real reactive apparent power
TRANSCRIPT
Power Engineering Foundation
By Fuad Latip 2006 1
Real, Reactive and Apparent Power in AC Circuit
P = V*I DC Circuit
AC circuit more complex Phase different between V and I
The Instantaneous Power supplied to an AC cct still
P = V*I
V = Instan. Voltage
I = Instan. Current
Average Power supplied to the load will affected by the θ (θ = phase angle between V
and I)
+ -
R
I
VDC
DC Voltage source supplying a load with resistance R
Power Engineering Foundation
By Fuad Latip 2006 2
Inductive Load Capacitive Load
θ = +ve θ = -ve
Current LAG the voltage by θ degrees. Current LEAD the voltage by θ degrees.
rms
peak
rms
IItIti
VV
VVtv
tVtv
=−=
=
===
)cos(2)(
load this toflowingcurrent The
2
voltageinstant)(cos2)(
load this toapplied voltageThe
θω
ω
rms
peak
rms
IItIti
VV
VVtv
tVtv
=+=
=
===
)cos(2)(
load this toflowingcurrent The
2
voltageinstant)(cos2)(
load this toapplied voltageThe
θω
ω
+ -
I AC = I∠-θo
VAC = V∠0o
AC Voltage source supplying a load with resistance Z =Z∠θo
Z =Z∠θo
θo I Lead V
V
I θo I Lag V
V
I
Power Engineering Foundation
By Fuad Latip 2006 3
Power in AC
2sin2sin1)12(coscos
sin2sin)12(coscos
)(coscos
)()()(
ComponenttVIComponenttVI
where
tVItVI
tttVI
titvtPLoadInductiveFrom
>−−>−−+
++=
−=
=
θωωθ
θωωθ
θωω
The components of power supplied to a single phase load versus time. The first Component represents the power supplied by the component of current in phase with the voltage, while the second term represents the power supplied by the component of current 90o out of phase with the voltage
Power Engineering Foundation
By Fuad Latip 2006 4
• The 1st term of the instantaneous power expression is always +ve, but it produces
pulses of power instead of a constant value.
• The average value of this term is
P = VICosθ
• This is the AVERAGE or REAL POWER (P) supplied to the load.
• The unit of real power are WATT (W)
1W = 1V * 1 A
• The 2nd term of instantaneous power expression is +ve half and –ve half so that
the average power supplied by this term is zero.
• This term represents power that is first transferred from source to the load, and
then returned from the load to the source.
• It is known as REACTIVE POWER (Q)
• Reactive Power represents the energy that is first stored and then released in the
magnetic field of an inductor or in the electric field of a capacitor.
• The Reactive Power of load is given by
Q = VISinθ
• The unit of Reactive Power is volt-amperes reactive (var).
1var = 1V * 1A
• θ is the impedance angle of the load for both cases (Active Power and
Reactive Power)
Power Engineering Foundation
By Fuad Latip 2006 5
Apparent Power (S) is the product of the voltage across the load and the current through
the load whereby the phase angle are ignored.
S = V*I
The unit of Apparent Power is volt-amperes (VA)
1VA = 1V * 1A
Relative Forms of the Power Equations
θθ
θ
θ
θθ
sincos
impedanceloadtheofmagnitud
sincos
(3)and(2)(1),tongsubstituti
constantloadaif
)3()2(sin)1(cos
2
2
2
jZjXRZ
ZZISZIQZIP
IZV
VISVIQVIP
+=+=
==
=
=
==
−−−−−−=−−−−=−−−−=
S
Q
P
θ
Power Engineering Foundation
By Fuad Latip 2006 6
Complex Power (S)
S = P + jQ
Complex Power, S supplied to a load can be calculated from the equation below
jQPjVIVIS
jVIVIVI
IV
IIVV
lett) I conjugaI* VI* (S
+=+=
−=
−+−=−∠=
−∠∠=
∠=∠=
==
θθβαθ
βαβαβα
βα
βα
sincos
)sin()cos()(
))((
The Relationship between Impedance Angle, Current Angle and Power
Inductive Load
• Has a +ve impedance angle ,θ since the reactance of an inductor is +ve
• The phase angle of the current flowing through the load will lag the phase angle
of the voltage across the load by θ.
θθ
−∠=∠∠
==ZV
ZV
ZVI 0
• Load is said to be consuming both real and reactive power from source.
Power Engineering Foundation
By Fuad Latip 2006 7
Capacitive Load
• Has a -ve impedance angle ,θ since the reactance of an inductor is -ve
• The phase angle of the current flowing through the load will lead the phase angle
of the voltage across the load by θ.
θθ
∠=−∠
∠==
ZV
ZV
ZVI 0
• Load is said to be consuming real power from the source and supplying reactive
power to the source.
+ -
I AC = I∠-θo
VAC = V∠0o Z =Z∠θo
P
Q
+ -
I AC = I∠-θo
VAC = V∠0o Z =Z∠θo
P
Q
Power Engineering Foundation
By Fuad Latip 2006 8
The Power Triangle
PQSQSP
=
=
=
θ
θ
θ
tan
sin
cos
• cosθ = cos(-θ), so the power factor produced by an impedance angle of +θ is
exactly same as the power factor produced by and impedance angle of –θ.
• So we cannot know whether a load is inductive or capacitive from the power
factor alone.
• Then the current leading or lagging have to know whenever a power factor is
quoted.
Power Engineering Foundation
By Fuad Latip 2006 9
Example
Figure below shows an AC voltage source supplying power to a load with impedance Z =
20∠-30o Ω. Calculate the current, I supplied to the load, the power factor of the load, and
the real, reactive, apparent, and complex power supplied to the load.
+ -
I
VAC = 120∠0o VoltZ =20∠-30o
Power Engineering Foundation
By Fuad Latip 2006 10