Sizing standards

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shortcircuit1

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Hello,

I am trying to see if we have a standard method where we can calculate feeder size and OCPD. For example:

1.We have a IWH(Instant water heater).Its wattage is 7kW@208V. Now to calculate the feeder size all i do is 7000/208=33.65A. So i choose as per 310.15(B)(16) #10AWG. Now for OCPD since IWH is not a continuous load i can choose 35A breaker.But according to 240.4(D) i cant choose 35A OCPD because of small conductor rule.Am i right on this?

2.Lets say you have a cassette@208v and Running nominal amps=0.25 and an MCU(Mode changing unit)@208v and running nominal amps=0.4.I want to hookup both to the same circuit since the amp draw is minimal.So i added up the ampacities of the motors and 0.56A@208v i can put them on a 15A,2P circuit.Is this procedure right?

3.If you have a 5HP motor how do you convert that to kVA?(1HP=765w,so 5HP=3825W or 3.825kVA).But if i go to table 430.250(Full load current for three phase motors) if i see FLA of 5HP@208V it gives me 16.7A so to get kva=208*1.732*16.7=6.016kVA. Which method is right?
 
shortcircuit1 said:
Hello,

I am trying to see if we have a standard method where we can calculate feeder size and OCPD. For example:

1.We have a IWH(Instant water heater).Its wattage is 7kW@208V. Now to calculate the feeder size all i do is 7000/208=33.65A. So i choose as per 310.15(B)(16) #10AWG. Now for OCPD since IWH is not a continuous load i can choose 35A breaker.But according to 240.4(D) i cant choose 35A OCPD because of small conductor rule.Am i right on this?

2.Lets say you have a cassette@208v and Running nominal amps=0.25 and an MCU(Mode changing unit)@208v and running nominal amps=0.4.I want to hookup both to the same circuit since the amp draw is minimal.So i added up the ampacities of the motors and 0.56A@208v i can put them on a 15A,2P circuit.Is this procedure right?

3.If you have a 5HP motor how do you convert that to kVA?(1HP=765w,so 5HP=3825W or 3.825kVA).But if i go to table 430.250(Full load current for three phase motors) if i see FLA of 5HP@208V it gives me 16.7A so to get kva=208*1.732*16.7=6.016kVA. Which method is right?
Click to expand...
My answer is "correct" for the first two questions.

Last question, 746 watts is one horsepower. But remember nameplate horsepower is the output power of the motor, you need to factor in power factor and efficiency to get actual kVA . NEC tables for motor full load currents typically represent the worst case PF and efficiency with the values that are in the tables.
 
My answer is "Incorrect"

The sum of the 2 small motor loads is .65, not .56. :p


JAP>
 
kwired said:
My answer is "correct" for the first two questions.

Last question, 746 watts is one horsepower. But remember nameplate horsepower is the output power of the motor, you need to factor in power factor and efficiency to get actual kVA . NEC tables for motor full load currents typically represent the worst case PF and efficiency with the values that are in the tables.
Click to expand...

For the 1st question if i use 30A as OCPD and if the heater is drawing FL current of 33.65A it is going to trip the breaker?
 
shortcircuit1 said:
For the 1st question if i use 30A as OCPD and if the heater is drawing FL current of 33.65A it is going to trip the breaker?
Click to expand...

It would be my guess that it might.

Generally the manufacturers have the MOCPD listed on the label.

At least the ones that I've installed have.

And they're never less than what the calculated load it.

JAP>
 
shortcircuit1 said:
1.We have a IWH(Instant water heater).Its wattage is 7kW@208V. Now to calculate the feeder size all i do is 7000/208=33.65A. So i choose as per 310.15(B)(16) #10AWG. Now for OCPD since IWH is not a continuous load i can choose 35A breaker.But according to 240.4(D) i cant choose 35A OCPD because of small conductor rule.Am i right on this?
Click to expand...

I would say you have this wrong. According to 240.4(D) you can't chose #10AWG conductors...According to 422.10(A) your OCPD has to be at least 35A.
 
shortcircuit1 said:
If i cant choose #10AWG for 35A breaker i have to go to next size which is #8?
Click to expand...

Yes. (echoing David who posted faster than I)

eta: also

shortcircuit1 said:
3.If you have a 5HP motor how do you convert that to kVA?(1HP=765w,so 5HP=3825W or 3.825kVA).But if i go to table 430.250(Full load current for three phase motors) if i see FLA of 5HP@208V it gives me 16.7A so to get kva=208*1.732*16.7=6.016kVA. Which method is right?
Click to expand...

1HP = 746W, not 765W (and I see kwired beat me to that correction! ninja'd again! haha)
 
Last edited:
shortcircuit1 said:
If i cant choose #10AWG for 35A breaker i have to go to next size which is #8?
Click to expand...
correct.

Minimum conductor and minimum overcurrent are both 33.65A, but the small conductor rule does limit 10 AWG to being protected at 30 amps, No exceptions for this kind of a load, if it were a motor it could be protected more then 30 amps, but then most motors you still need to multiply the 33.65 (if that were the FLA) by 1.25 and would end up needing 8 AWG anyway.
 
kwired said:
My answer is "correct" for the first two questions.

Last question, 746 watts is one horsepower. But remember nameplate horsepower is the output power of the motor, you need to factor in power factor and efficiency to get actual kVA . NEC tables for motor full load currents typically represent the worst case PF and efficiency with the values that are in the tables.
Click to expand...

So which one is the correct to calculate correct kVA?
 
shortcircuit1 said:
So which one is the correct to calculate correct kVA?
Click to expand...
As GD mentioned, NEC wants us to use the NEC tables for NEC calculations. They want to account for worst case efficiency motor when doing such calculations. Seldom does one run into a motor that actually has same FLA on it's nameplate as is in the NEC tables.
 
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