Class material on three phase system.pptx

Lecture 1
EEE1211:Electrical Circuit II Section: B
4/10/2024
Dr. Md. Abdur Rahman
Part 1 Even Semester Exam. 2021, Department of EEE, R.U. 1

Course outline
The course contains three modules to cover your course curriculum.
Module 1: 1-5 lectures, each of 45 minutes
• Introduction to power system structure
• Basic principle of polyphaser power generation
• Circuit analysis for YY, YΔ, ΔΔ connection with balanced load
• Circuit analysis for unbalanced 3 phase system
• Power calculation for 3 phase system
• Problem solving
• Applications
• Class test/quizzes/assignment
4/10/2024
Part 4 Batch 2, Even Semester Exam. 2019, Department of EEE,
R.U.
2

Module 2: 6-9 lecturs
• Analysis of two port network
• Determination of Z parameters
• Y parameters
• h parameters
• Problem solving for each parameter type
• Applications
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Module 3: 10-16 lectures
• Intro to electric filters and its classifications
• Understanding attenuation constant, cut off frequency, pass band and
stop band.
• Symmetric network and characteristic impedance
• Design principle of passive T section low pass filter
• Design principle of π section filter
• To understand M-derived filter
• Problem solving
• Applications
• Class test/quizzes/assignment
4/10/2024
R.U.
4

Reference sources
1. Alexander C. K. and Sadiku M. N. O, “Fundamental of Electric
Circuits”
2. Online resources if necessary.
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• Introduction to power system
• Introduction to poly phase circuits
• Why 3 phase circuits
• Balanced Y source and Δ source
• Principle of 3 phase power generation
• Balanced 3 phase load Y and Δ
• Line voltage and phase voltages for YY connection
• Line current and phase currents for YY connection
4/10/2024
Part 1, even and part 2, Batch 2, odd Exam. 2020,
Department of EEE, R.U.
6
Module 1:

Power system structure
4/10/2024 Part 1 Even Semester Exam. 2021, Department of EEE, R.U. 7

Poly phase circuits
Circuits or systems where more one source operates at same frequency
but maintains different phases then the circuits are said to be poly
phase circuits.
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Principle of 3 phase power generation
4/10/2024
Part 1 even and p 2 odd Batch 2, Even Exam. 2029,
9

Y and Δ source
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Y source
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_
_
_
+
+
+
n
a
b
c
c
b
a
V0
V-120
V-240
Wye Connected
Source

a
b
c
a
b
c
+
+
+
_
_
_
Delta
Source
Delta Source
Vab = | Vab |  0
Vbc = Vab  -120
Vca = Vab  -240

YY system
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Zl
Zl
Zl
ZL
ZL ZL
a
n
b
c
A
B C
N
(Alexander C.K and Sadiku M. N. O, chapter 12, pp509

Part 4 Batch 2, Even Semester Exam. 2019,
15
In
Ic
Ia
Ib
In
𝑣𝑎𝑏 = 𝑣𝑎𝑛 − 𝑣𝑛𝑏 = 𝑣𝑎𝑛 + 𝑣𝑏𝑛
𝑣𝑎𝑏 = 𝑣𝑝∠0 − 𝑣𝑝∠ − 120
𝑣𝑎𝑏=𝑣𝑝(cos0+jsin0-cos120+jsin120)
𝑣𝑎𝑏 = 3𝑣𝑝∠30

(Alexander C.K and Sadiku M. N. O, chapter 12, pp509
4/10/2024
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_
_
_
+
+
+
n
a
b
c
c
b
a
V0
V-120
V-240
A
B C
Z Z
Z
IaA ICA
IBC
IAB









Z
V
Z
V
V
I
V
Z
I
V
AB
bn
an
AB
bn
AB
an
,
0
CA
p
ca
BC
p
bc
AB
p
ab
V
V
V
V
V
V
V
V
V














210
|
|
3
90
|
|
3
30
|
|
3 ,

Power in 3 phase balanced system
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Zl
Zl
Zl
a
b
c
A
B C
ZL
Z
L
Z
L
+
+
+
_
_
_
Delta – Delta System

Zl
Zl
Zl
a
b
c
A
B C
+
+
+
_
_
_
ZL
ZL
ZL
Delta – Wye System

_
_
_
+
+
+
n
a
b
c
c
b
a
V0
V-120
V-240
A
B C
Z Z
Z
IaA ICA
IBC
IAB

Van
Vbn
Vcn
- Vbn
Vab
Vab = Van - Vbn
Vab = 3 Van 30o
30o

 

























30
|
|
3
2
3
2
1
|
|
|
|
)
120
sin
120
(cos
1
|
|
120
|
|
0
|
|
p
p
p
p
p
p
bn
an
ab
V
j
V
V
j
V
V
V
V
V
V








210
|
|
3
90
|
|
3
p
ca
p
bc
V
V
V
V
Voltage
Line

 |
|
3 p
L V
V

IAB
IBC
ICA
- ICA
IaA
IaA = 3 IAB -30o

Instantaneous power in 3 phase system

Comparisons of power loss between single and 3 phase systems
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Show that 33% more loss in single phase system

Complex power, real power and apparent power
We have average power 𝑃 =
1
2
𝑉
𝑚𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖

Average power, apparent power and power factor
• 𝑃 =
1
2
𝑉
𝑚𝐼𝑚 cos 𝜃𝑣 − 𝜃𝑖 which is average power in ac circuit. This can also be
written 𝑃 = 𝑉
𝑟𝑚𝑠𝐼𝑟𝑚𝑠 cos 𝜃𝑣 − 𝜃𝑖 watt, where S=𝑉
𝑟𝑚𝑠𝐼𝑟𝑚𝑠 is known as
apparent power i.e. VA.
• The average power in the circuit is the useful power absorbed by the load.
• Power facto pf is the dimensionless parameter which is the ratio between the
average power and the apparent power, therefore
𝑝𝑓 =
𝑃
𝑆
= cos 𝜃𝑣 − 𝜃𝑖
• The pf is the cosine of the phase difference between voltage and current in a
circuit.
• If current leads the voltage then pf is leading otherwise lagging.
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From previous lecture
𝑝 = 3𝑣𝑝𝑖𝑝cos(θ𝑣 − θ𝑖)
FOR Y load
P ?
For Delta load p ?
For single phase ?

3 phase power continued
Therefore, per phase
By analogy to single phase we can write
Alternatively
𝑆 = 𝑃 + 𝑗𝑄
Where
Average/ real power 𝑃 = √3𝑉𝐿𝐼𝐿 cos 𝜃𝑣 − 𝜃𝑖
Reactive power Q or VAR = √3𝑉𝐿𝐼𝐿sin 𝜃𝑣 − 𝜃𝑖
Apparent power =√3𝑉𝐿𝐼𝐿

Unbalanced line (home exercise)

Find line currents, phase currents and power
triangle

Power triangle [1]
4/10/2024 Part 1 even part 2 odd , 2020, Department of EEE, R.U. 36
S
P
Q
i
v 

 

Lagging
S
P
Q
i
v 

 

Leading

PF improvement inductive load

PF improvement (continued)

LO and home work
• LO 1?
• LO 2?
• LO3?
• An inductive load operating with pf=0.75, input rms voltage 220volt,
average power P= 0.25kwatt. Draw power triangle and determine
required circuit component for improving pf to 0.95.
[1] Reference textbook: Alexandar and Sadiku, “ Fundamental of Electric Circuits”, 5ed. Mc Graw Hill.

• Intro to two port network
• Why two port analysis
• Learning impedance method for two port analysis
Objectives

3/or 2 wattmeter method for balanced 3phase system[1]
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2 Wattmeter method for balanced YY connection
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Considering Y load and abc sequence
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)
1
.......(
2
1 P
P
PT 

)
30
cos(
)
30
cos(
1 


 
 L
L
a
ab I
V
I
V
P
)
30
cos(
)
30
cos(
2 


 
 L
L
c
cb I
V
I
V
P
Putting the values of P1 and P2 in Eq. (1) and after manipulation it reduces to

 cos
3
cos
3 p
p
L
L
T I
V
I
V
P 


Similarly for P1-P2
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)
(
3
sin
3 1
2 P
P
I
V
Q L
L
T 

 

sin
1
2 L
LI
V
P
P 


Class material on three phase system.pptx

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