Consider the 13.8 kV radial distribution feeder shown in the figure below. The s

Question

Consider the 13.8 kV radial distribution feeder shown in the figure below. The short-circuit MVA at node 2 is 450 MVA, the maximum load current through fuse A is 135 A and the maximum load current through fuse B is 42 A. The base values are 13.8 kV and 10 MVA for the feeder parameters. The pu values of the power and distribution transformers are in their own base values. Perform the following tasks (1) Calculate the fault current at node 4 for a bolted three-phase fault and a line-to-ground fault. (2) Select the protecting and protected fuses (i.e., fuses B and A, respectively) (3) Calculate the coordination limit for the selected fuses using (a) The Kearney Current vs. Time characteristics (or any other curves like those for Positrol fuse links from S&C; Electric Company). Please, provide the corresponding graphs similar to Fig. 10.19, and Coordination tables (see for example Tables 10.3 and 10.4 in the textbook) (b) (4) Repeat steps (2)-(3) assuming that the maximum load current through fuse A is now 80 A. (5) Present the results in the form of a written report. Table 1 Feeder parameters in entage (%) From 3 4 To 4 5 3.0 4.0 10 4.5 18 20 60 22.5

Explanation / Answer

1)Whenever the fault occur the current flow towards the fault location.current from 1-phase and 3-phase will flow towards node-4.

Three phase fault current will makes the zero sequence currents to flow through the neutral hence unbalance in the network results Ir+Iy+Ib is not equal to zero .hence zero sequence will flow..

Impedance in the line ZL=(r^2+jX^2)=3^2+10^2=10.44 ohms

Power factor is cos()=X/Z=10/10.44=0.957

Three phase fault current at 0.957 power factor is

If=10×10^6÷3×13.8×10^3×0.957=437.16A

3phase=3×437.16=1311.5A

Zero sequence current Io=3×Vsc÷Zsc=3×437/10.44÷62.64

Io=2A

Since Zsc is Ro^2+Xo^2=62.64 ohms

Hence total three phase fault current is 1313 A at node-4

Line to ground fault is calculated as

If=437 A will flow through the fault path I.e line to discharge into ground.

2)as we know fuse rating can be calculated by using 1.25 × making capacity

Hence at B we go with 55 amps fuse because 1.25×42= 52.5

At A Breaking capacity =1.25×135=168.75

Hence @ A we choose 170 amps fuse

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