Heater operation at 208v instead of 240v

[Fishn sparky]

Gents,
I don't like asking dumb questions but here I go...

Infrared Heater Nameplate:
Voltage: 240V/Single Phase
Wattage: 6000
Ampacity: 25A

The dumb question I have, wattage goes down when supplied by 208v/single phase, but how can I determine what wattage is if I don't know current? Current should go up with decrease in voltage or does current stay the same not dependent on voltage thus instead of 6000W it is now at 5200W.

Sorry about the dumbness, brain not working right and walking myself in circles.

 

[mayanees]

1) determine the resistance by 240/25=9.6 ohms
2) Power using a 208 Volt source is V squared/Resistance = (208*208)/9.6 = 4506 watts

 

[kwired]

Gents,
I don't like asking dumb questions but here I go...

Infrared Heater Nameplate:
Voltage: 240V/Single Phase
Wattage: 6000
Ampacity: 25A

The dumb question I have, wattage goes down when supplied by 208v/single phase, but how can I determine what wattage is if I don't know current? Current should go up with decrease in voltage or does current stay the same not dependent on voltage thus instead of 6000W it is now at 5200W.

Sorry about the dumbness, brain not working right and walking myself in circles.

The thing you need to know is resistance. That stays the same (or very close to the same anyway) on an item like a heating element regardless of what voltage is applied to it.

E=IxR

240 =25 x 9.6

208 = 21.66 x 9.6

V x A = W

208 x 21.66 = 4505

Current change is proportional with voltage change in pure resistance loads.

Current does go up with decrease in voltage in induction motor loads and similar loads - but only so much change can normally be tolerated before core saturation and overheating gets the best of the motor.

Mayanese gave you a shortcut for calculating, I tried to explain it with calculations that hopefully will help you why the result is what it is.

 

[charlie b]

I don't like asking dumb questions but here I go...

I subscribe to the theory that there is no such thing.

Current should go up with decrease in voltage or does current stay the same not dependent on voltage thus instead of 6000W it is now at 5200W.

That would only be true for a device that draws the same power regardless of the applied voltage (this might hold true only over a limited range). Motors fall into that category. For resistive loads, such as your heaters, the thing that is constant is resistance. Others have treated that topic.

I will add that a quick calculation is to divide "applied voltage" (in your case, 208) by "rated voltage" (240), then square that number. (208/240) squared is 75%. So the original 6000 watts becomes 75% of 6000, or 4506 watts.

There is a simple proof that this shortcut works. I will leave that for a homework problem.

 

[Jraef]

I subscribe to the theory that there is no such thing.

Careful, the Universe often regards that as a challenge...

 

[Adamjamma]

my wife, a math teacher, showed me that one, then I found it in one of my code textbooks as well..lol... think it was the ONSITE guide for BS7671...lol