Download Chapter 9 - AC Power (PowerPoint Format)

Steady-State AC Power
The power absorbed or delivered by any element is the product of the voltage and
current associated with the element.
i(t)
+
v(t)
-
If the current and voltage are functions of
time, then so is the power:
p(t) = v(t)i(t)
Specifically, if v(t) and i(t) are sinusoids, the power,
p(t), will also be a sinusoid:
v (t )  V cos( t   )
M
v
i (t )  I M cos( t   i )
p (t )  VM I M cos( t   v ) cos( t   i )
Using the trig identity: cos 1 cos 2  1 cos(1 2 ) cos(1 2 )
2
We can write:
Spring 2001
p (t ) 
VMIM
cos(v i )cos(2t v i )
2
EE202 Steady-State AC Power
1
v (t )  4 cos( t  60 ) volts , f  60 Hz    120
Example 9.1
i (t )  2 cos( t  30 ) amps
p (t )  v (t )  i (t )  8 cos( t  60 ) cos( t  30 ) watts
p (t )  4 cos(60 30 )cos(2 t 60 30 )
p (t )  4 cos(30 )cos(2 t 90 )
8
6
p(t)
4
2
0
-2
v(t)
-4
Spring 2001
i(t)
0
0.005
0.01
0.015
0.02
EE202 Steady-State AC Power
0.025
0.03
2
Average Power
The average value of any periodic waveform can be found by integrating (summing)
the function over one complete period and dividing by the length of the period.
1
X avg  T
t0  T

x (t ) dt
t0
The average value of a sinusoid such as a sinusoidal voltage or current, then, is zero:
1
Vavg  T
t0  T

VM cos( t  ) dt  0
t0
The average value of the instantaneous power function, however, is not zero:
t T
VM I M
1 0
VM I M
Pavg  T 
cos( v i )  cos(2 t  v i )dt 

2
2
t0
cos( v   i )
VM I M
Pavg 
cos( v   i )
2
Spring 2001
EE202 Steady-State AC Power
3
Average Power- Continued
+
I=I I
+
V=V V
-
VM I M
Pavg 
cos( v   i )
2
+-
Z=Z Z
-

I 
I

V
V V V

 Z (V
Z  Z
Z
V Z
and
 z )
 I  (V   z )   I  (V   z )
If Z is purely resistive Z = 0 and PAVG = VMIM/2 since cos(0) = 1.
If Z is purely reactive Z = 90 and PAVG =0 since cos(90) =0.
Reactive elements absorb (store) power during one half cycle and deliver
(release) that power during the next half cycle.
Spring 2001
EE202 Steady-State AC Power
4
Maximum Average Power Transfer
ZTH=ZT T
+
VTH = VOC V
-
+
-
ZL
+
VL=VL VL
-
PL , the power delivered to ZL , is maximized whenever ZL = Z*TH
(The load impedance is equal to the complex conjugate of the Thevenin
equivalent impedance).
Z L  RL  jX L  RTH  jX TH
*
or Z L  ZTH
Spring 2001
EE202 Steady-State AC Power
5
Effective (RMS) Value
The effective (or rms) value of a periodic signal is the dc equivalent value. For
example, household power is typically 110V, rms. This means, that a resistive
load connected to 110V ac source would absorb the same amount of power if it
were connected to a 110V dc source.
The term rms stands for root-mean-square and comes from the mathematical
definition of effective value:
X eff
1

T
It is easily shown that for a sinusoidal signal:
t0 T

x 2 (t )dt
t0
X eff 
XM
2
So, the 110Vrms household power is a sinusoid with a peak value of:
110  2  155.6V
The rms value of non-sinusoidal signals must be found by applying the definition.
Spring 2001
EE202 Steady-State AC Power
6
Effective (RMS) Values In Power Calculations
The average power formulas were written in terms of the peak values of the
sinusoidal voltage and current:
VM I M
Pavg 
cos( v   i )
2
But these can be re-written in terms of rms values:
VM I M
Pavg 

cos( v   i )
2
2
Pavg  VRMS I RMS cos( v   i )
For a resistor:
Spring 2001
2
V
2
Pavg  I RMS  R  RMS
R
EE202 Steady-State AC Power
7
The Power Factor
For a given load, the average ac power absorbed by the load is given by:
I=Irmsi
P = VrmsIrmscos (v -i)
+
V=Vrmsv
-
ZL
Notice that the passive sign convention is being
used: power absorbed (+) , power generated (-)
The product VrmsIrmsis called the apparent power, and is the power that would be
delivered if ZL were a purely resistive load.
The units for P, sometimes called the real power, is watts (W). However, apparent
power is specified in volt-amps (VA) or kilovolt-amps (kVA).
The power factor (pf) is the ratio of the real power to the apparent power and is
given by:
pf  cos(v  i )
Spring 2001
EE202 Steady-State AC Power
8
The Power Factor - Continued
I=Irmsi
P = VrmsIrmscos (v -i)
+
V=Vrmsv
-
ZL
For a purely resistive load, pf =1 since the voltage and current are in phase (v = i )
For a purely reactive load, pf =0 since the voltage and current are 90 out of phase.
The pf will range between 0 and 1 since the phase angle between the voltage and
current varies between 90 .
The power factor angle, (v - I), is determined by the angle of the impedance ZL,
since I = V/ZL.
If ZL is inductive (ZL = R + jX = Z +), then the pf angle will be negative and I will
lag V  lagging power factor.
If ZL is capacitive (ZL = R - jX = Z-), then the pf angle will be positive and I will
lead V  leading power factor.
Spring 2001
EE202 Steady-State AC Power
9
Example 9.12
RW = 0.08
+
VS
IL
P=88kW @
0.707pf lagging
480 V rms
-
An industrial load consumes 88kW at a pf of 0.707 lagging. The voltage at the load
is 480V (rms), find the total power supplied by the source if the load is connected to
the source via a transmission line that has a total resistance pf 0.08 .
P = 88kW = VrmsIrmspf
 Irms= 88kW/(480V  0.707)  Irms= 259.3 A rms.
Pwire = (259.3 A)2  0.08 = 5.38kW  Psource = 88kW + 5.38kW = 93.38 kW.
Suppose that the load power is maintained at 88 kW, but that the pf is improved to
0.9, determine the rms load current and the power delivered by the source.
P = 88kW = VrmsIrmspf
 Irms= 88kW/(480V  0.9)  Irms= 203.7 A rms.
PWire= (203.7 A)2  0.08 = 3.31kW  Psource = 88kW + 3.31kW = 91.31 kW.
Spring 2001
EE202 Steady-State AC Power
10
Complex Power
*
S  Vrms I rms
Complex power is defined as follows:
If I rms  I rms i

*
I rms
 I rms   i
Complex power has magnitude and phase and is found by multiplying the voltage phasor
with the complex conjugate of the current phasor.
*
S  Vrms I rms
 Vrms  v I rms    i  Vrms I rms ( v  i )
The expression above represents the complex power in polar form, but we can also
represent it is rectangular form.
S  Vrms I rms ( v  i )  Vrms I rms cos( v  i )  jVrms I rms sin( v  i )
P = real power
(Watts)
Q = reactive power
(volt-amps reactive or VARs)
S = P + jQ
Spring 2001
EE202 Steady-State AC Power
11
Complex Power - Continued
As stated previously, the complex power is composed of a real part (Watts) and an
imaginary part (VARS).
S = P + jQ
P = Re(S) and Q = Im(S)
The magnitude of S, is what we have called the apparent power (VA), and is the
hypotenuse of the power triangle.
S
z
Q
P
It can also be shown that:
Some observations can be made from the power triangle
using trig identities:
P = |S|cos z
Q = |S|sinz
|S| = [P2+Q2]0.5
z = tan-1(Q/P)
P  I 2rms  Re( Z)  I 2rms  R
Q  I 2rms  Im( Z)  I 2rms  X
Spring 2001
EE202 Steady-State AC Power
12
The Power Triangle
S = P + jQ = |S|Z
(inductive load)
S
Q
z
(Lagging pf)
P
If the load ZL is inductive, ZL= R + jX, then Z is positive and Q is also positive, as shown
in the diagram above. This is called a lagging pf since the current lags the voltage.
P
S = P - jQ = |S|-Z
(capacitive load)
z
S
-Q
(Leading pf)
If the load ZL is capacitive, ZL= R - jX, then Z is negative and Q is also negative, as shown
above. This is called a leading pf since the current leads the voltage.
If the load ZL is purely resistive, ZL= R, then Z =0 and Q =0. In this case, P = S.
Spring 2001
EE202 Steady-State AC Power
13
Power Factor Correction
The overall power factor of the typical industrial plant is lagging due to the inductive
nature of the devices used: motors and transformers. It is generally economically viable
to take some steps to improve (correct) the overall power factor. There are at least two
benefits to improving power factor:
(1) the same amount of real power can be delivered at a lower current;
(2) most power suppliers charge for the VAs used, not for Watts.
To improve power factor, a capacitive load is added to the system to offset the inductive
VARs generated by other equipment. Often, the only function of the capacitive load is to
improve power factor.
I
new
+
Power
Source
Iold
IC
VS
-
Spring 2001
EE202 Steady-State AC Power
14
Power Factor Correction - continued
Inew
Inew = Iold + IC
+
Power
Source
Iold
IC
Ic
VS
VS
Inew
-
Iold
• Since the capacitive current leads VS by 90, Inew = Iold + IC will be smaller in magnittde
than the original current.
• Also, the phase angle  between the voltage and current will be closer to zero (power
factor will be closer to unity).
Spring 2001
EE202 Steady-State AC Power
15
Power Factor Correction - continued
Power factor improvement can also be visualized using power triangles:
Pnew = Pold = P
Qnew = Qold - QC
Snew = P + jQnew
Inew
+
Power
Source
Iold
IC
new = tan-1[(Qold - QC)/P]
VS
Sold
QC
Qold
Snew
-
old
P
Spring 2001
EE202 Steady-State AC Power
new
16
Example 9.15 Plastic kayaks are manufactured using a process called rotomolding, which is
illustrated in the diagram below. Molten plastic is injected into a mold, which is then spun the
long axis of the kayak until the plastic cools, resulting in a hollow, one-piece craft. Suppose the
induction motors used to spin the molds consume 50kW of power at a 0.8 lagging pf. The power
source is 2200 , rms. Determine the rating of the capacitive reactance required to raise the pf
to 0.95 lagging. Also, calculate the new current demanded from the 220V source.
kayak mold
Solution: old = cos-1(0.8) = 36.87
Sold = 50k/0.8
Sold = 62.5kVA
old
Qold = S  sinold
Qold = 37.7kVAR
induction
motor
new = cos-1(0.95) = 18.2
Snew = 50k/0..95
Snew = 52.6kVA
P = 50kW
P = 50kW
Qnew = S  sinnew
Qnew = 16.4kVAR
new
Qc = Qnew - Qold = 21.1kVAR
Iold = S/V = 62.5kVA/220V = 284A
Spring 2001
Inew = S/V = 52.6kVA/220V = 239A
EE202 Steady-State AC Power
17