Minimum fuse required for feeder of a 3-Phase, Delta-Wye System

[julian padilla]

Another question for y'all:
Given a transformer diagram shows a configuration of 120/208 V, 3-Phase, Delta-Wye System. The system has three 14 kW, single-phase, continuous balanced loads, each connected 120 V line to neutral. Assuming each of the three loads are resistive, what would be the minimum fuse required to protect conductor A in the transformer feeder circuit?

 

[infinity]

That's 14,000 watts @ 120 volts on the A phase? What answer did you come up with?

 

[julian padilla]

I'm not 100% positive, as I solved it two different ways shown below:

Solution_1 (single power):
Step_1: Using Ohm's Law: I = P/(E x 1.732) = 14,000 W / (120 V x 1.732) = 67.36 amps
Step_2: Since loads are continuous, multiply by 125%: 67.36 amps x 1.25 = 84.2 amps
Step_3: Per Table 240.6(A) size up to 90 A.


- OR -


Solution_2 (total power):
Step_1: Using Ohm's Law: I = P/(E x 1.732) = 42,000 W / (208 V x 1.732) = 116.58 amps
Step_2: Since loads are continuous, multiply by 125%: 116.58 amps x 1.25 = 145.73 amps
Step_3: Per Table 240.6(A) size up to 150 A.

What do you think?

 

[augie47]

Looking at any one particular load, would you consider it to be 3 phase ?

 

[mivey]

use either 14000 ÷ 120

or

42000 ÷ sqrt(3) ÷ 207.85

Either gives the correct amps of 116.7

 

[infinity]

I'm not 100% positive, as I solved it two different ways shown below:

Solution_1 (single power):
Step_1: Using Ohm's Law: I = P/(E x 1.732) = 14,000 W / (120 V x 1.732) = 67.36 amps
Step_2: Since loads are continuous, multiply by 125%: 67.36 amps x 1.25 = 84.2 amps
Step_3: Per Table 240.6(A) size up to 90 A.


- OR -


Solution_2 (total power):
Step_1: Using Ohm's Law: I = P/(E x 1.732) = 42,000 W / (208 V x 1.732) = 116.58 amps
Step_2: Since loads are continuous, multiply by 125%: 116.58 amps x 1.25 = 145.73 amps
Step_3: Per Table 240.6(A) size up to 150 A.

What do you think?

I hinted at it in post #2. You're only looking for the 1Ø, 120 volt current as a continuous load so you end up:

14,000/120*125%=145.8 or #1/0 conductor and 150 amp OCPD.

 

[petersonra]

I wonder what number he is actually looking for. Is he looking to protect the primary or the secondary of the Transformer, or the Transformer conductors, or the loads on the secondary. The numbers would all be potentially different.

 

[electrofelon]

I wonder what number he is actually looking for. Is he looking to protect the primary or the secondary of the Transformer, or the Transformer conductors, or the loads on the secondary. The numbers would all be potentially different.

Right. He did say this: ".....conductor A in the transformer feeder circuit"

I assume you are talking about an OCPD after the transformer secondary, before the loads. You would need something there for 450.3(B) and 240.21(c). Then the conductors tapping to each load may be taps or need their own OCPD depending on what they are.

 

[Besoeker]

I'm not 100% positive, as I solved it two different ways shown below:

Solution_1 (single power):
Step_1: Using Ohm's Law: I = P/(E x 1.732) = 14,000 W / (120 V x 1.732) = 67.36 amps
Step_2: Since loads are continuous, multiply by 125%: 67.36 amps x 1.25 = 84.2 amps
Step_3: Per Table 240.6(A) size up to 90 A.


- OR -


Solution_2 (total power):
Step_1: Using Ohm's Law: I = P/(E x 1.732) = 42,000 W / (208 V x 1.732) = 116.58 amps
Step_2: Since loads are continuous, multiply by 125%: 116.58 amps x 1.25 = 145.73 amps
Step_3: Per Table 240.6(A) size up to 150 A.

What do you think?

The loads are all single phase, phase to neutral. You don't need the √3

 

[julian padilla]

Thanks gentleman!
Step_1: Using Ohm's Law: I = P/E = 14,000 W / 120 V = 116.67 amps.
Step_2: Since loads are continuous, multiply by 125%: 116.67 amps x 1.25 = 145.83 amps.
Step_3: Per 310.15(B)(16) 75 deg C column size up to 1/0 AWG.
Step_4: Per Table 240.6(A) size up OCPD to 150 A.
Julian

 

[topgone]

Thanks gentleman!
Step_1: Using Ohm's Law: I = P/E = 14,000 W / 120 V = 116.67 amps.
Step_2: Since loads are continuous, multiply by 125%: 116.67 amps x 1.25 = 145.83 amps.
Step_3: Per 310.15(B)(16) 75 deg C column size up to 1/0 AWG.
Step_4: Per Table 240.6(A) size up OCPD to 150 A.
Julian

If you use a fuse (as what you named this thread), a 150 fuse will not melt until about 300 amperes!

 

[infinity]

Thanks gentleman!
Step_1: Using Ohm's Law: I = P/E = 14,000 W / 120 V = 116.67 amps.
Step_2: Since loads are continuous, multiply by 125%: 116.67 amps x 1.25 = 145.83 amps.
Step_3: Per 310.15(B)(16) 75 deg C column size up to 1/0 AWG.
Step_4: Per Table 240.6(A) size up OCPD to 150 A.
Julian

Looks good.