Say I am running 120 through 3 heaters each rated 250, 375W. How would that impact the Watts for each heater? what is the relationship? If each section has a heater rated at 250V, 375Watts, what is the total current when operating the 250V rated heaters at 120VAC.
120V through a 250V Heater
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winnie
Senior Member
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Normal resistive heaters are nearly ideal resistors. Use simple Ohm's law relations, power = voltage^2/ resistance
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P1 = V1^2 / R.
P2 = V2^2 / R. (It'sthe same R)
Thus, P1 / P2 = (V1 / V2)^2
= (120 / 250)^2
I will let you finish the math.
P2 = V2^2 / R. (It'sthe same R)
Thus, P1 / P2 = (V1 / V2)^2
= (120 / 250)^2
I will let you finish the math.
LarryFine
Master Electrician Electric Contractor Richmond VA
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Welcome to the forum.
Simply put, for a resistive load, if you halve the voltage, the resultant current halves.
Since power is the product of voltage and current, with each being halved, the power quarters.
Thus, each 375w heater will instead produce 1/4 of 375, or 93.75w.
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I was hung up at running voltage through a heater, and if the heaters were in series or parallel.
texie
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Me too.
LarryFine
Master Electrician Electric Contractor Richmond VA
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I took it to mean paralleled.
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Hey, I resemble that remark... I have to think about it every single time.
I do it a lot actually, because it's a trick for sizing strip heaters for cabinets and I have to keep explaining it to the kids I work with when they see it. If you have a 6" strip rated 200W @ 250V, but apply 120V, the "watt density", which speaks to the SURFACE temperature if you happen to touch it, is 1/4 that of an equivalent 6" 50W 120V heat strip. So you get the same amount of heat, but lower surface temperature.
I do it a lot actually, because it's a trick for sizing strip heaters for cabinets and I have to keep explaining it to the kids I work with when they see it. If you have a 6" strip rated 200W @ 250V, but apply 120V, the "watt density", which speaks to the SURFACE temperature if you happen to touch it, is 1/4 that of an equivalent 6" 50W 120V heat strip. So you get the same amount of heat, but lower surface temperature.
winnie
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Huh? If you have two devices of the same dimension and the same outer case material, dissipating the same heat, how is the watt density lower? Is there some other dimension which changes when you switch between a 200W 250V 6" heater and a 50W 125V 6" heater?
-Jon
Eddie702
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I get 260 watts for 3 heater at 120volt.
LarryFine
Master Electrician Electric Contractor Richmond VA
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I get 3 x my 93.75, or 281.25. I quartered the 250v rating, so my numbers really apply to 125v.
There will be some mismatch because we're comparing applying 120v to a load rated at 250v.
There will be some mismatch because we're comparing applying 120v to a load rated at 250v.
~260W, 259.2W it is!:
Resistance per heater 250V, 375W) = E^2/P = (250^2)/375 = 166.67 ohms
@ 120V, the current per heater will be = 120/166.67 = 0.72 A
With 3 heaters, you have Amps = 2.16 A
Wattage = 2.16 X 120 = 259.2 W!
Rule of thumb if you halve the voltage to any electric heaters you quarter the wattage & BTU'S produced. Ex = a heater rated for 2,000 watts @ 240 volts would use 500 watts of power. Won a free lunch 40 some years ago with a fellow sparky who said that I was wrong.
kwired
Electron manager
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works the other way around as well. If you apply 240 volts to a 120 volt heater you get 4 times watts. Chances are the heater can not take that for very long though.
Fred B
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Interesting discussion.
Seems useful to calculate actual power for a residence heat that doesn't have the "rated" voltage available. A rated heater of 250V@5000W would only make you 4600W of heat when available voltage is only 230V.
Seems would be cumulative when calc. for an entire house load that is not being provided optimum voltages.
A house needing 45,000W of heat at a rated unit (250V@5000W) would require 9 such units, but It would require almost another whole unit (@230V) compared to if the optimum voltage was available.
Seems useful to calculate actual power for a residence heat that doesn't have the "rated" voltage available. A rated heater of 250V@5000W would only make you 4600W of heat when available voltage is only 230V.
Seems would be cumulative when calc. for an entire house load that is not being provided optimum voltages.
A house needing 45,000W of heat at a rated unit (250V@5000W) would require 9 such units, but It would require almost another whole unit (@230V) compared to if the optimum voltage was available.
wwhitney
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No, as per this discussion, for the fixed operating resistance assumption, it would make (230/250)^2 * 5000 = 4232W.
Cheers, Wayne
winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
I suspect that the OP is a spammer. New user, pops in, asks a simple question, lets the discussion build without any input, and ends up with a bit of cred that they use later to post spam or disinformation. With that said, they did start an interesting conversation amongst us real folk 
Once you get up to the scale of multiple heaters (say a house with a bunch of electric baseboards heaters) you end up having to deal with load diversity maths.
A single resistance heater rated at 250V will have lower wattage when run at 230V. But in a house these heaters are being cycled on and off because of thermostats, and the total power being delivered has to average out to the heat load of the building. Same power at lower voltage means _higher_ current.
How do we reconcile that lower voltage into a resistor means lower current, but lower voltage into the house means higher current (for resistance heating)?
For the house to be at the same temperature, the same total BTU needs to be delivered. For an individual heater at lower voltage, the heater draws lower current, operates at lower power, and needs to run longer to deliver the same total BTU.
When you average across a bunch of heaters, each cycling on and off on its own thermostat, in aggregate delivering the same total BTU, the load diversity gets changed, and the aggregate average current must go up. So instantaneous current to an individual heater goes down as voltage goes down, but average aggregate current to the entire house goes up to deliver the same BTU per hour.
-Jon
-Jon
Once you get up to the scale of multiple heaters (say a house with a bunch of electric baseboards heaters) you end up having to deal with load diversity maths.
A single resistance heater rated at 250V will have lower wattage when run at 230V. But in a house these heaters are being cycled on and off because of thermostats, and the total power being delivered has to average out to the heat load of the building. Same power at lower voltage means _higher_ current.
How do we reconcile that lower voltage into a resistor means lower current, but lower voltage into the house means higher current (for resistance heating)?
For the house to be at the same temperature, the same total BTU needs to be delivered. For an individual heater at lower voltage, the heater draws lower current, operates at lower power, and needs to run longer to deliver the same total BTU.
When you average across a bunch of heaters, each cycling on and off on its own thermostat, in aggregate delivering the same total BTU, the load diversity gets changed, and the aggregate average current must go up. So instantaneous current to an individual heater goes down as voltage goes down, but average aggregate current to the entire house goes up to deliver the same BTU per hour.
-Jon
-Jon
LarryFine
Master Electrician Electric Contractor Richmond VA
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- Henrico County, VA
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- Electrical Contractor
Yes, but unless you're replacing the heaters, you'll end up delivering less power over greater time.
See my above statement. You're applying theoretical math using the wrong variable.
For a resistive load, current (and the resultant power) varies proportionately with applied voltage.
For the power to stay constant, you would have to vary the load resistance as you vary the voltage.
Exactly.
Again, it's the duration that changes with voltage applied to a given (fixed resistance) heater.
kwired
Electron manager
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- NE Nebraska
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- EC
At same time if heating load calculation was done correctly you would only have trouble keeping up with heating demand during times when outside temperature is at or below the designed minimum outdoor temp.
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