I have a home that receives single phase 240 AC from the utility co.
Is there a way to wire and electric heater rated at 208 volts?
thanx
gary

 

Is it a simple heating filament? If yes then you can simply use a lamp dimmer to prevent the filament becomes overheated.
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What is the listed wattage of the heater?

 

It is a 10KW shop heater.

 

The light bulb idea is not so bright, I think...

P = E x I, so 10000 / 208 = 48 Amps (I hope you have a spare circuit in your breaker box.)

240V - 208V = 32V across the lightbulb

E = I x R, so R = E / I. Light bulb resistance needed is 32 / 48 = 0.67 Ohm

P = E^2 / R, so light bulb rating = 240 x 240 / 0.67 = 86 kiloWatts.

 

you can perhaps use a variable autotransformer a.k.a dimmerstat to bring down the voltage to 208 volts.

btw Mr. Thingmaker3 i'm confused why is the rating not 32*48 = 32^2/0.67? m i missing something (hopefully not a brain ). will that bulb be on when the heater is off? (i've little idea abt practical concepts)

 

The light bulb idea is not so bright, I think...

P = E x I, so 10000 / 208 = 48 Amps (I hope you have a spare circuit in your breaker box.)

240V - 208V = 32V across the lightbulb

E = I x R, so R = E / I. Light bulb resistance needed is 32 / 48 = 0.67 Ohm

P = E^2 / R, so light bulb rating = 240 x 240 / 0.67 = 86 kiloWatts.

Ahem,.... the voltage across the "lightbulb" is 32 V, not 240 V and the power would be about 1.5 KW. I doubt such a lightbulb exists but this resistor could be of any type and could be custom built.

'tis strange that a heater would be built for 208 V.

If it is just a resistive load (no motors) then a triac "dimmer" might be a good idea (except that you need a specially large triac, a regular dimmer will not do). The harmonics generated by such a large load might be too much and would have to be dealt with accordingly.

OTOH, a load desighned for 10 KW @ 208 V would generate 13.3 KW @ 240 V and it may be able to withstand that indefinitely or cycling. With a 75% duty cycle it would be generating 10 KW. A zero crossing switching device with a 75% duty cysle might be a better idea.

I still don't understand where such a heater came from.

 

i'm confused why is the rating not 32*48 = 33^2/0.67

You are not confused, Recca, I simply expressed myself poorly. Required resistance is indeed 0.67 Ohm.

Ahem,.... the voltage across the "lightbulb" is 32 V, not 240 V

Most bulbs in 240 Volt nations are rated for 240 Volts, not 32 Volts. Most bulbs are rated by wattage, not resistance. To find bulb resistance, one must use the given ratings.

Either way, my point is that a series resistance will not be practical. (Unless perhaps another resistive heater is chosen as the series resistance.)

I apologize for my ambiguity.

'tis strange that a heater would be built for 208 V. --- I still don't understand where such a heater came from.

The heater is for industrial use. 208 Vac is one leg of a 240 Vac 3-phase wye system taken to ground.

 

Most bulbs in 240 Volt nations are rated for 240 Volts, not 32 Volts. Most bulbs are rated by wattage, not resistance. To find bulb resistance, one must use the given ratings.

actually that answers my doubt
so it was me wrong all along.
thanks.

 

That answere's it. Thanx. I should be looking for a 240v heater. The confusion for me came as the 208v was recommended by a licensed HVAC installer who came out and looked at my application. I will look for a different HVAC person since he obviously doesn't understand I am residential single phase and not 3 phase.

 

Most bulbs in 240 Volt nations are rated for 240 Volts, not 32 Volts. Most bulbs are rated by wattage, not resistance. To find bulb resistance, one must use the given ratings.

Even after reading this several times I was confused but I think I now understand what you are trying to communicate which is that if we could find a bulb rated at 240 V, 86 KW then the resistance would be 0.67 ohm and we could use it for the intended purpose. But that is not the case because the resistance varies greatly with the power being generated and such a bulb connected in such a way would have a *much* lower resistance. Way much lower. It would only have a resistance of 0.67 ohm when connected to 240 V and dissipating 86 KW. What you *really* can use is a lightbulb (or other resistance) rated 32 V, 1.5 KW.

The heater is for industrial use. 208 Vac is one leg of a 240 Vac 3-phase wye system taken to ground.

Can you explain that? According to my calculations if you have 240 V between two phases then you have 240 / SQRT(3) = 139 V between a phase and neutral. Am I missing something?

 

most probably that 208 voltage is line to line voltage between 120 volts supply at 120 phase difference that is phase to phase voltage for a 120v three phase combination.
maybe this can put some light on this matter.

 

Well, on a three phase system the line to line voltage is always the line to neutral voltage multiplied by sqrt(3) so no need to go into that. In Europe the standard used to be 127/220 and was gradually changed to 220/380. The UK had their own 240 V just to show off. Now it seems the voltage will be gradually standarised at 230 V in Europe AND in the UK. Most people will not even notice the change. People still say 220 V when they should be saying 230 V.

I guess 120/208 must be an American thing. Never heard of it. But it does make sense.

 

Well, on a three phase system the line to line voltage is always the line to neutral voltage multiplied by sqrt(3)

true,
was trying to be descriptive

here (in my country) it is 400/sqrt(3)=230.9
ppl are always confused since ratings for boards generally read 240.

 

It seems 230/400 V has become the international standard replacing 220/380. Most appliances will not notice the difference.

 

My bad. I really, really, should not post when tired.

208 \(\sqrt{3}\) =120.

In any event, a 208V heater should not be run on 240V. If obtaining a 240V heater is an option, by all means do it.