CHAPTER 4 THREE-PHASE POWER

4.1       GENERAL

FIGURE 4.1 3-PHASE SYSTEM -STAR

Consider a 3-phase system supplied from a generator as shown at the top of Figure 4.1.  Suppose the generator is star-connected and that each terminal, R, Y and B, is connected to the corresponding terminal of a 3-phase, star-connected load.

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 V_R, V_Y and V_B 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. V_R) developed in it.  If the line current is I_L, then, if the loading on that phase is purely resistive, the power transmitted by that phase alone is:

V_R x I_L

Similarly the power transmitted by the two other phases alone is:

                                                                  V_Y x I_L

                                                           and V_B x I_L

Since V_R = V_Y = V_B = phase voltage, the total power in all three phases is:

                                                                  3V_R I_L

If the load were not purely resistive but had a power factor cos j, then the active power transmitted by each phase would be:

V_R I_L cos

and the total power for all three phases would then be:

                                                                  3V_R I_L cos                                                    ….(i)

If V_L is the line-to-line voltage (always indicated by switchboard voltmeters), it has already been shown (the manual ‘Fundamentals of Electricity 2’) that V_L = √ 3V_R, or V_R  =

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 I_RY, I_YB and I_BR, and the voltages developed in each generator phase are V_RY, V_YB and V_BR, then, considering the winding RY alone, there is a voltage V_RY causing a current I_RY 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 I_RY and a return current I_BR 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 I_RY – I_BR, which we will call ‘I_R’.  Similarly the single wire from terminal Y carries the difference current I_YB – I_RY, which will be ‘I_Y’ and that from terminal B carries I_BR – I_YB, which will be ‘I_B’.

The vector diagram at the bottom of Figure 4.2 shows the generator’s three balanced phase currents I_RY, I_YB and I_BR.  The differences are obtained, as explained in the manual ‘Fundamentals of Electricity 2’, by joining the arrow-tips, so that line AC equals I_RY – I_BR, which, as explained above, is the net line current I_R leaving terminal R.  Similarly CB represents the net line current I_Y, leaving terminal Y, and BA the net line current I_B 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 I_R, I_Y and I_B.  Since these are all numerically equal, they will be called I_L (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:

                                                I_L (=I_R = I_Y = I_B) = √ 3 x I_RY (or I_YB or I_BR)

                                  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 I_RY lagging by an angle j on the corresponding voltage V_RY).  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:

With a delta connection any line-to-line voltage (V_L) is the same as the corresponding generator phase voltage, since both are in parallel, so that V_L = V_RY. 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 V_L (volts) and the line current I_L (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

   \    Total power  = √ 3 x 6 600 x 200 x 0.8W

= 285 780W

= 286kW approximately.