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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