# Electrical supply system

1. May 5, 2016

### topcat123

1. The problem statement, all variables and given/known data
FIGURE 3 shows part of an electrical supply system.

Draw the impedance diagrams reducing to a single impedance.

Calculate the fault MVA levels in each unit for a fault on the left 3.3 kV bus with the bus section
switch B open and A closed. Use a base MVA of 10 and show
the fault MVA levels on a diagram.

Comment on the result and suggest a method of improving the system to limit the fault MVA through T1 to approximately 5 × FL current. (It is not necessary to recalculate.)

2. Relevant equations

3. The attempt at a solution
The fault MVA and pu impedances
$$G_1 and G_2\text{fault MVA} = \frac{16}{0.8}=20MVA$$
$$G_1 and G_2\text{ pu impedance} = \frac{10}{20}\frac{20}{100}=0.1pu$$
$$G_3\text{fault MVA} = \frac{1.6}{0.8}=2MVA$$
$$G_3\text{ pu impedance} = \frac{10}{2}\frac{20}{100}=1pu$$
$$T_1,T_2and T_3\text{ pu impedance} = \frac{10}{10}\frac{5}{100}=0.05pu$$
So we need to work out the total impedance and fault current.
$$T_2||T_3=0.025pu$$
$$G_3+T_2||T_3=1.025pu$$
$$(G_3+T_2||T_3)||G_1||G_2=0.0477pu$$
$$[(G_3+T_2||T_3)||G_1||G_2]+T_1=0.0977pu$$
So the fault MVA
$$\frac{10}{0.0977}=102.4MVA$$
So the fault MVA through T1 is 102.4
and through the rest
$$\text{voltage MVA}(G_3+T_2||T_3)||G_1||G_2=102.4*0.0477=4.88MVA$$
Fault through G1 and G2
$$\frac{4.88}{0.1}=48.88MVA$$
Fault through G3+T2||T3
$$\frac{4.88}{1.025}=4.76$$
$$\text{voltage MVA G3}=4.76*1=4.76MVA$$
voltage MVA over T2||T3
$$=0.025*4.76=0.119MVA$$
Fault through T2 and T3
$$\frac{0.110}{0.05}=2.38MVA$$

I think this is correct.
The part I am struggling with is
"Comment on the result and suggest a method of improving the system to limit the fault MVA through T1 to approximately 5 × FL current. (It is not necessary to recalculate.)"

Any help would be appreciated.
Thanks

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2. May 10, 2016