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hands
11-05-2007, 08:05 PM
I have an electric heating element that is rated for 18 kw at 200 volts 3 Ph, can anyone tell me what amp draw I should have.

thanks

jemawalton
11-05-2007, 08:48 PM
51.96 amps

(Watts / Volts) X 1.732

herrerao
11-05-2007, 08:50 PM
Hey Hands.
I think you probably working on 208V, what you do take the power 18000 Watts and divided by (208V times 1.73 a 3 phase constant number). then you get your answer.

P
-------------
(Volts x 1.73)

jayguy
11-05-2007, 11:08 PM
jema got it.

but keep in mind that this is if you have 200 VAC.

51.96 "rated" amps at 200 VAC

if you actually have 208 VAC, then the amp draw would be: 54.04 amps @ 19.47 kw. if it were to climb to 215 VAC (i don't know...i just pulled this number out of my arse to make a point) then you would have: 55.86 amps @ 20.80 kw.

have fun.
good luck.

sudarshan
11-06-2007, 02:04 AM
its
Watts/(voltage x PF)

PF=power factor ,0.8

CommtechinVA
11-06-2007, 07:13 PM
its
Watts/(voltage x PF)

PF=power factor ,0.8

Sudarshan, check out this link for some useful formulas. The .8 PF is not applicable for three phase. best of luck!!

http://www.elec-toolbox.com/Formulas/Useful/formulas.htm
Bill

hands
11-06-2007, 08:06 PM
Thanks for the help guys, if I wasn't so lazy I could of got the books out and looked it up myself. By the way, the voltage I was working with is 200 VAC. It's a Japanese Temper Oven, all of the machining equipment in the building that is from Japan is 200 VAX.

Shophound
11-07-2007, 09:35 AM
its
Watts/(voltage x PF)

PF=power factor ,0.8

Power factor comes into play if the load is inductive. Since the OP stated electric heating element, that is a resistive load, where current and voltage are in phase. Power factor references the phase angle between voltage and current...in a resistive load it is considered to be 1. Inductive loads will see power factors below 1, indicating a ratio of true power to apparent power.

Perhaps you meant to write your formula as:

Watts/(volts x 1.732) = amps

hvacker
11-09-2007, 02:20 PM
Three phase electric heat strips can be hooked up eather as delta or wye.
Delta draws a lot more amps than wye and develops a lot more heat. Check the wiring diagram to be sure how to wire to get the heat you require.

There is a 4 wire wye. The load between any legs will be for instance 208 single phase and the ground wire is bonded to the buss where all the phases join. This gives a 120 volt single phase availability between any hot phase and ground. This is often used for 208 ceiling light fixtures and 120 volt for other uses.

That 120 volt leg in your case might be for the control circuit.

Other than that with a resistive load it's straight ohms law. And 3414 btu's per kw.

herschmidt
11-10-2007, 06:44 PM
Thanks for the help guys, if I wasn't so lazy I could of got the books out and looked it up myself. By the way, the voltage I was working with is 200 VAC. It's a Japanese Temper Oven, all of the machining equipment in the building that is from Japan is 200 VAX.

Hands:

It doesn't matter where the equipment is caming from, what is important is the voltage that is available at the place where you install the equipment or the voltage that the heaters will be connected to.

I hope this help.

Hernan

hands
11-10-2007, 08:14 PM
Hands:

It doesn't matter where the equipment is caming from, what is important is the voltage that is available at the place where you install the equipment or the voltage that the heaters will be connected to.

I hope this help.

Hernan

Yes, I realize that. The building voltage is 208 and 460 depending on what section of the building you are in. They use transformers to drop down the voltage to 200 VAC at each machine.

Airmechanical
11-11-2007, 07:56 AM
basic (easy way to figure amp draw)

i been using this SIMPLE formula for 27 years

if its a (230 volt system), take k.w multiply x 4

example, 5 kw at 230 volts is (approximately) 20 amps

keep it simple, this (easy) equation will get you thinking in the ballpark

without having to get out the calculator:)

.

beenthere
11-11-2007, 07:59 AM
Delta draws a lot more amps than wye and develops a lot more heat.

And 3414 btu's per kw.

18 KW strip heat is the same heat weather wye or delta.

hvacker
11-12-2007, 05:56 PM
18 KW strip heat is the same heat weather wye or delta.

Think about it. For the same applied voltage, say 240, In a wye connected all phases run through 2 windings where in a delta they run through 1 winding.
A wye connected strip has twice the resistance that a delta has.
A manufactuer can have the same heater supply 2 different needs.

Say one strip has a resistance of 10 ohms. Then 240 / 10 would = 24 amps. But in a wye reisitance would be 20 ohms and that strip would draw 12 amps.

beenthere
11-13-2007, 03:18 AM
18 KW is still the same amount of heat. Weather delta or wye.
If the heat changes, then so did the KW.

hvacker
11-15-2007, 04:46 PM
18 KW is still the same amount of heat. Weather delta or wye.
If the heat changes, then so did the KW.

Yup, your right. Had to do some math and the problem can be misleading in 3 phase.
Thanks.