240V to 208V Water Heater

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Trey4U

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I have a 240v water heater I need to connect to a 208v panel. Based on the wattage on the nameplate the current at 208v works out to 43A. All I have is #6 and a 60A breaker. Can I use this to feed the water heater?

Thanks!
 
Trey4U said:
I have a 240v water heater I need to connect to a 208v panel. Based on the wattage on the nameplate the current at 208v works out to 43A. All I have is #6 and a 60A breaker. Can I use this to feed the water heater?

Thanks!
Click to expand...
That is pretty much what you do need to run. Minimum conductor ampacity needs to be 125% of rated load - this means you nedd 53.75 amp conductor. @ 75C #8 is only good for 50A

Overcurrent protection also needs to be at least 53.75 - next standard size up is 60 amp.
 
Trey4U said:
I have a 240v water heater I need to connect to a 208v panel. Based on the wattage on the nameplate the current at 208v works out to 43A. All I have is #6 and a 60A breaker. Can I use this to feed the water heater?

Thanks!
Click to expand...

The wattage at 208V will be 75% of that at 240V.
 
dkidd said:
The wattage at 208V will be 75% of that at 240V.
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I was presuming he had that figured out, but yes if the 43 amps is the rating at 240 volts then it will draw less on 208.

43 amps either way is larger than typical household storage tank type water heater, but I suppose might be expected from an instantaneous water heater. I don't think you need to apply 125% factor to an instantaneous type heater.
 
kwired said:
I was presuming he had that figured out, but yes if the 43 amps is the rating at 240 volts then it will draw less on 208.

43 amps either way is larger than typical household storage tank type water heater, but I suppose might be expected from an instantaneous water heater. I don't think you need to apply 125% factor to an instantaneous type heater.
Click to expand...

The wattage is 9,000 so I figured it at 208 and got 43 and change.
 
infinity said:
The current at 240 volts is 9000/240=37.5 amps. At 208 volts it will be considerably less current.
Click to expand...

And IF the 9000 is made up of 2 - 4500 watt elements, a standard water heater control setup will only energize one at a time!

But regardless, your proposal (#6 on a 60A) will work. You may have difficulty landing the 6s on the terminals provided.
 
retirede said:
And IF the 9000 is made up of 2 - 4500 watt elements, a standard water heater control setup will only energize one at a time!

But regardless, your proposal (#6 on a 60A) will work. You may have difficulty landing the 6s on the terminals provided.
Click to expand...

The label on the water heater says simultaneous wattage is 9000. (4500 each element) It's a commercial water heater so I guess both elements can come on together when needed.
 
Trey4U said:
Uhhh, if you lower the voltage doesn't the current go up? 9,000/208=43.26
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No your formula is incorrect. You need to find the resistance of the heater at 240 volts and then use that with 208 volts in your calculation. Lower voltage equals lower current with a resisitive load.
 
Trey- resistive heat works backwards from motor loads etc

Here is an example of how resisitance heat works using ohms law
Lets take a baseboard heater that is rated 2000 watts at 240V. What is the wattage rating @208 volts?
Our intuition says as the voltage increases the amps go down however that is not the case with resistance heat.


R= V^2 / P
R= 240^2/2000= 28.8
Since the resistance does not change we can now use this formula


P=V^2 / R
P= 208^2/ 28.8= 1502 watts


So a 2000 watt heater at 240V will be rated 1502 watts at 208V

75% of 2000 watts is 1500 watts-- pretty close
 
Trey4U said:
Uhhh, if you lower the voltage doesn't the current go up? 9,000/208=43.26
Click to expand...
Not for a constant-resistance load; the current varies with the voltage. A motor, on the other hand, tends to be a constant-power load, and the current would increase as the voltage lowers.

Now, if you designed and built a new element to maintain the same heating power (wattage) on the lower voltage, its resistance would be lower, resulting in the expected increase in current.
 
Trey4U said:
Uhhh, if you lower the voltage doesn't the current go up? 9,000/208=43.26
Click to expand...
In a resistive load where the resistance remains the same, the lower voltage results in a lower wattage.

P (watts) = E (voltage)2 / R (ohms)

Since you already have the heater, the R is already existing and stays the same.

If you bought a NEW heater designed as 9000W at 208V, then the R would be lower in order to attain the same watts and the current would be higher.
 
Same if you replace the 240V elements in the heater with 208V elements (if available) or change the tap connection on dual voltage elements.

Sent from my XT1585 using Tapatalk
 
I always use a variation of P= E²R to solve these problems.
(new voltage/old voltage)² (original wattage)/(new voltage)
(208/240)² (9000)/208
0.75111(9000)/208
6760/208
32.5 amps
 
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