July 8, 2004
10- Sinusoidal Steady-State Power Calculations
•
•
All electrical energy is supplied in the form of sinusoidal voltages and currents.
Our interest is the average power delivered to or supplied from a pair of terminals as a
result of sinusoidal voltages and currents
Instantaneous Power
i
+
v
-
Back
Box
Instantaneous Power is the power at any instant of time is
p = vi
where
v = Vm cos(ωt + θ v )
i = I m cos(ωt + θ i )
θi
v = Vm cos(ω t + θ v − θ i )
i = I m cos(ωt )
Let’s shift both voltage and current by
Therefore, instantaneous power becomes
p = vi = Vm I m cos(ωt + θ v − θ i ) cos ωt
(1.1)
To simplify this, we can use the following trigonometric identity
cos α cos β =
p=
We know that
1
1
cos(α − β ) + cos(α + β )
2
2
Vm I m
V I
cos(θ v − θ i ) + m m cos(2ωt + θ v − θ i )
2
2
(1.2)
cos(α + β ) = cos α cos β − sin α sin β
Therefore, instantaneous power
p=
Vm I m
V I
V I
cos(θ v − θ i ) + m m cos(θ v − θ i ) cos 2ωt − m m sin(θ v − θ i ) sin 2ωt
2
2
2
(1.3)
ECE309-05--1
In this equation,
• the first term is constant,
• the 2nd and 3rd show that instantaneous power has twice frequency of the voltage or
current.
• It has negative for some portion of cycle. That means, energy stored in the inductors o
capacitors is now being extracted.
Average and Reactive Power.
We can divide the instantaneous power (last equation) in three terms.
p = P + P cos 2ωt − Q sin 2ωt
(1.4)
where P is called the average power or real power
P=
Vm I m
cos(θ v − θ i ) [W]
2
(1.5)
Q is called the reactive power
Q=
Vm I m
sin(θ v − θ i ) [VAR]
2
(1.6)
The average power be represented in the following form too
1
P=
T
t 0 +T
∫ pdt
t0
T is period of the sinusoidal function.
ECE309-05--2
Power for Purely Resistive Circuits
If the circuit between the terminal is purely resistive, there is no phase different between voltage
and current, which means θ v = θ i , so that instantaneous power will be
p=
Vm I m
(1 + cos 2ω t )
2
This is called as the instantaneous real power.
• It has some average value
• Frequency is 2ω
• Can never be negative, that means power can not be extracted from purely resistive
circuit
Power for Purely Inductive Circuits
If the circuit between the terminal is purely inductive, there is phase different between voltage and
current, The current lags the voltage by 90° which means
θ i = θ v − 90 D ( θ v − θ i = +90 D ), so
that instantaneous power will be
p=−
Vm I m
sin 2ω t
2
This is called as the instantaneous reactive power.
• The average will be zero.
• Frequency is 2ω
• When p is positive, energy is being stored in the magnetic fields associate with the
inductive elements,
• When p is negative, energy is being extracted from the magnetic fields.
ECE309-05--3
Power for Purely Capacitive Circuit
If the circuit between the terminal is purely capacitive, there is phase different between voltage
and current, The current leads the voltage by 90° which means
θ i = θ v + 90 D ( θ v − θ i = −90 D ),
so that instantaneous power will be
p=
Vm I m
sin 2ω t
2
This is called as the instantaneous reactive power.
• The average will be zero.
• Frequency is 2ω
• When p is positive, energy is being stored in the capacitive elements
• When p is negative, energy is being extracted from the capacitive elements.
ECE309-05--4
The Power factor
The angle
θ v − θ i is the power factor angle.
The power factor is
pf = cos(θ v − θ i )
(1.7)
The reactive factor is
rf = sin(θ v − θ i )
Depends on the phase, the current lags the voltage by 90° ,
we call lagging power factor pf = cos(θ v − θ i )
and the current laeads the voltage by 90° ,
we call leading power factor
pf = cos(θ i − θ v )
ECE309-05--5
Example:
v = 100 cos(ωt + 15 D ) V.
i = 4 sin(ωt − 15 D ) A
Let’s convert current i(t) in cos function
i = 4 cos(ω t − 15D − 90D ) = 4 cos(ω t − 105D )
the average power or real power will be
Vm I m
cos(θ v − θ i )
2
(100)(4)
P=
cos(15 − (−105)) = −100 W
2
P=
(Network inside the box is delivering average power to the terminal)
The reactive power
Vm I m
sin(θ v − θ i )
2
(100)(4)
Q=
sin(15 − (−105)) = 173.21 VAR
2
Q=
(Inside the box is absorbing magnetizing vars at its terminal.)
The instantaneous power will be
p = −100 − 100 cos 2ω t − 173.21sin 2ω t
ECE309-05--6
The RMS Value and Power Calculation
The RMS value of a periodic sinusoidal function is defined as a square root of the mean value of
the square sinusoidal function. Also, the RMS value can be called as effective value of the
Ex: The RMS value of
v(t ) = Vm cos(ω t + θ )
is
(1.8) Vrms =
V
1 t0 + T 2
Vm cos 2 (ω t + θ )dt = m
∫
T t0
2
(1.9)
Let’s look at the average power delivered to the resistor
+
Vmcos(ωt+θ)
R
-
P=
1  1 t0 +T 2

Vm cos 2 (ω t + θ )dt 
∫

t
0
R T

P=
2
Vrms
V2
= m
R
2R
If we know the current on the resistor, we can find the average power delivered to the resistor as
P=I
2
rms
I m2
R=
R
2
The previous pages we give the average power as
P=
Vm I m
cos(θ v − θ i )
2
(1.10)
This can be written in terms of RMS value
P=
Vm I m
cos(θ v − θ i )
2 2
P = Vrms I rms cos(θ v − θ i )
(1.11)
ECE309-05--7
The reactive power
Vm I m
sin(θ v − θ i )
2
Q = Vrms I rms sin(θ v − θ i )
Q=
Example:
Vrms = 110 V , R = 50 Ω ,
Find Vm and the Average power delivered to the resistor
Vrms =
P=
Vm
2
Æ Vm = Vrms 2 = 110 2 = 155.56 V
2
Vrms
1102
=
= 242W
50
R
ECE309-05--8