Pages

Wednesday, December 5, 2012

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 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

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  =
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 IRYIBR, which we will call ‘IR’.  Similarly the single wire from terminal Y carries the difference current IYBIRY, which will be ‘IY’ and that from terminal B carries IBRIYB, 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 IRYIBR, 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:

                            

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

   \    Total power  = 3 x 6 600 x 200 x 0.8W
= 285 780W
= 286kW approximately.

No comments:

Post a Comment