Circuit theory. Power triangle.

1. Example 10.6
Calculating Power in Parallel Loads
Load 1 absorbs an average power of 8 kW at a leading power factor of 0.8.
Load 2 absorbs 20 kVA at a lagging power factor of 0.6.
1
ECE 201 Circuit Theory 1

2. ECE 201 Circuit Theory 1 2
Determine the power factor of the two loads in parallel.
The total complex power absorbed by the two loads is
2
1
*
2
*
1
*
2
1
*
)
250
(
)
250
(
)
)(
250
(
)
250
(
S
S
S
I
I
S
I
I
S
I
S S








3. Power Triangle for Load 1
ECE 201 Circuit Theory 1 3
8 kW at a leading power factor of 0.8.
cos-1(0.8) leading = -36.87°
S1 = 8,000 –j6,000 VA

4. Power Triangle for Load 2
ECE 201 Circuit Theory 1 4
20 kVA at a lagging power factor of 0.6.
cos-1(0.6) lagging = 53.13°
S2 = 12,000 +j16,000 VA

5. ECE 201 Circuit Theory 1 5
S = S1 + S2 = 8,000 – j6,000 = 12,000 +j16,000
S = 20,000 +j10,000 VA
The resulting Power Triangle

6. ECE 201 Circuit Theory 1 6
A
I
A
j
I
A
j
I
j
I
S
S
S
S










57
.
26
44
.
89
40
80
40
80
250
000
,
10
000
,
20
*
*
The power factor of the two loads in parallel is
cos(26.57°) = 0.8944 lagging
(net reactive power is positive)

7. ECE 201 Circuit Theory 1 7
Determine the apparent power required to supply the
loads, the magnitude of the current IS, and the average
power loss in the transmission line.
The apparent power which must be supplied to these
loads is 22.36 kVA as shown in the power triangle
below.

8. ECE 201 Circuit Theory 1 8
The magnitude of the current that supplies the
required 22.36 kVA is
A
j
IS 44
.
89
40
80 


The average power lost in the line is due to the
current flowing through the line resistance.
W
P
R
I
P
line
S
line
400
)
05
.
0
(
)
44
.
89
( 2
2



Note: the total power supplied is equal to 20,400 W,
even though the loads require only 20,000 W.

9. ECE 201 Circuit Theory 1 9
Given that the frequency of the source is 60 Hz, compute
the value of capacitor that would correct the power factor
to 1 if placed in parallel with the two loads.
As seen from the power triangle, the capacitor would
have to supply 10 kVAR of magnetizing reactive power.






25
.
6
000
,
10
)
250
( 2
2
C
eff
C
X
Q
V
X

10. ECE 201 Circuit Theory 1 10
Calculation of the capacitance from the required reactance
F
C
X
C
C
X
C
C




4
.
424
)
25
.
6
)(
60
)(
2
(
1
1
1









11. ECE 201 Circuit Theory 1 11
Look at the resulting power triangle
(a) is the sum of the power triangles for loads 1 and 2.
(b) is the power triangle for the 424.4 µF capacitor at 60 Hz.
(c) is the “corrected” power triangle (for a power factor equal to 1).

12. ECE 201 Circuit Theory 1 12
Once the power factor has been corrected,
W
R
I
P
A
I
kVA
P
S
S
line
S
320
)
05
.
0
(
)
80
(
80
250
000
,
20
20
2
2







The total power supplied is now
20,000 + 320 = 20,320 W
The line loss has been reduced by correcting the
power factor.