BEF 23803 - Lecture 9 - Three-Phase Power Calculations.ppt

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 VandI 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 = V
RNI
Rcos
R+ V
YNI
Ycos
Y+ V
BNI
Bcos
B
If the load is balanced, then
V
RN= V
YN= V
BN = V
ph
I
R= I
Y= I
B= I
ph
and

R= 
Y= 
B = 
where V
RN, V
YN, V
BN, I
R, I
Y, and I
Bare phase quantities.
Three-phase load
R
Y
B
I
R
I
Y
I
B
POWER CONSUMED BY THREE -PHASE LOADS

5. If the load is star-connected, then
I
L= I
phandLph VV
3
1

Therefore, total power
giving cos
3
1
3cos3 
LLphph IVIVP
Therefore, total power,
= 3V
phI
phcos
P = V
RNI
Rcos
R+ V
YNI
Ycos
Y+ V
BNI
Bcos
Bcos3
LLIV P
POWER CONSUMED BY THREE -PHASE LOADS

6. If the load is delta connected, then
andphL
VV
from which it follows that total power, phL
II3 cos3
LLIVP
Thus, for all methods of connection, the power in a three-phase
balanced load is given bycos3
LLIVP
If the load is unbalanced, then total power is obtained by summing
the all the phase powers, that is,
P = V
RI
Rcos
R+ V
YI
Ycos
Y+ V
BI
Bcos
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 isBYR
SSSS 
whereRRNR IVS YYNY IVS BBNB IVS
V
RN= V
YN= V
BN= V
ph
I
R= I
Y= I
B= I
ph
and
Therefore, total apparent powerphph
IVSS 33
ph

POWER CONSUMED BY THREE -PHASE LOADS
If the load is star-connected and balanced, thenphBYR
SSSS 

8. SinceLph
3
1
VV
Then, total apparent power supplied to the load isLLLLphph IVIVIVSS 3
3
1
333
ph 
I
ph= I
Land
POWER CONSUMED BY THREE -PHASE LOADS
In complex notation, *
3 LLIVS
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 sin3sin
3
1
3sin33
phphph LLLLph IVIV θIVQQ  θIVQ
Y YYY sin BYR QQQQ 
where θIVQ
R RRR sin θIVQ
B BBB sin
POWER CONSUMED BY THREE -PHASE LOADS
are phase quantities. If the load is star-connected, then
I
L= I
phandLph VV
3
1

Thus,
Likewise, it can be shown that for a delta-connected load, the total
reactive power consumed is given by sin3
LLIVQ

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 Example20
3/400
400V
400V
400V
N
0.5H
0.5H0.5H
20 20
Solution
First of all calculating the
impedance of the coils.
Given:20R
P H 5.0
PL
Therefore

157
5.0502
P
X

11. 
22
PPPPP
XRjXRZ 
831588315720
22

p
Z 
83
20
157
tantan
11

















P
P
R
X
 where
Solution
Coil impedance is
Therefore,
Since it is a balanced loadV 231
3
400

PV

12. cos3
LLIVP W1281264.046.14003 P Total power absorbed by the load is1264.083coscos 


where
Therefore,A 46.1
158
231

P
P
LP
Z
V
II
Current taken by the load,
Solution

13. Alternative Solution20
3/400
400V
400V
400V
N
0.5H
0.5H0.5H
20 20
First of all calculating the
impedance of the coils.
Given:20R
P H 5.0
PL
Therefore

157
5.0502
P
X   15720jjXRZ
PPP
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

PV
R
NRNV RI RNZ
Phase voltage is
Choosing V
RNas the reference phasor, we can writeV 0 231RNV
Therefore,A 74.8246.1
15720
0231




jZ
V
I
RN
RN
R
From which we obtain the line currentA 46.1R
L
II

15. R
NRNV RI RNZ
Complex power consumed by Z
RNis
S
RNVA 74.822.337
74.8246.10231
*


 RRNRN IVS VA 82.741012
74.822.3373
3


 RNSS
Total complex power consumed by the
three load impedances is
Alternative Solution

16.    W5.12774.82cos1012Re SP Therefore, total power absorbed by the load is
Alternative Solution

17. Abalancedthreephaseloadconnectedinstar,eachphaseconsists
ofresistanceof100paralleledwithacapacitanceof31.8F.The
loadisconnectedtoathreephasesupplyof415V;50Hz.
Calculate:
415
Worked Example
(a)the line current;
(b)the total power absorbed;
(c)total kVA;
(d)power factor .

18. V240
3
415
3
V
V
L
P  Solution
Admittance of the loadCj
1
X
P


wherePP
P
XR
Y
11
 Cj
R
1
P
 6
108.31502j
100
1

 S)01.0j01.0(

19. )01.001.0(240 jYVII PPPL  
4539.34.24.2  j *
PIVS Pp 
454.8144539.3240    W57645cos4.814Re SP kW 728.15763 P Line current
Volt-ampere per phase
Active power per phase
Total active power
Solution

20.   kVAR 728.157633 
p
QQ Reactive power per phase VAR 57645sin4.814 

p
Q
Total reactive powerkVA 44.24.81433 
p
SS
Total volt-ampere
Power Factor = cos= cos 45= 0.707 (leading)
Solution

21. Worked Examplerms
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
is3 204 326 VA
A j  SS
Alsorms rms1.16 177 A , 1.16 62 A
bB cC     II 68 109 VA
BC j  SS
Solution

23. A balanced 3-wire circuit hasrms
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
Solution3 204 326 VA
A j  SS

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. Z
L
n
V
a V
cV
b
N
I
A
I
B I
C
Z
A Z
B Z
C
Z
L Z
L
V
Nn
Solution
Determine V
NnV 15156
8.262.49

 jV
Nn BYA
C
c
B
b
A
a
Nn
ZZZ
Z
V
Z
V
Z
V
V
111



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. Solution1.71 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
Solution20
400V400V
400V
0.5H
0.5H
0.5H
20
20

30. Solution
From previous Worked Example we found coil impedance
831588315720
22

p
Z
Since it is a balanced loadV 400
LPVV 20
400V400V
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 currentA 38.453.233 
PL II

31. cos3
LLIVP W6.2211264.053.24003 P Total power absorbed by the load is1264.083coscos 


where
Therefore, total power absorbed by the load is
Solution

32. Exercise

33. END