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ELEC1300
Electrical Engineering 1
Friday Revision Lecture 11
Maximum Power Transfer theorems in a.c.
Circuits
And Power in a.c. Circuits
Semester 2, 2014
1
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
IS
VS
Load A
Load B
Load C
• 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?
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Reactive Power Compensation Example
IS
A:
500kVA
0.6pf
11kV
B:
300kW
0.8pf
C
Item
P
(kW)
Q
(kVAR)
|S|
(kVA)
pf
cos(θ)
θ
A
300
400
500
0.6
53.13°
B
300
225
A+B
600
625
375
0.8
36.87°
Method:
1. Draw up power
table
2. Compute all
elements in table
pf  cos( )
S  P  jQ
| S | P 2  Q 2
P | S | cos( )
866.39 0.6925 46.17°
Q | S | sin( )
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Reactive Power Compensation Example
Item
P
(kW)
Q
(kVAR)
|S|
(kVA)
pf
cos(θ)
θ
A
300
400
500
0.6
53.13°
B
300
225
375
0.8
36.87°
A+B
600
625
866.39 0.6925
46.17°
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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Method:
1. Draw up power
table
2. Compute all
elements in table
3. Derive further
quantities needed
pf  cos( )
S  P  jQ
| S | P 2  Q 2
P | S | cos( )
Q | S | sin( )
| S || V || I |
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Reactive Power Compensation Example
Item
P
(kW)
Q
(kVAR)
|S|
(kVA)
pf
cos(θ)
θ
A
300
400
500
0.6
53.13°
B
300
225
375
0.8
36.87°
A+B
600
625
C
0
-334.41
0
-90°
A+B+C
600
290.59 666.67
0.9
25.84°
866.39 0.6925
46.17°
Method:
1. Draw up power
table
2. Compute all
elements in table
3. Derive further
quantities needed
4. Extend table to
include
compensation
Q | S | sin( ) P | S | cos( ) pf  cos( )
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|S |
| I S |
 60.61 A
|V|
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Reactive Power Compensation Example
IS
500kVA
0.6pf
11kV
300kW
0.8pf
C
Q  334.41kVAR
  V * 
Q  imag VC I C*   imag VC  C  
 Z C  


 VCVC* 
 imag 
*
1
/
(
j

C
)

 Q   V 2 C
C
C
Q
2 f VC
2
C
334.41k
2 50 11k 
2
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Method:
1. Draw up power
table
2. Compute all
elements in table
3. Derive further
quantities needed
4. Extend table to
include
compensation
5. From Qc and line
voltage, find C
 8.80  F
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RMS Values
Suppose v(t) is a periodic “sawtooth”
waveform. v(t)
12V
5mS
10mS
15mS
t
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?
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Maximum Power Transfer Theorem
in 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 Exam
1. 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.
a
5F
4cos(200t+30°)A
10H
2k
b
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