Electrical supply system

  • Thread starter topcat123
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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 T1 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 G1 and G2
[tex]\frac{4.88}{0.1}=48.88MVA[/tex]
Fault through G3+T2||T3
[tex]\frac{4.88}{1.025}=4.76[/tex]
[tex]\text{voltage MVA G3}=4.76*1=4.76MVA[/tex]
voltage MVA over T2||T3
[tex]=0.025*4.76=0.119MVA[/tex]
Fault through T2 and T3
[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
 

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Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 
  • #3
Hi,
I have attempted the above question (before looking for help on here) and have the same answer as topcat123 for the first part, so assume I have calculated correctly.
I am to struggling on the second part of limiting the fault current through T1 to 5 x FL current.
I think I have found a way of doing this but this would be by changing the Voltage Impedance of T1 and also making use of a Reactor.
I am unsure if this is correct as changing the Voltage Impedance of T1 would effectively be changing the Transformer??
Any help would be appreciated.
Matt
 
  • #4
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126
The calculation, it seems to me to be correct. However I did the calculation according to IEC 60909 also-using only reactances- and the difference is 2.5% less.
Since the system is isolated-that means no connection with the outside system ,even the generator voltages will be 11[3.3] kV the voltage on 11 kV bus decrease up to 5.365 kV[1.61 kV on 3.3 system] in the case of short-circuit when the short-circuit current will be Isc=102.4/sqrt(3)/3.3 =17.92 kA. and then the MVA on T1 will be sqrt(3)*17.92*1.69=52.424 MVA only.
So now the T1 load is 5.24 times the rated-at 17.92 kA.
If the transformer T1 impedance stays unchanged you have to reduce the current flowing through by 5/5.24[17.50 kA] by inserting a reactor in the T1 circuit [my opinion a series reactor of 0.00252 ohm/17.5 kA/3.3 kV ]
 

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