Three-Phase Power Calculations

1. Lecture 9:
Three-Phase Power
Calculations

2. Lesson Outcomes
i. calculate power consumed by a star-connected load,
ii. calculate power consumed by a delta-connected load,
After completing this unit and doing the exercises givenyou will be
able to:

3. Recall
The power (p) supplied at any instant in a single-phase circuit is
given by
POWER CONSUMED BY THREE-PHASE LOADS
p = vi
The total power for a single-phase circuit is
P = VIcos
where V and I are rms values of voltage and current respectively,
and  is the phase angle between them.
v(t)
i(t)

4. The power consumed by a three-phase load is thus given by the
sum of the powers in each phase:
P = VRNIRcosR + VYNIYcosY + VBNIBcosB
If the load is balanced, then
VRN = VYN = VBN = Vph
IR = IY = IB = Iph
and
R = Y = B = 
where VRN, VYN, VBN, IR, IY, and IB are phase quantities.
Three-phase load
R
Y
B
IR
IY
IB
POWER CONSUMED BY THREE-PHASE LOADS

5. If the load is star-connected, then
IL = Iph
and
L
ph V
V
3
1

Therefore, total power
giving

 cos
3
1
3
cos
3 



 L
L
ph
ph I
V
I
V
P
Therefore, total power,
= 3VphIphcos
P = VRNIRcosR + VYNIYcosY + VBNIBcosB

cos
3 L
L I
V
P 
POWER CONSUMED BY THREE-PHASE LOADS

6. If the load is delta connected, then
and
ph
L V
V 
from which it follows that total power,
ph
L I
I 3


cos
3 L
L I
V
P 
Thus, for all methods of connection, the power in a three-phase
balanced load is given by

cos
3 L
L I
V
P 
If the load is unbalanced, then total power is obtained by summing
the all the phase powers, that is,
P = VRIRcosR + VYIYcosY + VBIBcosB
POWER CONSUMED BY THREE-PHASE LOADS

7. The total apparent power supplied to the load is equal to the sum of
apparent powers supplied to all the three phases, that is
B
Y
R S
S
S
S 


where
R
RN
R I
V
S  Y
YN
Y I
V
S 
B
BN
B I
V
S 
VRN = VYN = VBN = Vph
IR = IY = IB = Iph
and
Therefore, total apparent power ph
phI
V
S
S 3
3 ph 

POWER CONSUMED BY THREE-PHASE LOADS
If the load is star-connected and balanced, then
ph
B
Y
R S
S
S
S 



8. Since L
ph
3
1
V
V 
Then, total apparent power supplied to the load is
L
L
L
L
ph
ph I
V
I
V
I
V
S
S 3
3
1
3
3
3 ph 





Iph = IL
and
POWER CONSUMED BY THREE-PHASE LOADS
In complex notation,
*
3 L
L I
V
S 
It can be shown that this result also holds for a balanced delta-
connected load.
So, providing we know the line current and phase-to-phase (line)
voltage and the impedance angle, we can calculate the power of the
load without having to know how it is connected.

9. Similarly, the total reactive power supplied to the load is the sum of
reactive powers consumed by each phase; that is

 sin
3
sin
3
1
3
sin
3
3 ph
ph
ph L
L
L
L
ph I
V
I
V
θ
I
V
Q
Q 






θ
I
V
Q Y Y
Y
Y sin

B
Y
R Q
Q
Q
Q 


where
θ
I
V
Q R R
R
R sin
 θ
I
V
Q B B
B
B sin

POWER CONSUMED BY THREE-PHASE LOADS
are phase quantities. If the load is star-connected, then
IL = Iph
and
L
ph V
V
3
1

Thus,
Likewise, it can be shown that for a delta-connected load, the total
reactive power consumed is given by

sin
3 L
L I
V
Q 

10. Three identical coils, each having a resistance of 20  and an
inductance of 0.5 H are connected in star to a three phase supply of
400 V; 50 Hz. Calculate the current and the total power absorbed
by the load.
Worked Example
20
3
/
400
400V
400V
400V
N
0.5H
0.5H 0.5H
20 20
Solution
First of all calculating the
impedance of the coils.
Given: 
 20
RP
H
5
.
0

P
L
Therefore





157
5
.
0
50
2
P
X

11. 





2
2
P
P
P
P
P X
R
jX
R
Z

83
158
83
157
20 2
2






p
Z

83
20
157
tan
tan 1
1

















 

P
P
R
X

where
Solution
Coil impedance is
Therefore,
Since it is a balanced load
V
231
3
400


P
V

12. 
cos
3 L
L I
V
P 
W
128
1264
.
0
46
.
1
400
3 




P
Total power absorbed by the load is
1264
.
0
83
cos
cos 
 

where
Therefore,
A
46
.
1
158
231




P
P
L
P
Z
V
I
I
Current taken by the load,
Solution

13. Alternative Solution
20
3
/
400
400V
400V
400V
N
0.5H
0.5H 0.5H
20 20
First of all calculating the
impedance of the coils.
Given: 
 20
RP
H
5
.
0

P
L
Therefore





157
5
.
0
50
2
P
X
 



 157
20 j
jX
R
Z P
P
P
Coil impedance is

14. Alternative Solution
Since it is a balanced load, we can analyse it on a per phase basis
as follows:
V
231
3
400


P
V
R
N
RN
V
R
I
RN
Z
Phase voltage is
Choosing VRN as the reference phasor, we can write
V
0
231 


RN
V
Therefore,
A
74
.
82
46
.
1
157
20
0
231









j
Z
V
I
RN
RN
R
From which we obtain the line current
A
46
.
1

 R
L I
I

15. R
N
RN
V
R
I
RN
Z
Complex power consumed by ZRN is
SRN
VA
74
.
82
2
.
337
74
.
82
46
.
1
0
231
*










 R
RN
RN I
V
S
VA
82.74
1012
74
.
82
2
.
337
3
3








 RN
S
S
Total complex power consumed by the
three load impedances is
Alternative Solution

16.     W
5
.
127
74
.
82
cos
1012
Re 


 S
P
Therefore, total power absorbed by the load is
Alternative Solution

17. A balanced three phase load connected in star, each phase consists
of resistance of 100  paralleled with a capacitance of 31.8 F. The
load is connected to a three phase supply of 415 V; 50 Hz.
Calculate:
415
Worked Example
(a) the line current;
(b) the total power absorbed;
(c) total kVA;
(d) power factor .

18. V
240
3
415
3
V
V L
P 


Solution
Admittance of the load
C
j
1
XP


where
P
P
P
X
R
Y
1
1


C
j
R
1
P



6
10
8
.
31
50
2
j
100
1 






S
)
01
.
0
j
01
.
0
( 


19. )
01
.
0
01
.
0
(
240 j
Y
V
I
I P
P
P
L 




45
39
.
3
4
.
2
4
.
2 


 j
*
P
I
V
S P
p  

45
4
.
814
45
39
.
3
240 






    W
576
45
cos
4
.
814
Re 



 S
P
kW
728
.
1
576
3 


P
Line current
Volt-ampere per phase
Active power per phase
Total active power
Solution

20.   kVAR
728
.
1
576
3
3 





 p
Q
Q
Reactive power per phase
  VAR
576
45
sin
4
.
814 


 
p
Q
Total reactive power
kVA
44
.
2
4
.
814
3
3 


 p
S
S
Total volt-ampere
Power Factor = cos = cos 45 = 0.707 (leading)
Solution

21. Worked Example
rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50 80
50 80
A
B
C
j
j
j
  
  
  
Z
Z
Z
A balanced four-wire circuit has
Calculate the total complex power supplied to the load

22. rms
110 0
1.16 58 A
50 80
a
aA
A j
 
     

V
I
Z
*
68 109 VA
A aA a j
  
S I V
The total complex power delivered to the three-phase load
is
3 204 326 VA
A j
  
S S
Also
rms rms
1.16 177 A , 1.16 62 A
bB cC
      
I I
68 109 VA
B C
j
  
S S
Solution

23. A balanced 3-wire circuit has
rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50 80
50 80
A
B
C
j
j
j
  
  
  
Z
Z
Z
Calculate the total apparent power supplied to the circuit
Example

24. rms
110 0
1.16 58 A
50 80
a
aA
A j
 
     

V
I
Z
*
68 109 VA
A aA a j
  
S I V
The total complex power delivered to the three-phase load is
Solution
3 204 326 VA
A j
  
S S

25. rms
rms
rms
110 0 V
110 120 V
110 120 V
a
b
c
  
   
  
V
V
V
50 80
50
100 25
A
B
C
j
j
j
  
 
  
Z
Z
Z
An unbalanced 3-wire star-star connected system has
Calculate the total apparent power supplied to the circuit
Example

26. ZL
n
Va Vc
Vb
N
IA
IB IC
ZA ZB ZC
ZL ZL
VNn
Solution
Determine VNn
V
151
56
8
.
26
2
.
49






 j
VNn
B
Y
A
C
c
B
b
A
a
Nn
Z
Z
Z
Z
V
Z
V
Z
V
V
1
1
1





Therefore
, , and
a Nn b Nn c Nn
aA bB cC
A B C
  
  
V V V V V V
I I I
Z Z Z

27. Solution
1.71 48 , 2.45 3 , and 1.19 79
aA bB cC
        
I I I

28. Exercise

29. Three identical coils, each having a resistance of 20  and an
inductance of 0.5 H connected in delta to a three phase supply of
400 V; 50 Hz. Calculate the current and the total power absorbed
by the load.
Worked Example
Solution
20
400V
400V
400V 0.5H
0.5H
0.5H
20
20

30. Solution
From previous Worked Example we found coil impedance

83
158
83
157
20 2
2






p
Z
Since it is a balanced load
V
400

 L
P V
V
20
400V
400V
400V 0.5H
0.5H
0.5H
20
20
A
53
.
2
158
400



ph
L
P
Z
V
I
Load phase current
Therefore, line current
A
38
.
4
53
.
2
3
3 


 P
L I
I

31. 
cos
3 L
L I
V
P 
W
6
.
221
1264
.
0
53
.
2
400
3 




P
Total power absorbed by the load is
1264
.
0
83
cos
cos 
 

where
Therefore, total power absorbed by the load is
Solution

32. Exercise

33. END