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Short Circuit Force on Bus Bars for Low Voltage Connection

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kevc70

Mechanical
Jul 15, 2009
37
Hello,

I am wondering how to compute the short circuit force that would be exerted on (3) aluminum bus bars within a 3 phase transformer. Here are the specs:

2750 kVA 3 phase transformer
HV voltage is 13200Y/7620
LV voltage is 600 Delta
LV current is 1528A
(3) LV bushings on right side of transformer
# of bus bars = 3
Bus bar dimensions are : 42.5" long x 6.00" wide x 0.5" thk

I would use Inventor to simulate the force of stabilizing stiffeners to see if I need to see if a change to the stiffener shape is needed.
 
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Percent impedance voltage?
X/R ratio?
Peak asymmetrical current?
Temperature conversion factor? (The above factors are typically measured or calculated on a transformer at rated working temperature. The values will be higher on a cold transformer.)
or
The cold resistances of the windings.
Once you know the peak asymmetrical current on the cold transformer, you will need to know the spacing of the buss bars.
And;
There are some phase interactions and switching transients that are unique to a wye:delta transformer.
And;
Why a wye:delta?
They tend to be problematic in distribution service which your ratings imply.
A wye:wye or a delta:wye would be more common with those ratings.
Is this a Generator Step Up transformer?

--------------------
Ohm's law
Not just a good idea;
It's the LAW!
 
See
IEC Std 60865-1
"Short-circuit currents –
Calculation of effects –
Part 1:
Definitions and calculation methods "
chapter 2.2 Rigid conductor arrangements 2.2.1 Calculation of electromagnetic forces
 
From IEC 60076-5/ 2000 standard Table 1 – Recognized minimum values of short-circuit impedance, for transformers with two separate windings of 2501 to 6300 kVA ukr=7%. So, transformer L.V. impedance will be:
ZT=ukr%/100*Urt^2/SrT where Urt=0.6 kV and SrT=2.75 MVA.
ZT=7/100*0.6^2/2.75=0.009164 ohm.
Ik=600/sqrt(3)/0.009164=37802.7 A
From IEC IEC 61936-Part 1 Table 1 – Minimum clearances in air – Voltage range I (1 kV < Um ≤ 245 kV) for 3 kV it is 60 mm [0.06 m].
According to Schneider brochure no. 162 electrodynamic forces on busbars in LV systems:
F/lng = 2* 10-7* I1* I2 (k/d)
lng=42.5”= 1.0795 m
d=0.06 m
k=1.28 vertical bars 0.7 horizontal bars
F=1.0795*2/10^7*37802.7^2*1.28/0.06=6581.987 N
 
 https://files.engineering.com/getfile.aspx?folder=97ebfdfb-9eef-4bab-aad9-5e988fc47c6c&file=Schneider_162_k_for_short-circuit.jpg
Two corrections:
k for vertical bars is 0.7 and for horizontal 1.28
But d=0.060+0.0127=0.0727m
Then F=1.0795*2/10^7*37802.7^2*0.7/0.0727=2970.7 N
 
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