POWER TRIANGLE

A Power triangle, or using this triangle, is a standard practice of representing the relationship between S, P, and Q – the complex power, real power, and reactive power, respectively. This is similar to the impedance triangle that shows the relationship between Z, R, and X. The power triangle has four items: the apparent/complex power, real/activepower, reactive power, and the power factor angle. Given two of these four items, you’ll be able to easily solve for the other two with the triangle, by simply applying the basic trigonometric identities or the well known mnemonic SOH-CAH-TOA.

This is what a (a) power triangle looks like and the (b) impedance triangle as mentioned:

Comparing a power triangle with a lagging and leading power factors:

When S lies in the first quadrant, it means that the circuit has an inductive load and the power factor or pf is lagging. If S lies in the fourth quadrant, the load is capacitive and the pf is leading. It is also possible for the complex power S to lie in the second or third quadrant, though it requires the circuit with a load impedance to have a negative resistance. In summary, if the pf is lagging, the angle is positive, and so is the reactive power. Otherwise, the angle and the reactive power are in the negative if the pf is leading.

Power factor or pf is the cosine of the difference between the phase angles of the voltage and current. It is also the cosine of the angle of the load impedance. The power factor may also be regarded as the ratio of the real power dissipated in the load to the apparent power of the load.

pf = P/S = cos(θv – θi)

Now with the power P. The real power P is the average power, measured in watts W, delivered to a load, and totally depends on the load’s resistance R. According to a reference, it is the only useful power and the actual power dissipated by the load. Average power P can be obtained by this formula:

P = Vrms · Irms cos(θv – θi)

The root-mean-square or rms value, is the effective value of a periodic signal. The effective value of a periodic current is the DC current that delivers the same average power to a resistor as the periodic current. The idea of effective value arises from the need to measure the effectiveness of a voltage or current source in delivering power to a resistive load. Therefore:

Vrms = Vm/√(2)

Irms = Im/√(2)

The reactive power Q is a measure of the energy exchange between the source and the reactive part of the load. The unit of Q is the volt-ampere-reactive VAR to distinguish it from the real power, whose unit is wattW. Q depends on the load’s reactance X and is called the reactive power.

Q = Vrms · Irms sin(θv – θi)

And finally, the apparent power S and the complex power S. S for apparent power and S for complex power. The only difference is that the symbol for complex power is a bolded S. The apparent power S is the product of the rms values of the voltage and the current. It is so called because it seems apparent that the power should be the product of Vrms and Irms, by analogy with DC resistive circuits. It is measured in voltage-ampere VA.

S = Vrms · Irms ∠(θv – θi)

S² = P² + Q²

While the complex power S is, also measured in voltage-ampere VA, the product of the rms voltage phasor and the complex conjugate of the rms current phasor. S contains all power information of a load as the real part of S is the real or active power P, its imaginary part is the reactive power Q, its magnitude is the apparent power S, and the cosine of its phase angle is the power factor pf.

S = P + jQ

we could say that apparent power S is the polar form of complex power S, and that complex power S is the rectangular form of apparent power S.

example:

Assuming we are to find the remaining items of a power triangle with a load of 16 kW @ 0.8 pf lagging.

Judging from the unit of the given load, we can already say that it is a real power P. All we have to solve for now is the power factor angle θ, the reactive power Q, and the complex power S. We have to calculate for its phase angle then from the given power factor.

pf = cosθ

θ = cos¯¹ (pf)

θ = cos¯¹ (0.8)

θ = 36.87º

Now that we have the phase angle, we can get the reactive power Q by using the trigonometric identity TOA. Since, in a power triangle, the real power P is in the adjacent line while the reactive power Q is in the opposite line.

tanθ = O/A

O = A tanθ

Q = 16 kW tan(36.87º)

Q = 12 kVAR

And lastly, is the complex power S. As stated above, real power P is the real part of S, while reactive power Q is the imaginary part. So, all you have to do is to substitute them to the formula:

S = P + jQ

S = 16 + j12 kVA

To get the apparent power S, you just have to apply the Pythagorean Theorem