EE 204
Lecture 29
Three Phase Circuits
Introduction of three phase circuit concept
•
The three phase circuit is the main circuit structure used in power system.
•
The electrical power generation is accomplished through three-phase circuit.
•
The three phase connection lines transmit power over long distance.
•
The energy is distributed and consumed at the load level through three phase and
single phase.
•
The basic structure of a three phase system consists of voltage source,
transformers, transmission line and connected loads.
Balance Three Phase Voltages
•
Three sinusoidal voltages form a set of balanced voltages when they have the
same amplitudes and frequency.
•
These voltages are shifted in phase by 120o with each other.
•
The standard practice is to name those phases by a, b and c and use phase a as
reference.
•
These voltages represent phase a voltage, phase b voltage and phase c voltage.
Balanced Y-connected Load
•
Large blocks of power are generated, transmitted, distributed and used with three
phase circuits.
•
For economic reasons, three phase systems are usually designed to operate in the
balanced state. Figure shows a balanced Y-connected load.
•
A balanced Load is composed of three identical impedances ZL.
•
For a Y-connected load the voltages across the loads are the phase voltages,
regardless of whether neutral is attached.
Balanced ∆-connected Load
•
Another common way of interconnecting the individual three phase loads is the
delta connection.
•
In ∆ -connected load there is no neutral connection.
•
For a ∆ -connected load the voltages across the loads are the line to line voltages.
Example 1:
Consider a balanced wye-connected load where each load impedance is
Zˆ L = 50 + j 50 Ω and the phase voltages are 120 V. Determine the total average
power delivered to the load.
Solution:
The line currents are
o
ˆI = 120∠0 = 1.7∠ − 45o A
a
50 + j 50
o
ˆI = 120∠ − 120 = 1.7∠ − 165o A
b
50 + j 50
o
ˆI = 120∠120 = 1.7∠75o A
c
50 + j 50
Hence, the average power delivered to each load is:
(
)(
)
Pav = Re ⎡⎣ 120∠0o 1.7∠45o ⎤⎦
( )
Pav = 120 × 1.7 × cos 45o
= 144 W
The total average power delivered to the load is:
PavTot = 3 × 144 = 432 W
Example 2:
If the line voltage of a balanced, wye-connected load is 208 V and the total average
power delivered to the load is 900 W, determine each load if their power factors are 0.8
leading.
Solution:
From equation:
PavTot = 3 ×
VL
I L cos θ = 3VL I L cos θ
3
IL =
then:
900 = 3VL I L cos θ
900
= 3.12 A
3(208)(0.8)
The phase voltage is:
VL
3
= 120 V
Vp =
Thus the magnitude of the individual load impedance is:
ZL =
Vp
IL
= 38.43 Ω
Since the power factor is 0.8 leading (current leads voltage; voltage lags current),
θ = − cos −1 0.8 = −36.87o
Thus the individual loads are:
Z L = 38.43 ∠ − 36.87o = 30.74 − j 23.06 Ω
or
Example 3:
A balanced ∆ -connected load is 208 V, and the total average power delivered to the load
is 600 W. Determine the individual loads if they have a lagging power factor of 0.7.
Solution:
The average power delivered to each load is 200 W, so that
200 = 3 ×
IL =
IL
VL cos θ
3
or
200 3
= 2.38 A
(208)(0.7)
Thus the magnitude of the individual load impedance is:
ZL =
VL
IL
3
= 151.42 Ω
Since the power factor is 0.7 lagging,
θ = cos −1 0.7 = 45.57o Thus the individual loads are:
Z L = 151.42 ∠45.57 o = 106.74 + j108.14 Ω
Example 4:
Three-phase power is supplied to a balanced, ∆ -connected load. The line-to-line voltage
is 208 V, and the load consumes a total power of 15 kW at a lagging power factor of 0.6.
Determine the transmission-line currents and the individual loads.
Solution:
From equation:
PavTot = 3 ×
IL
VL cos θ = 3VL I L cos θ
3
then: 15000
= 3VL I L cos θ
Therefore the transmission-line current is:
IL =
15000
= 69.39 A
3(208)(0.6)
For delta connection, VP =VL, thus the magnitude of the individual load impedance is:
ZL =
Vp
IL
= 5.19 Ω
Since the power factor is 0.6 lagging,
θ = cos −1 0.6 = 53.13o
Thus the individual loads are:
Z L = 5.19 ∠53.13o = 3.12 + j 4.15 Ω