
Accountable Engineer
 Cambridge, Cambridgeshire
 Competitive
A professional engineer providing technical and project management delivery across a portfolio of projects
 Recruiter: AstraZeneca

 Recruiter: Atkins

Engineering Manager
 Crich, Derbyshire
 £40,000 PA
The post holder will be a qualified electrical engineer and have experience in running a workshop and extensive project management skills.
 Recruiter: Crich Tramway Village
IET 
search :
help :
home


Latest News:


Topic Title: Transformer details given on Amtech Topic Summary: Created On: 09 May 2013 11:03 AM Status: Post and Reply 
Linear : Threading : Single : Branch 
Search Topic

Topic Tools

09 May 2013 11:03 AM


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


bump




10 May 2013 07:43 AM


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


Hi Geoff, thanks for responding.
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


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


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


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


Thank you



IET
» Wiring and the regulations
»
Transformer details given on Amtech

Topic Tools

FuseTalk Standard Edition v3.2  © 19992016 FuseTalk Inc. All rights reserved.