Download Dr. Sahar Abd El Moneim Moussa

Electrical Power & Machines
“Electrical Engineering Dept”
Prepared By:
Dr. Sahar Abd El Moneim Moussa
Dr. Sahar Abd El Moneim Moussa
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Balanced Three-Phase System
Balanced three-phase voltage consists of three sinusoidal
voltage having the same amplitude & frequency but are out
of phase with each other exactly by 120o
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3 Phase Voltages in Time Domain
 Va = Vm Sin ωt
 Vb = Vm Sin (ωt-120)
 Vc = Vm Sin (ωt-240)
Phase
(a)
Phase
b
Dr. Sahar Abd El Moneim Moussa
Phase
(c)
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3-Phase Voltages in Terms of Phasors
 Va = Vm ∠0
 Vb = Vm ∠-120
 Vc = Vm ∠-240 = Vm ∠120
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Types of Connections in 3-phase system
Wye”Y”
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Delta”∆”
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Wye Connection “Y”
 Wye Connection: “Y”
For Y circuit:
Iline = Iphase
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Delta Connection “∆”
 For Delta Circuit:
Eline = Ephase
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Relationship between three-phase delta-connected and
wye connected impedance
Wye connected load
Delta connected
load
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Four Different Configurations for the three-phase source
and loads Connections
Source
Load








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Power in 3-φ System
• P(total) = 3𝑉𝑙𝑖𝑛𝑒 𝐼𝑙𝑖𝑛𝑒 𝐶𝑜𝑠 𝜃
• Q(total) = 3𝑉𝑙𝑖𝑛𝑒 𝐼𝑙𝑖𝑛𝑒 𝑆𝑖𝑛 𝜃
 S(total) = 3𝑉𝑙𝑖𝑛𝑒  𝐼𝑙𝑖𝑛𝑒   − 𝜃
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Example 1:
A balanced three-phase Y-connected generator with positive
sequence has an impedance of 0.2 +j0.5 / and internal voltage
120V/ feeds a -connected load through a distribution line
having an impedance of 0.3 +j0.9 /. The load impedance is
118.5+ j85.8 /. Use the a phase internal voltage of the generator
as a reference.
A. Construct the single-phase equivalent circuit of the 3- system.
B. Calculate the line currents IaA , IbB and IcC.
C. Calculate the phase voltages at the load terminals.
D. Calculate the phase currents of the load.
E. Calculate the line voltages at the source terminals.
F. Calculate the complex power delivered to the -connected
load.
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Solution:
A. The load impedance of the Y equivalent is
118.5+𝑗85.8
3
= 39.5 + 𝑗28.6 /
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B. The a-phase line current is
𝐼𝑎𝐴
1200
=
0.2 + 0.3 + 39.5 + 𝑗(0.5 + 0.9 + 28.6)
=
120 < 0
40+𝑗 30
= 2.4
− 36.87 A.
Therefore,
IbB=2.4-156.87 A.
IcC= 2.483.13 A.
C. because the load is - connected, the phase voltages are the
same as the line voltages. To calculate the line voltages,
VA=(39.5 + j28.6)(2.4-36.87) = 117.04-0.96
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The line voltage VAB is
𝑉𝐴𝐵 = 3 𝑉𝐴  + 30
= 202.72 29.04V
Therefore,
VBC=202.72 -90.96 V
VCA= 202.72 149.04 V
D. The phase currents of the load will be,
𝐼𝐴𝐵 =
1
𝐼𝑎𝐴  + 30
3
= 1.39 -6.87 A.
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Therefore,
IBC=1.39-126.87 A
ICA=1.39113.13 A
E. The line voltage at the source terminals will be,
Va=(39.8 + j29.5) (2.4-36.87)
=118.9 -0.32 V.
The line voltage will be
𝑉𝑎𝑏 = 3 𝑉𝑎  + 30
= 205.9429.68 V.
Therefore ,
Vbc=205.94 -90.32 V.
Vca= 205.94149.68 V.
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F. The total complex power delivered to the load will be,
V=VAB= 202.72 29.04 V.
I=iAB=1.39-6.87 A.
Therefore,
ST= 3 (202.72 29.04) (1.396.87)
= 682.56 +j 494.21 VA
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