Download The complex conjugate of

EE301 – POWER FACTOR CORRECTION
Learning Objectives
a. Perform AC power calculations using the complex form of Apparent Power
b. Define power factor
c. Define unity, leading and lagging power factors
d. Define unity power factor and unity power factor correction
e. Calculate the inductor or capacitor value required to correct AC series parallel networks to the
desired apparent power
f. Compare currents, voltages, and power in AC series parallel networks before and after power factor
correction
Complex Numbers in Power Calculations AC power can be calculated using complex equations. The
resultant can be used to determine real and reactive power by changing it to rectangular form.
I*is complex conjugate of I

 *
 V 2
2
S  P  jQ  E IT  I Z   *
Z
NOTE! The complex conjugate
of current is used to make the
power angle the same as the
impedance angle!
Power Factor The power factor (FP) tells us what portion of the apparent power (S) is actually real
power (P). The power factor is a ratio given by
FP = P / S
Power factor is expressed as a number between 0 to 1.0 (or as a percent from 0% to 100%)
From the power triangle it can be seen that
FP = P / S = cos 
The power factor angle is thus given as
= cos-1(P / S)
For a pure resistance,  = 0.
For a pure inductance,  = 90
For a pure capacitance,  = -90
NOTE:  is the phase angle of the total impedance, Z T , not the current or voltage.
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EE301 – POWER FACTOR CORRECTION
Unity Power Factor (FP = 1)
A unity power factor implies that all of a load’s apparent power is
real power (S = P). Now, if FP = 1, then  = 0º.
It could also be said that the load looks purely resistive. The load current and voltage are in phase.
Lagging Power Factor ( > 0º)
The load current lags the load voltage, which implies that the load looks inductive.
Leading power factor ( < 0º)
The load current leads the load voltage which implies that the load looks capacitive.
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EE301 – POWER FACTOR CORRECTION
Example: For the circuit shown:
a. Determine PT and QT by determining individual
component real and reactive powers and summing them.
Use PT and QT to find Apparent Power for the circuit.
b. Determine the complex form of Apparent Power by
∙ . Change the complex S
using the equation
into rectangular form to determine real and reactive total
powers.
Solution:
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EE301 – POWER FACTOR CORRECTION
Why is Power Factor Important? Consider the following
example: A generator is rated at 600 V and supplies one of
two possible loads.
Load 1: P = 120 kW, FP = 1
Load 2: P = 120 kW, FP = 0.6
How much current (I) is the generator required to supply in
each case?
Consider Load 1. Since FP = 1, it is the case that S = P.
Thus I = 120,000/600 = 200 A.
Now consider Load 2. In this case FP = 0.6, so S = P/0.6 =
200,000.
Thus I = 200,000/600 = 333 A.
What does this mean? Answer: The source feeding Load 2 has
to provide 133 A more current for the same real power P. The
larger current means larger equipment (wires, transformers,
generators) which cost more. Larger current also means larger
transmission losses (think I2R).
Because of the wide variation in possible current requirements due
to power factor, most large electrical equipment is rated using
apparent power (S) in volt-amperes (VA) instead of real power
(P) in watts (W).
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.
In practice, almost all loads (commercial, industrial and
residential) look inductive (due to motors, fluorescent lamp
ballasts, etc.).
Hence, almost all power factor correction consists of adding
capacitance.
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EE301 – POWER FACTOR CORRECTION
Solution steps:
1. Calculate the reactive power (QLD) of the load
2. Insert a component in parallel of the load that will cancel out that reactive power. For example, if
the load has QΦ=512 VAR, insert a capacitor with QΦ=-512 VAR.
3. Calculate the reactance (X) that will give this value of Q. Normally the Q=V2/X formula will work.
4. Calculate the component value (F or H) required to provide that reactance.
Transmission lines and generators must be sized to handle the larger current requirements of an
unbalanced load.
Industrial customers are frequently fined by the utility if their power factor deviates from the prescribed
value established by the utility.
Example: A generator, shown below, is rated for 600 V, 120 KVA.
a. Determine the power factor of the load.
b. Determine the size (in VAR) of capacitive load (QC)
required to correct the power factor to unity.
Solution:
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EE301 – POWER FACTOR CORRECTION
Example: For the circuit shown below:
a. Determine the value of the capacitance (in Farads)
required to bring the power factor up to unity (at 60 Hz).
b. Determine generator current before and after correction.
Solution:
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EE301 – POWER FACTOR CORRECTION
Example: For the system shown below:
a. Determine the value of S, PT, QT
and FP. Draw the power triangle.
b. Determine the value of the
capacitance (in F) required to
correct the power factor to unity
(frequency of 60 Hz).
c. Determine generator current before
and after correction
Solution:
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