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Topic Title: Transformer details given on Amtech
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Created On: 09 May 2013 11:03 AM
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 09 May 2013 11:03 AM
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ofsdairw

Posts: 7
Joined: 25 April 2013

Transformer 11kV/400V

Under the Fault Level Calculator tab we have

Primary Fault Level (MVA) 250
Rating (kVA) 400
Impedance (%) 5

the accompanying calculation gives:

250 MVA Transformer Primary supply network: r = 0.000064, x = 0.000637

Transformer winding: r = 0.004, x = 0.019596

I'm trying to work out how these inpedances relate to the 5% given for the transformer.

I don't normally use the p.u.system so any help would be appreciated

Saul

Edited: 09 May 2013 at 11:12 AM by ofsdairw
 10 May 2013 12:24 AM
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ofsdairw

Posts: 7
Joined: 25 April 2013

bump
 10 May 2013 07:43 AM
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GeoffBlackwell

Posts: 3588
Joined: 18 January 2003

I shall ignore the primary fault level and related impedance as your question appears to be concerned with the transformer.

The quoted 5% impedance Z% (or 0.05 p.u Zpu) is not really an impedance it is a ratio derived from a voltage drop .

Specifically it is the ratio of the voltage drop across a phase winding to the rated phase voltage.

Zpu = Z*Ifl/Vphase

(Note that Z*Ifl is a voltage.)

Where: Z is the impedance of the winding in ohms; Ifl is the rated full load current; Vphase is the rated phase voltage.

Z in terms of Zpu can be found from:

Z = Zpu*Vline/Ifl*root3

Zpu may also be given as Ifl/Isc

Where: Isc is the current that would flow in the winding if shorted when supplied at rated phase voltage.

This form is used to calculate Isc from the rated full load current and Zpu.

Isc = Ifl/Zpu

Now for your homework

Show that Z = Vphase/Isc can be derived from Z = Zpu*Vline/Ifl*root3

Regards

Geoff Blackwell
 10 May 2013 02:12 PM
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ofsdairw

Posts: 7
Joined: 25 April 2013

Hi Geoff, thanks for responding.

I shall ignore the primary fault level and related impedance as your question appears to be concerned with the transformer.


I asked about the transformer as the transformer information was given and I couldn't see how it related to the 5% but I do now thanks. If i've got this right

Using the original figures of:

Transformer winding: r = 0.004, x = 0.019596

gives

Z (Transformer winding) = √(r2+x2) = √[(0.004)(0.019596)+(0.000637)(0.000637)]= 0.02 Ohms

this verifies with Z = Vp*(0.05)/Ifl = 230*(.05)/577.35 = 0.02

since Ifl = rated kVA / (Vp*√3) = 400,000 / (230*√3) = 577.35 amps


Now, I assume (rightly or wrongly) that the

250 MVA Transformer Primary supply network: r = 0.000064, x = 0.000637

is needed for calculating the maximum phase fault current, so

I max phase fault (kA) = V(line) / [Z (250 primary network) + Z (winding) + other Z values]

Edited: 10 May 2013 at 02:18 PM by ofsdairw
 10 May 2013 02:16 PM
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hpcompaq

Posts: 78
Joined: 27 September 2007

Hello Saul.

Another take on the answer.

The Tx winding figure is of course referred to the secondary.

The square root of [(0.004) squared + (0.019596) squared] = 0.02ohms

Tx secondary voltage = 400V, Rating 400kVA, Impedance 5% so;
(400V) squared divided by 400kVA = 0.4 ohms
5% of 0.4 ohms = 0.02ohms, as above.

Best regards.
 10 May 2013 02:58 PM
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ofsdairw

Posts: 7
Joined: 25 April 2013

Originally posted by: hpcompaq

5% of 0.4 ohms = 0.02ohms, as above.

Best regards.


Thanks. as I said in my first post, I'm not used to the p.u. system but I read the basic formulas in a book and now have Z = Zb*Zpu and have enough to go on with I think as all the parameters have been stated.

Referring to the transformer, the turns ratio is clearly 11kV/400 = 27.5 so the primary current will be around 21A.

Is there anything else we can work out from the primary network?
 10 May 2013 03:21 PM
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hpcompaq

Posts: 78
Joined: 27 September 2007

Hello Saul.

One minor point. You need to use V(phase) and not V(line) to calculate the "I max phase fault" which is of course the 3 phase symmetrical short circuit current.

You will no doubt note that ignoring the resistive element altogether introduces only a minor error - this is often done in practice at these levels.
 10 May 2013 03:28 PM
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ofsdairw

Posts: 7
Joined: 25 April 2013

Thank you
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