7.1 VECTOR DIAGRAMS
In alternating current practice the
quantities such as voltages, currents, impedances, fluxes and so on have not
only numerical values but also a time or phase relationship with each
other. Although their numerical values
may follow a pure sine-wave form, the instants they reach their peaks may be
different, one following (or ‘lagging’) on the other. This time difference is expressed not in
seconds but as a fraction of the time for one period, or ‘cycle’, which is
taken to be 360°. Thus a quantity whose peak
value follows one-quarter of a cycle behind another’s is said to lag 90° on it. (If there is likely
to be confusion between any other degrees - say the rotation of a rotor - the
lag may be expressed in ‘electrical degrees’.)
A lag of half a cycle is 180°, and of
one-tenth of a cycle 36°.
If the peak value of a quantity occurs before that of the reference quantity, it is said to ‘lead’ on it,
the amount of lead being similarly expressed in (electrical) degrees. The difference angle, usually called ‘
‘, is the ‘phase relationship’ between the quantities.
‘, is the ‘phase relationship’ between the quantities.
FIGURE 7.1
VECTOR DIAGRAM
This relationship is easily
expressed in diagram form. In any
relationship one of the partners is considered to be the ‘reference’ or datum
against which the other is measured. In Figure 7.1(a) suppose the line OA
represents the reference quantity, which may be an alternating voltage at a
certain point or any other quantity. Its
length represents the numerical peak value of the reference voltage, and an
arrow is added to show the direction
which is taken as the reference
datum. OA is said to be a ‘vector’,
having both length and direction. (Note: the term ‘phasor’ is now often used in
such cases instead of ‘vector’. However
in this series of manuals the word ‘vector’ will be used exclusively.)
The line OA is by convention
considered to be-rotating anti-clockwise at a speed which will cause it to
complete one full revolution in the time of one period or cycle; thus its speed
is determined solely by the system frequency.
Figure 7.1(a) is simply a flash ‘photograph’ of the line as it passes
through the datum direction, here taken to be 12 o’clock (though some prefer to
use 3 o’clock as datum; it makes no difference which is used).
FIGURE 7.2
CURRENT/VOLTAGE RELATION
If now another alternating quantity of a different peak value and lagging in phase on OA exists (it might be another voltage at some other point, or a current, or a flux), it could be fully represented by another vector line OB, where the length OB represents the numerical peak value of the second quantity, and the angle
between OB and OA is the angle in
electrical degrees by which OB lags on OA, as shown in Figure 7.1(b). Since both are considered to be rotating
anti-clockwise, OB is following OA
and therefore truly lagging on it, and Figure 7.1(c) is the flash photograph of
both vectors at the same instant when OA is passing through the 12 o’clock
datum direction. If OB were leading on
OA, it would be on the left of OA and so rotating ahead of it
A particular and important case is
when OA represents the voltage applied to a circuit with an impedance having
both resistance and reactance (see Chapters 8, 9 and 10), and OB represents the
current that consequently flows in it.
In general the current will lag on the
voltage by an angle which depends on
the relative amounts of resistance and reactance (see Figure 7.2(a)). As explained in Chapter 9, the angle j of lag is given by the expression
where X and R are respectively
the reactance (= 2pfL) and the resistance in ohms.
It can be seen at once that if there is only resistance and no
reactance, X = 0 and so tan j = 0, which makes j = 0. This means that there
is no angle between current and voltage, and the current is said to be ‘in
phase’ with the voltage (see Figure 7.2(b)).
If on the other hand there were only
reactance and no resistance, R = 0
and tan j becomes infinite, which makes j = 90°. In that case the current
lags 90° on the voltage and is said to be ‘in quadrature’ with the voltage
(see Figure 7.2(c)). This is a very
common situation when dealing with generators and transformers under fault
conditions, as discussed in the manual ‘Electrical Protection’.
Both the above, R = 0 or X = 0, are
extreme cases. The general case is that
shown in Figure 7.2(a), where the phase angle j is somewhere between 0° and 90°.
7.2 ADDING AND SUBTRACTING VECTORS
Vectors represent quantities, such as voltages and currents, and they can be added and subtracted just like ordinary numbers.
FIGURE 7.3
VECTORS IN PHASE
OA or OB are two vector quantities P and Q in phase with each other.
In Figure 7.3(a) it is required to add them together. It is clear that, if OB were picked up and
put down over OA, the base (O) of OB being placed on the arrow-tip of OA to the
position O’B’, then the combined length OB’ represents the sum of OA and OB and
so is the quantity P + Q.
If OA were 4 inches and OB 3 inches, P
+ Q would be 7 inches.
In Figure 7.3(b) it is required to subtract
OB from OA. If a line were drawn joining
the two arrow-tips, its length would be BA; it represents the difference
between OA and OB and so is the quantity P
- Q.
The direction of this difference line is from the arrow-tip of
the subtracted quantity to that of the quantity it was taken from - that
is from B to A.
Where the quantities OA and OB were in phase, as above, the process was simple, as shown in Figure 7.3.
FIGURE 7.4
VECTORS NOT IN PHASE
The process is
exactly the same however where the quantities are not in phase. To add the two vectors OA and OB in Figure
7.4(a), simply pick up vector OB and place it, without altering its direction,
on top of OA in position O’B’. OA and
O’B’ are added together as in Figure 7.4(a), but the route now goes round a
corner. The line OB’ joining O to the
new point B’ is then the vector sum
of OA and OB; it is written P + Q to distinguish it from the numerical
sum P + Q. It is clear that the
vector sum P + Q is in general less than the numerical sum P + Q.
Some prefer to use
the construction where the parallelogram on OA or OB is completed. The diagonal through O is then the addition
vector representing the sum P + Q.To subtract two
vector quantities which are not in phase, the process is exactly similar to
that shown in Figure 7.3(b). Simply join
the tips of the two vector arrows, as shown in Figure 7.4(b). The vector BA is then the difference between
the vectors OA and OB and is written P - Q.
Since OB is being subtracted from
OA, the direction of the difference vector is from B to A, as
shown in Figure 7.4(b). If OA had been
subtracted from OB, the direction would have been from A to B.
If the parallelogram on OA and OB is
drawn as before, the difference vector is then the other diagonal - that
is the diagonal across O.
7.3 EXAMPLE: ADDITION OF ACTIVE AND REACTIVE ELEMENTS
Chapters 8 to 10 show that a current in a purely resistive circuit is in phase with the applied voltage, whereas that in a purely inductive circuit lags 90° on the voltage. Vector methods can be used to handle such currents - for example to add them.
FIGURE 7.5
EXAMPLE OF THE VECTOR
ADDITION OF CURRENTS
In Figure 7.5 an a.c. generator feeds a resistance and a reactance in parallel which draw 40A and 30A respectively. The resistance current IR is in phase with the applied voltage, but the reactance current Ix lags 90° on the voltage (see Chapter 9). The combined current is now not 70A but is 50A
because the currents
add vectorially.
FIGURE 7.6
STAR CONNECTION VOLTAGES
7.4 PHASE AND LINE VOLTAGES AND CURRENTS
An important application of vectors
is for determining the relationship between the phase and the line-to-line
voltages and currents in a 3-phase system.
FIGURE 7.7
VOLTAGE BETWEEN LINES
Consider a 3-phase system supplied from a generator as shown
at the top of Figure 7.6. Suppose that
the generator is star-connected. Then,
since the three 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 as developed in the three windings of the generator. They are the voltages between each of the
three terminals and the star (or neutral) point.
Consider the simple system of Figure
7.7. Two lines A and B carry voltages VA
and VB measured with respect to a common line (or earth)
C. Clearly the voltage between lines A
and B is the difference between the individual line voltages VA
and VB.
Reverting to Figure 7.6, by the same
token the voltage between any two lines, say R and Y, is the difference between
their individual voltages VR and VY
measured with respect to their common neutral point.
VR and VY, are
vectors, and it has already been shown that the difference between two vector
quantities is obtained by joining their arrow-tips. So the voltage between lines R and Y, which
we will call ‘VRY‘, is the difference between vectors VR
and VY, and is therefore represented by the line AB. Similarly line BC represents the voltage
between lines Y and B, which is VYB, and line CA represents
the voltage between lines B and R, which is VBR.
Thus in Figure 7.6 the sides of the
triangle represent the line-to-line voltages, and the ‘spokes’ represent the
three phase (or line-to-neutral) voltages.
FIGURE 7.8
PHASE AND LINE VOLTAGES
Consider more closely the geometry
of the triangle OAB. This is reproduced
in Figure 7.8, and an equilateral triangle OAP is constructed on line OA. Let OP cut AB at point N.
If OA (= OB) is assumed to be one
unit of length, then ON (which is half OP) will be one-half unit of
length. By Pythagoras:
This means that the triangle’s sides
are √ 3 times as long as the spokes.
Hence, with a star connection:
line
voltage = √ 3 (phase voltage)
This is a very important result
which comes into numerous system calculations, especially the calculation of
3-phase power which is explained in the manual ‘Fundamentals of Electricity 3’.
It should also be noticed from
Figure 7.6 that the current in each line is the same as the current in
the corresponding generator phase winding, since both are in series. So, with a star connection:
line current = phase current
A similar method is used to
determine the individual phase currents (that is, the actual machine currents)
in a delta-connected machine or transformer when only the line currents are
known.
Figure 7.9 (top) shows a
delta-connected generator. Suppose it is
feeding a balanced load - that is where each of the three load elements
has the same impedance and the same power factor. This load itself may be star- or
delta-connected, as indicated. If IRY, IYB and IBR
are the currents in the three phase windings of the machine, and if IR, IY and IB
are the three line currents, then applying Kerchoff’s Law to the machine
terminal R (which says that the total current entering any point is balanced by
the current leaving it):
IBR (in) = IRY + IR (out)
or IR = IBR - IRY ….(i)
where all these are vector
quantities.
FIGURE 7.9
DELTA CONNECTION
CURRENTS
The three balanced currents are
shown at the bottom of Figure 7.9. The
vectors IBR and IRY are two of them, and
their difference (see equation (i)) is obtained, as usual, by joining the
arrow-tips, A and C. The line AC then
represents the line-current vector IR
from equation (i).
Using the same geometry as was used
in Figure 7.8 for the voltages, it is clear that IR =√ 3 times IRY. Similarly for IY and IB
Thus, with a delta connection:
line
current = √ 3 (phase current)
It should also be noticed from
Figure 7.9 that the voltage across each pair of lines is the same as that of
the corresponding generator phase winding, since both are in parallel. So with a delta connection:
line
voltage = phase voltage
7.5 SUMMARY
A. Star Connection
line
voltage = √ 3 (phase voltage)
line
current = phase current
B. Delta Connection
line
current = √ 3 (phase current)
line
voltage = phase voltage
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