Open Delta Single Phase Fault Calc

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O'Dell

Member
Location
Forest Grove, Oregon
Hello everyone,

I have been requested to provide a fault calculation for a new 400 Amp single phase service. The service will feed from an Open Delta Bank consisting of a 100kVA lighter pot and a 50kVA power pot both with an impedance of 1.5%. The different calculators that I have available give a fault current of approx. 42,000 Amps at the secondary bushings. But since the service length will be around 75 feet of 1/0 overhead triplex with a #2 reduced neutral, I know that the fault current will be less at the meter base. I don't want to give the fault current at the transformer because it will force the contractor to install higher rated gear and cost more money. Any help would be appreciated on how to figure out the fault current at the meter base. Thank you.
 
T

topgone

Senior Member
O'Dell said:
Hello everyone,

I have been requested to provide a fault calculation for a new 400 Amp single phase service. The service will feed from an Open Delta Bank consisting of a 100kVA lighter pot and a 50kVA power pot both with an impedance of 1.5%. The different calculators that I have available give a fault current of approx. 42,000 Amps at the secondary bushings. But since the service length will be around 75 feet of 1/0 overhead triplex with a #2 reduced neutral, I know that the fault current will be less at the meter base. I don't want to give the fault current at the transformer because it will force the contractor to install higher rated gear and cost more money. Any help would be appreciated on how to figure out the fault current at the meter base. Thank you.
Click to expand...

We need to be clear on what you wanted. Which is which? Single-phase or three-phase?
Open-delta --> three-phase, fyi.:happysad:
 
O

O'Dell

Member
Location
Forest Grove, Oregon
It is a single phase service feeding from a three phase Open-Delta bank. This is why I am unsure of how to calculate it. I assume that I need to use the total KVA because the transformers are tied together. Do I factor in 1.732 into my calculation. How do I figure the secondary wire into the formula?
 
T

topgone

Senior Member
O'Dell said:
It is a single phase service feeding from a three phase Open-Delta bank. This is why I am unsure of how to calculate it. I assume that I need to use the total KVA because the transformers are tied together. Do I factor in 1.732 into my calculation. How do I figure the secondary wire into the formula?
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Base your calcs on the 3-phase ratings. Line amps will be kVA/(1.732 x secondary volts). Use that as the current reference when choosing the conductor size. Then treat the fault at your service as a line to line fault of the transformer. Three-phase fault cannot happen since you only brought out two of the three lines.
 
kwired

kwired

Electron manager
Location
NE Nebraska
O'Dell said:
Hello everyone,

I have been requested to provide a fault calculation for a new 400 Amp single phase service. The service will feed from an Open Delta Bank consisting of a 100kVA lighter pot and a 50kVA power pot both with an impedance of 1.5%. The different calculators that I have available give a fault current of approx. 42,000 Amps at the secondary bushings. But since the service length will be around 75 feet of 1/0 overhead triplex with a #2 reduced neutral, I know that the fault current will be less at the meter base. I don't want to give the fault current at the transformer because it will force the contractor to install higher rated gear and cost more money. Any help would be appreciated on how to figure out the fault current at the meter base. Thank you.
Click to expand...
The high leg pot isn't contributing anything to what you are connected to, use the 100kVA pot as though it were stand alone for your calculations - that is all you are connected to. No different then calculating the available line to ground fault current - you disregard the other half of the winding as it doesn't contribute anything in that instance (you do only have half the kVA and half the full winding impedance though).
 
kwired

kwired

Electron manager
Location
NE Nebraska
Phil Corso said:
O'Dell...

Search forum archives for:

http://forums.mikeholt.com/showthread.php?t=179180

Regards, Phil Corso
Click to expand...
Seems you already did the searching;)

OP is only utilizing single phase and connected to one of the "pots" The other "pot" is not going to contribute any energy into any fault he may have on this single phase feed. He has essentially the same thing at his point of use as if the 100 kVA "pot" were the only thing on the pole.
 
Julius Right

Julius Right

Senior Member
Occupation
Electrical Engineer Power Station Physical Design Retired
It depends on high voltage and low voltage.
Let's say the high-voltage is 480 V and the low-voltage 240 A.
3*1/0 awg-according to NEC CH.9 Tb.9 :0.13 ohm/kft=0.13/1000*75=0.00975 ohm[75 ft.]
The cable reactance as seen at low voltage side is 0.00975/4=0.002438 ohm[(480/240)^2=4].
There are two currents in series: Ibc= Iab-Ica.
Iab=Vab/Z1 and Ica=Vca/Z2
Zab=240^2/100/1000*1.5/100=0.00864 ohm
Zca=240^2/50/1000*1.5/100=0.01728 ohm
Total impedance Z1=0.00868+0.002438=0.0110775 ohm and Z2=0.0197175 ohm.
Iab=240/Z1*[cos(0)-isin(0)]=21665.54 A
Iac=-Ica=240/0.0197175=12172 A and the angle will be -240+180=-60.
Iac=240/Z2*[cos(-60)-isin(-60)]=6086+10541i
If you'll neglect the cable impedance the current will be:36747 A
If you'll take the cable impedance the current will be: 29686 A
For a more accurate calculation we have to separate resistance and reactance but in my opinion the difference is negligible.
 
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