Transformer Question

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.

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

Regards

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

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

Before the circuit protective device operation.

What is the circuit protective device set to trip at?

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.

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

Primary current = 160A

Secondary current = 1928A

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

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

regards

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 .

thanks.Please explain your answer in details.

regards

% 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

Why would anyone want to do this?

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.

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

Regards

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.