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Example of power including reactive power
Example of power including reactive power
With DC the instantanesous values of voltage and current are
With DC the instantanesous values of voltage and current are
constant with respect to time, hence the power is constant.
constant with respect to time, hence the power is constant.
In contrast to this the instantaneous value of power of AC or
In contrast to this the instantaneous value of power of AC or
AC + DC signals will fluctuate, its amplitude and polarity will
AC + DC signals will fl uctuate, its amplitude and polarity will
periodically change. If the phase angle is zero this is the special
periodically change. If the phase angle is zero this is the special
case of pure active power which remains positive (exclusively
case of pure active power which remains positive (exclusively
directed from source to load) at all times.
directed from source to load) at all times.
If there is a reactive component in the circuit there will be a
If there is a reactive component in the circuit there will be a
phase difference between voltage and current. The inductive
phase difference between voltage and current. The inductive
or capacitive element will store and release energy periodically
or capacitive element will store and release energy periodically
which creates an additional current component, the reactive
which creates an additional current component, the reactive
part. The product of voltage and current will therefore become
part. The product of voltage and current will therefore become
negative for portions of a period which means that energy will
negative for portions of a period which means that energy will
flow back to the source.
fl ow back to the source.
Apparent power (unit VA)
3.6.3 Apparent power (unit VA)
The apparent power is equal to the product of voltage and
The apparent power is equal to the product of voltage and cur-
current. The apparent power is further equal to the geometric
rent. The apparent power is further equal to the geometric sum
sum of active and reactive power as shown in this diagram:
of active and reactive power as shown in this diagram:
With the designations:
With the designations:
S
S
= apparent power
= apparent power
P
P
= active power
= active power
Q
Q
= reactive power
= reactive power
V
V
= rms voltage
= rms voltage
rms
rms
I
I
= rms current
= rms current
rms
rms
the apparent power is derived:
the apparent power is derived:
S = P
2
+ Q
2
= V
rms
S =
P
2
+ Q
2
= U
Power factor
3.7
Power factor
In general the power factor PF is derived:
In general the power factor PF is derived:
P
PF = – – – –
P
S
PF = – – – –
S
PF
= power factor
PF
S
= power factor
= apparent power
S
P
= apparent power
= active power
P
= active power
In the very special case of sinusoidal voltage and
In the very special case of sinusoidal voltage and
current the power factor equals
current the power factor equals
PF = cos ϕ
PF = cosϕ
HINT
x J
rms
x J
eff
eff
B a s i c s o f P o w e r M e a s u r e m e n t
B a s i c s o f P o w e r M e a s u r e m e n t
If e.g. the current is rectangular while the voltage is sinusoidal
If e.g. the current is rectangular while the voltage is sinusoi-
the power factor will be P/S. Also in such case the reactive power
dal the power factor will be P/S. Also in such case the reacti-
can be determined as demonstrated in the following example:
ve power can be determined as demonstrated in the following
example:
û = 325,00 V
î = 12,25 A
û = 325,00 V
î = 12,25 A
How to calculate the power factor (example):
rms voltage is:
û
U
= — — = 229,8 V ≈ 230 V
eff
√
2
The rms current is given by:
2π
1
∫
I
=
î
2
· dϕ
——
eff
2π
0
î
2
π
[(
=
·
π –
——
— —
2π
3
2
=
î
2
·
——
= î ·
3
2
I
= 12,25 A ·
——
= 10,00 A
eff
3
The apparent power S:
S = V
· I
= 230 V · 10,0 A = 2300 VA
rms
rms
The active power is derived from:
1
π
∫
P =
——
û · î sin ϕ · dϕ =
π
π
3
û · î
[(
)
P
=
– (-1)
– (-0,5
———
π
1,5
=
· 325 V · 12,25 A = 1900 W
——
π
The power factor thus becomes:
P
1900 W
PF = —— = —————— = 0,826
S
2300 VA
Obviously there is a reactive power component as the
apparent power exceeds the active power:
Q =
S
2
– P
2
= (2300 VA)
4π
)
(
)]
+
2π –
— —
3
2
——
3
û · î
[
]
π
———
– cos ϕ
π
π
3
1,5
)]
=
· û · î
——
π
2
– (1900 W)
2
= 1296 var
Subject to change without notice
Subject to change without notice
31
31
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