# 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 **= **Z**L**Q* where Z = *J*R^{2} + X^{2} 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 V_{L} and V_{G} 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/V_{c}), 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 V_{G} 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 mm^{2} line at 275 kV.)

**Figure 2.25 **Phasor diagram when V_{G} 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 V_{G} 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 V_{G} is specified and V_{L} is required, the phasor diagram in Figure 2.25 is used.

From this, If