Improving Electrical Supply System for Lower Fault MVA: Analysis and Solutions

In summary: A practically 5 times the rated current [at fault] and the system stays balanced and the voltage will be 11*sqrt(3)*5.365=100.8 kV.In summary, the conversation discusses the calculation of fault MVA levels in an electrical supply system and suggests a method of improving the system to limit the fault MVA through T1 to approximately 5 times the full load current. This can be achieved by changing the voltage impedance of T1 and using a reactor in the T1 circuit.
  • #1
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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
 

Attachments

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  • #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
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 ]
 

1. What is an electrical supply system?

An electrical supply system is a network of interconnected components that provide electricity to buildings, homes, and other structures. It includes power plants, transmission lines, distribution lines, and transformers.

2. How does an electrical supply system work?

An electrical supply system works by generating electricity at power plants, which is then transmitted at high voltages through transmission lines. The voltage is then reduced at substations and distributed through distribution lines to homes and buildings. The electricity is then used to power appliances and devices.

3. What are the main components of an electrical supply system?

The main components of an electrical supply system include power plants, transmission lines, substations, distribution lines, transformers, and meters. These components work together to generate, transmit, and distribute electricity to end-users.

4. How is electricity measured in an electrical supply system?

Electricity is measured in an electrical supply system using a unit called kilowatt-hour (kWh). This measures the amount of energy consumed over a period of time. Meters are installed in homes and buildings to track the amount of electricity used.

5. What are the safety considerations in an electrical supply system?

Safety is a critical aspect of an electrical supply system. Some of the safety considerations include proper installation and maintenance of equipment, grounding and earthing systems, and following safety protocols when working with electricity. It is also important to have safety features, such as circuit breakers and fuses, in place to prevent electrical accidents.

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