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## Homework Statement

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.)

## Homework Equations

## The Attempt at a Solution

The fault MVA and pu impedances

[tex]G_1 and G_2\text{fault MVA} = \frac{16}{0.8}=20MVA[/tex]

[tex]G_1 and G_2\text{ pu impedance} = \frac{10}{20}\frac{20}{100}=0.1pu[/tex]

[tex]G_3\text{fault MVA} = \frac{1.6}{0.8}=2MVA[/tex]

[tex]G_3\text{ pu impedance} = \frac{10}{2}\frac{20}{100}=1pu[/tex]

[tex]T_1,T_2and T_3\text{ pu impedance} = \frac{10}{10}\frac{5}{100}=0.05pu[/tex]

So we need to work out the total impedance and fault current.

[tex] T_2||T_3=0.025pu[/tex]

[tex]G_3+T_2||T_3=1.025pu[/tex]

[tex](G_3+T_2||T_3)||G_1||G_2=0.0477pu[/tex]

[tex][(G_3+T_2||T_3)||G_1||G_2]+T_1=0.0977pu[/tex]

So the fault MVA

[tex]\frac{10}{0.0977}=102.4MVA[/tex]

So the fault MVA through T

_{1}is 102.4

and through the rest

[tex]\text{voltage MVA}(G_3+T_2||T_3)||G_1||G_2=102.4*0.0477=4.88MVA[/tex]

Fault through G

_{1}and G

_{2}

[tex]\frac{4.88}{0.1}=48.88MVA[/tex]

Fault through G

_{3}+T

_{2}||T

_{3}

[tex]\frac{4.88}{1.025}=4.76[/tex]

[tex]\text{voltage MVA G

_{3}}=4.76*1=4.76MVA[/tex]

voltage MVA over T

_{2}||T

_{3}

[tex]=0.025*4.76=0.119MVA[/tex]

Fault through T

_{2}and T

_{3}

[tex]\frac{0.110}{0.05}=2.38MVA[/tex]

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