CHAPTER 7 VECTORS (OR PHASORS)

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.

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 I_R is in phase with the applied voltage, but the reactance current I_x 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 V_R, V_Y and V_B 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 V_A and V_B 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 V_A and V_B.

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 V_R and V_Y measured with respect to their common neutral point.

V_R and V_Y, 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 ‘V_RY‘, is the difference between vectors V_R and V_Y, and is therefore represented by the line AB.  Similarly line BC represents the voltage between lines Y and B, which is V_YB, and line CA represents the voltage between lines B and R, which is V_BR.

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 I_RY, I_YB and I_BR are the currents in the three phase windings of the machine, and if I_R, I_Y and I_B 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):

                                                            I_BR (in) =     I_RY + I_R (out)

                                                        or    I_R       =     I_BR - I_RY                                        ….(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 I_BR and I_RY 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 I_R from equation (i).

Using the same geometry as was used in Figure 7.8 for the voltages, it is clear that I_R =√ 3 times I_RY.  Similarly for I_Y and I_B

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