3-phase power and network equations

Spyros Chatzivasileiadis

31730: Electric Power Engineering, fundamentals

3-phase power and Network Equations

12 September 2017

If not otherwise indicated, all figures are taken from the course textbook: J. D. Glover, T. J. Overbye, M. S. Sarma, Power System Analysis and Design, Sixth Edition - SI, Cengage Learning, 2016

DTU Electrical Engineering, Technical University of Denmark

Reviewing previous lecture

• For 5 minutes discuss with the person sitting next to you about:

– Three main points we discussed in the last lecture

– One topic or concept that is not so clear and you would like to hear again about it

2 Spyros Chatzivasileiadis


The goals for today• Line-to-neutral and line-to-line voltage

• Δ-loads and Y-loads

• Single-line diagram

• Power in three-phase balanced systems

• Network equations

• Bus Admittance Matrix



Balanced three-phase systems

• Voltage line-to-neutral

• Voltage line-to-line

• Question:What are the phasors for the line to neutral voltages?(for each phase)



Balanced three-phase systems• Voltage line-to-neutral

• Voltage line-to-line

• Assuming a 𝑉"#$ = 10for each phase:

• Question:What are the phasors for the line-to-line voltages?(for each phase)

S = V · I∗

ZY =Z∆

3

p3φ(t) = P3φ = 3VLNIL cos(δ − β)

Ean = 10 0◦

Ebn = 10 −120◦

Ecn = 10 +120◦

7 DTU Electrical Engineering 31730: Electric Power Engineering, Fundamentals Sep 8, 2017



Line-to-neutral and line-to-line voltages



Delta-Wye Loads and transformation

S = V · I∗

ZY =Z∆

3

p3φ(t) = pa(t) + pb(t) + pc(t) =

3VLNIL cos(δ − β) + VLNIL[cos(2ωt+ δ + β)

cos(2ωt+ δ + β − 240◦)+ cos(2ωt+ δ + β + 240◦)]



• Delta vs Wye

• Delta-Volt. = 3� × Wye-Volt.

• Delta-Curr. = 3� ×Wye-Curr.

If I want to transform:

• Ι- = 3� ./0/= 3� 1� .2

0/⇒45

65= 3 42

6/


Interesting fact: (outside the scope of this course)

Delta-Wye switch for induction motor start

Spyros Chatzivasileiadis8

• Many induction motors use a delta-wye switch when they start• wye-start-delta-run• Reason:

• Induction motors have usually very high currents during the first seconds they start (up to 6x the nominal current)

• Apply line-neutral voltage instead of line-line à lower voltage• In this case, impedance values in the wye-formation are the same as in

the delta-formation (it is just a different connection)• è Reduces the inrush current, i.e. less current when the motor starts!

Video: https://www.youtube.com/watch?v=km8MSWm39Z0


Three-phase to single line diagram



Three-phase to single line diagram• Three main points:

1. Voltages must be RMS values

2. Transform line-to-line voltages to line-to-neutral

3. Transform all Δ-loads to Y-loads



Power balanced in three-phase systems

• The three-phase instantaneous power remains constant!

• Relationship between line-to-line and line-to-neutral differs by √3

• Similar for Q

S = V · I∗

ZY =Z∆

3

p3φ(t) =pa(t) + pb(t) + pc(t) =

3VLNIL cos(δ − β) + VLNIL[cos(2ωt+ δ + β)+

cos(2ωt+ δ + β − 240◦)+

cos(2ωt+ δ + β + 240◦)]


p3φ(t) = P3φ =√3VLLIL cos(δ − β)


S = V · I∗

ZY =Z∆

3

p3φ(t) =pa(t) + pb(t) + pc(t) =


cos(2ωt+ δ + β − 240◦)+

cos(2ωt+ δ + β + 240◦)]




S = V · I∗

ZY =Z∆

3

p3φ(t) =pa(t) + pb(t) + pc(t) =


cos(2ωt+ δ + β − 240◦)+

cos(2ωt+ δ + β + 240◦)]




S = V · I∗

ZY =Z∆

3



p3φ(t) = Q3φ = 3VLNIL sin(δ − β)




Why did three-phase systems prevail?



Three-phase vs. single-phase systems• Benefits

1. Neutral current and voltage is zeroa. need for less conductorsb. Lower costs for transmission

2. Total instantaneous power remains constanta. Single-phase would create shaft

vibration and noise in motors and generators

b. Shaft failures in large generators c. Gens>5 kVA are three-phase



Network equations

How much is 𝐼:̅?

Assuming 𝑉<:, 𝑉<>, 𝑉<1 and 𝑦<:>, 𝑦<:1,𝑦<>1 known

Bus Reactance Matrix

V 1 V 2

V 3

I1 y12 I2

y13 y23

How much is P1?

1 DTU Electrical Engineering 31765: Optimization in modern power systems Sep 12, 2017



Network equationsHow much is 𝐼:̅?


Step 1:

How much is 𝐼:̅>?

I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦



V 1 V 2

V 3

I1 y12I12 I2

I13

y13 y23

How much is P1?




Network equationsHow much is 𝐼:̅?


Step 1:

Step 2:

I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


⟹I12 =

V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦



V 1 V 2

V 3

I1 y12I12 I2

I13

y13 y23

How much is P1?




I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


Network equationsHow much is 𝐼:̅?Assuming 𝑉<:, 𝑉<>, 𝑉<1 and 𝑦<:>, 𝑦<:1,𝑦<>1 known

Step 1:

Step 2:

Step 3:

I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦



V 1 V 2

V 3

I1 y12I12 I2

I13

y13 y23

How much is P1?


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3) ⇒I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13


I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦





Step 1:

I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13

I1 = y12(V 1 − V 2) + y13(V 1 − V 3)

I1 = (y12 + y13)V 1 − y12V 2 − y13V 3

I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦



V 1 V 2

V 3

I1 y12I12 I2

I13

y13 y23

How much is P1?


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13


I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13


I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13


I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦

2 DTU Electrical Engineering 31765: Optimization in modern power systems Sep 12, 201718 Spyros Chatzivasileiadis




V 1 V 2

V 3

I1 y12I12 I2

I13

y13 y23

How much is P1?


I12 =V 1 − V 2

z12I12 = y12(V 1 − V 2)

I1 = I12 + I13


I1 =[y12 + y13 −y12 −y13

]⎡

⎣V 1

V 2

V 3

⎤

⎦


⎡

⎣I1I2I3

⎤

⎦ =

⎡

⎣y12 + y13 −y12 −y13−y12 y12 + y23 −y23−y13 −y12 y13 + y23

⎤

⎦

⎡

⎣V 1

V 2

V 3

⎤

⎦

I = YV



⎡

⎣I1I2I3

⎤

⎦ =

⎡

⎣y12 + y13 −y12 −y13−y12 y12 + y23 −y23−y13 −y23 y13 + y23

⎤

⎦

⎡

⎣V 1

V 2

V 3

⎤

⎦

I = YV



Bus Admittance Matrix 𝒀B

• Diagonal Elements 𝑌<DD–𝑌<DD = ∑ 𝑌<DG�

G

• Off-diagonal elements 𝑌<DH:– Are nodes 𝑖 and 𝑗 connected? Yes: 𝑌<DH = −𝑦<DH– Are nodes 𝑖 and 𝑗 connected? No : 𝑌<DH = 0


Y =

⎡

⎣y12 + y13 −y12 −y13−y12 y12 + y23 −y23−y13 −y23 y13 + y23

⎤

⎦


3-phase power and network equations

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