Lesson 36
AC Three Phase
Power
Learning Objectives

Compute the real, reactive and apparent power in
three phase systems

Calculate currents and voltages in more
challenging three phase circuit arrangements.
Apply the principles of Power Factor Correction to a
three phase load.

Review
AC Power Summary
Real Power
P = VI (W)
P = I2R =V2/R
P = 0 (W)
P = 0 (W)
Reactive
Power
Q = 0 (VAR)
Q = I2XL =V2/XL
= I2XC =V2/XC
Resistance
Reactance
R
XL = L
XC = 1/C
Review
Power Triangle

The power triangle graphically shows the
relationship between real (P), reactive (Q) and
apparent power (S).
P

S
QL

QC
S
P
P  VI cos  S cos
Q  VI sin   S sin 
(W)
(VAR)
Active Power to Wye (Y) Load
Y-load
Single phase of Y-load
Z  Z   R  X  j phase impedance
2
V
P  V I cos =I 2 R = R
R
phase power
Active Power (P) to Wye (Y) Load

Because we are considering a balanced system, the
power per phase (P) is identical and the total active
power (PT) is simply PT = 3 P.
PT  Pan  Pbn  Pcn  3P

Using line voltage (VL 
3V ) and line current (IL=I):
 VL 
PT  3P  3V I cos  3 
 I L cos
 3
 3VL I L cos
(W)
Example Problem 1a
EAN = 277-30 V . Compute PΦ, PT.
Reactive Power (Q) to Wye (Y) Load

The reactive power per phase (Q) is given
Q  V I sin 
(VAR)
2
V
 I 2 X   X
X
S

P
(VAR)
Q = V I sin 
Reactive Power (Q) to Wye (Y) Load

Because we are considering a balanced system, the
power per phase (Q) is identical and the total reactive
power (QT) is simply QT = 3 Q.
QT  Qan  Qbn  Qcn  3Q

Using line voltage (VL ) and line current (IL):
QT  3VL I L sin 
(VAR)
Example Problem 1b
EAN = 277-30 V . Compute QΦ, QT.
Apparent Power (S) to Wye (Y) Load

The apparent power per phase (S) is given
S  V I
(VA)
 I  2 Z 
V 2
Z
ST  3VL I L
(VA)
(VA)
S = V I
Q

P
Power Factor (FP)

The power factor (FP) is given
PT P
FP 

 cos
ST S
S
Q

P
Example Problem 1c
EAN = 277-30 V . Compute SΦ, ST, and FP.
Power to a Delta () Load
 -load
PT  Pab  Pbc  Pca  3P
Single phase of -load
Z  Z  phase impedance
P  V I cos
phase power
Active Power (P) to Delta () Load

Total active power (PT) is simply PT = 3 P.
PT  Pab  Pbc  Pca  3P

Using line voltage (VL=V) and line current (I L  3I ):
 IL 
PT  3P  3V I cos  3VL 
 cos
 3
 3VL I L cos

(W)
Which was the EXACT same equation as for Y loads
Reactive and apparent power to Delta (Δ) Load

The equations for calculating total reactive and apparent
power are also identical to the Wye load versions:
QT  3VL I L sin 
ST  3VL I L
(VA)
(VAR)
Example Problem 2a
EAN=120-30 V.
Determine per phase and total power (active, reactive, and
apparent).
Determine total powers (active, reactive, and apparent) by
multiplying the per-phase powers by 3.
Example Problem 2b
EAN=120-30 V.
Determine total powers (active, reactive, and apparent) by
using these formulas: S  3V I
T
L L
PT  ST cos
QT  ST sin 
Power in Advanced 3 phase




You must pay attention to the problem statement!
Does it ask for total or per-phase power?
What kind of power? S, P, or Q?
Where is the power?
Pline=?
 Generator
Qline =?
 Line Impedances
 Load
Sgen =?
Pgen =?
Qgen =?
Sload =?
Pload =?
Qload =?
Review
Power Factor

Power factor (FP) tells us what portion of the
apparent power (S) is actually real power (P).
FP = P / S = cos 

Power factor angle
 = cos-1(P / S)=cos-1(FP)



For a pure resistance,  = 0º
For a pure inductance,  = 90º
For a pure capacitance,  = -90º
S
Q
NOTE:  is the phase angle of ZT, not the
current or voltage.

P
Review
Power Factor Correction

In order to cancel the reactive component of
power, we must add reactance of the opposite
type. This is called power factor correction.
Three Phase Power Correction

Capacitors will be connected in parallel with
each load phase
Power Factor Correction Solution Steps
1.
2.
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.
3.
4.
Calculate the reactance (X) that will give this value of Q
Normally the Q=V2/X formula will work
Calculate the component value (F or H) required to provide
that reactance.
Example Problem 3
EAB=4800 V. Frequency 60 Hz.
Determine value of capacitor which must be
placed across each phase of the motor to correct
to a unity power factor.