AC Power Analysis 3-Phase

Prepared for

Electrical Engineering Laboratory II, ECE L302

by

Mohammed Muthalib

Center for Electric Power Engineering

Drexel University

(http://power.ece.drexel.edu) 1

Outline

Power Generation Fundamental Concepts (recap)

1-Phase: V, I, S,P, Q, PF

Balanced 3-phase Line-line vs. line-neutral voltage 3-phase power

S3Φ, P3Φ, Q3Φ,

Why 3-phase? This weeks experiment Power lab safety

2

Generation

3

Nuclear plant Hydro plant

Coal plant Fossil fuel plant

Generation

Conservation of energy

4

Generation

Fossil Fuel: Burn fuel Steam Turbine Generator

Nuclear: Nuclear reaction Steam Turbine Generator

Hydro: P.E. of reservoir K.E. of water Turbine Generator

Wind: K.E. of wind rotational K.E. of blades in turbine Generator

Solar: Solar radiation semiconductor energy transfer Electron flow

Solar and wind require storage and AC/DC conversion

5

Generation

Turbine – Generator Turbine provides mechanical torque to the rotor of the generator Stator has coils that react to the electro-magnetic field variations

6 http://dcacmotors.blogspot.com/2009/03/basic-generator.html

Generation

3-phase generation Stator has 3 coils wound 120° out of phase

7

ECE Department, University of Minnesota http://www.ece.umn.edu/users/riaz/animations/alternator.html

Fundamental Concepts

AC voltage and current Sinusoidal time varying waveforms

Phasor representation (time invariant)

8

( ) ( )( ) ( )

peak V

peak I

V

I

v t V sin t

i t I sin t

voltage phase anglecurrent phase angle

= ω + θ

= ω + θ

θ −

θ −

V

I

jV

jI

peak

V V e

I I e

Where V and I are rms values

V (amplitud ) V

V

I

e 2

θ

θ

= ∠θ =

= ∠θ =

=

Voltage and current phasors (polar coordinates)

Fundamental Concepts

Voltage and current waveforms

9

Voltage and current phasors (polar coordinates)

Fundamental Concepts

Instantaneous power: Product of v(t) and i(t)

10

( ) ( )( ) ( )( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( )

V

I

V I

V I

V I V I

v t 2 V sin t

i t 2 I sin t

p t v t * i t

2 V I sin t sin tuse trigonometric identity2sin usin v cos u v cos u vu t , v tp t V I cos V I cos 2 t

= ω + θ

= ω + θ

=

= ω + θ ω + θ

= − − +

= ω + θ = ω + θ

= θ − θ − ω + θ + θ

time invariant (constant)

time varying with frequency 2ω

Fundamental Concepts

Instantaneous power

11

( ) ( ) ( )p t v t * i t=

Fundamental Concepts

Complex power: Is a representation of power in a complex vector space Denoted S with units of VA

|S| - apparent power

12

2 2S P Q V I= + =

( )( ) ( )

( ) ( )

V I

*

jV I

V I V I

S VI *denotes complex conjugate

S V I V I eusing Euler's formulaS V I cos j V I sin

P jQ

θ −θ

=

= ∠ θ − θ =

= θ − θ + θ − θ

= +

Fundamental Concepts

Real power: Real power = Average power

Denoted P with units of Watts Power absorbed by the resistive components of the system

Reactive power: Denoted Q with units of VAR Power absorbed by the reactive components of the system Example, Inductance and Capacitance

13

( )AVG V IP P V I cos= = θ − θ

( )V IQ V I sin= θ − θ

Fundamental Concepts

Power Factor: Is a measure of how effectively a system component draws real

power. It is the ratio between real power and apparent power

PF is presented as a real number between 0 and 1 with a leading/lagging denotation for the PF angle o Lagging - current angle lags the voltage angle, θV>θI

o Leading - current angle leads the voltage angle, θV<θI

14

( )V IPPF cosS

= = θ − θ

Fundamental Concepts

Power Factor: Which figure represents a load with a lagging PF?

o What kind of load has a lagging power factor? Which figure represents a load with a leading PF?

o What kind of load has a leading power factor?

15

Fundamental Concepts

PF=1 indicates that all the power consumed in the system is real power. A load with PF=1 emulates a resister Reactive power draw is zero

Is PFC necessary? Why?

16

Fundamental Concepts

PFC by injecting reactive power

Reactive power injection is done by adding capacitive and inductive loads to the system. Capacitive – supply reactive power Inductive – consume reactive power

17

Balanced 3-phase

Balanced 3-phase circuit All components, loads, lines, etc. are assumed to be identical

Balanced 3Φ circuits are analyzed through a single phase equivalent analysis

18

Balanced 3-phase

Balanced 3-phase voltages

19

( )( )( )

an an

bn an

cn an

V 2 V sin t 0 Volts

V 2 V sin t 120 Volts

V 2 V sin t 120 Volts

= ω + °

= ω − °

= ω + °

( )( )( )

a a I

b a I

c a I

I 2 I sin t 0 Amps

I 2 I sin t 120 Amps

I 2 I sin t 120 Amps

= ω + θ + °

= ω + θ − °

= ω + θ + °

Line-line voltage

Line-line voltage Voltage measure between phases, Vab, Vbc, Vca

20

( )( ) ( )

ab an nb

a

ab an b

n bn

n

an

ab an

V V cos 30 V cos 30

3 V

V V VV

V V 30

V= +

= +

= ° + °

=

∠ = ∠ +

−

°

an

bn

ab

V 120 0 VV 120 120 VV 208 30 V

= ∠ °

= ∠− °

= ∠ °

Example:

3-phase power

Real Power P3Φ= PA+ PB+ PC

21

( )( )( )

( ) ( )( )( )

A A A V,A I,A

B B B V,B I,B

C C C V,C I,C

V,B V,A I,B I,A

V,C V,A I,C I,A

A B C A B C

B A A V,A I,A

A A V,A I,A

A

P V I cos Watts

P V I cos Watts

P V I cos Watts

120 , 120120 , 120

V V V , I I I

P V I cos 120 120

V I cos

P

= θ − θ

= θ − θ

= θ − θ

θ = θ − ° θ = θ − °

θ = θ + ° θ = θ + °

= = = =

= θ − ° − θ − °

= θ − θ

=

Balanced 3-phase voltage phasor

3-phase power

Real Power

Reactive power

Complex Power

22

( )( )

3 A B C

3 an a V,A I,A

3 LL L V,A I,A

P P P P Watts

P 3 V I cos Watts

P 3 V I cos Watts

Φ

Φ

Φ

= + +

= θ − θ

= θ − θ

( )( )

3 A B C

3 an a V,A I,A

3 LL L V,A I,A

Q Q Q Q VAr

Q 3 V I sin VAr

Q 3 V I sin VAr

Φ

Φ

Φ

= + +

= θ − θ

= θ − θ

Balanced 3-phase voltage phasor ( )*3 A A

3 3 3

S 3V I VAS P jQ VA

Φ

Φ Φ Φ

=

= +

LL ab anV V 3 V= =

Why 3-phase?

Electrical energy has to be transmitted/distributed over long distances Cables are heavy (engineering problem to string up in the air) and

lossy

23

https://www.osha.gov/SLTC/etools/electric_power/illustrated_glossary/transmission_lines.html

Why 3-phase?

Neutral Current (balanced 3-Φ)

24

( )( )( )

( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )( ) ( ) ( )( )

a a I

b a I

c a I

n a b c

a I I I

a I I I

a I I

a I I

I 2 I 0 Amps

I 2 I 120 Amps

I 2 I 120 Amps

I I I I

2 I cos cos 120 cos 120

j 2 I sin sin 120 sin 120

2 I cos 2cos cos 120

j 2 I sin 2sin cos 120

0

= ∠ θ + °

= ∠ θ − °

= ∠ θ + °

= + +

= θ + θ − ° + θ + °

+ θ + θ − ° + θ + °

= θ + θ − °

+ θ + θ − °

=

Why 3-phase?

Transmission of power in balanced 3-phase can be done without a neutral conductor

More efficient and economical since: less losses from conductor Less cost with respect to material and construction

NOTE: Distribution systems are unbalanced and require a neutral

conductor Connect back to local electrical substations Supplied through distribution transformers

25

Electrical Supply

Transmission/Distribution

26 http://www.avalon-energy.com/sample.aspx?C1=28

Electrical Supply

Residential connection

27

http://waterheatertimer.org/See-inside-main-breaker-box.html

Lab Experiment

Simulating a 2-bus power system (Transmission) Measure voltages and currents; calculate power Observe effects of loading

28

Lab Experiment

Test system with 3 load types Resistive Load (4 – 20 bulbs) RL (20 bulbs + 0 - 5 inductors) RC (20 bulbs + 0 - 4 capacitors) Nonlinear Load

29

Lab safety

You will be dealing with voltages and currents of 120V and 25A

Adherence to safety and conduct guidelines is imperative Please read the Power Lab Safety document Please watch lab safety video

http://power.ece.drexel.edu/videos/

30

31

Thank you

Questions are most welcome