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Transformer Question

04/29/2010 7:18 AM

A transformer of 2 MVA , 6.6/0.433 KV and percentage impedence of 6.5 % is supplied with 0.433 KV on primary side(i.e 6.6 KV side0 and secondary side(i.e 0.433 KV side) shorted.Then what will be the exact current in primary and secondary side?

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#1

Re: Electrical,transformer

04/29/2010 7:22 AM

Whatever it is, it should be enough to operate the circuit protective devices recommended in the transformer manufacturer's manuals. After that time, it will be zero.

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#2
In reply to #1

Re: Electrical,transformer

04/29/2010 7:26 AM

Sir,i want to know about the current.So please calculate and send it.

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#3

Re: Electrical,transformer

04/29/2010 7:27 AM

6,6 kV x 6,5 / 100 = short circuit voltage and at with this voltage while supplied nominal current will be circulate from winding of transformer

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#4
In reply to #3

Re: Electrical,transformer

04/29/2010 7:39 AM

Is that before or after the circuit protective device has operated?

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#5
In reply to #4

Re: Electrical,transformer

04/29/2010 7:53 AM

Before the circuit protective device operation.

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Guru
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#16
In reply to #5

Re: Electrical,transformer

05/04/2010 9:05 AM

What is the circuit protective device set to trip at?

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#6

Re: Transformer Question

04/29/2010 8:18 AM

Dear Sir,

As I have understood your wordings, for an iuput voltage of .433 Kv

on primary side , the voltage ( open circuit) will be 28.47 V and the % impedance

being 6.5 the short circuit current ( secondary side) will be 437.04 A.On primary

side this will be 28.67 A. Please correct me if I am wrong.

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#7
In reply to #6

Re: Transformer Question

04/29/2010 11:18 AM

Once again i am writting down the question.

A transformer with following details is given.

2 MVA ,6.6 KV/0.433 KV(no load)

% impedance(Z)=6.25 %

primary current=175 amp

secondary current=2666.7 amp

Now on primary side(i.e on 6.6 KV side),0.433 KV is applied and secondary side is shortcircuited.then with this condition what will be the primary and secondary current.please explain your answer with details.

regards

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#8
In reply to #7

Re: Transformer Question

04/29/2010 2:52 PM

Primary current = 160A

Secondary current = 1928A

NOTE - very dangerous when attempting to do as a practical test!!!

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#10
In reply to #8

Re: Transformer Question

04/30/2010 1:13 AM

Thanks sir,but will you please explain,how you got these answer?

regards

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#9
In reply to #7

Re: Transformer Question

04/29/2010 11:49 PM

Transformer % impedance is 6.25%

Means at 6.25 % of short ckt voltage full load current will flow in transformer

short ckt voltage = 6600*6.25/100=412.5

at 412.5 volt secondary current will be 2666.7amp

for 433v it will be (2666.7/415) x 433=2782a

never do this in practical .

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#11
In reply to #9

Re: Transformer Question

04/30/2010 1:16 AM

thanks.Please explain your answer in details.

regards

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#12
In reply to #11

Re: Transformer Question

04/30/2010 5:03 AM

% impedance refer to the pecentage of voltage applied to transformer atwhich full load current will flow in the transformer with one winding short circuited.

hope it will solve your query

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Guru
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#17
In reply to #7

Re: Transformer Question

05/04/2010 9:06 AM

Why would anyone want to do this?

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#13

Re: Transformer Question

04/30/2010 10:00 AM

Simple approximations:-

Full Load Current = Tx VA / (SQRT(3) x Voltage) assuming normal 3 phase system

FLC = 2,000,000 / 1.73 x 400

= 2890 A (Approx. for 400V nominal)

= 2669 A (Approx. for 433V)

You can choose, but if you want to know approx. short circuit current, then divide FLC by impedance = 2890/0.065 = 44461A (50KA rated board required).

Stand well back when you switch it on to test, just in case you don't believe me.

If you supply primary side (6.6kV) with 433V, secondary will be about 28V. Don't short that out either.

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#14
In reply to #13

Re: Transformer Question

05/01/2010 10:59 AM

Thanks, but i want to know about the value of primary and secondary current.

Regards

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#15
In reply to #14

Re: Transformer Question

05/03/2010 11:48 AM

The maximum currents are the same whatever voltage is used, due to the design of the transformer i.e. copper size used, or in this case around 2500A secondary, 160A primary, before overheating. If you mean value of current under short circuit conditions, again this will be limited by the copper etc. Perhaps you could explain your purpose behind the question.

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