 # X/R Ratio

The range of X/R values for typical voltage class (Canadian) are as follows: 735 kV, 18.9-20.4; 500 kV, 13.6-16.5; 220 kV, 2-25; 110 kV, 3-26. The X/R value decreases with separation of the fault point from the source and can be substantially decreased by the fault resistance.

For distribution circuits, X/R is lower, and, although data is limited, typical values are 10 (at source point) and 2-4 on a medium voltage overhead line.

Example 7.4

A synchronous machine A generating 1 p.u. voltage is connected through a star-star transformer, reactance 0.12 p.u., to two lines in parallel. The other ends of the lines are connected through a star-star transformer of reactance 0.1 p.u. to a second machine B, also generating 1 p.u. voltage. For both transformers, X1 = X2 = X0.

Calculate the current fed into a double-line-to-earth fault on the line-side terminals of the transformer fed from A.

The relevant per unit reactances of the plant, all referred to the same base, are as follows:

For each line: X1 = X2 = 0.30, X0 = 0.70.

For generators:

 X1 X2 X0 Machine A 0.30 0.20 0.05 Machine B 0.25 0.15 0.03

The star points of machine A and of the two transformers are solidly earthed.

Solution

The positive-, negative-, and zero-sequence networks are shown in Figure 7.21. All per unit reactances are on the same base. From these diagrams the following equivalent reactances up to the point of the fault are obtained: Z1 = /0.23 p.u. Z2 = /0.18 and Z0 = /0.17 p.u.

The red phase is taken as reference phasor and the blue and yellow phases are assumed to be shorted at the fault point. From the equivalent circuit for a line-to-line fault,  Figure 7.21 Line diagram and sequence networks for Example 7.4  The correctness of the first part of the solution can be checked as Example 7.5

An 11 kV synchronous generator is connected to a 11/66 kV transformer which feeds a 66/11/3.3 kV three-winding transformer through a short feeder of negligible impedance. Calculate the fault current when a single-phase-to-earth fault occurs on a terminal of the 11 kV winding of the three-winding transformer. The relevant data for the system are as follows:

Generator: X1 = /0.15 p.u., X2 = /0.10 p.u., X0 = /0.03 p.u., all on a 10 MVA base; star point of winding earthed through a 3 V resistor.

11/66kV Transformer: X1 = X2 = X0 = /0.1 p.u. on a 10 MVA base; 11 kV winding delta connected and the 66 kV winding star connected with the star point solidly earthed.

Three-winding transformer: A 66 kV winding, star connected, star point solidly earthed; 11 kV winding, star connected, star-point earthed through a 3 V resistor; 3.3 kV winding, delta connected; the three windings of an equivalent star connection to represent the transformer have sequence impedances,

• 66 kV winding X1 = X2 = X0 = /0.04 p.u.,
• 11 kV winding X1 = X2 = X0 = /0.03 p.u.,
• 3.3 kV winding X1 = X2 = X0 = /0.05 p.u.,

all on a 10 MVA base. Resistance may be neglected throughout.

Solution:

The line diagram and the corresponding positive -, negative -, and zero-sequence networks are shown in Figure 7.22. A 10 MVA base will be used. The 3 V earthing resistor has the following p.u. value: Much care is needed with the zero-sequence network owing to the transformer connections. For a line-to-earth fault, the equivalent circuit shown in Figure 7.13 is used, from which Hence  Figure 7.22 Line diagram and sequence networks for Example 7.5 