The Long Line (above 240 km, 150 miles)

Here the treatment assumes distributed parameters. The changes in voltage and current over an elemental length Dx of the line, x metres from the sending end, are determined, and given below:

where

z = impedance/unit length y = shunt admittance/unit length

If Dx ! 0, then

By differentiating (3.2) and substituting from (3.3)

Similarly,

Solution to equations (3.4) and (3.5) takes the form Vx = A1 cosh Px + A2 sinh Px and Ix = B1 cosh Px + B2 sinh Px. When x = 0, as Vx = VS and Ix = IS, the voltage and current at x metres from the sending end are given by

where

and

where

R = resistance/unit length L = inductance/unit length G = leakage/unit length C = capacitance/unit length

Z0 is the input impedance of an infinite length of the line; hence if any line is terminated in Z0 its input impedance is also Z0.

The propagation constant P represents the changes occurring in the transmitted wave as it progresses along the line; a measures the attenuation, and b the angular phase-shift. With a lossless line, where R = G = 0, P = jv/LC and b = л/LC. With a velocity of propagation of 3 x 105km/s the wavelength of the transmitted voltage and current at 50 Hz is 6000 km. Thus, lines are much shorter than the wavelength of the transmitted energy.

Usually conditions at the load are required when x = l in equations (3.6). Alternatively,

The parameters of the equivalent four-terminal network are thus, where

Z = total series impedance of line Y = total shunt admittance of line

The easiest way to handle the hyperbolic functions is to use the appropriate series.

Usually not more than three terms are required, and for (overhead) lines less than 500 km (312 miles) in length the following expressions for the constants hold

Equivalent circuit to represent accurately the terminal conditions of a long line

Figure 3.29 Equivalent circuit to represent accurately the terminal conditions of a long line

approximately:

An exact equivalent circuit for the long line can be expressed in the form of the p section shown in Figure 3.29. The application of simple circuit laws will show that this circuit yields the correct four-terminal network equations. Figure 3.29 is only for conditions at the ends of the line; if intermediate points are to be investigated, then the full equations must be used.

If only the first term of the expansions are used, then

that is, the medium-length p representation.

Example 3.2

The conductors of a 1.6 km (1 mile) long, 3.3 kV, overhead line are in horizontal formation with 762 mm (30 in) between centres. The effective diameter of the conductors is 3.5 mm. The resistance per kilometre of the conductors is 0.41 V. Calculate the line-to- neutral inductance of the line. If the sending-end voltage is 3.3 kV (50 Hz) and the load is 1 MW at a lagging p.f. of 0.8, estimate the voltage at the load busbar and the power loss in the line.

Solution

The equivalent equilateral spacing is given by de — •Зd12 x d23 x d31 In this case de — -3762 x 762 x 1524

The inductance (line to neutral)

The total inductance of 1.6 km The inductive reactance Resistance of line Impedance of the line

An estimate of the voltage drop is required and so the distribution approximation will be used.

Choosing 1MVA and 3.3 kV as the bases for the calculation Using the distribution approximation in per unit

The voltage drop is therefore 347 V (3.3 kV line-line) and 201V (1.9 kV line-neutral)

The apparent power of the load is 1.25 MVA or 1.25 per unit. Assuming the load voltage to be 1 per unit, this is also the current in per unit. The line loss is 1.252 x 0.0602 = 0.094 per unit or 94 kW for all three phases.

As the sending end voltage and receiving end power are specified, a more precise iterative calculation may be undertaken with Equation (2.13)

Example 3.3

A 150 km long overhead line with the parameters given in Table 3.2a for 400 kV, quad conductors is to be used to transmit 1800 MW (normal weather loading) to a load with a power factor of 0.9 lagging. Calculate the required sending end voltage using three line representations and compare the results.

Solution

From Table 3.2 a

Choosing a base of 2000 MVA and 400 kV.

For a 150 km line

Short-line representation:

Load power, S = 1800 + j870 MVA S = VI* and as the receiving end voltage is at 400 kV,

Hence

Medium-line representation:

Hence

Long-line representation:

(Note: This is an extremely high voltage for normal operation and would not be tolerated. A 400 kV system would be designed for only about 10% steady-state overvoltage, that is 400 + 40 = 440 kV. In practice, either the reactance of the line would be reduced by series capacitors and/or the power factor at the receiving end would be raised to at least 0.95 lag by the use of shunt capacitors or synchronous compensators).

 
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