Post on 24-Dec-2015
ELEC1300
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Electrical Engineering 1
Friday Revision Lecture 11Maximum Power Transfer theorems in a.c.
CircuitsAnd Power in a.c. Circuits
Semester 2, 2014
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Announcements• Week 12: Lab 6 (AC Power, Reactive Power)– Complicated lab– Read notes– Watch video
• Pre Lab 6 Quiz, due 2pm Mon 27 Oct
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Outline• Further notes on Reactive Power
Compensation• RMS Values• Maximum Power Transfer• Real, Reactive and Complex Power• Lab 6: Power in AC Circuits
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Reactive Power Compensation Example
• Vs (Voltage supply, 11kVrms)• Load A: Induction furnace, 500kVA, PF= 0.6 lagging• Load B: 300kW Motor, PF=0.8 lagging• Load C: Capacitor for ‘reactive power compensation’Questions:1. Without compensation (Load C absent) find the overall real
power, reactive power, power factor and |Is|?
2. Find the value for the capacitance, as load C, that will improve the power factor to 0.9. What is |Is|in this case?
IS
VS Load A Load B Load C
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Reactive Power Compensation Example
Method:1. Draw up power
table2. Compute all
elements in table
IS
11kVA:500kVA0.6pf
B:300kW0.8pf
C
Item P (kW)
Q (kVAR)
|S|(kVA)
pfcos(θ)
θ
500 0.6A
300 0.8B
cos( )pf S P jQ
2 2| |S P Q | | cos( )P S
| | sin( )Q S
53.13°300 400
36.87°375225
A+B 600 625 866.39 0.6925 46.17°
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Reactive Power Compensation ExampleMethod:1. Draw up power
table2. Compute all
elements in table3. Derive further
quantities needed
Item P (kW)
Q (kVAR)
|S|(kVA)
pfcos(θ)
θ
500 0.6A
300 0.8B
cos( )pf S P jQ
2 2| |S P Q | | cos( )P S
| | sin( )Q S
53.13°300 400
36.87°375225
A+B 600 625 866.39 0.6925 46.17°
| | | || I |S V
Without compensation:• real power = 600kW• reactive power = 625kVAR• power factor = 0.6925 (lagging)• |Is|= 866.39 (kVA)/11kV = 78.76 A
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Reactive Power Compensation ExampleMethod:1. Draw up power table2. Compute all
elements in table3. Derive further
quantities needed4. Extend table to
include compensation
Item P (kW)
Q (kVAR)
|S|(kVA)
pfcos(θ)
θ
500 0.6A
300 0.8B
53.13°300 400
36.87°375225
A+B 600 625 866.39 0.6925 46.17°
0 0C -90°
A+B+C 600 0.9 25.84°290.59 666.67
| | cos( )P S cos( )pf | | sin( )Q S
-334.41
| || I | 60.61
| V |S
SA
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Reactive Power Compensation ExampleIS
11kV 500kVA0.6pf
300kW0.8pf
C
Method:1. Draw up power table2. Compute all
elements in table3. Derive further
quantities needed4. Extend table to
include compensation
5. From Qc and line voltage, find C
334.41Q kVAR
*C CQ imag V I
*
CC
C
Vimag V
Z
*
*1/ ( )C CV V
imagj C
2
CQ V C
22 C
QC
f V
2
334.41
2 50 11
kC
k
8.80 F
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RMS Values
Suppose v(t) is a periodic “sawtooth” waveform.
What is the period of v(t)?What is the r.m.s. value of v(t)?If v(t) is the voltage across an 8 resistor, what is the average power?
12Vv(t)
t5mS 10mS 15mS
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Maximum Power Transfer Theoremin a.c. Circuits
• Maximum power will be delivered to a load when the load impedance is the complex conjugate of the Thévenin or Norton impedance of the a.c. circuit.So if:
ZTh = (3 + j4) then ZL =
ZTh = (-j5) then ZL =
ZTh = 5 then ZL =
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Example: 2009 Final Exam1. Find the impedance, which when connected across
points a and b, will give maximum power transfer
2. Find the real power in the load under this condition.
2kW10H
5mF
4cos(200t+30°)A
a
b