Powerpoint slides to_chapter_12_three_phase_circuits_and_systems-corrected3

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BASIC ELECTRICAL ENGINEERING
PowerPoint Slides
D. C. KULSHRESHTHA,
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Chapter 12
Three-Phase Circuits
and Systems
 D.C. Kulshreshtha
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Wednesday, April 6, 2022 Ch. 12 Three-Phasae Circuits and Systems 3
Thought of The DAY
The best way
out of a difficulty
is
through it.
Next

Topics to be Discussed
 Three-Phase System.
 Advantages.
 Concept.
 Generation.
 Unbalanced Three-Phase
System.
 Voltages And Currents
Relations.
 (1) Star-Connected
System.
 (2) Delta-Connected
System
 Power In Three-phase
System.
 Power Measurement.
 Two-Wattmeter
Method.
 Power Factor
Measurement.
 Important Points.
Next

Advantages of Three-Phase
System
 Transmission lines require much less conductor
material.
 A three-phase machine gives a higher output.
 A three-phase motor develops a uniform (not a
pulsating) torque.
 The three-phase induction motors are self-starting.
 Can be used to supply domestic as well as
industrial (or commercial) power.
 The voltage regulation is better.
Next

Concept of Three-phase Voltages
Next
• The phase order
or phase sequence or
phase rotation is
abc.

• Whether the coils rotate anti-
clockwise or the magnet on the rotor
rotates clockwise, the effect is the
same.
• But the latter is safer and easier to
make external connections to
stationary coils.
Next

Three windings connected to three loads using
six line conductors
Next

 Thus, the terminals on the periphery appear
in the order : R, B’, Y, R’, B, Y’.
 The three emfs generated eR, eY and eB
connected to three respective loads L1, L2
and L3.
 This necessitates the use of six line
conductors.
 Obviously, it is cumbersome and expensive.
 Let us now consider how it may be
simplified.
Next

Three-phase Loads
There are two kinds of three-phase systems :
(i) Star or wye (Y) connection, and
(ii)Delta (Δ) or mesh connection
If all the three impedances are equal, the load
is said to be a balanced load.
Next

Star (Y) Connected Three-phase System
Next

 Practically, it may not be possible always to make
this star-connected domestic load balanced.
 However, as per KCL, the sum of the three line
currents must still be zero.
 Hence, the voltages across the three loads get
adjusted, resulting in neutral shift or floating
neutral.
 This situation is definitely undesirable.
 Therefore, we use Four-Wire Three-Phase
Voltage System.
Next

Delta (Δ) Connected Three-phase System
Note that the ‘finish’ of one phase is connected to
the ‘start’ of another phase.
Next

Voltages And Currents Relations
in 3-φ Systems
(1) Star-Connected System
Next

 In a three-phase system, there are two sets of
voltages :
 the set of phase voltages, and
 the other is the set of line voltages.
 VRN, VYN and VBN denote the set of three
phase voltages.
 The term ‘line voltage’ is used to denote the
voltage between two lines.
 VRY represents line voltage between the lines
R and Y.
Next

RY RNY RN NY
RN YN RN YN
( )
  
    
V V V V
V V V V
 
RY RN
L ph ph
2( cos30 )
or 2 3 / 2 3
V V
V V V

 
L ph
3
V V

L ph
and I I

VRN
VBN
VYN
-VYN
VRY
30°
VRN cos 30°
Next
Click

Star-connected System
Next

(2) Delta-Connected System
R'R Y'Y B'B ph (say)
I
  
I I I
Next

R R'R B'B
 
I I I
R ph ph
2( cos30 ) 3
I I I
 
L ph
3
I I
 L ph
and V V

IR’R
IB’B
IY’Y
-IB’B
IR
30°
IR’R cos 30°
Next
Click

Delta-connected System
Next

Important Points about
Three-Phase Systems
1. It is normal practice to specify the values
of the line voltages and line currents.
2. The current in any phase can be
determined by dividing the phase voltage
by its impedance.
Next

Example 1
 A 400-V, 3- supply is connected across
a balanced load of three impedances
each consisting of a 32-Ω resistance and
24-Ω inductive reactance. Determine the
current drawn from the power mains, if
the three impedances are
(a) Y-connected, and
(b) Δ-connected.
Next

Solution : Z = R + jX = (32 + j24) Ω.
2 2 2 2
32 24 40
Z R X
      
(a) Y-connection :
ph
L
ph ph
400 400/ 3 10
V A
40
3 3 3
V
V
V I
Z
     
L ph
10
.
3
I I
    5 78 A
(b) For Δ-connection :
ph
ph L ph
400
400V 10 A
40
V
V V I
Z
     
L ph
3 3 10
I I
    17.32 A
Next
Click
Click
Click

Power In Three-phase System With A
Balanced Load
Consider one phase only, 1 ph ph cos
P V I 

Hence, the total power consumed,
1 ph ph
3 3 cos
P P V I 
 
For a star-connected system, L ph L ph
3 and
V V I I
 
L L L L
3( / 3) cos 3 cos
P V I V I
 
 
For a delta-connected system, L ph L ph
and 3
V V I I
 
L L L L
3 ( / 3)cos 3 cos
P V I V I
 
 
Thus, for any balanced load,
L L
3 cos
P V I 
 Next
Click
Click
Click
Click

• In three-phase system with balanced
load (such as a three-phase motor), there is
no variation of power at all.
• This is the reason why for driving heavy
mechanical loads we prefer a three-phase
motor rather than a single-phase motor.
Next

Example 2
 A 400-V, 3- supply is connected to a balanced
network of three impedances each consisting of a
20-Ω resistance and a 15-Ω inductive reactance.
 If the three impedances are (a) star-connected,
and (b) delta-connected, in each case determine
 (i) the line current,
 (ii) the power factor, and
 (iii) the total power in kW.
Next

(a) For star-connected load : L ph L ph
3 and
V V I I
 
L
ph
400
231 V
3 3
V
V   
2 2
ph
and 20 15 25
Z    
(i) ph
L ph
ph
231
25
V
I I
Z
    9.24 A
(ii) ph
ph
20
cos
25
R
Z
    0.8 (lagging)
(iii)
L L
3 cos 3 400 9.24 0.8
P V I 
      5.12 kW
Next
Click
Click
Click
Click

(b) For delta-connected load : L ph L ph
and 3
V V I I
 
L ph 400 V
V V
  
(i) ph
ph
ph
400
16 A
25
V
I
Z
  
L ph
3 3 16
I I
     27.71 A
(ii) The power factor is same as above,
pf = 0.8 (lagging)
(iii) L L
3 cos 3 400 27.71 0.8
P V I 
      15.36 kW
Note that power consumed has become 3 times.
Next
Click
Click
Click
Click

No. Star-Connected System Delta-Connected System
1.
2.
3.
4.
5.
6.
Similar ends are joined together.
Neutral wire available.
4-wire, 3- system possible.
Both domestic and industrial
loads can be handled.
By earthing the neutral wire,
relays and protective devices can
be provided in alternators for
safety.
Dissimilar ends are joined.
Neutral wire not available.
4-wire, 3- system not
possible.
Only industrial loads can be
handled.
Due to absence of neutral
wire, it is not possible.
L ph L ph
3 and
V V I I
  L ph L ph
and 3
V V I I
 
Next

Measurement of Power
(i) Three-Wattmeter Method : This is simplest
and straight forward method.
(ii)Two-Wattmeter Method : This can be used
for any balanced or unbalanced load, star- or
delta-connected.
(iii)One-Wattmeter Method : This can be used
only for a star-connected balanced load.
Next

The sum of the wattmeter readings gives the
average value of the total power absorbed by the
three phases
Next
Proof :
R RN Y YN B BN
Total instantaneous power
.
i v i v i v
  
1
1 R RN YN
2
2 B BN YN
The instantaneous power measured by ,
( )
The instantaneous power measured by ,
( )
W
p i v v
W
p i v v
 
 

1 2 R RN YN B BN YN
R RN B BN R B YN
( ) ( )
( )
p p i v v i v v
i v i v i i v
     
   
R Y B R B Y
By KCL, 0 ( )
i i i i i i
      
1 2 R RN B BN Y YN
total instantaneous power
p p i v i v i v
    

Since we did not assume a balanced load or a
sinusoidal waveform, it follows that the sum of the
two wattmeter readings gives the total power under
all conditions.
Next

Power Factor Measurement by Two-
Wattmeter Method
Concept of ‘power factor’ is meaningful only if
the load is balanced.
Next

Since the load is balanced,
R Y B L (say)
I I I I
  
Since VRB = VRN  VBN, and VYB = VYN  VBN,
we can determine the line voltages VRB and VYB by
phasor method,
RN YN BN ph
and (say)
V V V V
  
Next

VRN
VBN
VYN
IR
IB
IY -VBN
VYB
VRB
Φ
Φ
Φ
Next

• It is seen that the line voltage VRB lags the
phase voltage VRN by 30° and VYB leads VYN by
30°.
• Thus, the phase angle between VRB and IR is
(30°  ).
• Similarly, the phase angle between VYB and
IY is (30° + ).
• Therefore, the readings of the two wattmeters
are
Next

1 RB R L L
cos(30 ) cos(30 )
P V I V I
 
   
2 YB Y L L
cos(30 ) cos(30 )
P V I V I
 
   
1
2
cos(30 )
cos(30 )
P
P


 
 
 
Next
Click
Click

By applying componendo and dividendo,
1 2
1 2
cos(30 ) cos(30 )
cos(30 ) cos(30 )
2sin30 sin
tan30 tan
2cos30 cos
P P
P P
 
 



   

   

  

1 2
1 2
tan 3
P P
P P

 

   

 
Calculate the phase angle , and then determine
the power factor.
Next
Click
Click

Important Points :
 If Ф = 0°, P1 = P2.
 If Ф < 60°, both P1 and P2 are positive; and
P = P1 + P2
 If Ф = 60°, P2 is zero.
 If Ф > 60°, (i.e., if pf < 0.5), P2 is negative.
In such case, the connection of either the current
coil or the potential coil has to be reversed to
make positive deflection.
The value P2 should then be taken as negative
while calculating the power factor or the total
power.
Next

Example 3
 Two-wattmeter method was used to
determine the input power to a three-phase
motor. The readings were 5.2 kW and -1.7
kW, and the line voltage was 415 V.
Calculate
(a) the total power,
(b) the power factor, and
(c) the line current.
Next

Solution :
(a) The total power,
1 2 5.2 kW 1.7 kW
P P P
     3.5 kW
(b) 1 2
1 2
5.2 ( 1.7)
tan 3 3 3.41
5.2 ( 1.7)
P P
P P

   
  
  
   
  
 
 
1
tan 3.41 73 39'
 
   
cos cos73 39'
pf 
     0.281
(c) 3500 3 415 0.281
L
L
I
I
   
  17.3 A
Next
Click
Click
Click

Review
 Three-Phase System.
 Advantages.
 Concept.
 Generation.
 Unbalanced Three-Phase
System.
 Voltages And Currents
Relations.
 (1) Star-Connected
System.
 (2) Delta-Connected
System
 Power In Three-phase
System.
 Power Measurement.
 Two-Wattmeter
Method.
 Power Factor
Measurement.
 Important Points.
Next

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