Download Chapter 5 Steady-State Sinusoidal Analysis

ECE 3183 – EE Systems
Chapter 5 – Part D
AC Power, Power Factor
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D1
EFFECTIVE OR RMS VALUES
i (t )
Instantaneous power
R
p( t )  i ( t ) R
2
The effective value is the equivalent DC
value that supplies the same average power
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D2
EFFECTIVE OR RMS VALUES
If current is DC (i (t )  I dc ) then
Pdc 
2
RI dc
I eff : Pav  Pdc
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D3
EFFECTIVE OR RMS VALUES
If current is periodic with period T
1
Pav 
T
2
I eff
 1 t0 T 2



p
(
t
)
dt

R
i
(
t
)
dt




T
t0
t


0
t 0 T
1

T
I eff 
Definition is valid
for ANY periodic
signal with period T
t 0 T
1
T
2
i
 (t )dt
t0
t 0 T
2
i
 (t )dt
t0
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D4
EFFECTIVE OR RMS VALUES
If the current is sinusoidal the average power is
1 2
Pav  I M R
2
2
 I eff
ECE 3183 – Chapter 5 – Part D
1 2
 IM
2
CHAPTER 5 - D5
LEARNING EXAMPLE
Compute the rms value of the voltage waveform
T 3
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D6
LEARNING EXAMPLE
Compute the rms value of the voltage waveform
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D7
LEARNING EXAMPLE
Compute the rms value of the voltage waveform and use it to
determine the average power supplied to the resistor
i (t )
i(t )  5 cos20000  t  Amps
ECE 3183 – Chapter 5 – Part D
R
R  2
CHAPTER 5 - D8
LEARNING EXAMPLE
Compute the rms value of the voltage waveform and use it to
determine the average power supplied to the resistor
T  4( s)
i (t )
R
ECE 3183 – Chapter 5 – Part D
R  2
CHAPTER 5 - D9
LEARNING EXTENSION
Compute rms value of the voltage waveform
T 4
v  2t
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D10
LEARNING EXTENSION
Compute the rms value for the current waveforms and use
them to determine average power supplied to the resistor
i (t )
X rms
R
1

T
t 0 T
2
x
 (t )dt
t0
2
Pav  I rms
R
T 6
R  4
T 8
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D11
LEARNING EXTENSION
Compute the rms value for the current waveforms and use
them to determine average power supplied to the resistor
i (t )
X rms
R
T 6
2
I rms
4
6

1 2
   4dt  16dt   4dt  
6 0

2
4
8  32  8
8
6
1

T
t 0 T
2
x
 (t )dt
t0
2
Pav  I rms
R
R  4
P  8  4  32(W )
T 8
2
I rms
6

1 2
   16dt   16dt   8
8 0

4
ECE 3183 – Chapter 5 – Part D
P  32(W )
CHAPTER 5 - D12
STEADY-STATE POWER ANALYSIS
LEARNING GOALS
Instantaneous Power
For the special case of steady state
sinusoidal signals
Average Power
Power absorbed or supplied during one cycle
Maximum Average Power Transfer
When the circuit is in sinusoidal steady state
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D13
STEADY-STATE POWER ANALYSIS
LEARNING GOALS
Effective or RMS Values
For the case of sinusoidal signals
Power Factor
A measure of the angle between current and
voltage phasors
Power Factor Correction
How to improve power transfer to a load by
“aligning” phasors
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D14
STEADY-STATE POWER ANALYSIS
LEARNING GOALS
Complex Power
Measure of power using phasors
Single Phase Three-Wire Circuits
Typical distribution method for households
and small loads
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D15
INSTANTANEOUS POWER
Instantane ous
Power Supplied
to Impedance
p(t)  v(t)i(t)
v(t )  4 cos t  60 Volts
Z  2  30

FIND i(t) and p(t)
We know that:
p(t )  v(t )i(t )  Vm I m cos(t  v ) cos(t  i )
where
1
cos(1 ) cos( 2 )  cos(1   2 )  cos(1   2 )
2
yielding:
Vm I m
cos( v   i )  cos(2t   v   i )
p (t ) 
2
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D16
INSTANTANEOUS POWER
V 460
I 
 230( A)
Z 230
LEARNING EXAMPLE
Assume : v (t )  4 cos( t  60),
Z  230
Find : i (t ), p(t )
i (t )  2 cos( t  30)( A)
VM  4, v  60
I M  2, i  30
p(t )  4 cos 30  4 cos(2 t  90)
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D17
INSTANTANEOUS POWER
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D18
AVERAGE POWER
For sinusoidal (and other periodic signals) we
compute averages over one period
1
P
T
t 0 T
 p(t )dt
t0
T
2

VM I M
cos( v   i )  cos(2 t   v   i )
p( t ) 
2
VM I M
P
cos( v   i )
2
ECE 3183 – Chapter 5 – Part D
DOES IT MATTER WHO LEADS?
CHAPTER 5 - D19
AVERAGE POWER
If voltage and current are in phase
1
 v   i  P  VM I M
2
Purely
resistive
(cos(0)=1)
If voltage and current are in quadrature
v   i  90  P  0
ECE 3183 – Chapter 5 – Part D
Purely
inductive or
capacitive
(cos(90)=0)
CHAPTER 5 - D20
AVERAGE POWER
LEARNING EXAMPLE
Find the average power absorbed
by the impedances

VR

1060 1060
I

 3.5315( A)
2  j 2 2 245
VM  _____, I M  _____,
 v  _____ , i  _____ 
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D21
AVERAGE POWER
LEARNING EXAMPLE
Find the average power
absorbed
by the impedances

VR

Since an inductor does
not absorb power, one
can use voltages and
currents for the resistive
part only.
2
VR 
1060  7.0615(V )
2  j2
1
P  7.06 V  3.53 A  12.48 W
2
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D22
LEARNING EXAMPLE
Determine the average power absorbed by each resistor,
the total average power absorbed and the average power
supplied by the source
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D23
LEARNING EXTENSION
I1 
Vs
Find the AVERAGE power absorbed by each PASSIVE
component and the total power supplied by the source
I2
1
P4   4  4.122 (W )
2
Pj 2  0(W )

I1 
4  j2
1030
3  4  j2
4.4726.57
I1 
1030  6.1440.62( A)
7.2815.95
1 2 1
P3  RI M
  3  6.142 (W )
2
2
I 2  1030  6.1440.62
Power supplied by source
Method 1. Psupplied  Pabsorbed
Psupplied  P3  P4  90.50W
Method 2: P  1 V I cos(   )
M M
v
i
2
Vs  3I1  18.4240.62
1
P   18.42  10  cos(40.62  30)
2
3
3030
1030 
3  4  j2
7.2815.95
 4.1214.05( A)
I2 
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D24
LEARNING EXAMPLE
Determine average power absorbed or supplied by each
element
1230  60 10.39  j 6  6

 6  j 4.39
j1
j
 7.43  36.19( A)
1
P60   6  7.43 cos(0  36.19)  18W
2
I3 
Passive sign convention
I2 
1230
 630( A)
2
I1  I 2  I 3  5.20  j 3  6  j 4.39  11.2  j1.39( A)
1 2 1
P2  RI M
  2  62  36(W )
2
2
Pj1  0
To determine power absorbed/supplied
by sources we need the currents I1, I3
Average Power
1
P  VM I M cos( v   i )
2
ECE 3183 – Chapter 5 – Part D
 11.28  7.07
1
P1230    12  11.28  cos(30  7.07)
2
 54(W )  (36  18)
For resistors
1 2 1 VM2
P  RI M 
2
2 R
CHAPTER 5 - D25
Determine average power absorbed/supplied by each
element
LEARNING EXTENSION
V
I1
I2
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D26
Determine average power absorbed/supplied by each
element
LEARNING EXTENSION
V
I1
I2
V  14.77  j1.85 Volts
240  V 24  14.77  j1.85
I1 

 4.62  j 0.925
2
2
I1  4.71  11.32( A)
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D27
Determine average power absorbed/supplied by each
element
LEARNING EXTENSION
V
I1
I2
V  14.77  j1.85 Volts
120  V 12  14.77  j1.85  j
I2 


j2
j2
j
 1.85  j 2.77
I2 
 0.925  j1.385( A)  1.67123.73( A)
2
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D28
Determine average power absorbed/supplied by each
element
LEARNING EXTENSION
V
V  14.77  j1.85 Volts
I1
I2
I1  4.71  11.32( A)
I 2  1.67123.73( A)
For resistors
1 2 1 VM2
P  RI M 
2
2 R
1
P2   2  4.712  22.18(W )
2
2
1 14.88
P4  
 27.67(W )
2
4
Check :
Pabsorbed  22.18  27.67  5.565(W )
Psupplied  55.42(W )
1
P120    12  1.67 cos(0  123.73)  5.565(W )
2
1
P240    24  4.71 cos(0  11.32)  55.42(W )
2
ECE 3183 – Chapter
5 – Part D
CHAPTER 5 - D29
THE POWER FACTOR

Z L V 
M
v
I M  i

V  ZI  V  Z  I
v   z  i
V
v
pf 
Papparent
 cos( v   i )  cos z
I
i
Papparent  Vrms I rms
1
P  VM I M cos( v   i )  Vrms I rms cos( v   i )
2
P
z
 90   z  0
current leads
(capacitiv e)
P  Vrms  I rms  pf
0   z  90
current lags
(inductive )
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D30
V
THE POWER FACTOR
z
0
 90
0  pf  1  90   z  0
1
0
0  pf  1 0   z  90
0
90
pf
ECE 3183 – Chapter 5 – Part D
purely capacitive
leading or capacitive
resistive
lagging or inductive
purely inductive
CHAPTER 5 - D31
LEARNING EXAMPLE
Find the power supplied by the power company.
Determine how it changes if the power factor is changed to 0.9
Power company
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D32
LEARNING EXAMPLE
Find the power supplied by the power company.
Determine how it changes if the power factor is changed to 0.9
P  Vrms  I rms  pf
 cos z  0.707  z  45
Current lags the voltage
Power company
480(V )rms
259.3  45( A)rms
To solve for z, take the arccos of p.f.
z = cos-1(0.707) =  45°. Which one to choose?
Since the load is lagging (inductive), we know 0 < z < 90°
So pick z = + 45°.
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D33
LEARNING EXAMPLE
Find the power supplied by the power company.
Determine how it changes if the power factor is changed to 0.9
P  Vrms  I rms  pf
 cos z  0.707  z  45
Current lags the voltage
480(V )rms
259.3  45( A)rms
Power company
If pf=0.9
I rms
88  103 (W )

 259.3( A) rms
480  0.707
2
Plosses  I rms
R  259.32  0.08  5.378kW
PS  Plosses  88,000(W )  93.378(kW )
ECE 3183 – Chapter 5 – Part D
I rms 
Plosses
88,000
 203.7( A)rms
480  0.9
2
 I rms
R  3.32kW
Losses can be reduced by 2kW!
CHAPTER 5 - D34
LEARNING EXAMPLE
Find the power supplied by the power company.
Determine how it changes if the power factor is changed to 0.9
P  Vrms  I rms  pf
 cos z  0.707  z  45
Current lags the voltage
480(V )rms
259.3  45( A)rms
Power company
I rms
88  103 (W )

 259.3( A) rms
480  0.707
2
Plosses  I rms
R  259.3 2  0.08  5.378kW
PS  Plosses  88,000(W)  93.378(kW)
ECE 3183 – Chapter 5 – Part D
VSrms  0.08 I rms  VL
VSrms
 0.08  259.3  45  480
 0.08  (183.4  j183.4)  480
 494.7  j14.7  495  1.7(V )
CHAPTER 5 - D35
LEARNING EXTENSION
P  Vrms  I rms  pf
PL  100kW ,VL  480(V )rms , pf  0.707
Rline  0.1
Determine the power savings if the power factor can be increased to 0.94
I rms
P

Vrms  pf
2
P
Rline
1
2
Plosses  I rms
Rline 

2
Vrms
pf 2
1010  0.1
1
Plosses ( pf  0.707) 

(W )
4802
0.707 2
 2  4.34kW
1010  0.1
1
Plosses ( pf  0.94) 

(W )
4802
0.942
 1.13  4.34kW
Psaved  0.87  4.34kW  3.77kW
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D36
COMPLEX POWER
Definition of Complex Power
1
S  Vrms I*rms  V I*
2
S  P  jQ
S
complex
P
real
jQ imaginary
ECE 3183 – Chapter 5 – Part D
KVA
KW
KVAR
Apparent Power
Real Power
Reactive Power
CHAPTER 5 - D37
COMPLEX POWER
S  Vrms I
*
rms
S  Vrms V  I rms  I 
*
S  Vrms I rms V   I 
S  Vrms I rms  cosV   I   j Vrms I rms  sinV   I 
S  P  jQ
P=Active Power
ECE 3183 – Chapter 5 – Part D
Q=Reactive Power
CHAPTER 5 - D38
COMPLEX POWER
Let a load be ZL = R + jX.
Then Q is:
•Positive if the load is inductive (X>0)
•Negative if the load is capacitive (X<0)
•Zero if the load is resistive (X=0)
inductive
capacitive
ECE 3183 – Chapter 5 – Part D
CHAPTER 5 - D39