It's a 3 phase hookup. Sheet shows 3 hp 3 ph (2.24 kW)

The school has 3 ph 208.

Am I correct in finding amps by figuring 2,240/208?

I came up with 10.7

Seems I'm forgetting something.

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It's a 3 phase hookup. Sheet shows 3 hp 3 ph (2.24 kW)

The school has 3 ph 208.

Am I correct in finding amps by figuring 2,240/208?

I came up with 10.7

Seems I'm forgetting something.

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8,054 Posts

Yeah you need to divide it by the square root of three. 1.732.

6.33

6.33

So, 2,240/6.33 then / by 208?ponyboy said:Yeah you need to divide it by the square root of three. 1.732. 6.33

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Nope. 2240/208/1.732sparky723 said:So, 2,240/6.33 then / by 208?

Gotcha! Geez, messing with 240 all the time gettin me rusty on calcs.ponyboy said:Nope. 2240/208/1.732

Got 6.23 amps

Thx

The voltage/KW calcs don't include power factor.

Ok, thank youmicromind said:

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8,054 Posts

True. But when hooking up industrial type machinery wouldn't you just go off the kW listed on the nameplate?micromind said:

When motors are rated in KW, the KW rating is what appears at the shaft, not the input.True. But when hooking up industrial type machinery wouldn't you just go off the kW listed on the nameplate?

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Could you explain further please? When I hook up machines (like a laser) that has maybe 15 motors in it I'm not even looking at the motor nameplates. Usually there is one nameplate with a total kW on it and that's what I figure it for.micromind said:When motors are rated in KW, the KW rating is what appears at the shaft, not the input.

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7,984 Posts

This k.w. thing now isn't burned into my brain like amps on nameplates were that I grew up with.

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In places OTHER than North America, the kW rating on a motor nameplate is used as a MECHANICAL rating in the same way we show HP on a motor nameplate. It's only relevant to shaft torque and speed, not ELECTRICAL kW. The ELECTRICAL kW rating would be what they refer to as the "Absorbed power".This k.w. thing now isn't burned into my brain like amps on nameplates were that I grew up with.

But if you are looking at a nameplate for a machine or system that has multiple loads connected to the same main incoming terminals, and the kW rating is referring to the entire machine, then that is the ELECTRICAL kW rating of the machine.

Now if you are wanting to size anything for your punch and this information is coming from the single motor on it, the other stuff is actually somewhat irrelevant anyway. The NEC does not ask you to calculate FLA. You have to size the conductors, OCPD etc. based on the motor HP rating. That's easier; there are 746W / HP. So 2.24kW = 2240W / 746 = 3.002HP, or 3HP. You then use the charts in Article 430 to size things based on 3HP.

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So that's why the engine in my Italian tractor is rated in KW.

So are we going with 6.23A, 8A, no Amps, or what?

I love this forum, but where's the final answer??

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1) 430.6(A)(1) states that table values are to be used, or the nameplate, whichever is greater.

2) Table 430.250 lists a 3HP 3ø motor operating on 208 as 10.6 amps.

3) 430.22(A) states that for a single motor, the table value must be multiplied by 1.25 to get the minimum wire size.

4) 110.14(C)(1)(a)(4) states that wire is to be sized based on 75ºC, unless any part of it is rated less.

5) 240.4(D) and (G) state that the 15, 20, and 30 amp limitations to 14s, 12s, and 10s do not apply to motors.

6) 10.6 X 1.25 = 13.25; Table 310.16 lists #14 at 20 amps, therefore this is the minimum wire size.

7) Table 430.52 lists the maximum size of a basic circuit breaker as 250% of the value in Table 430.250. 10.6 X 2.5 = 26.5 430.52(C)(1) Exception #1 allows the use of the next standard breaker if the calculated value does not correspond to a standard value, so the maximum breaker would be 30 amp.

8) Yes, you can have a 30 amp breaker and #14s. But 14s are the minimum (there's no maximum), and a 30 is the maximum (there's no minimum). Personally, I'd use 14s or 12s and a 15 or a 20.

9) 430.6(A)(2) states that the overloads need to be based on nameplate current.

There, wasn't that easy; 9 steps and 4 different articles just to hook up a dumb motor.....

The EU only uses HP as a secondary rating to watts, as of 2010.So that's why the engine in my Italian tractor is rated in KW.

The concept of how James Watt originally calculated horsepower is fairly ridiculous, and reads a little like the witch scene in

No doubtmicromind said:

I was gonna throw it on a 20A/#10

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FLC for a 3 phase motor (as opposed to the nameplate FLA, since you apparently don't have it) can be ESTIMATED using the following formula:

So are we going with 6.23A, 8A, no Amps, or what?

I love this forum, but where's the final answer??

A = (746 x HP) / (1.732 * Volts * Efficiency * Power Factor

You don't know the efficiency or power factor of your motor, otherwise you would have the FLA as well. So the best you can do is guess by ASSuming 90% efficiency (if it is less than 10 years old, 80% if it is older) and a .8 power factor.

A = (746*3)/(1.732 * 208 * .9 * .7)

A = 2238 / 259.38

A = 8.63A

Your right, apparently-I don't have it, JrJRaef said:FLC for a 3 phase motor (as opposed to the nameplate FLA, since you apparently don't have it) can be ESTIMATED using the following formula: A = (746 x HP) / (1.732 * Volts * Efficiency * Power Factor You don't know the efficiency or power factor of your motor, otherwise you would have the FLA as well. So the best you can do is guess by ASSuming 90% efficiency (if it is less than 10 years old, 80% if it is older) and a .8 power factor. A = (746*3)/(1.732 * 208 * .9 * .7) A = 2238 / 259.38 A = 8.63A

Who's ASS..? uming

It's a brand new unit, I'm going by info given on the spec. sheet.

I bet I'll be fine with #10 on a 20A.

APPARENTLY

I appreciate help, but don't appreciate sarcastic attitude or being called an ass.

Frustrated I can't get a clear answer? Yes

Ticked that someone's getting mad about it? APPARENTLY!

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