EE 205 Dr. Chokri. Belhaj Ahmed
Electric Circuits II
Balanced Three Phase Circuits
Lecture # 03
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EE 205 Dr. Chokri. Belhaj Ahmed
Objective of this lecture is as follows:
™ Transformation from Delta to Why.
™ Transformation from Why to Delta.
™ The single-phase equivalent circuit.
™ The line and phase current quantities for a Delta circuit.
™ Complete Analysis of Y-∆ circuit.
After finishing this lecture, you should be able to:
™ Transform the three-phase circuit Y to ∆ and vice versa.
™ Extract the single-phase equivalent circuit for Y-∆ three-phase circuit.
™ Deduce the line and phase current quantities for ∆ connection.
™ Identify the difference between delta and why currents and voltage relations.
™ Conduct a complete analysis for Y-∆ circuit.
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EE 205 Dr. Chokri. Belhaj Ahmed
Transformation from Delta to Why
• Starting from a Delta connection to reach Wye Connection
Figure 13 Delta to Wye Transformation circuits
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EE 205 Dr. Chokri. Belhaj Ahmed
Transformation from Delta to Why (Cont)
Z AB .Z AC
ZA =
Z AB + Z BC + Z AC
Z AB.ZBC
ZB =
Z AB + ZBC + Z AC
ZC =
™ In case where
™ Then
™ So we can deduce that
Z A C .Z B C
Z AB + Z BC + Z AC
ZAB =ZAC =ZBC =Z
2 1
Z
ZA =
= Z
3Z 3
ZY =
1
3
Z∆
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EE 205 Dr. Chokri. Belhaj Ahmed
Transformation from Why to Delta
• Starting from a Why connection to reach Delta Connection
Figure 14 Wye to Delta to Transformation circuits
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EE 205 Dr. Chokri. Belhaj Ahmed
Transformation from Why to Delta (Cont)
Z Z +Z Z +Z Z
ZAB = A B B C A C
ZC
Z AZB + ZB ZC + Z AZC
ZBC =
ZA
Z Z +Z Z +Z Z
Z AC = A B B C A C
ZB
™ In case where
™ Then
™ So we can deduce that
ZA = ZB = ZC = Z
2
3
Z
Z AB =
= 3Z
Z
2
Z AB = 3Z = 3ZY
Z
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EE 205 Dr. Chokri. Belhaj Ahmed
The single-phase equivalent circuit
™ The source is Y connected and the load is Delta connected and balanced. So
Z AB = Z BC = Z CA = Z ∆
™ the conversion of or transformation of ∆ to Y transformation gives
Z
Z ∆
=
Y
3
™ After replacing the data connected load by the equivalent Y connected load, the circuit can be
analyzed by the equivalent single phase equivalent circuit .
™ After replacing the delta connected load by the equivalent Y connected load, the circuit can be
analyzed by the equivalent single phase equivalent circuit.
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EE 205 Dr. Chokri. Belhaj Ahmed
The single-phase equivalent circuit (cont)
Figure 15 Delta to Wye Load Transformation Single Phase Equivalent Circuit
™ This circuit is used to calculate phase voltages, phase currents, line voltages and line current.
™ Generators phase and line voltages, in each of the original delta load.
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EE 205 Dr. Chokri. Belhaj Ahmed
The line and phase current quantities for a Delta circuit
™ When positive phase sequence is assumed and
Iφ
is taken for the phase reference current.
Figure 16 Delta balanced load connections
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EE 205 Dr. Chokri. Belhaj Ahmed
The line and phase current quantities for a Delta circuit (cont)
™ Then
I AB = Iφ ∠00 A
I BC = Iφ∠−1200 A
I CA = Iφ∠+1200 A
™
By KCL at node A we get
I aA = I AB − I CA = Iφ∠00 − Iφ∠+1200 = 3Iφ∠−300 = 3∠−300I AB
™
By KCL at node B we get
IcB =I BC −I AB = 3∠−300I BC = 3Iφ∠−1500
™
By KCL at node C we get
I cC = I CA − I BC = 3∠− 300 I CA = 3Iφ ∠900 A
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EE 205 Dr. Chokri. Belhaj Ahmed
The line and phase current quantities for a Delta circuit (cont)
Figure 17 Phasor Diagram for Delta balanced currents and voltages
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EE 205 Dr. Chokri. Belhaj Ahmed
Complete Analysis of Y-∆ circuit.
™
The analysis is conducted through the following application example.
Example:
™ Y connected source load feeds ∆ connected load through a distribution line having impedance of
0.3+ j 0.9Ω/φ,Z∆ =118.5+ j 85.8Ω/φ
™ The internal per phase impedance is
a)
Zy =
et
Van' =120∠0V /φ
is taken as reference.
0.2 + j0.5Ω /φ .
Construct the single-phase equivalent circuit.
Z∆
= (118.5 + j85.8) /3 = 39.5 + j 28.6 =146.3∠35.9/3
3
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EE 205 Dr. Chokri. Belhaj Ahmed
Figure 18 Single Phase Equivalent Converted Circuit .
b)
Calculate the line currents
I aA =
I aA , I b B , I c C
.
120∠0V
120∠0V
=
(0.2+0.3+39.5) + j(0.5+ 0.9+ 28.6) 40+ j30A
I aA = 2.4∠−36.870 A, I bB = 2.4∠−156.870 A, I cC = 2.4∠−83.130 A
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EE 205 Dr. Chokri. Belhaj Ahmed
Example(cont)
c)
V AN
Calculate the phase voltages at the load terminals.
= (39.5 + J 28.6)(2.4∠ − 36.87) = 117.04∠ − 0.960V , V
= 117.04∠−120.960V
BN
V CN = 117.04 = ∠−119.04 A = 117.04 = ∠− 240.960V
V AB = 3∠300V AN = 202.72∠29.040V
V BC = 202.72∠− 90.960V ,V CA = 202.72∠149.040V
d)
Calculate the phase current of the load.
Figure 19 Three Phase Delta Connected Load.
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EE 205 Dr. Chokri. Belhaj Ahmed
Example(cont)
I BC
I aA = 1 ∠30 I AB , A = 1.39∠30∠ − 36.87 = 1.39∠ − 6.87 A
3
0
= 1.39∠− 6.87 −120 = 1.39∠−126.870 A
I CA = 1.39∠113.130 A
In an other way
I AB = V AB / Zφ∆ = 202.72∠29.04 = 1.39∠− 6.870 A
118.5 + J 85.8
e)
Calculate the line voltage at the source terminals.
Van = (Zl + Z y )I aA = (39.8 + j 29.5)(2.4∠− 36.870 )
Van = 118.90∠− 0.320V ,Vbn = 118.9∠− 0.32 −120
V an = 118.90∠−120.32 −120 = V cn = 118.9∠− 240.320V
V ab = 3∠30V an = 205.94∠29.680V
V bc = 205.94∠− 90.320V ,V ca = 205.94∠149.680V
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EE 205 Dr. Chokri. Belhaj Ahmed
Self test:
To analyse a Y-∆ connected balanced circuit we have to ?
a) Include the Delta phase load without any changes in the single-phase equivalent circuit.
b) Include the transformed Delta to Y phase load in the single-phase equivalent circuit.
c) Include three times the Delta phase load value in the single phase equivalent circuit.
Answer is b)
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EE 205 Dr. Chokri. Belhaj Ahmed
Self test:
Which options are true for a set of balanced delta variables?
a) Each phase voltage leads its corresponding line voltage by 30o.
b) Each phase current leads its corresponding line current by 30o.
c) Each phase current is higher in magnitude than its corresponding line current.
d) For the single-phase equivalent circuit, the line load voltage has to be used between A and N.
f)
Each line voltage magnitude is equal each phase voltage magnitude.
Answer b) and f)
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