Download 2300NotesSet23v05

Dave Shattuck
University of Houston
© University of Houston
ECE 2300
Circuit Analysis
Lecture Set #23
Complex Power –
Background Concepts
Dr. Dave Shattuck
Associate Professor, ECE Dept.
[email protected]
713 743-4422
W326-D3
Dave Shattuck
University of Houston
© University of Houston
Overview of this Lecture Set
Complex Power – Background
Concepts
In this set of lecture notes, we will cover
the following topics:
• Review of Sinusoidal Sources and
Phasors
• Review of RMS
• Power with Sinusoids
Dave Shattuck
University of Houston
© University of Houston
Textbook Coverage
This material is introduced in different ways in
different textbooks. Approximately this same
material is covered in your textbook in the
following sections:
• Electric Circuits 6th Ed. by Nilsson and Riedel:
Sections 10.1 through 10.3
Dave Shattuck
University of Houston
© University of Houston
Power in the Sinusoidal Steady
State (Complex Power)
We studied Phasor Transforms. Using
these transforms, we can find things we
want to know, more quickly and more
easily.
Now we are going to do a similar thing
with power absorbed or delivered in circuits
with sinusoidal sources. Again, we will only
consider what is happening in the steady
state. We will find that:
• The use of phasors and transforms can
be used for power calculations, and
• Some very useful new concepts will help
us deliver more power to where we want it,
with fewer losses at the same time.
The power we use is
often sinusoidal. That
is, the wall plugs, and
some other sources,
are voltages and
currents that vary as a
sine wave. Thus, this
subject is very useful
to us.
AC Circuit Analysis Using
Transforms
Dave Shattuck
University of Houston
© University of Houston
Let’s remember first and foremost that the end goal is to
find the solution to real problems. We will use the
transform domain, and discuss quantities which are
complex, but obtaining the real solution is the goal.
Solutions Using Transforms
Problem
Transform
Solution
Real, or time
domain
Complicated and difficult
solution process
Inverse
Transform
Transformed
Transformed
Problem
Problem
Relatively simple
solution process, but
using complex numbers
Transformed
Transformed
Solution
Solution
Complex or
transform domain
Dave Shattuck
University of Houston
© University of Houston
Power with Sinusoidal
Voltages and Currents
• It is important to remember that nothing has
really changed with respect to the power
expressions that we are looking for. Power is
still obtained by multiplying voltage and
current.
• The fact that the voltage and current are sine
waves or cosine waves does not change this
formula.
Dave Shattuck
University of Houston
© University of Houston
Some Review – Sinusoids
The figure below is taken from Figure 6.2 in Circuits by A. Bruce
Carlson. A general sinusoid has the equation given below. Note
that in this equation there are three parameters, the amplitude (Xm),
the frequency (w), and the phase (f). The time, t, is the
independent variable. The sine function is just as good as the
cosine function, but in electrical engineering the cosine function is
used more often.
x(t )  X m cos(w t  f )
Dave Shattuck
University of Houston
© University of Houston
Definition of RMS – Review
We are now going to review an
important term, the rms value of a
voltage or current. This was covered
before. The rms value, also called
the effective value, has the most
meaning in terms of power
calculations.
It is so useful, that we will redefine our
phasor transforms in terms of rms
values, to make our formulas simpler
and easier to use. Thus, it is worth
the time to review rms concepts.
Derivation of RMS – 1
Dave Shattuck
University of Houston
© University of Houston
We want the effective value that could be used in power
calculations, for average power, in the formula below.
Pave  Vrms I rms .
We will do the derivation for a resistance, since we want the
formula to work with the resistance power formulas. Let’s
arbitrarily choose to work with the voltage. What we want to
get is a value that will work in the formula,
Pave
Vrms 


R
2
.
With T as the period, the average value
of the power is obtained by the formula,
1 t0 T  v 2 (t ) 
Pave   
dt.

t
T 0  R 
Derivation of RMS – 2
Dave Shattuck
University of Houston
© University of Houston
Now, to get the formula, we simply set the two equations
from the previous slide equal to each other,
Pave
Vrms 


R
2
1 t0 T  v 2 (t ) 
  
dt.

T t0  R 
Now, we need to simplify. The resistance is assumed to
be a constant, and so it can be taken out of the integral.
When we multiply both sides by R, we get
Vrms 
2
1 t0 T 2
   v (t )  dt.
T t0
Dave Shattuck
University of Houston
© University of Houston
Derivation of RMS – 3
Finally, we can solve for the rms value of the voltage, by
taking the square root of both sides,
Vrms
1 t0 T 2

v (t )  dt .


t
T 0
This is the result that we have been working
toward. We only need to interpret this
result. We have taken the periodic voltage,
v(t), and squared it. Then, by integrating it
over a period and dividing by the period, we
are taking the mean value of the squared
function. Finally, we take the square root of
the mean value of the squared function.
We call this rms.
Dave Shattuck
University of Houston
© University of Houston
RMS Value of a Sinusoid
The rms value for a general periodic function, x(t), is
X rms
1 t0 T 2

x (t )  dt .


t
T 0
Now, this was derived for any periodic function. The
function must be periodic for the formula for the mean value to
apply.
If we perform the calculus to get the rms value for a
sinusoid, we find the rms value is equal to the zero-to-peak
value (or amplitude) divided by the square root of 2, or
X rms
Xm

.
2
Remember, this only holds
for sinusoids!
Dave Shattuck
University of Houston
© University of Houston
Power with Sinusoids – 1
If we have sinusoidal voltages and currents, we can get the power by
multiplying the two. We could plot the power as shown below. This
curve was obtained by taking a couple of arbitrary sinusoids for voltage
and current. If you change the magnitudes and phases, this curve will
change.
Power with Sinusoids
voltage [Volts]
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Dave Shattuck
University of Houston
© University of Houston
Power with Sinusoids – 2
Let’s approach this same issue using equations. Let’s
assume that our voltage and current have the formulas,
v(t )  Vm cos(w t   v ), and
i (t )  I m cos(w t   i ).
Now, we can use the formula from trigonometry, which
most of us learned but forgot,
cos(a) cos(b)  1 cos(a  b)  1 cos(a  b).
2
2
Power with Sinusoids
Applying this here, we get
voltage [Volts]
current [Amps]
p(t )  v(t )i (t ) 
6
4
2
Vm I m
Vm I m

cos( v   i ) 
cos(20 w t   v   i ).
-2
2
2
-4
8
-6
time
power [Watts]
Dave Shattuck
University of Houston
© University of Houston
Power with Sinusoids – 3
In the last slide we found that
p(t )  v(t )i (t ) 
Vm I m
Vm I m

cos( v   i ) 
cos(2w t   v   i ).
2
2
Now, we can use another formula from trigonometry,
cos(a  b)  cos(a) cos(b)  sin(a ) sin(b).
Applying this here in
the second term, with
a  2w t , and
b   v  i ,
Vm I m
p(t )  v(t )i (t ) 
cos( v   i ) 
2
8
V64m I m
we get 
cos( v   i ) cos(2w t ) 
2
2
0
V-2-4m I m
 -6 sin( v   i )sin(2w t ).
2
time
Power with Sinusoids
voltage [Volts]
current [Amps]
power [Watts]
Dave Shattuck
University of Houston
© University of Houston
Power with Sinusoids – 4
So, we have that
Vm I m
p(t )  v(t )i (t ) 
cos( v   i ) 
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t ).
2
Now, we can look at
the plot that we had,
and understand it
somewhat better.
Let’s note some
special properties in
the slides that follow.
voltage [Volts]
Power with Sinusoids
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Power with Sinusoids –
Note 1
cos(   ) 
Dave Shattuck
University of Houston
© University of Houston
Vm I m
p(t )  v(t )i (t ) 
v
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t ).
2
First, note that
even though
v(t) and i(t)
were zeromean
sinusoids, the
product, p(t),
does not
generally have
a zero mean.
i
voltage [Volts]
The power curve is not
centered at zero, so the
average is not zero.
Power with Sinusoids
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Power with Sinusoids –
Note 2
cos(   ) 
Dave Shattuck
University of Houston
© University of Houston
Vm I m
p(t )  v(t )i (t ) 
v
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t ).
2
In fact, the
mean, or
average value,
of the power is
equal to the first
term of this
equation, since
the average of
the sinusoids is
zero.
i
voltage [Volts]
p AVERAGE
Vm I m

cos( v   i ).
2
Power with Sinusoids
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Power with Sinusoids –
Note 3
cos(   ) 
Dave Shattuck
University of Houston
© University of Houston
Vm I m
p(t )  v(t )i (t ) 
v
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t ).
2
We use a
capital letter P
to represent the
average value
of the power,
p(t). The
average power
is very useful to
know.
i
p AVERAGE
voltage [Volts]
Vm I m
P
cos( v   i ).
2
Power with Sinusoids
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Power with Sinusoids –
Note 4
cos(   ) 
Dave Shattuck
University of Houston
© University of Houston
Vm I m
p(t )  v(t )i (t ) 
v
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t ).
2
This average
power P is a
function of the
magnitudes of
the voltage and
current, but also
of the difference
in phase
between voltage
and current.
i
p AVERAGE
voltage [Volts]
Vm I m
P
cos( v   i ).
2
Power with Sinusoids
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Power with Sinusoids –
Note 5
cos(   ) 
Dave Shattuck
University of Houston
© University of Houston
Vm I m
p(t )  v(t )i (t ) 
v
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t ).
2
The power
varies with time,
and is in fact
sinusoidal with a
frequency twice
that of the
voltage and
current.
i
p AVERAGE
voltage [Volts]
Vm I m
P
cos( v   i ).
2
Power with Sinusoids
current [Amps]
8
6
4
2
0
-2
-4
-6
time
power [Watts]
Dave Shattuck
University of Houston
© University of Houston
Shifting the Time Axis
For notational reasons, electrical engineers take the general case, which
we have been considering, and then shift the time axis, so that the phase of
the current is zero. The new phase of the voltage is now reduced by the
phase of the current, and now is v-i. We redefine this phase of the voltage
as , and get a new set of formulas, below.
v(t )  Vm cos(wt   v ) and General
case
i(t )  I m cos(wt  i )
Vm I m
p(t )  v(t )i (t ) 
cos( v   i ) 
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t )
2
p AVERAGE
Vm I m
P
cos( v   i )
2
v(t )  Vm cos(w t    ) and Shifted
case
i(t )  I m cos(w t )
p(t )  v(t )i (t ) 
Vm I m
cos( ) 
2
Vm I m

cos( ) cos(2w t ) 
2
Vm I m

sin( )sin(2w t )
2
p AVERAGE
Vm I m
P
cos( )
2
Dave Shattuck
University of Houston
© University of Houston
Shifting the Time Axis – Note 1
Some of you may be disturbed by the relationship between
the two cases below. It may look like we have used  = v + i
in some cases, and  = v - i in other cases. We have not.
Remember that the t in the General case is different from the t'
in the Shifted case.
v(t )  Vm cos(wt   v ) and General
case
i(t )  I m cos(wt  i )
Vm I m
p(t )  v(t )i (t ) 
cos( v   i ) 
2
Vm I m

cos( v   i ) cos(2w t ) 
2
Vm I m

sin( v   i )sin(2w t )
2
p AVERAGE
Vm I m
P
cos( v   i )
2
v(t )  Vm cos(w t    ) and Shifted
case
i(t )  I m cos(w t )
p(t )  v(t )i (t ) 
Vm I m
cos( ) 
2
Vm I m

cos( ) cos(2w t ) 
2
Vm I m

sin( )sin(2w t )
2
p AVERAGE
Vm I m
P
cos( )
2
Dave Shattuck
University of Houston
© University of Houston
Shifting the Time Axis – Note 2
Since we shifted the time axis, we changed the t to a t' in the
previous slide. However, the original choice of t was arbitrary,
and there is no reason to keep the prime any longer.
Therefore, for the rest of this material, we will use the notation
below. Remember that  is the phase of the voltage with
respect to the phase of the current. This way of expressing the
phases will be useful to us. We will generally get  by
subtracting the phase of the current from the phase of the
voltage.
Vm I m
p(t )  v(t )i(t ) 
cos( ) 
2
Vm I m
i(t )  I m cos(wt )

cos( ) cos(2w t ) 
2
Vm I m
p AVERAGE  P 
cos( )  Vm I m sin( )sin(2w t )
2
2
Given that:
v(t )  Vm cos(wt   ) and
Dave Shattuck
University of Houston
© University of Houston
Power with Sinusoidal Sources
The formulas below are important, and are the beginning of
the concepts that follow in the next two parts. We have found
two things. First, the average power is a function of the product
of the magnitudes of the voltage and current, and also a
function of the difference between the phase of the voltage and
current, . Second, we found that the expression for the power
as a function of time has a constant term, which is that average
value, and terms at twice the frequency of the voltage and
current.
Vm I m
p(t )  v(t )i(t ) 
cos( ) 
2
Vm I m
i(t )  I m cos(wt )

cos( ) cos(2w t ) 
2
Vm I m
p AVERAGE  P 
cos( )  Vm I m sin( )sin(2w t )
2
2
Given that:
v(t )  Vm cos(wt   ) and
Dave Shattuck
University of Houston
© University of Houston
So what is the point of all this?
• This is a good question. First, our premise is
that since electric power is usually distributed
as sinusoids, the issue of sinusoidal power is
important.
• In addition, there are some significant
problems that will arise when we connect
loads to our power lines that act like inductors.
This problems can be addressed using phasor
analysis, and some additional concepts that
we will lay out in the next set of lecture notes.
These concepts involved quantities called real
and reactive power.
Go back to
Overview
slide.