Voltage regulation is often the limiting factor
in the choice of either conductor or type of insulation. While the heat loss in the cable
determines the maximum current it can safely carry without excessive deterioration, many
circuits will be limited to currents lower than this in order to keep the voltage drop
within permissible values. In this connection it should be remembered that the high
voltage circuit should be carried as far as possible so that the secondary runs, where
most of the voltage drop occurs, will be small.
The voltage drop of a feeder may be calculated from the
following formulae:
V = 100 (VS
VL) / VL
V = Voltage regulation in percent
VL = Voltage across load
VS = Voltage at source
VS =![VS = square root [(VL Cos theta + Rl)2 + (VL sin theta + Xl)2]](images/Eqn12-1.gif)
= is the angle by which the load current lags the voltage
across the load
Cos = Power factor of
load
R = Total a-c resistance of feeder
X = Total reactance of feeder
I = Load current
Approximate formula for voltage drop:
(VS VL)
= RI cos + XI sin  |
This above formula is
satisfactory where the power factor angle is nearly the same as the impedance angle. It is
exact when they are equal.
That is: tan = X / R
Above values apply directly for single phase
lines when resistance and reactance are loop values and voltage is voltage between lines.
For 3-phase circuits, use voltage to neutral
and resistance and reactance of each conductor to neutral. This gives voltage drop to
neutral. To obtain voltage drop line-to-line, multiply voltage drop by . (The percent
voltage drop is of course the same between conductors as from conductor to ground and
should not be multiplied by ).
Example: 3 single coated copper conductors
600 volt cables in non-metallic conduit.
Size conductor =4/0, Awg Copper .080 insulation, .045 jacket.
O.D. = .810
Voltage = VS = 440 volts 3 phase
Current = I= 250 amperes
Power Factor = cos = 0.8
Length = 750 ft.
Resistance
Per conductor = R
= .0525 ohms 1000 feet at 25°C
= .047 ohms for 750 feet at 75°C
Reactance
Per conductor = X
= .031 ohms 1000 feet (see table)
= .028 ohms for 750 feet
(including 20% for random lay)
VS =![VS = square root [(VL Cos theta + Rl)2 + (VL sin theta + Xl)2]](images/Eqn12-1.gif)
![440 / square root 3 = square root [(.8VL + .047 x 250)2 + (.6VL + .028 x 250)2]](images/Eqn12-4.gif)
Solving for VL ; VL = 240.4
Line-to-line voltage = 240.4 = 417
Voltage drop = 440 417 = 23 volts
![V = [(440 - 417) / 417 ] (100) = 5.52%](images/Eqn12-5.gif)
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Approximate Formula:
Voltage drop = line to
neutral
= Rl cos
+ Xl sin 
= 0.047 X 250 X .08 + 0.028 X 250 X 0.6
= 9.4 + 4.2 = 13.6
Line-to-line voltage drop = 13.6 = 23.5 volts |
Conductor Reactance
The following table shows a nomogram for
determining the reactance of any solid or concentric stranded conductor. This covers
spacings encountered for conduit wiring as well as for open wire circuits. Various
modifications necessary for use under special conditions are covered in notes on the
nomogram. The reactances shown are for 60-Hertz operation.
Where regulation is an important
consideration several factors should be kept in mind in order to obtain the best operating
conditions.
Open wire lines have a high reactance. This
may be improved by using parallel circuits but is much further reduced by using insulated
cable. Three conductors in the same conduit have a lower reactance than conductors in
separate conduits.
Single conductors should not be installed in
individual magnetic conduit because of the excessive reactance.
Three conductors in magnetic conduit will
have a somewhat higher reactance than cables in non-magnetic conduit.
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