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cos cos sin sin
P VI S Q VI S
(W) (VAR)
2 P V I cos I^2 R VR (^) R = = phase power
PT Pan Pbn Pcn 3 P
VL 3 V
2 2 Q V I sin I X VX Q X (VAR) =
QT 3 V I (^) L L sin (VAR)
Learning Objectives a. Compute the real, reactive and apparent power in three phase systems b. Calculate currents and voltages in more challenging three phase circuit arrangements c. Apply the principles of Power Factor Correction to a three phase load
Recall that the power triangle graphically shows the relationship between real ( P ), reactive ( Q ) and apparent power ( S ).
We will first examine three-phase power in the context of a wye-load; then we’ll examine a delta load.
Power to the Wye-Load Active (Real) Power. Suppose that each phase has impedance.
Then the active (real) power per phase ( P ) is given
Because we are considering a balanced system, the power per phase ( P ) is identical in all three phases, and thus the total active power ( PT ) is simply PT = 3 P .
Using line voltage ( ) and line current ( I (^) L =I ), we have
Reactive Power The reactive power per phase ( Q ) is given by
The total reactive power can be calculated similar to the total active power:
Apparent Power
The apparent power per phase ( S ) is given
The power factor ( FP ) is given T cos P T
P P
F
S S
3 3 cos 3 cos 3 cos
L T L L L
V
P P V I (^) I V I
(W )
2 2
T^3 L L
V S V I I Z Z S V I
(VA)
(VA)
Z Z (^) R X (^) j
PT Pab Pbc Pca 3 P
I L 3 I
3 3 cos 3 cos 3 cos
L T L L L
I
P P V I V V I
(W)
Power to the Delta ( ) Load
Active (Real) Power.
Total active power ( PT ) is simply PT = 3 P
Using line voltage (VL=I ) and line current ( ):
Which was the EXACT same equation as for Y loads
Reactive and Apparent Power The equations for calculating total reactive and apparent power are also identical to the Wye load versions:
The applicable portion of the equation sheet:
QT 3 V I (^) L L sin (VAR)
ST 3 V IL L (VA)
ST 3 V I L L PT ST cos QT ST sin
Example: In the circuit shown E AN = 120-30 V
a. Determine per phase powers (active, reactive, and apparent) b. Determine total powers (active, reactive, and apparent) by multiplying the per-phase powers by 3 c. Determine total powers (active, reactive, and apparent) by using these formulas:
Solution:
Example: In the circuit shown, E AB = 208 0 V a. Determine the line currents b. Determine total real power delivered by the generator c. Total real power dissipated by the load d. Determine the load phase voltage Van
Solution:
Power Factor Correction Recall: In order to cancel the reactive component of power, we must add reactance of the opposite type. This is called power factor correction.
In a three phase circuit, capacitors are connected in parallel with each load phase (presuming the actual load is inductive, which is usually the case)
Solution steps:
- Calculate the reactive power (Q) of ONE PHASE of the load
- Insert a component in parallel of the load that will cancel out that reactive power e.g. If the load has QΦ=512 VAR, insert a capacitor with QΦ= - 512 VAR
- Calculate the reactance (X) that will give this value of Q Normally the Q=V^2 /X formula will work
- Calculate the component value (F or H) required to provide that reactance
Example: In the system shown we have E AB = 480 0 V. The frequency is 60 Hz.
Determine value of capacitor which must be placed across each phase of the motor to correct to a unity power factor.
Solution:
