Download average power - Department of Electrical Engineering

Chapter 11
AC Steady-State Power
Matching Network for Maximum Power Transfer
Cellular Telephone
Design the matching network to transfer maximum
power to the load where the load is the model of
an antenna of a wireless communication system.
George Westinghouse, 1846-1914
The greatest engineer of his day,
George Westinghouse modernized
the railroad industry and established
the electric power system.
Nikola Tesla, 1856-1943
Tesla was responsible for many inventions,
including the ac induction motor, and was a
contributor to the selection of 60Hz as the
standards ac frequency in the United States.
Instantaneous Power and Average Power
Instantaneous Power
p (t )  v (t )  i (t )
A circuit element
If v(t) is a periodic function
v (t )  v (t  T )
Then for a linear circuit i(t) is also a periodic function
i (t )  i (t  T )
 p (t )  v (t  T )  i (t  T )
Instantaneous Power and Average Power(cont.)
Average Power
1
P
T
t0 T

p(t ) dt
t0
Arbitrary point in time
If v(t) is a sinusoidal function
v(t )  Vm (cos t  V )
For a linear circuit i(t) is also a sinusoidal function
i (t )  I m (cos t   I )
p(t )  Vm I m (cos t  V )(cos t   I )
Vm I m
p (t ) 
cos(V   I )  cos(2 t  V   I )
2
T
1 Vm I m
P 
cos(V   I )  cos(2 t  V   I ) dt
T 0 2
T
T
1 Vm I m
1 Vm I m
 
cos(V   I ) dt  
cos(2 t  V   I ) dt
T 0 2
T 0 2
T
T
Vm I m
1 Vm I m

cos(V   I )  dt  
cos(2 t  V   I ) dt
2T
T 0 2
0
Vm I m

cos(V   I )  0
2
Vm I m

cos(V   I )
2
average value of the cosine
function over a complete
period is zero
Example 11.3-1
P=?
Using the period from
t = 0 to t = T
Im
i  t ;0t T
T
i(t) through a resistor R
The instantaneous power is
2
Im R 2
pi R 2 t
T
;0t T
2
The average power is
2
1 Im R 2
P 
t dt
2
T 0 T
T
2
2
3
2
Im R
Im R T
Im R
2
P  3  t dt  3

0
T
T
3
3
T
W
Example 11.3-2
PL = ? PR = ?
i (t )  721cos(100t  41) mA
The element voltages are
vs (t )  20cos(100t  15) V
vR (t )  18cos(100t  41) V
vL (t )  8.66cos(100t  49) V
The average power delivered by the voltage source is
(20)(0.721)
Ps 
cos( 15  ( 41))  6.5 W
2
The average power delivered to the voltage source is 6.5 W
Example 11.3-2 (cont.)
The average power delivered to the resistor is
(18)(0.721)
PR 
cos( 41  ( 41))  6.5 W
2
The average power delivered to the inductor is
(8.66)(0.721)
PL 
cos(49  ( 41))  0 W
2
WHY the average power delivered to the inductor = 0 ?
The angle of vL always be 90larger than the angle of iL
and cos(90)  0
Effective Value of a Periodic Waveform
The goal is to find a dc voltage, Veff
(or dc current, Ieff),
for a specified vs(t) that will deliver the same average power
to R as would be delivered by the ac source.
The energy delivered in a period T is
W  PT
The average power delivered to the resistor by a periodic
current is
1 T
P

T
0
i 2 Rdt
Effective Value of a Periodic Waveform (cont.)
The power delivered by a direct current is
PI R
2
eff
1
P 
T
Solve for Ieff

T
0
i Rdt  I R
2
2
eff
1 T 2
I eff 
i dt

T 0
 I rms rms = root-mean-square
The effective value of a current is the steady current (dc)
that transfer the same average power as the given time
varying current.
Example 11.4-1
Ieff = ?
Express the waveform over
the period of t = 0 to t = T
Im
i  t ;0t T
T
i(t) = sawtooth waveform
I eff
1 T 2
1 T I m2 2

i dt 
t dt
2


T 0
T 0 T
T
I t 



3
T  T 0
2
m
3
3
 I eff
2
m
I
3
Im

3
Complex Power
A linear circuit is excited by a sinusoidal input and the circuit
has reached steady state.
The element voltage and current can be represented in
(a) the time domain
or
(b) the frequency domain
Complex Power (cont.)
To calculate average power from frequency domain
representation of voltage and current i.e. their phasors
I( )  I m I
and V( )  VmV
The complex power delivered to the element is defined to be
VI
( I m   I )(VmV )
S

2
2
I mVm

(V   I )
2
*
Apparent power
where I  complex conjugate of I
*
Complex Power (cont.)
The complex power in rectangular form is
I mVm
I mVm
S
cos(V   I )  j
sin(V   I )
2
2
P
Q
or
S  P  jQ
real or average power
reactive power
Units S : VA, P:W, Q:VAR
Volt-Amp
Volt-Amp Reactive
Complex Power (cont.)
The impedance of the element can be expressed as
V ( ) VmV Vm
Z( ) 


(V   I )
I( ) I m I I m
In rectangular form
Vm
Vm
Z( )  cos(V   I )  j sin(V   I )
Im
Im
R
X
or
Z( )  R  jX
resistance
reactance
Complex Power (cont.)
The complex power can also be expressed in terms of
the impedance
I mVm
I mVm
S
cos(V   I )  j
sin(V   I )
2
2
2
2
 I m  Vm
 I m  Vm
   cos(V   I )  j   sin(V   I )
 2  Im
 2  Im
I

 2
2
m

 Re( Z) 

P
I
j
 2
2
m

 Im( Z)

Q
Complex Power (cont.)
The impedance triangle
The complex power triangle
The complex power is conserved

all
elements
*
k k
VI
0
2
The sum of complex power absorbed by all elements
of a circuit is zero.
Complex Power (cont.)
The complex power is conserved implies that both
average power and reactive power are conserved.

*
VI
V
I

k k

0

Re



2
2
all
all

elements
 elements

*
V
I

k k
Re
0


2
all
 elements

*
V
I
 Im   k k
 all
2
 elements

*
V
I

k k
Im
0


2
all
 elements
*
k k
and
or

all
elements
Pk  0 and

all
elements
Qk  0
Example 11.5-1
S is conserved ?
vs  100cos1000t
V
=1000
Solving for the mesh current
Vs
I( ) 
 7.07  45
1
R  j L  j
C
Use Ohm’s law to get the element voltage phasors
VR ( )  RI( )  70.7  45
VL ( )  j LI( )  141.445
j
VC ( ) 
I( )  70.7  135
C
Example 11.5-1 (cont.)
Consider the voltage source
Vs I*
supplied by the source
SV 
2
 353.545 VA
For the resistor
VR I*
SR 
absorbed by the resistor
2
 2500 VA
For the inductor
VLI* delivered to the inductor
SL 
2
 50090 VA
Example 11.5-1 (cont.)
For the capacitor
VC I*
delivered to the capacitor
SC 
2
 250  90 VA
The total power absorbed by all elements (except source)
S R  S L  SC  2500  50090  250  90
 353.545  SV
For all elements

all
elements
Vk I*k
0
2
Example 11.5-2
P is conserved ?
vs  100cos1000t
V
=1000
The average power for the resistor, inductor, and capacitor is
 I m2 
P    Re( Z)
2 

 I m2 
PR    R  250 W PL  PC  0
 2 
The average power supplied by the voltage source is
 Vs I* 
PV  Re  SV   Re 

2


 Re(353.545)  250 W
Power Factor
The ratio of the average power to the apparent power is
called the power factor(pf).
I mVm
average power P 
cos(V   I )
2
S apparent power
 pf  cos (V   I )
pf angle
Therefore the average power
I mVm
P 
pf
2
Power Factor (cont.)
The cosine is an even function cos(  )  cos( )
 pf  cos(V   I )  cos( I  V )
Need additional information in order to find the angle
Ex
pf  0.8 leading for V   I  0  36.87
and
pf  0.8 lagging for V   I  0  36.87
Ex The transmission of electric power
Time
domain
Power Factor (cont.)
Frequency domain
We will adjust the power factor by adding compensating
impedance to the load. The objective is to minimize the
power loss (i.e. absorbed) in the transmission line.
R1
L1 R1
L1
The line
Z LINE ( )   j   j
impedance
2
2 2
2
 R1  j L1
Power Factor (cont.)
The average power absorbed by the line is
PLINE
I m2
I m2
 Re( Z LINE ) 
R1
2
2
The customer requires average power delivered to the load P
at the load voltage Vm
Solving for Im
Vm I m
P
pf
2
2P
Im 
Vm pf
2
 PLINE
 P 
 2
 R1
 Vm pf 
max pf =1
Power Factor (cont.)
compensating
impedance
A compensating impedance has been attached across the
terminals of the customer’s load.
pf  1
pfc  cos c
corrected
The load impedance is Z  R  jX and the
compensating impedance is ZC  RC  jX C
We want ZC to absorb no average power so ZC  jX C
Power Factor (cont.)
The impedance of the parallel combination ZP
ZZC
ZP 
 RP  jX P  Z P P
Z  ZC
The power factor of the new combination
 1 X P 
pfc  cos P  cos  tan

RP 

Calculate for R and X
P
P
ZZC
( R  jX ) jX C
ZP 

Z  ZC ( R  jX )  jX C
RX C2  j  R 2 X C  ( X C  X ) XX C 

R 2  ( X  X C )2
RX C2
R 2 X C  ( X C  X ) XX C
 2
j
2
R  ( X  XC )
R 2  ( X  X C )2
Power Factor (cont.)
X P R2  ( X C  X ) X

RP
RX C
 1 X P 
From pfc  cos  tan

RP 

XP
 tan(cos 1 pfc)
RP
Solving for XC
R2  X 2
XC 
R tan(cos1 pfc)  X
Typically the customer’s load is inductive  ZC = capacitive
j
ZC 
 jX C
C
1
R2  X 2

C R tan(cos1 pfc)  X
Power Factor (cont.)
Solving for C
X  R tan(cos1 pfc)
C 
R2  X 2
R X

1
 2

tan(cos
pfc
)

2 
R X R

Let   tan 1  X 
R
R
C  2
tan   tan  C 
2 
R X
where
  cos1  pf  and C  cos1  pfc 
Example 11.6-1
I and pf = ?
Load = 50 kW of heating (resistive) and motor 0.86 lagging pf
Load 1 50 kW resistive load
S1  P1  50 kW
Load 2 motor 0.86 lagging pf 2  0
1
1
2  cos ( pf 2 )  cos (0.86)  30.7 
P
Q
S2  S2 2  10030.7  86  j51 kVA
Example 11.6-1 (cont.)
S  S1  S2  136  j51  145.220.6 kVA
pf  cos(20.6)  0.94
To calculate the current
Vm I m
S
 Vrms I rms
2
S
 I rms 
Vrms
145200

4
10
 14.52 Arms
Example 11.6-2
pf ==> 0.95, 1 C = ?
  377 rad/s
Z=100+j100 
cos  cos 45  0.707
pf lagging
We wish to correct the pf to be pfc
pfc  0.95 lagging
R2  X 2
XC 
R tan(cos1 pfc)  X
 297.9
1
C
 8.9 μF
 XC
Example 11.6-2 (cont.)
pfc  1
R2  X 2
XC 
R tan(cos1 pfc)  X
 200
1
C
 13.3 μF
 XC
Or use
R
C  2
tan   tan  C 
2 
R X
C  13.3 μF
The Power Superposition Principle
i  i1  i2
p  i R  (i1  i2 ) R  (i  i  2i1i2 ) R
2
2
2
1
2
2
1 T
R T 2 2
P   pdt   (i1  i2  2i1i2 )dt
T 0
T 0
R T 2
R T 2
2R T
  i1 dt   i2 dt 
i1i2dt

T 0
T 0
T 0
2R T
0
 P1  P2 
i1i2dt

T 0
The Power Superposition Principle (cont.)
2R T
i1i2dt  0 ?

T 0
Let the radian frequency of the 1st source = m and
the radian frequency of the 2nd source = n
integer
i  I cos(m t   )
1
1
i2  I 2 cos(n t   )
2R T
P12 
i1i2dt

T 0
2R T

I1 I 2 cos(m t   ) cos(n t   )dt

T 0
The Power Superposition Principle (cont.)
2 RI1 I 2 T
 P12 
cos(m t   ) cos(n t   )dt

0
T
2 RI1 I 2 T

(cos(( m  n ) t  (   ))  cos(( m  n ) t  (   )))dt

0
T
;m  n
0

  RI1 I 2 cos(   )
;m  n

2
For the case that m and n are not integer
for example m = 1, n = 1.5     0
1 2t
1 2t
P12  lim  t pdt  lim  t 2 RI1I 2 cos  t cos(1.5 t )dt
t  T 
t  T 
2
2
1 2t
 lim  t 2 RI1 I 2 (cos0.5 t  cos 2.5 t )dt  0
t  T 
2
The Power Superposition Principle (cont.)
The superposition of average power
The average power delivered to a circuit by several
sinusoidal sources, acting together, is equal to the sum
of the average power delivered to the circuit by each
source acting alone, if and only if, no two of the source
have the same frequency.
If two or more sources are operating at the same frequency
the principle of power superposition is not valid but the
principle of superposition remains valid.
For N sources I  I1  I 2  I3 
 IN
I m2 R
P
2
Example 11.7-1
P=?
(1) v A (t )  12cos3t
V and iB (t )  2cos 4t A
(2) v A (t )  12cos 4t
V and iB (t )  2cos 4t A
Example 11.7-1(cont.)
Case I I1 ( )  1.414  45 and
I 2 ( )  1.6  143
These phasors correspond to different frequencies and
cannot be added.
i1 (t )  1.414cos(3t  45) and i2 (t )  1.6cos(4t  143)
Using the superposition
i (t )  1.414 cos(3t  45)  1.6cos(4t  143)
The average power can be calculated as
R T
P   (1.414 cos(3t  45)  1.6cos(4t  143)) 2 dt
T 0
Since the two sinusoidal sources have different frequencies
1.4142
1.62
P  P1  P2 
6
6  13.7 W
2
2
Example 11.7-1(cont.)
Case II I1 ( )  1.2  53.1 and I 2 ( )  1.6  143
Both phasors correspond to the same frequency and
can be added.
I ( )  1.2  53.1+1.6  143  2.0  106.3
The sinusoidal current is
i (t )  2.0cos(4t  106.3)
The average power can be calculated as
2.02
P
6  12 W
2
Power superposition cannot be used here because
Both sources have same frequencies
The Maximum Power Transfer Theorem
Zt  Rt  jX t
and
ZL  RL  jX L
Vt
I
( Rt  jX t )  ( RL  jX L )
2
2
m
Vt RL
I
P  RL 
2
( Rt  RL )2  ( X t  X L )2
We wish to maximize P set X L   X t
2
Vt RL
P
2
( Rt  RL )
dP
For
 0 we get RL  Rt
dRL
 Z L  Rt  jX t  Z*t  Maximum Power Transfer
Coupled Inductors
di1
di2
v1  L1
M
dt
dt
di2
di1
v2  L2
M
dt
dt
(a) both coil currents enter
the dotted ends of the coils
di1
di2
v1  L1
M
dt
dt
di2
di1
v2  L2
M
dt
dt
(b) one coil current enters the
dotted end of the coil, but
the other coil current enters
the undotted end
Summary
Instantaneous Power and Average Power
Effective Value of a Periodic Waveform
Complex Power
Power Factor
The Power Superposition Principle
The Maximum Power Transfer Theorem
Coupled Inductor and Transformer