Solving for Fault Current using Square Root 3

[Brytonnegahban]

I am having a tough time wrapping my head around when square root 3 is used or not used when solving for fault current. Both of the NCEES Practice Exam questions below solved for fault current, but one used sqrt(3) and the other didn't.

NCEES Practice Exam #110
The only load on a 3-phase, 4-wire system is placed between Phase B and Phase C. The phase-to-phase voltage is 13.2 kV. The load is 500 kVA at 0.85 lagging power factor. The magnitude of the line current (amperes) in Phase C is most nearly:

Answer
I_c = 500 kVA/13.2 kV = 37.9 A

NCEES Practice Exam #132
Consider the 60-kV transmission system below. Transmission line impedances are: Z_1 = 16.75 <71deg Ohms, Z_2 = 13.4 <71deg Ohms. With a system impedance of Z_system = 13.25 <81 deg Ohms, the 3-phase fault current (amperes) at Station C is most nearly:

Answer
I_fault = [60,000/sqrt(3)]/(16.75 <71 + 13.4 <71 + 13.25 <81)

 

[akyip]

For 110: the given load is only a 1-phase load, connected between 2 phases. You do not need to use the square root of 3 for a single phase load.

For 132: you are asked to find the fault current resulting from a 3-phase fault. The given voltage is a line-to-line voltage, and the division by square root of 3 is used to convert that voltage into a line-to-neutral voltage for a standard per-phase circuit analysis.

 

[kris7o2]

Hi everyone, is this the correct formula to use for questions 110 and 132?

110) Sp=ILL*VLL and

132) S3p = sqrt(3) * ILL*VLL

 

[alex93]

For 110: the given load is only a 1-phase load, connected between 2 phases. You do not need to use the square root of 3 for a single phase load.

For 132: you are asked to find the fault current resulting from a 3-phase fault. The given voltage is a line-to-line voltage, and the division by square root of 3 is used to convert that voltage into a line-to-neutral voltage for a standard per-phase circuit analysis.

The question asks for the 3 phase fault current though. What context clues tell you that the fault current they're looking for is per phase?

 

[yaoyaodes]

The question asks for the 3 phase fault current though. What context clues tell you that the fault current they're looking for is per phase

The question asks for the 3 phase fault current though. What context clues tell you that the fault current they're looking for is per phase?

want to ask the same question.

 

[Zach Stone P.E.]

The question asks for the 3 phase fault current though. What context clues tell you that the fault current they're looking for is per phase?

@yaoyaodes:

want to ask the same question.

NCEES #110:

This is a three-phase system with the "only load placed between Phase B and Phase C".

This is a single-phase load connected phase to phase.

Think of a two-pole breaker in your garage panel that provides 240V to your clothes dryer or HVAC equipment.

The question asks for the magnitude of the line current in Phase C, which is equal to the magnitude of the line current in Phase B, since the two voltages are in series with each other by placing the load across the two phases (instead of connecting a load line to neutral).


NCEES #132:

This is a three-phase system shown with three-phase line impedances.

To calculate the current on a three-phase system, you have to use Ohm's law with the phase voltage (not line voltage).

The square root of three in the solution is dividing the line voltage to get the phase voltage, in order to use Ohm's law to solve for the line current flowing through each of the three conductors of the three-phase system.

Electrical PE Review - Avoid These Three Phase Power Formula Mistakes!

 

[yaoyaodes]

NCEES #110:

This is a three-phase system with the "only load placed between Phase B and Phase C".

This is a single-phase load connected phase to phase.

Think of a two-pole breaker in your garage panel that provides 240V to your clothes dryer or HVAC equipment.

The question asks for the magnitude of the line current in Phase C, which is equal to the magnitude of the line current in Phase B, since the two voltages are in series with each other by placing the load across the two phases (instead of connecting a load line to neutral).


NCEES #132:

This is a three-phase system shown with three-phase line impedances.

To calculate the current on a three-phase system, you have to use Ohm's law with the phase voltage (not line voltage).

The square root of three in the solution is dividing the line voltage to get the phase voltage, in order to use Ohm's law to solve for the line current flowing through each of the three conductors of the three-phase system.

Electrical PE Review - Avoid These Three Phase Power Formula Mistakes!

Hi Zack. I think the question is the three phase fault current = single phase current?

 

[Zach Stone P.E.]

Hi Zack. I think the question is the three phase fault current = single phase current?

It's line current.

 

[aliwaqar]

It's line current.

But Zach in the question no where its mentioned that it is a Wye system, for a Delta system we might not need to convert. Thoughts?
Thanks, Ali

 

[Zach Stone P.E.]

But Zach in the question no where its mentioned that it is a Wye system, for a Delta system we might not need to convert. Thoughts?
Thanks, Ali

When using Ohm's law to solve for the line current drawn by a three-phase system, you need to use line-to-neutral voltage of the system, regardless if the source or load are delta or wye connected.

If you have a delta connected load, you can solve for the phase current in each phase of the delta connection, and convert it to line current. You'll get the same value for line current as the previous approach (try it!).