7.4. Power of Alternating Current
When loads carry current, a voltage drop is caused. The product
of the instantaneous values of current and voltage is called instantaneous
power. Normally, the instantaneous power changing from time to time is of less
interest than its mean value.
In an effective resistance, current and voltage are in phase.
The electrical power becomes completely, i.e. effectively, utilisable. It is
called effective power Pe (active power). It can be determined
on the basis of the effective values according to the relations derived in
Section 4.1.
[Pe] = W (watt)
In a reactance (ideal coil, ideal capacitor), current and
voltage are subjected to a phase shift of 90°. The electrical power is
required for the duration of a quarter of a cycle for the building up of the
magnetic field (in a coil) or of the electrical field (in a capacitor) and
delivered in the subsequent quarter of a cycle. There is no power conversion in
a temporal mean. This power is called reactive power
Pr.
[Pr] = Var (voltampere -
reactive)
In an impedance, the phase shift between current and voltage is
between 0° and 90°. The electrical power is converted partly as
effective power and partly as reactive power. This power is called apparent
power Pa.
[Pa] = VA (voltampere)
Since effective resistances and reactances are made up at right
angles, this analogously applies to the powers. Consequently, effective and
reactive powers are the legs and the apparent power is the hypotenuse of a
right-angled triangle having the phase angle (Fig. 7.17.).
Fig. 7.17. Correlation between
effective power, reactive power and apparent power
Pw = Pe, Pb =
Pr, Ps = Pa
From this follows
Pa = U ·
I
(7.14)
where:
|
Pa |
apparent power |
|
U |
effective value of voltage |
|
I |
effective value of current |
From Fig. 7.17. we can cerive
(7.15.)
Pe = Pa cos j
(7.16.a)
Pr = Pa sin j
(7.16.b)
The expression cos in equation (7.16.a) is called power
factor.
From equation (7.16.a) we have
cos j =
Pe/Pa
(7.16.c)
The power factor cos can be between 0 and 1;
With a low power factor, the reactive power is high. Since
reactive power unnecessarily loads the generators of the power stations and the
distribution network, the reactive power must be kept as small as possible for
economical reasons; in other words, cos j must
be as high as possible. A power factor of cos j
= 1 is an ideal case.
In networks of power electrical engineering, an inductive phase
shift occurs always because of the necessary transformers and connected motors;
this phase shift always worsens the power factor.
The power factor can be improved up to the value of cos j = 1 by means of an additional capacitive component.
In practice, capacitors are connected in parallel having a total capacity of
C = Pr/(w·U2)
(7.17.)
where
|
C |
capacity required to attain a cos j = 1 |
|
Pr |
reactive power |
|
w |
angular frequency |
|
U |
alternating voltage (effective value) |
Of the energy or work has to be determined., then
the product of power times time must be formed according to Section 4.1. In
accordance with the various types of power, there are effective work, reactive
work and apparent work.
Example 7.6.
An enterprise is connected to a 380 V network (50 Hz). A current
of 66 A passes through the loads with a power factor of cos j = 0.5. Determine the apparent power, effective power
and reactive power and the capacity of the capacitor necessary to improve the
power factor to cos j = 1!
Given:
U = 380 V
I = 66 A
f = 50 Hz
cos j = 0.5
To be found:
Pa, Pe, Pr,
C
Solution:
Pa = U I
Pa = 380 V · 66 A = 25,000 VA
Pa = 25
kVA
Pe = Pa cos j
Pe = 25 kVA · 0.5
Pe = 12.5 kW
Pr = Pa sin j
Pr =
25 KVA · 0.865
Pr = 21.6
kVar
Proof with equation (7.15.)
C = Pr/(wU2)
C » 500
µF
With a capacitor connected in parallel with the loads having a
capacity of C 500 µF, the bad power factor of cos j = 0.5 can be increased to the ideal value of cos
j = 1.
With alternating current, there are three types of
powers, namely, apparent power, effective power and reactive power. Effective
power and reactive power are made up at right angles. The apparent power is
always greater than effective power or reactive power but it is always smaller
than the algebraic sum of the latter two. Efforts are always made to achieve a
high effective power and a reactive power which is as small as possible. The
ratio of effective power to apparent power is called power factor. It can reach
the value of 1 in the most favourable case. The power factor can be improved
when a capacitor is connected in parallel with the loads.
Questions and Problems:
1. What are the differences between effective power,
reactive power and apparent power?
2. What are the values of the effective power and reactive power
when the apparent power is 23 kVA with a phase angle of j = 30°?
3. Why is a high power factor desired in energy supply systems?
Explain measures by means of which the power can be improved!
4. Two motors are connected, in parallel and to a supply system
of 220 V. One motor has a power factor of 0.65 and carries a current of 2 A; the
other motor having a power factor of 0.85 carries a current of 5.5 A. Calculate
for the total circuit the effective power, reactive power and apparent power and
the total power
factor!
