# Fundamental Concepts of Electrical Energy Systems

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```					Fundamental Concepts                       MTE 320          Spring 2006                           E.F. EL-Saadany

Fundamental Concepts of Electrical Energy Systems

1. Introduction
Electricity is only one of many forms of energy used in industry, homes, businesses and transportation.
It has many desirable features; it is clean, convenient, relatively easy to transfer from point of source to
point of use and highly flexible in its use. In some cases, it is an irreplaceable source of energy.

Power system engineers are more interested in defining power rather than current. In steady state, most
power system quantities (voltage and current) are sinusoidal functions of time and all with the same
frequency. Therefore, it is essential to understand the sinusoidal steady state analysis using phasor,
impedances, admittance and different forms of power.

2. Impedance Concept

Average value
1T
If v(t ) = Vm cos(ω t + θ v ) , then the average value of the voltage is given by: VAVE =      ∫ v(t )dt , For pure
T0
sinusoidal waveform this value is equal to zero.

Effective value

1T
If v(t ) = Vm cos(ω t + θ v ) , then the effective value of the voltage is given by: VRMS =     ∫ v(t ) dt
2
T0

100
If v(t ) = 100 cos(ω t + 30) then Vrms =
2

In the circuit shown in Fig. 1, If the terminal voltage is given as: v(t ) = Vm cos(ω t + θ v )

This can be converted to the phasor format
(frequency domain) as: v = VRMS ∠θ v

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Fundamental Concepts                      MTE 320                 Spring 2006                   E.F. EL-Saadany

I (t)

V (t)                                Element

Fig. 1

If the current through the load is given by: i (t ) = I m cos(ω t + θ I ) And in phasor format is: i = I RMS ∠θ I

Then, the impedance of the element (load) can be defined as the phasor voltage divided by the phasor
current;
VRMS ∠θ v V
Z=              = ∠θ v − θ I = Z ∠θ
I RMS ∠θ I I

Since Z is a phasor, then it can be defined by a magnitude and angle or in rectangular format as real
and imaginary components.
Z = Z∠θ = R + jX

Where: (Resistance) R = Z cos θ and (Reactance) X = Z sin θ

3. Power Concept
Before presenting the power concept, it is essential to remind with the current reference direction. The
passive sign convention is normally used for this purpose.

Passive Sign Convention
If the current is entering the positive terminal of the element, then this current is positive and vice
versa. If the product of the voltage and the current (the power) is positive, then this element is
ABSORBING power and if the product is negative, then the element is DELIVERING power.

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Fundamental Concepts                          MTE 320               Spring 2006                       E.F. EL-Saadany

In order to illustrate the concept of power, we will use the cosine representation of the waveform as a
reference.

Assume any load as shown in the Fig. 2. If the voltage and current in the load are given by:
v(t ) = Vm cos(ω t + θ v )

i (t ) = I m cos(ω t + θ I )

I (t)

+     V (t)       -

Fig. 2

Then the instantaneous power will be given as p(t ) = v(t ) i(t ) , but usually we are interested in the

1T
average power over one complete cycle, where Pave =                     ∫ p(t ) dt , and T is the periodic time of the
T0
sinusoidal waveform.

If we take the current as the reference phasor and shift both the voltage and the current by θ I , then:
v(t ) = Vm cos(ω t + θ v − θ I )

i(t ) = I m cos(ω t )
and
p(t ) = Vm I m cos(ω t + θ v − θ I ) cos(ω t )

Vm I m
p (t ) =          [ cos (θv − θ I ) + cos(2ω + θv −θ I ) ]                                        (*)
2

From this equation, we can realize that the instantaneous power has double the frequency of the
voltage and current.

Now we have three different cases depending on the element type:

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Fundamental Concepts                          MTE 320              Spring 2006                                 E.F. EL-Saadany

•   Resistance: For resistors, the voltage and current are in-phase and θ v − θ I = 0 . Moreover, the

1T
average value of the component cos(2ω + θ v −θ I ) is equal to zero,                     ∫ cos(2ω + θ v −θ I ) dt = 0 .
T0

Vm I m
Therefore, for resistive elements, the average power is equal to                         = VRMS I RMS .
2

•   Inductance: For inductances, θ v − θ I = 90 . Therefore, the average power in this case will be
equal to zero.

•   Capacitance: For capacitances, θ v − θ I = −90 . Therefore, the average power in this case will
be equal to zero.

4. Complex Power Concept
Vm I m
Recall that     p (t ) =          [cos(θ v − θ I ) + cos(2ω + θ v −θ I )]
2
Vm I m
Then            p (t ) =          [cos(θ v − θ I ) + cos(θ v −θ I ) cos(2ω t ) − sin (θ v −θ I )sin (2ω t )]
2
Vm I m                                  V I
p (t ) =          cos(θ v − θ I )[1 + cos(2ω t )] − m m sin(θ v − θ I ) sin 2ω t
2                                       2
p(t ) = P[1 + cos(2ω t )] − Q sin 2ω t
Vm I m
P=            cos(θ v − θ I ) = Real (Active) power (WATT) and
2
Vm I m
Q=            sin(θ v − θ I ) = Reactive power (VAR)
2

If we compare the above two formulas with the expression for power in R, L and C circuits, then:

P = Real Power = Associated with power dissipated in resistive portion of the circuit.
Q = Reactive Power = Associated with the power in the reactive (or L & C) portion of the circuit.

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Fundamental Concepts                             MTE 320                Spring 2006   E.F. EL-Saadany

The term cos(θ v − θ I ) is known as the power factor (PF) and it controls the amount of active power

supplied and the term sin(θ v − θ I ) is known as the reactive factor (rf).

pf = cos(θ v − θ I )

rf = cos(θ v − θ I )

Two important terminologies have to be known. The first is lagging power factor that happens in
inductive circuits where the angle θ v − θ I > 0 and the current lags the voltage. The second is the

leading power factor case that happens in capacitive circuits when the current leads the voltage and
θv −θ I < 0 .

The complex power (or known as the apparent power) is an abstract concept, where:
S = P + jQ (VA)

Vm I m
Recall that:      P=          cos(θ v − θ I )
2
Vm I m
and               Q=           sin(θ v − θ I )
2
Vm I m
then:             S=          [cos(θ v − θ I ) + j sin(θ v − θ I )]
2
Vm I m j (θ v −θ I ) 1
S=         e             = Vm e j (θ v ) I m e − j (θ v )
2                  2
1      *           *
S = V m I m = V rms I rms
2

Important Remarks

2
2           Vrms
•   P = VI cos(θ v − θ I ) = I rms R =          = S × pf
R
2
2           Vrms
•   Q = VI sin(θ v − θ I ) = I rms X =          = S × rf
X
2
2          Vrms     2
•   S = VI = I rms Z =         = VrmsY
Z

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Fundamental Concepts                     MTE 320            Spring 2006                  E.F. EL-Saadany

2
Vrms
•    S=      *
2                     2
= Vrms × Y * = P + jQ = I rms Z
Z
•   Upon computing S, we can obtain P and Q as the real and imaginary parts of S.
•   The units for S is (VA), P is (watt) and Q is (VAR).
•   The practical significance of the apparent power is as a rating unit for generators and
transformers.

Example
The equivalent induction motor load that is connected to a 79.7kV busbar can be represented by
impedance Z = 80 + j 60 Ω . Determine the real and reactive power absorbed by this load. Calculate the

Solution

1     1
Y=       =         = 0.008 − j 0.006 moh
Z 80 + j 60

Y * = 0.008 + j 0.006 moh
2
S = Vrms × Y * = P + jQ

S = (0.008 + j 0.006) × (79.7 kV ) 2 = (50.8 + j 38.1) × 10 6
∴P = 50.8 MW
Q = 38.1 MVAR

P         Q     ⎛        38.1 ⎞
PF = cosθ =       = tan −1 = cos⎜ tan −1      ⎟ = 0 .8
S         P     ⎝        50.8 ⎠

5. Power in Inductive and Capacitive Circuits

5.1 Inductive Circuits
ωL
Consider an inductive load with Z = R + jωL = Z∠φ Ω , where φ = tan −1               . If the voltage is:
R
V = V∠0 , then

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Fundamental Concepts                   MTE 320                 Spring 2006                       E.F. EL-Saadany

V V∠0 V
I=      =    =  ∠ −φ
Z Z ∠φ Z

The current flowing in the inductive load is then said to be lagging the voltage. The apparent power S
absorbed by the impedance is given by:
*
S = V rms I rms = P + jQ

P = VI cos φ
Q = VI sin φ

Q

V
P

I

Fig. 3

5.2 Capacitive Circuits
1                                        1
Consider a capacitive load with Z = R +                 = Z∠ − φ Ω , where φ = tan −1       . If the voltage is:
jω c                                     ωRC
V = V∠0 , then

V   V∠0   V
I=       =     =   ∠φ
Z Z∠ − φ Z

The current flowing in the capacitive load is then said to be leading the voltage. The apparent power S
absorbed by the impedance is given by:

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Fundamental Concepts                    MTE 320         Spring 2006       E.F. EL-Saadany

*
S = V rms I rms = P + jQ

P = VI cos φ
Q = −VI sin φ

I

P
V

Q

Fig. 4

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Fundamental Concepts                   MTE 320          Spring 2006                     E.F. EL-Saadany

Three-Phase Systems

1. Sources
An ideal three-phase power source may be visualized as consisting of three independent voltage
sources, each producing a sinusoidal voltage displaced in phase by 120° from the other two sources.
The peak magnitude and the frequency of all three sources must be the same.

Va = Vm cos ω t               Va = Vm cos(ω t − 120)           Va = Vm cos(ω t − 240)

Va = Vrms ∠0                  Va = Vrms ∠ − 120                Va = Vrms ∠ − 240

Fig. 5 Three-phase systems representations

Vm
Where Vrms is the phase RMS voltage and is equal to Vrms =            .
2

Important Remarks;

•   The magnitude and frequency of each single-phase source is the same and the angular
•   The phasors are defined using the RMS quantities of their magnitude.

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Fundamental Concepts                  MTE 320                  Spring 2006                        E.F. EL-Saadany

If the three sources are connected to one common point, this point is called the neutral of the system
and it is said that the system is three-phase system. Two different arrangements for three-phase
systems are possible; the STAR (Wye) or the DELTA connection.

2. Star (Wye) Connection
A star connection is achieved by connecting the three individual sources to one common point known
as the neutral. The three-phase terminals will be the three remaining terminals of the sources. A star
connection is shown in Fig. 6.

A

Neutral Point
Va

V ab

Vc                                Vb

B
V bc
Physical Connection                                                                      C

Circuit Representation

Va

0
0                     120
120

0
120

Vc                                                             Vb

Phasor Representation

Fig. 6 Three-phase star connection

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Fundamental Concepts                            MTE 320            Spring 2006                         E.F. EL-Saadany

Where:
•   Va , Vb and Vc are the line to neutral voltages and are called the phase voltages V ph .

•   Vab , Vbc and Vca are the line-to-line voltages and are called the line voltage VL .

3. Phase and Line Voltages Relationship

The magnitude and the phase angle of the line-to-line voltage (line values) can be directly calculated
from the phasor diagram of the three-phase voltages as:
Vab = Va − Vb
Vbc = Vb − Vc
Vca = Vc − Va

These relations can be seen in phasor representation as follows:
-Va

Vab                                       Vca
Vc
Vca                                       -Vb                       Vc
30                                          Va
Vbc
Va                                     Vb
Vb               120

Vab
Vbc

-Vc

Vb
120                 Va

30
b                                                      a
Vab

Fig. 7 Phase and line voltages relationships

3
If               ab = 2 cm ab = 2V ph cos 30° = 2V ph            = 3 V ph
2

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Fundamental Concepts                    MTE 320              Spring 2006                       E.F. EL-Saadany

∴ab = 3 V ph = VL

The line-to-line magnitude is equal to         3 the line to neutral voltage.

Line voltage = 3 phase voltage.

From the figure above, the angle between Vab and Va is 30° , then:

Phasor Representation
Vab = 3V ph ∠30

Vbc = 3V ph ∠ − 90

Vca = 3V ph ∠150

where V ph is the RMS voltage

Time-domain representation
( ) (
Vab (t ) = 3 2V ph cos ωt + 30°  )
Vbc (t ) =   3 ( 2V )cos(ωt − 90 )
ph
°

Vca (t ) =   3 ( 2V )cos(ωt + 150 )
ph
°

Using KVL:
Va + Vb + Vc = 0
Vab + Vbc + Vca = 0

NOTE The line voltage LEADS the phase voltage by 30 degrees, regardless of the sequence and the
type of load (resistive, inductive or capacitive).

4. Three-phase Four-wire System
Another important and useful connection for three-phase systems is the three-phase four-wire
connection, where a fourth wire is utilized as a neutral wire or a return path. In a balanced three-phase
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Fundamental Concepts                    MTE 320                 Spring 2006                     E.F. EL-Saadany

system, the neutral current will be equal to ZERO, but in case of unbalanced systems or systems that
are polluted with harmonics, this wire can carry currents that may reach or exceed the phase current
magnitude.

5. Balanced Delta Connected Sources
Frequently, three phase voltages are generated by connecting voltage sources directly to the lines
rather than to a neutral point. In this case, the connection is said to be DELTA (or π) connection. The
circuit connection as well as the phasor representation of the Delta connection is shown in Fig. 8.

a

Vca                        Vab

b

Physical                                     Vbc
connection                                                       c

Circuit
representation

Vca                       Vab

Vbc

Phasor
representation

Fig. 8 Three-phase delta connection

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Fundamental Concepts                                MTE 320             Spring 2006                  E.F. EL-Saadany

When it comes to circuit connection, loads are treated as sources. Loads can be connected in single-
phase or three-phase connections. In three phase circuits, loads are normally connected in three-phase
configuration either in star or delta connection. As will be discussed latter, in three-phase systems,
especially in distribution systems, loads are often connected in single-phase configuration.

1.1 Connection
Balanced STAR loads are connected as shown in Fig. 9.

Ia
a

Va
Za
Balanced System
Three-Phase

Vc                        Vb
n

Zc             Zb
Ic       c                                  b

Ib

Fig. 9 Balanced star load connection.

1.2 Balancing condition
The following condition has to be satisfied for balanced star load system:
Z a = Z b = Z c = Z y ∠θ

In the above circuit, if we apply Ohm’s law in any branch we get:
Va
Ia =
Za

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Fundamental Concepts                            MTE 320             Spring 2006                E.F. EL-Saadany

For a balanced Y-load, the line current is equal to the phase current and the line voltage is equal to      3
of the phase voltage. Therefore,
I L = I ph

and               VL = 3 V ph .

1.4 Phasor Diagram of balanced Y-load

Va

Ia
Vca                θ                  Vab

120 0
120 0
Ic
120 0
Ib
Vc                                     Vb

Vbc

Fig. 10 Phasor representation for balanced star load

•   Vca = 3 Va

•   Angles between I a , I b and I c is 1200.

•   Angles between Va , Vb and Vc is 1200.

•   Angle between Vab , Vbc and Vca is 1200.

1.5 Real and reactive power in Y-loads
The average real and reactive power delivered to each individual phase of the star connected load may
be calculated as:
P / phase = V ph I ph cos θ

VL           V    V
P / phase =      I L cosθ = L ⋅ L cosθ
3             3 3Z y

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Fundamental Concepts                       MTE 320              Spring 2006               E.F. EL-Saadany

VL2
P / phase =        cosθ
3Z y

VL2
Also           Q / phase =         sin θ
3Z y

We have to realize that the total power is given by the three phases together
VL2
Ptotal = 3 × P / phase = 3V ph I ph cosθ = 3VL I L cosθ =         cosθ
Zy

VL2
and            Qtotal = 3 × Q / phase = 3V ph I ph sin θ = 3VL I L sin θ =        sin θ
Zy

The apparent power (the magnitude of the complex power) is given by:

2        2                     VL2
S = Ptotal + Qtotal = 3VL I L =
Zy

The power factor of the three-phase system is given by:
Ptotal
PF =          = cosθ
S

Example
A three-phase voltage generator with line voltage of 120V (rms) is used to drive a balanced y-
connected load consisting of three balanced impedances of Z L = 4 + j 3 Ω through a three-phase four-
wire power feeder with per phase impedance of Z f = 0.1 + j 0.1Ω . Calculate the line voltage at the

Solution
The impedance seen by the source/phase is:
Z T = Z f + Z L Ω = (0.1 + j 0.1) + (4 + j3) = 4.1 + j 3.1Ω

The current flowing in the load is:

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Fundamental Concepts                                       MTE 320                  Spring 2006                    E.F. EL-Saadany

V ph             120 / 3
I ph = I L =                    =                          = 13.48 A
ZT             (4.1) + (3.1)
2         2

Zf                Ia                            a

ZL       Va
Balanced System
Three-Phase

Vb
n

ZL              ZL
Zf           Ic        c              Vc                       b

Zf

Ib

Fig. 11 Three-phase circuit diagram for the example

The line voltage at the load terminal is given as:
VL = 3 I ph Z L = 3 × 13.48 × 5 = 116.73V

The total power delivered to the load is

PLoad = 3I 2 (Re Z ) = 3 × (13.48)2 × 4 = 2180Watt

2.1 Connection
Balanced delta loads are connected as shown in Fig. 12.

•   Vab is the voltage at point a with respect to point b.

•    I ab is the current from a to b.

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Fundamental Concepts                             MTE 320               Spring 2006                        E.F. EL-Saadany

Ia
a

I ca                               I ab

Balanced System
Z ca                 Z ab
Vac

Three-Phase

Ic        c                        ZL                   b

I bc
Vcb

Ib

Fig. 12 Balanced delta load connection.

Applying KCL at the nodes a, b, and c we get:
I a = I ab − I ca
I b = I bc − I ab
I c = I ca − I bc

2.2 Balancing condition
The following condition must be satisfied for a balanced Delta connected load.
Z ab = Z bc = Z ca = Z Δ ∠θ

2.3 Current in balanced Δ –load
In the above circuit, if Vab is taken as reference and applying Ohm’s law in any branch we get:

Vab V ph ∠0 V ph
I ab =       =      =     ∠ − θ = I ph ∠ − θ = the phase current
Z ab Z Δ ∠θ Z Δ

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Fundamental Concepts                         MTE 320                Spring 2006                   E.F. EL-Saadany

2.4 Phasor Diagram of balanced Δ -load

Vca

-Ibc
Ica
θ
Ic
120 0
120 0
Ibc
120 0
Iab

Vbc                                                   Vab

Fig. 13 Phasor representation for balanced delta load

2.5 Phase currents
V ph
I ab =          ∠ − θ = I ph ∠ − θ
ZΔ

I bc = I ph ∠ − θ − 120

I ca = I ph ∠ − θ − 240

2.6 Line currents
I L = 2 × I ph cos 30 = 3I ph

I a = 3I ph ∠ − θ − 30
I b = 3I ph ∠ − θ − 150
I c = 3I ph ∠ − θ − 270

NOTE The line current LAGS the phase current by 30 degrees

2.7 Real and reactive power in Δ-loads
The average real and reactive power delivered to each individual phase of the delta-connected load
may be calculated as:

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Fundamental Concepts                          MTE 320            Spring 2006               E.F. EL-Saadany

P / phase = V ph I ph cos θ

I L2ZΔ
P / phase   = VL
IL
(
cosθ = I ph Z Δ ×
IL
)cosθ =
3
cosθ
3                    3
2
I LZΔ
P / phase =         cosθ
3

2
I LZΔ
Also           Q / phase =          sin θ
3

We have to realize that the total power is given by the three phases together
2
Ptotal = 3 × P / phase = 3V ph I ph cosθ = 3VL I L cosθ = I L Z Δ cosθ
2
and            Qtotal = 3 × Q / phase = 3V ph I ph sin θ = 3VL I L sin θ = I L Z Δ sin θ

The apparent power (the magnitude of the complex power) is given by:
2        2                  2
S = Ptotal + Qtotal = 3VL I L = I L Z Δ

The power factor of the three-phase system is given by:
Ptotal
PF =            = cosθ
S

3. Star-Delta Transformation
Loads connected in balanced Star can be converted to balanced delta connection and vice versa.

3.1 Star to Delta Transformation
Za Zb + Zb Zc + Zc Za
Z ab =
Zc
Za Zb + Zb Zc + Zc Za
Z bc =
Za

Za Zb + Zb Zc + Zc Za
Z ca =
Zb

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Fundamental Concepts                             MTE 320            Spring 2006          E.F. EL-Saadany

Z ab

Za                   Zb

n
Z ca                             Z bc
Zc

Fig. 14 Star delta transformation

3.2 Delta to Star Transformation
Z ab Z ca
Za =
Z ab + Z bc + Z ca
Z bc Z ab
Zb =
Z ab + Z bc + Z ca

Z ca Z bc
Zc =
Z ab + Z bc + Z ca

Let us consider the case of a balanced star-connected source delivering power to a balanced delta-
connected load as shown in Fig. 15.

Note that:
• I a = I A , I b = I B and I c = I C

• Vab = V AB , Vbc = VBC and Vca = VCA

If we take VAB as the reference phasor and given that the load is Z = Z∠θ , then starting from the load
V AB
side, I ab =        and the phasor diagram will be given as shown in Fig. 16.
Z

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Fundamental Concepts                 MTE 320             Spring 2006                             E.F. EL-Saadany

Z                       Z

Z

Fig. 15 Balanced star-connected source delivering power to a balanced delta-connected load

VCA
Vc

I CA

V AB
θ
30 °
I AB

Vb                                               Va
IA
− I CA

VBC

Fig. 16 Phasor diagram

Example
The three winding of the source is connected in wye. The source produces a line voltage of 440 v, and
supplies two resistive loads. One load contains resistors with a value of 4Ω each and connected in wye.
The second load contains resistors with a value of 6Ω each and connected in delta. Find:
a) Line and phase voltage of load 2.
b) Line and phase current of load 2.

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Fundamental Concepts                        MTE 320           Spring 2006                 E.F. EL-Saadany

c) Line and phase current of load 1.
d) Line current of the source.
e) Power delivered by the source.

Solution
Since both loads are connected directly to the output of the source, then the line voltage of both loads
will be the same as the line voltage of the source.
VL1 = VL2 = 440 V

Since load 1 is connected in star and load 2 is connected in delta, then
VL1
V ph1 =         = 254.04V
3
V ph2 = VL2 = 440

V ph
The phase currents in the two loads are equal to
Z ph

254.04
For the star load,        I ph1 =          = 63.51 A
4

440
For the delta load,       I ph2 =       = 73.33 A
6

For star connection       I L1 = I ph1 = 63.51 A

For delta connection I L2 = 3I ph2 = 127.01 A

The supply line current is,
I LT = I L1 + I L2 = 63.51 + 127.01 = 190.52 A

The supply power is
P = 3V ph I ph cosθ = 3VL I L cosθ = 3 × 440 × 190.52 × 1 = 145.2 kW

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Fundamental Concepts                      MTE 320           Spring 2006                      E.F. EL-Saadany

5. Power Factor Correction
It is highly desirable to have a unity power factor for industrial and residential loads. The
consequences of power factor that is less than unity are numerous including loading the supply with
undesirable reactive current and increasing losses. This implies that the current drawn from the supply
is greater than the minimum current required to transmit the needed power. A common method to
improve the power factor is to install an inductive element (capacitive or inductive depending on the
circuit) in parallel or series with the load.

Example
A balanced inductive load is connected to a 550 v, 60 Hz, three-phase supply. The line current is 20A
and the total power delivered to the load is 10kW. It is desired to improve the power factor to 0.9
lagging. Find:
a) The power factor before correction
b) The line current after correction
c) The KVAR rating of the capacitor bank.
d) The per phase capacitance.

Solution
P    10,000
a) The power factor before correction is PF =         =              = 0.52 lagging
S   3 × 550 × 20
b) The line current before correction is 20A
After correction, the power factor will be 0.9 and the line current in this case will be:
P         10,000
PF = 0.9 =      =                   and I L2 = 11.66 A
S2      3 × 550 × I L2

c) The reactive power for the original load is:

d)   Q1 = S12 − P 2 =     (                 )
2
3 × 550 × 20 − (10000)2 = 16.22kVAR

The reactive power after compensation should be:

2
Q2 = S 2 − P 2 =     (                )2
3 × 550 ×11.66 − (10000)2 = 4.84kVAR

The necessary capacitive reactive power to correct the power factor to 0.9 is:

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Fundamental Concepts                        MTE 320        Spring 2006                E.F. EL-Saadany

Qc = Q1 − Q2 = 11.38kVAR

The capacitors should be connected in parallel with the load to have the advantage of the
known applied voltage. Assume that the capacitors are connected in Delta connection, then:
VL2
Qc phase =       = ωCVL2
Xc

Qc = 3Qc phase = 3ωCVL2 = 3 × 2π × 60 × 5502 C = 11.38 ×103

∴ C phase = 33.26 μF

If the capacitors were connected in Star, the capacitance per phase would have been three times
larger.

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Fundamental Concepts                  MTE 320              Spring 2006                    E.F. EL-Saadany

Three-Phase Power Measurements

One of the main tasks that are needed in power systems is to measure the power associated with any
load. Different circuit topologies are used to measure the power delivered by a supply or absorbed by a
load. These connections depend mainly on the circuit connection and the load characteristics (balanced
or unbalanced). Given the fact that we are interested in active (real) power measurements, then the
wattmeters are utilized as the power-measuring device. Two main configurations are utilized for power
measurements, three wattmeter or two wattmeter methods.

1. Three Wattmeter Method
1.1 Four-wire system

W1
A                                               Z1

W2
N
B                                               Z2

W3
C                                               Z3

N

Fig. 17 Three wattmeter methods for four-wire systems

In this case, we are measuring the power of the load by measuring the power in each phase and adding
them.
PT = P1 + P2 + P3

The wattmeter consists of two coils, the current coil and the potential coil. In this case the current coil
is measuring the phase current and the potential coil is measuring the phase voltage. Each wattmeter is
producing the phase power.

26
Fundamental Concepts                        MTE 320             Spring 2006              E.F. EL-Saadany

Remarks

•   The neutral point N is not available for delta connection. Therefore, this method is not practical
when the loads are connected in Delta.

•   This method is suitable for balanced and unbalanced loads since each wattmeter is measuring
the power in one phase independent on the other.

1.2 Three-wire system

Ia
Z1

Ib
Z2

Ic
Z3

Fig. 18 Three wattmeter methods for three-wire systems

Let x be the common point of the three wattmeter voltage coils. If T is the periodic time of the source
voltage and I a , I b and I c are the time-domain line current entering the load, then the total power Px
indicated by the three wattmeters is given by;
T
1
Px = ∫ (Vax I a + Vbx I b + Vcx I c ) dt
T0

Regardless of the point x, which is completely arbitrary, we have;
Vax = VaN + VNx
Vbx = VbN + VNx
Vcx = VcN + VNx

Substituting in the power equation we obtain:

27
Fundamental Concepts                          MTE 320              Spring 2006               E.F. EL-Saadany

T                                      T
1                                      1
Px =
T0∫ (VaN I a + VbN I b + VcN I c )dt + T ∫VNx (I a + I b + I c )dt
0

But I a + I b + I c = 0 for balanced three phase system. Therefore,
T
1
(VaN I a + VbN I b + VcN I c )dt
T∫
Px =
0

Thus, the sum of the three wattmeters reading is precisely the total average power delivered to the

NOTE For delta loads, point x can be connected to any point. Further, this method is suitable for either
star or delta, balanced or unbalanced.

2. Two Wattmeter Method
Since the common point x in the previous method is arbitrary, then we may place it on one of the lines.
The wattmeter whose current coil is placed in that line will read zero because the voltage across its
potential coil is zero and thus this wattmeter can be removed.

Ia
Z1

Ib
Z2

Ic
Z3

Fig. 19 Two wattmeter method

Let us assume that the sum of the two wattmeters measurements is P \ and is given by:

28
Fundamental Concepts                          MTE 320               Spring 2006          E.F. EL-Saadany

T
1
(Vab I a + Vcb I c )dt
T∫
P\ =
0

Vab = VaN + VNb
Where:
Vcb = VcN + VNb

Substituting in the power equation we get:
T
1
(VaN I a + VcN I c ) +VNb (I a + I c )dt
T∫
P\ =
0

Also             Ia + Ib + Ic = 0

and              I a + Ic = −Ib

Therefore,
T
1
(VaN I a + VcN I c ) +VNb (− I b )dt
T∫
P\ =
0

T
1
P = ∫ (VaN I a + VbN I b + VcN I c ) dt
\

T0

The above equation is the equation of the average power in three-phase system

∴ P \ = P , the measured value P \ gives the three-phase power in balanced systems.

3. Power Measurement Summary

1. In case of three-phase, four-wire systems, the three-wattmeter method is suitable for power
measurements in both balanced and unbalanced systems. The sum of the three wattmeters will
give the total power absorbed by the load. The potential coil terminals for the three wattmeters
should be connected to the neutral wire.

29
Fundamental Concepts                 MTE 320            Spring 2006                      E.F. EL-Saadany

2. In case of three-phase, three-wire systems, the three-wattmeter method is also suitable for
power measurements in both balanced and unbalanced systems. For balanced systems, the
potential coils of the wattmeters can be connected to any arbitrary point. However, in case of
unbalanced systems, the potential coils must be connected to the neutral point.

3. For Delta connected loads, three-wattmeter method is suitable for balanced and unbalanced
system power measurements.

4. Two-wattmeter method is suitable for balanced or unbalanced load systems either in delta or in
three-wire star connection. The common point of the two potential coils must be connected to
the phase that does not have a wattmeter.

30
Fundamental Concepts                      MTE 320        Spring 2006                   E.F. EL-Saadany

Three-Phase vs. Single-Phase Operation

A big question mark was always raised as which is the favorable system to deal with; is it single-phase
or three-phase system? The answer to this question was always three-phase systems, and the reasons
behind that were mainly related to the system performance, efficiency and economics.

1. Generation point of view
Three-phase systems always provide better utilization of the generating material. For the same size of
generator, the power capacity of three-phase exceeds single-phase. Moreover, the efficiency of three-
phase generators is higher than single-phase generators.

2. Transmission and distribution point of view
For this analysis, a comparison between the most common three-phase system and single-phase system
will be conducted. The most common three-phase configuration is the three-phase, four-wire system.
This is compared to the 1-phase, 2-wire system. Different factors such as the current transmitted,
power losses and copper weight are compared.

1.3 The value of current transmitted
Assume that the same power P is transmitted by the 1-phase and 3-phase systems, then:

For 3-phase     P3− ph = 3VI 3− ph cosθ

For 1-phase     P − ph = VI1− ph cos θ
1

Therefore,
P
I 3− ph =
3V cosθ
P
and           I1− ph =
V cosθ

31
Fundamental Concepts                              MTE 320        Spring 2006               E.F. EL-Saadany

In conclusion, for the same power transmitted and transmission line voltage, the 3-phase current is
1
only 1         of the 1-phase current, I 3− ph = I1− ph . This fact affects the size of the cables used in
3                                    3
transmission lines since less conductor material is needed for the 3-phase system to transmit the same
power in 1-phase system. Moreover, since the current is lower, the cost for the protection devices
will be less.

1.4 The power losses in transmission system
Assume that the same power P is to be transmitted and the transmission voltage is the same. The
power losses equations are given by:
2
⎛    P ⎞
PL3− ph = 3⎜         ⎟ R3− ph
⎝ 3V cosθ ⎠
2
⎛ P ⎞
PL1− ph = 2⎜        ⎟ R1− ph
⎝ V cosθ ⎠

where R3− ph and R1− ph are the 3-phase and 1-phase line resistance, respectively.

PL3− ph    1 R3− ph
= ×
PL1− ph    6 R1− ph

In conclusion, if the 3-phase and 1-phase resistances are equal, then the losses in the three-phase

system are 1         that in the single-phase system. We have to note that since the three-phase system is
6
balanced, then no current flow through the neutral wire and there will be no losses in the fourth wire.
This indicates that the 3-phase systems are 6 times more efficient than the 1-phase systems.

It is important to note that if the power losses in the 3-phase and 1-phase systems are the same, then
R3− ph
= 6 which means that for the same type of material and for the same transmission length, the
R1− ph

conductors of the three-phase system have smaller cross-section area and hence, they are more
economical.

32
Fundamental Concepts                            MTE 320             Spring 2006                     E.F. EL-Saadany

1.5 The copper weight
Before we analyze this item, let us consider the following facts:

•    Fact #1: For three-phase balanced systems there is NO neutral current (NO losses in neutral
conductor). Therefore, the cross-section area (CSA) of the neutral conductor needs not to be
the same as the other three-phases. Normally, the CSA of the neutral conductor is 1/2 of the
CSA of the line conductors.

1           ρL
•    Fact #2: The copper CSA for any system is proportional to                     since R =    , so for the
R            A
1
same length L and conductor material ρ , we have R ∝ l                  .
A

•    Fact #3: The copper weight for a given conductor is proportional to its CSA.

Given the previous three facts then, the copper weight for the three-phase, fore-wire system, which is
1
2
copper weight = K × 3.5 × A
1
A1− ph ∝ al
R1− ph

1
A3− ph ∝ al
R3− ph

A1− ph         R3− ph
=              =6
A3− ph         R1− ph

Therefore, the ratio of the copper weight of the two systems is:
Wcopper3− ph        K × 3.5 × A3− ph       3.5 A3− ph 3.5 1 7
=                      =      ×      =   × =
Wcopper1− ph         K × 2 × A1− ph         2 A1− ph   2 6 24

In conclusion, the saving in the cost of the material, which is proportional to the transmission and
distribution line cost is about 71% in favor of 3-phase, 4-wire systems (material cost in 3-phase
systems are 29% of the material cost in 1-phase systems having the same conductor material and

33
Fundamental Concepts                  MTE 320            Spring 2006                    E.F. EL-Saadany

transmitting the same power at the same voltage level). R3− ph is the resistance per phase of the 3-phase

system and R1− ph is the resistance per line of the 1-phase system.

34

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