Lecture Notes
ELE A6
Ramadan El-Shatshat
Three Phase circuits
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ELE A6 Three-phase Circuits
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Three-phase Circuits
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ELE A6 Three-phase Circuits
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Advantages of Three-phase
Circuits
• Smooth flow of power (instantaneous power
is constant).
• Constant torque (reduced vibrations).
• The power delivery capacity tripled
(increased by 200%) by increasing the
number of conductors from 2 to 3
(increased by 50%).
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ELE A6 Three-phase Circuits
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Three-phase Circuits
Wye-Connected System
•
The neutral point is grounded
•
The three-phase voltages have
equal magnitude.
•
The phase-shift between the
voltages is 120 degrees.
Van = V ∠0 ° = V
Vbn = V ∠ − 120 °
Vcn = V ∠ − 240 °
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ELE A6 Three-phase Circuits
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Three-phase Circuits
Wye-Connected System
•
Line-to-line voltages are the
difference of the phase voltages
Vab = Van - Vbn = 3 V ∠30
Vbc = Vbn - Vcn = 3 V ∠ - 90
Vca = Vcn - Van = 3 V∠150
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Three-phase Circuits
Wye-Connected System
•
•
•
•
•
Phasor diagram is used to visualize
the system voltages
Wye system has two type of
voltages: Line-to-neutral, and
line-to-line.
The line-to-neutral voltages are
shifted with 120o
The line-to-line voltage leads the
line to neutral voltage with 30o
The line-to-line voltage is 3 times
the line-to-neutral voltage
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Three-phase Circuits
Wye-Connected Loaded System
• The load is a balanced load and each one = Z
• Each phase voltage drives current through the
load.
• The phase current expressions are:
Vcn
Van
Vbn
Ia =
, Ib =
, Ic =
z
z
z
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Three-phase Circuits
Wye-Connected Loaded System
• Since the load is balanced (Za = Zb =
Zc) then: Neutral current = 0
Van
Za
a
Ia
• This case single phase equivalent
circuit can be used (phase a, for
instance, only)
• Phase b and c are eliminated
n
Io
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Three-phase Circuits
Wye-Connected System with balanced load
•
A single-phase equivalent circuit is used
•
Only phase a is drawn, because the magnitude of currents and voltages
are the same in each phase. Only the phase angles are different (-120o
phase shift)
•
The supply voltage is the line to neutral voltage.
•
The single phase loads are connected to neutral or ground.
V
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ln
Load
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Three-phase Circuits
Balanced Delta-Connected System
• The system has only one voltage :
the line-to-line voltage ( VLL )
• The system has two currents :
– line current
– phase current
• The phase currents are:
Vab
Ia =
z
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Three-phase Circuits
Delta-Connected System
The line currents are:
Ia= Iab − Ica
Ib= Ibc − Iab
Ic = Ica − Ibc
•
In a balanced case the line
currents are:
I line =
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3I phase ∠ − 30
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Three-phase Circuit
Delta-Connected System
•
The phasor diagram is used to
visualize the system currents
•
The system has two type of
currents: line and phase currents.
•
The delta system has only line-toline voltages, that are shifted by
•
The phase currents lead the line
currents by 30 °
•
The line current is 3 times the
phase current and shifted by 30
degree.
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ELE A6 Three-phase Circuits
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Three-phase Circuit
• Circuit conversions
– A delta circuit can be converted to an equivalent wye
circuit. The equation for phase a is (will not be used for
this course):
Za =
Z ab Z ca
Z ab + Z bc + Z ca
– Conversion equation for a balanced system is:
Za =
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Z ab
3
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Three-phase Circuit
Power Calculation
•
The three phase power is equal the sum of the phase powers
P = Pa + Pb + Pc
•
If the load is balanced:
P = 3 Pphase = 3 Vphase I phase cos (φ )
•
•
Wye system: Vphase = VLN
I phase = I L VLL =
P = 3 Vphase I phase cos (φ ) =
3 VLL I L cos (φ )
Delta system: I Line =
3 I phase VLL = Vphase
P = 3 Vphase I phase cos (φ ) =
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3 VLN
3 VLL I L cos (φ )
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Three-phase Circuit
Power measurement
• In a four-wire system (3 phases and a neutral)
the real power is measured using three singlephase watt-meters.
• In a three-wire system (three phases without
neutral) the power is measured using only two
single-phase watt-meters.
- The watt-meters are supplied by the line
current and the line-to-line voltage.
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Three-phase Circuit
Load
Watt meter 1
Power measurement
a
b
• The total power is
the algebraic sum of
Wattmeter 2
the two watt-meters
reading.
W1 = Vab I a cos(θ vab − θ Ia )
c
W2 = Vcb I c cos(θ vcb − θ Ic )
PT = W1 + W2 = 3VL I L cos(θ )
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Three-phase Circuits
Example 1
A 345 kV, three phase transmission line delivers 500 MVA, 0.866 power
factor lagging, to a three phase load connected to its receiving end
terminals. Assume the load is Y connected and the voltage at the receiving
end is 345 kV, find:
•
The load impedance per phase.
•
The line and phase currents.
•
The total real and reactive power.
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ELE A6 Three-phase Circuits
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(a ) Zφ =
Vφ
Iφ
345kV
S
500 MVA
Vφ =
∠0V , Iφ =
=
= 836.7 A
3
3VL
3 345kV
Iφ = 836.7∠ − cos −1 (0.866) = 836.7∠ − 30
Zφ = 238∠30 = 206 + j119
(b) I L = Iφ = 836.7∠ − 30
(c) P = 3VL I L cos(θ ) = 433 MW
Q = 3VL I L sin(θ ) = 249.9 MVAR
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Three-phase Circuits
Example 2
Repeat example 2 assuming the load is Delta connected.
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(a ) Vφ = VL = 345kV∠0V ,
S
500MVA
Iφ =
=
= 483.1A
3VL 3 * 345kV
Iφ = 483.1∠ − cos −1 (0.866) = 483.1∠ − 30
Zφ =
Vφ
Iφ
= 714∠30 = 618.3 + j 357
(b) I L = 3Iφ ∠ − 30 = 836.7∠ − 60
(c) P = 3VL I L cos(θ ) = 433 MW
Q = 3VL I L sin(θ ) = 249.9 MVAR
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ELE A6 Three-phase Circuits
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Power factor correction
• Power factor (p.f.) correction is the process of
making P.f. = 1.
• In order to correct the power factor in any system,
a reactive (either inductive or capacitive) will be
added to the load.
• If the load is inductive, then a capacitance is
added.
– Note: Correcting the P.f. WILL NOT affect the active
power, why?
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Three-phase Circuits
Example 3
A 3-phase load draws 120 kW at a power factor of 0.85 lagging from a
440 V bus. In parallel with this load, a three phase capacitor bank that
is rated 50 kVAR is inserted, find:
•
The line current without the capacitor bank.
•
The line current with the capacitor bank.
•
The P.F. without the capacitor bank.
•
The P.F. with the capacitor bank.
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3V L I L cos( θ )
(a ) P =
IL =
120 * 10 3
= 185 . 25 A
3 440 ( 0 . 85 )
I L = 185 . 25 ∠ − cos
( b ) Q capacitor =
−1
( 0 . 85 ) = 185 . 2 ∠ − 31 . 8 ,
3V L I L sin( θ )
Since we have pure capacitor
sin( θ ) = 1
Q capacitor =
I L ( capacitor
I L ( capacitor
3V L I L ( capacitor
)
50 * 10 3
=
= 65 . 6 A
)
3 440
= 65 . 6 ∠ 90
)
I L ( new ) = I L + I L ( capacitor
( c ) P .F. (no capacitor)
= 160 . 6 ∠ − 11 . 49
)
= 0.85
( d ) P .F. (with capacitor)
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load
= c os (11 . 49 ) = 0 . 98
ELE A6 Three-phase Circuits
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Home Work
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Home Work
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