Download EGN 3373 Week 12b – AC Power

EGN 3373 Introduction
to Electrical Systems I
A Systems Approach to Electrical Engineering
Graphics Adapted from “Physical, Earth, and Space Science”, Tom Hsu, cpoScience.
Let’s Look
at Power in AC Circuits
Emphasis is placed on Power in AC circuits, and to
facilitate the discussion the following voltage and current
signals will be used along with impedance Z.
P = VRMS I RMS [ cos(! v ! ! i )]
Power in AC Circuits
Based on the definition for the average power:
of For
sines
and cosines over a full cycle (period) is zero, the
a Resistive Load the phase angle is zero
ns after
integration.
and therefore
the average power becomes:
The residential voltage of 110 V is actually the RMS value of
the sinusoid signal delivered by the power company, which
ad the
phase
angle is value
zero of
(i.e.
! v ! ! i ) )isand
therefore the
means
the maximum
the(sinusoid
approximately
155 volts.
2
VRMS
2
P = VRMS I RMS =
= I RMS R
R
ge of 110 V is actually the RMS value of the sinusoid signal d
Effective RMS Values
The value of an AC voltage is continually
oscillating from the positive peak to the
negative peak through zero. This
makes the voltage value less than the
peak value during most of the cycle.
We use the root mean square (rms)
value for sine waves (V and I) to
represent the effective value of a varying
voltage or current.
VRMS is 0.7 of the peak voltage Vpeak
In addition, effective or RMS (root mean square) values of a periodic signal
are defined in terms of the average power delivered to a resistive load.
Effective RMS Values
In addition, effective or RMS (root mean
square) values of a periodic signal are
defined in terms of the average power
delivered to a resistive load.
Applying a periodic voltage v(t) with
period T to a resistive load R, then
power delivered is defined as
P(t) = v2(t) / R
Furthermore, the average power delivered
to the resistance is the energy delivered in
one cycle divided by the period T
Pavg = ET/T = Vrms2/ R
where
and demonstrates that resistors dissipate energy while inductors and capacitors store energy.
The terms “leading” and “lagging” are also introduced.
RESISTIVE!LOAD:!
Z R = Z R ! 0°
Power in AC Circuits
Im
I
v(t) = VM cos(! t + ! ) = VM ! !
i(t) = I M cos(! t + ! ) = I M ! !
I, V, and P for a Resistive Load
V
φ
For a purely resistive load
the current and voltage will be in phase.
0
The instantaneous power is given by:
p(t) = v(t)²i(t) = VM I M cos2 (! t + ! )
INDUCTIVE!LOAD:!
Z L = j! L = ! L! 90°
Im
v(t) = VM cos(! t + ! ) = VM ! !
V
i(t) = I M cos(! t + ! ² 90°) = I M ! (! ² 90°)
φ
For a purely inductive load the current and voltage
will be out of phase by 90°.
The current is lagging the voltage.
The instantaneous power is given by:
0
Re
I
CAPACITIVE!LOAD!
Power in AC ZCircuits
1
j
=
=!
=
C
j! C
!C
I, V, and P for a Capacitive Load
1
² ! 90°
!C
v(t) = VM cos(! t + ! ) = VM ² !
i(t) = I M cos(! t + ! + 90°) = I M ² (! + 90°)
Im
I
φ
0
Re
V
For a purely capacitive load the current and voltage
will be out of phase by 90°.
The current is leading the voltage.
The instantaneous power is given by:
p(t) = v(t)#i(t) = !
VM I M
sin 2(! t + ! )
2
The above derivations indicate that the average power for inductors and capacitors will be zero,
simply because this quantity is a sinusoid in both cases, and therefore its average value over
one cycle will be zero. On the other hand, the average power for a resistor is the square of the
cosine, the integral of which over one cycle is not zero; these results simply suggest that
resistances dissipate power while inductors and capacitors are (ideally) storage elements.
The phase difference between the current and voltage for an inductor and a capacitor results
in the current lagging the voltage for an inductor, while for a capacitor the current leads the
The instantaneous power is given by:
p(t) = v(t)²i(t) = VM I M cos2 (! t + ! )
INDUCTIVE!LOAD:!
Power in AC Circuits
Z L = j! L = ! L! 90°
I, V, and P for a Inductive Load
v(t) = V
M
Im
cos(! t + ! ) = VM ! !
V
i(t) = I M cos(! t + ! ² 90°) = I M ! (! ² 90°)
φ
For a purely inductive load the current and voltage
will be out of phase by 90°.
0
Re
I
The current is lagging the voltage.
The instantaneous power is given by:
p(t) = v(t)#i(t) =
Prepared'by'C.'S.'Ferekides'
VM I M
sin 2(! t + ! )
2
83
•Three-Phase Systems
•Why not DC?
•
No easy way to change voltage levels
•
Complicated rotating machine design and construction
•Why not single-phase AC?
•
Pulsating torque
•
Efficiency issue: kVA per MCM of copper
•Two-Phase Power
•Two-Phase Power
•Two-Phase Power
•Two-Phase Power
•AC Power
•v(t) = Vm cos (t + v)
•i(t) = Im cos (t + i)
•s(t) = v(t) x i(t)•= Vm Im cos (t + v) cos (t + i)
1
cos   cos    cos      cos    
2
•s(t) = (Vm Im)/2 [cos (v  i) + cos (2t + v + i)]
•AC Power
•s(t) = (Vm Im)/2 [cos (v  i) + cos (2t + v + i)]
•2t + v + i = 2(t + v) – (v – i)
Vm Im
s(t) 

 cos v  i   cos2  t  v   v  i  
2
2
•cos ( – ) = cos  cos  + sin  sin 
s(t)  Vrms Irms cosv  i   cos2 t  v  cosv  i   sin 2  t  v  sin v  i  
•AC Power
s(t)  Vrms Irms cosv  i   cos2  t  v  cosv  i   sin2  t  v  sin v  i  
•Constant component of real power
•Oscillating component of real power
•Reactive power
•Let P = Vrms Irms cos (v – i)
•and Q = Vrms Irms sin (v – i)
•s(t) = P + P cos [2(t + v)] + Q sin [2(t + v)]
•AC
Power
•AC Power
•AC Power
•s(t) = P + P cos [2(t + v)] + Q sin [2(t + v)]
•P = Vrms Irms cos (v – i)
•Q = Vrms Irms sin (v – i)
•S = P + j Q•= Vrms Irms [cos (v – i) + j sin (v – i)]
•P = P + P cos [2(t + v)]
•Q = Q sin [2(t + v)]
•AC Power
•S = P + j Q•= Vrms Irms [cos (v – i) + j sin (v – i)]
cos   j sin   e j 
•S = Vrms Irms e j(v –
 i)
•= Vrms Irms e j(v) e j(–
 i)
•S = Vrms e j(v) Irms e j(–
 i)
•= Vrms v Irms –
i
•S = V I*
•Complex Conjugate
•Power Triangle
•Our electrical system must
handle this component.
•Total or Apparent Power
•S
•kVA
•
•Real Power
•P
•kW
•This component of the
power furnishes the energy
required to establish and
maintain
electric
and
magnetic fields.
•Reactive Power
•Q
•kVAR
•THEREFORE, VARS REDUCE
THE “USEFULNESS” OF THE
TOTAL ELECTRIC POWER.
•cos   Power Factor
•This is the only “useful” work done.
•Power Triangle
•
•Power triangle in first quadrant (positive )
•indicates INDUCTIVE load.
•Power Triangle
•
•Power triangle in fourth quadrant (negative )
•indicates CAPACITIVE load.
•Power Triangle
•Z
•S
•Q
•
•X
•R
•R = resistance
•P
•X = reactance
•Z = impedance
•Power triangle and impedance triangle are
•SIMILAR TRIANGLES.
•
•Three-Phase Systems
•Three-phase systems
•
Most common: 3-wire delta and 4-wire wye
•Mathematical representation
•
Time domain (oscilloscope view)
•
Frequency domain (phasor diagram)
•Three-Phase Systems
Positive Sequence (A-B-C)
A
B
C
A
B
C
A
B C
A
A

time
C
Time Domain
B
Frequency Domain
•Three-Phase Systems
Negative Sequence (A-C-B)
A
C
B
A
C
B
A
C B
A
A

time
B
Time Domain
C
Frequency Domain
•Three-Phase Systems
•Delta-connected loads
•
Line-to-line voltage only
•
No neutral, so best for balanced loads
•
Ideal for motors
•Wye-connected loads
•
Line-to-line AND line-to-neutral voltage
•
Better for imbalanced loads, particularly mix of
three-phase and single-phase
•Mathematical Relationships
•P = |V|  |I|  cos 
•Q = |V|  |I|  sin 
•S = V  I*
•* denotes COMPLEX CONJUGATE
•Recommendation:
•For three-phase systems, use LINE-TO-LINE
voltages and THREE-PHASE POWER values.
•Alternatively, LINE-TO-NEUTRAL voltages and
POWER PER PHASE.
A Simple Derivation
A Simple Derivation
Based on the definition for the average power:
Power
inT AC Circuits
efinition for the average power:
T
1
1
P = ! p(t)dt = ! VM I M cos(! t + ² v )cos(! t + ² i )dt =
T 0
Based on the definition
forTthe
average power:
0
1
P=
T
1
=
T
T
!
0
T
11T 1
cos(!² vt ²+ ²²iv))cos(
+ cos(2
p(t)dt== ! VVMMIIMM [cos(
! t! +t +² i²)dt
v + ²=i )] dt =
TT 0 02
!
1
1
= VM I M cos(! v ² ! i )
! 2 VM I M [ cos(
2 ² v ² ² i ) + cos(2! t + ² v + ² i )] dt =
0
T
1
= VM I M cos(! v ² ! i )
2
Therefore:
1
P = VM I M [ cos(! v ! ! i )]
2
Therefore:
1
using
P = VM I M [or
cos(
! v ! the
! i )]RMS values:
2
P = VRMS I RMS [ cos(! v ! ! i )]
Based on the definition for the average power:
Power
inT AC Circuits
efinition for the average power:
T
1
1
P = ! p(t)dt = ! VM I M cos(! t + ² v )cos(! t + ² i )dt =
T 0
Based on the definition
forTthe
average power:
0
1
P=
T
1
=
T
T
!
0
T
11T 1
cos(!² vt ²+ ²²iv))cos(
+ cos(2
p(t)dt== ! VVMMIIMM [cos(
! t! +t +² i²)dt
v + ²=i )] dt =
TT 0 02
!
1
1
= VM I M cos(! v ² ! i )
! 2 VM I M [ cos(
2 ² v ² ² i ) + cos(2! t + ² v + ² i )] dt =
0
T
1
= VM I M cos(! v ² ! i )
2
Therefore:
1
P = VM I M [ cos(! v ! ! i )]
2
Therefore:
1
using
P = VM I M [or
cos(
! v ! the
! i )]RMS values:
2
P = VRMS I RMS [ cos(! v ! ! i )]
EXAMPLES