- Thread starter mbrooke
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Sure and in some cases a 30 amp breaker would violate 422.11(E)Is a 25 amp breaker permitted for a residential water heater?

My thinking is this: 4500 watt per element none simaltanous. 4500/240=18.75x1.25=23.4, making that a 25 amp breaker. Is this to code?

Sure and in some cases a 30 amp breaker would violate 422.11(E)

Awsome!

so I have to ask, if the supply is 208 volts to a residential condo apartment:

4500watts@208=3380/208=16.25x1.25=20amps.

So in theory 12 guage and a 20amp breaker would work for 208?

And now you have nailed the situation where a 30 amp breaker would be a violation.

Awsome!

so I have to ask, if the supply is 208 volts to a residential condo apartment:

4500watts@208=3380/208=16.25x1.25=20amps.

So in theory 12 guage and a 20amp breaker would work for 208?

Yeah I guess 12 with a 20 would be compliant. Most likely I would have run 10 AWG during the rough so I might go with a 25 amp breaker anyway.

If the water heater is rated 4500 watts at 240 then your formula is incorrect at 208.

Awsome!

so I have to ask, if the supply is 208 volts to a residential condo apartment:

4500watts@208=3380/208=16.25x1.25=20amps.

So in theory 12 guage and a 20amp breaker would work for 208?

Why multiply by 1.25? A water heater is not a continuous load.Is a 25 amp breaker permitted for a residential water heater?

My thinking is this: 4500 watt per element none simaltanous. 4500/240=18.75x1.25=23.4, making that a 25 amp breaker. Is this to code?

Basically with these voltage the 4500 watt heat at 208v will be about 75% OF THE 240V WATTAGE.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

4500 x .75= 3375 actually 3380

3380/208 = 16.25 * 1.25

It can be if the tank is 120 gal or less. section 422.13Why multiply by 1.25? A water heater is not a continuous load.

Yes that formula works providing that the heater is actually 4500 watts @ 208 volts and not the same heater in the example in post #1.

Awsome!

so I have to ask, if the supply is 208 volts to a residential condo apartment:

4500watts@208=3380/208=16.25x1.25=20amps.

So in theory 12 guage and a 20amp breaker would work for 208?

:huh:Yes that formula works providing that the heater is actually 4500 watts @ 208 volts and not the same heater in the example in post #1.

If the heater is actually 4500 watts @ 208 volts it will draw 21.6 amps when connected to 208 volts, not 16.25.

16.25 is what a 4500 watt @ 240 draws when connected to 208.

Awsome!

so I have to ask, if the supply is 208 volts to a residential condo apartment:

4500watts@208=3380/208=16.25x1.25=20amps.

So in theory 12 guage and a 20amp breaker would work for 208?

If the water heater is rated 4500 watts at 240 then your formula is incorrect at 208.

Check this out

Basically with these voltage the 4500 watt heat at 208v will be about 75% OF THE 240V WATTAGE.

4500 x .75= 3375 actually 3380

3380/208 = 16.25 * 1.25

The OP did have the right wattage listed in his formula. He just shouldn't have put in the first part saying "4500 @ 208". He did have 3380 in his formula. He couldn't have got 16.25A without using that.Yes that formula works providing that the heater is actually 4500 watts @ 208 volts and not the same heater in the example in post #1.

For the record, I just looked at my new water heater that I put in a few weeks ago. The nameplate says:

4500W at 240V

3375W at 208V

So the 3380 is very close!

Thanks Bill

Awsome!

so I have to ask, if the supply is 208 volts to a residential condo apartment:

4500watts@208=3380/208=16.25x1.25=20amps.

So in theory 12 guage and a 20amp breaker would work for 208?

I just realized you did convert 4500 watt to 3380 watt. My apologizes. Of course it should have stated 4500 watt at 240 v not 208 otherwise it would be incorrect. also 16.25 * 1.25= 20.13 not 20 amps.

Oops you're correct, I missed the 3380 part.:huh:

If the heater is actually 4500 watts @ 208 volts it will draw 21.6 amps when connected to 208 volts, not 16.25.

16.25 is what a 4500 watt @ 240 draws when connected to 208.

Ohms do stay the same (or at least very close to the same) with a fixed resistance type of load like we have here.My bad, I was just trying to find a way to say 75% reduced wattage. I did it by assuming ohms stayed the same.

So if it is 4500 watts @ 240 volts then current is 18.75. E = I x R so that means 240/18.75 = 12.8 ohms.

That resistance remains same no matter what voltage is applied, so apply 208 instead and we have 208 = I x 12.8 >>> I = 16.25 amps. 16.25 amps x 208 volts = 3380 watts.

75% reduced wattage is not exact, but is close enough for most quick calculations when reducing voltage from 240 to 208 on a fixed resistance load - 75% of 4500 is actually 3375.

Ohms do stay the same (or at least very close to the same) with a fixed resistance type of load like we have here.

So if it is 4500 watts @ 240 volts then current is 18.75. E = I x R so that means 240/18.75 = 12.8 ohms.

That resistance remains same no matter what voltage is applied, so apply 208 instead and we have 208 = I x 12.8 >>> I = 16.25 amps. 16.25 amps x 208 volts = 3380 watts.

75% reduced wattage is not exact, but is close enough for most quick calculations when reducing voltage from 240 to 208 on a fixed resistance load - 75% of 4500 is actually 3375.

Very true, but I would assume the above because unlike a light bulb the elements are in a medium with exceptional thermal inertia so the coil temperature will not change that much between 208 and 240 imo.

Ohms do stay the same (or at least very close to the same) with a fixed resistance type of load like we have here.

So if it is 4500 watts @ 240 volts then current is 18.75. E = I x R so that means 240/18.75 = 12.8 ohms.

That resistance remains same no matter what voltage is applied, so apply 208 instead and we have 208 = I x 12.8 >>> I = 16.25 amps. 16.25 amps x 208 volts = 3380 watts.

75% reduced wattage is not exact, but is close enough for most quick calculations when reducing voltage from 240 to 208 on a fixed resistance load - 75% of 4500 is actually 3375.

Echo?The OP did have the right wattage listed in his formula. He just shouldn't have put in the first part saying "4500 @ 208". He did have 3380 in his formula. He couldn't have got 16.25A without using that.

For the record, I just looked at my new water heater that I put in a few weeks ago. The nameplate says:

4500W at 240V

3375W at 208V

So the 3380 is very close!