and it is measured in a unit called Volt-Amps-Reactive (VAR), rather than watts. The mathematical symbol for reactive power is (unfortunately) the capital letter Q. The actual amount of power being used, or dissipated, in a circuit is called true power, and it is measured in watts (symbolized by the capital letter P, as always). The combination of reactive power and true power is called apparent power, and it is the product of a circuit's voltage and current, without reference to phase angle. Apparent power is measured in the unit of Volt-Amps (VA) and is symbolized by the capital letter S.

As a rule, true power is a function of a circuit's dissipative elements, usually resistances (R). Reactive power is a function of a circuit's reactance (X). Apparent power is a function of a circuit's total impedance (Z). Since we're dealing with scalar quantities for power calculation, any complex starting quantities such as voltage, current, and impedance must be represented by their polar magnitudes, not by real or imaginary rectangular components. For instance, if I'm calculating true power from current and resistance, I must use the polar magnitude for current, and not merely the “real” or “imaginary” portion of the current. If I'm calculating apparent power from voltage and impedance, both of these formerly complex quantities must be reduced to their polar magnitudes for the scalar arithmetic.