4.1 GENERAL
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Although it has been shown that three wires are
sufficient in this instance, nevertheless for the sake of this description the
neutral point of the generator is shown connected to the neutral point of the
load; it is known that, under these balanced conditions, no current will
actually flow in this fourth wire (see manual ‘Fundamentals of Electricity 2’,
Chapter 2).
Then, since the three phase voltages are equal in
magnitude and are spread 120° apart in time, the three vectors VR,
VY and VB in the lower part of the figure
represent the three phase voltages of the system - that is, the voltages
generated between the neutral point and each of the three terminals R, Y and B.
4.2 3-PHASE POWER
If in Figure 4.1 the three phases are considered quite
separately, each generator winding has phase voltage (e.g. VR)
developed in it. If the line
current is IL, then, if
the loading on that phase is purely resistive, the power transmitted by that
phase alone is:
VR x IL
Similarly the power transmitted by the two other phases alone is:
VY x IL
and
VB x IL
Since VR
= VY = VB = phase voltage, the total
power in all three phases is:
3VR IL
If the load were not purely resistive but had a power
factor cos j, then the active power transmitted by each phase would be:
VR IL cos ![](data:image/png;base64,iVBORw0KGgoAAAANSUhEUgAAABQAAAASCAIAAADUsmlHAAAAcklEQVQ4jdXSyRHAIAgF0F8XBVEP1dAMxZBDjMsoCZlcDCcOPpUF/iGwETYhgMRMCABAYklsQgDrmbG6u3LsR6xcqLtyyUwo0gPuz1Xc3/j65TRuJY/fXtup25UXfNev1ZyvGeFpUvGStJrj2BBn4qf4ACB10u9PktHqAAAAAElFTkSuQmCC)
and the total power for all three
phases would then be:
3VR IL cos
….(i)
If VL
is the line-to-line voltage (always indicated by switchboard voltmeters), it
has already been shown (the manual ‘Fundamentals of Electricity 2’) that VL
= √ 3VR, or VR = ![](data:image/png;base64,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)
Therefore from equation (i) the total 3-phase active power (W)
is:
FIGURE 4.2
3-PHASE SYSTEM – DELTA
Figure 4.2
shows a 3-phase system feeding a balanced load, but this time the generator is
delta-connected. As before, the
generator terminals, R, Y and B, are connected to the corresponding terminals
of a balanced 3-phase, delta-connected load, but for the sake of this
description suppose that each generator winding is independently connected to
its corresponding load element, so that six wires are used, as shown in the
diagram.
If the currents in the generator windings (i.e. the
phase currents) are IRY, IYB and IBR, and the voltages developed in each generator phase
are VRY, VYB and VBR, then, considering the winding RY alone, there is a
voltage VRY causing a
current IRY to flow from
terminal R, through one element of the load, returning to terminal Y and
through the generator phase winding YR back to terminal R. Similarly for the other two generator
windings.
Thus the two wires leaving terminal R carry respectively
an outward current IRY and
a return current IBR from
a different load element.
In practice, of course, there are not two wires from
each terminal, but each pair is commoned into a single wire at each terminal R,
Y and B. The single wire from terminal R
thus carries the difference current IRY
– IBR, which we will call
‘IR’. Similarly the single wire from terminal Y
carries the difference current IYB
– IRY, which will be ‘IY’ and that from terminal B
carries IBR – IYB, which will be ‘IB’.
The vector diagram at the bottom of Figure 4.2 shows the
generator’s three balanced phase currents IRY,
IYB and IBR. The differences are obtained, as explained in
the manual ‘Fundamentals of Electricity 2’, by joining the arrow-tips, so that
line AC equals IRY – IBR, which, as explained
above, is the net line current IR
leaving terminal R. Similarly CB
represents the net line current IY,
leaving terminal Y, and BA the net line current IB leaving terminal B.
That is to say, the ‘spokes’ of the diagram represent the three phase
currents in the generator, and the three sides of the triangle represent the
three line currents IR, IY and IB. Since these
are all numerically equal, they will be called IL (for line current).
It has been shown, from the geometry of Figure 7.6 in
the manual ‘Fundamentals of Electricity 2’, that the length of the triangle’s
sides is √ 3 times the length of its spokes, so that:
IL (=IR = IY
= IB) = √ 3 x IRY (or IYB or IBR)
that is, line
currents = √ 3 x any phase
current
The active power delivered by any one phase (say phase
RY) alone is:
where cos j is the power factor of the load (indicated on the vector diagram by
IRY lagging by an angle j on the corresponding voltage VRY). But it has just been shown that
Therefore, from equation (iii), the power delivered by
phase RY alone is:
and the power delivered by all three phases (P) is three times this, namely:
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With a delta connection any
line-to-line voltage (VL) is the same as the corresponding
generator phase voltage, since both are in parallel, so that VL
= VRY. Therefore, from equation (iv), the total active power
transmitted is:
If this is compared with equation (ii) for a
star-connected generator, it will be found to be exactly the same.
Therefore, if in a balanced 3-phase system the line
voltage is VL (volts) and the line current IL (amperes), and if the load
power factor is cos j, then the active power (P)
delivered is:
whether the power source is star or delta connected.
Example
In a 6.6kV system a 3-phase load draws 200A line current
at a power factor
![](data:image/png;base64,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)
\ Total power
= √ 3 x 6 600 x 200 x 0.8W
= 285 780W
= 286kW approximately.
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