Power Transfer and Reactive Power
The circuit shown in Figure 2.23 represents the simplest electrical model for a source (with voltage Vg) feeding into a power system represented by a load of P + jQ. It can also represent the power flows in a line connecting two busbars in an interconnected power system.
The voltage at the source end and load end are related by:
Figure 2.23 Power transfer between sources
If R + jX = ZLQ where Z = JR2 + X2 and U = tan 1 (X/R), then the current can be obtained as:
Therefore the apparent power at the source end is given by:
If the load end voltage is chosen as the reference and the phase angle between the load end and source end is d:
Substituting for VL and VG from equation (2.10) into (2.9) yields:
Therefore
Similarly,
The power output to the load is a maximum when cos(U — d) = 1, that is, U = d
Calculation of Sending and Received Voltages in Terms of Power and Reactive Power
The determination of the voltages and currents in a network can obviously be achieved by means of complex notation, but in power systems usually power (P) and reactive power (Q) are specified and often the resistance of lines is negligible compared with reactance. For example, if R = 0.1X, the error in neglecting R is 0.49%, and even if R = 0.4X the error is 7.7%.
From the transmission link shown in Figure 2.23:
For the load:
The voltage at the source and load are related by:
As Vl = VL = Vl (in this case):
Equation (2.14) can be represented by the phasor diagram shown in Figure 2.24.
Figure 2.24 Phasor diagram for transmission of power through a line
and
If d is small (as is usually the case in Distribution circuits) then
then
and
Hence the arithmetic difference between the voltages is given approximately by
In a transmission circuit, R ' 0 then
that is, the voltage magnitude depends only on Q.
The angle of transmission d is obtained from sin-1 (Д Vq/Vc), and depends only on P.
Equations (2.15) and (2.16) will be used wherever possible because of their great simplicity.
Example 2.6
Consider a 275 kV line of length 160 km (R — 0.034 V/km and X = 0.32 V/km). Obviously, R < X.
Compare the sending end voltage VG in Figure 2.25 when calculated with the accurate and the approximate formulae. Assume a load of 600 MW, 300MVAr and take a system base of 100 MVA. (Note that 600 MVA is nearly the maximum rating of a 2 x 258 mm2 line at 275 kV.)
Figure 2.25 Phasor diagram when VG Is specified
Solution
The base impedance is
For line of length 160 km,
and let the received voltage be
then
and
, 0.406
Hence S — tan 1 — 18
1.203
This is a significant angle between the voltages across a circuit and so the use of the approximate equations for this heavily loaded, long transmission circuit will involve some inaccuracy.
Using equation (2.14) with R neglected.
An approximate value of VG can be found from equation (2.17) as
that is, an error of 5.6%.
With a shorter, 80 km, length of this line at the same load (still neglecting R) gives
and
The accurate formula gives
and the approximate one gives
that is, an error of 1.7%.
In the solution above, it should be noted that p.u. values for power system calculations are conveniently expressed to three decimal places, implying a measured value to 0.1% error. In practice, most measurements will have an error of at least 0.2% and possibly 0.5%.
If VG is specified and VL is required, the phasor diagram in Figure 2.25 is used.
From this,
If
