Chapter-7
Electrical Energy & Its Derivation From Power.
Whenever a battery is required to supply a relatively large amount of power,
secondary cells must be used. Such batteries may supply power in telephone exchanges,
emergency lighting for hospitals etc. They also form portable sources of power for starting,
ignition and lighting of motor vehicles and as motive power for electrically propelled
vehicles.
Most of the readers are familiar with secondary cells or batteries. Batteries have
specific characteristic charge rate and discharge rate. Discharge rate is expressed in amperehour and battery capacity is also represented in ampere-hours. i.e. 100 Ah capacity battery
can supply 01A for almost 100 hours, 10 A for 10 hours and 100 Amps for 01 hour at the
battery terminal voltage. Actually terminal voltage drops-off rapidly with heavy discharge
rates (hence proper continuous charging is required).
Historically from 19th centenary, when A.C system was not present, experiments
started on D.C systems and battery capacity was described as amper-hour rating at specific
terminal voltages as 6V D.C, 12V D.C were common at earlier stage of development. Now a
days 24V D.C, 48V D.C or 110V D.C and 220V D.C. are common at Utility Grid Stations
and power houses for protection purposes and black-starts.
As per law of conservation of energy, it can not be generated, instead it can be
converted from one source to another one.
Actually chemical energy is stored in a battery and electrical energy can be converted
by chemical reaction in the battery, so some sort of approach must be incorporated to work
out the energy supplied by a D.C battery to D.C system.
Ampere-hour capacity of the Battery = Ampere X Hours at specific terminal
voltage.
A X h
……………………………( 7.1)
That is,
i X t
at specific terminal voltage / or system voltage of D.C system ( as
the battery terminal voltage is constant if proper charging of the battery is maintained).
7-1
If above expression is jointly expresses, i.e. taking into account voltage,
We have,
i
or
X t
X
V
V X i X t
…………………………..(7.2)
Where the product V X i is the power in a d.c circuit. Its units are Volt-ampere.
And V . i . t is the energy in a d.c circuit. The units are Volt-ampere-hour.
Equation (7.2) is only valid for d.c. circuits, where voltage and current are in-phase.
However, for a.c. circuits, the phase angle or angle displacement between
voltage and current brought about another consideration.
In direct-current circuits power may be measured either by a watt-meter or by an
ammeter and voltmeter, the product of whose readings gives the power in the circuit.
In alternating-current circuits the power at any instant is given by :
p = e i…………………………………… (11.6.1)
Where p = instantaneous power
e = instantaneous voltage
i = instantaneous current
Thus, if both the current and voltage waves are sinusoidal, the current is lagging
the voltage, in phase by an angle Φ, then
e = Emax Sin wt
i = Imax Sin (wt-Φ)
The instantaneous power p is therefore given by
p=ei
= Emax Imax Sin wt Sin (wt-Φ)
Or, writing θ for wt
= Emax Imax Sin θ Sin (θ -Φ)
7-2
The mean power is :
P= 1
∫02 Π
Emax Imax Sin θ Sin (θ -Φ) d θ……….(11.6.2)
2Π
P = Emax Imax
∫02 Π
2Π
Cos Φ Cos (2θ -Φ) d θ
2
P = Emax Imax
[
θ Cos Φ - Sin (2θ -Φ) ] 02 Π
4Π
2
P = Emax Imax
Cos Φ
2
P= E I
Cos Φ ………………………………..(11.6.3)
Where E and I are r.m.s. values of voltage and current.
The fact that the power factor (Cos Φ) is involved in the expression for the power
means that a wattmeter must be used instead of merely an ammeter and voltmeter, since the
latter method takes no account of power factor.
It may be stated generally that the measurement of energy is essentially the same
process as the measurement of power, except that the instrument used must not merely
indicate the power, or rate of supply of energy, but must take into account also the length of
time for which this rate of supply is continued.
Actually, energy or “supply” meters do not indicate power directly. For a given
amount of energy supplied to a circuit, their registrations should always be the same, no
matter what the instantaneous values of the power during the time in which the energy is
supplied.
7-3
As the rate of supply of energy is called as Power, hence converse is true other
way round i.e. if we integrate the mean power over a specific time interval, we will get mean
amount of energy to be supplied to a circuit.
Hence,
E = ∫ P. dt
= ∫ E I
Cos Φ
i.e Energy is integral of power over specific time interval.
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7-4