277 Volt single phase water heater on 240 volts

Merry Christmas
I think we've gone the long way around on the physics of the situation and the relative merits of the voltage systems in different countries. IMHO the focus should go back to the legality of the situation and what the OP should do.

We know that the water heater as designed is a 12.2 KW unit, and will draw a nameplate 44A at 277V. We also know that it is essentially a constant resistance, and will draw about 38A at 240V.

My question: For purpose of installing a code compliant circuit to feed this water heater, should we use the calculated 38A or the nameplate 44A? If the appliance manufacturer provided different ratings at different voltages, then the 240V rating would be 38A. But absent the manufacturer providing this rating, are we allowed to calculate it?
 
winnie said:
I think we've gone the long way around on the physics of the situation and the relative merits of the voltage systems in different countries. IMHO the focus should go back to the legality of the situation and what the OP should do.

We know that the water heater as designed is a 12.2 KW unit, and will draw a nameplate 44A at 277V. We also know that it is essentially a constant resistance, and will draw about 38A at 240V.

My question: For purpose of installing a code compliant circuit to feed this water heater, should we use the calculated 38A or the nameplate 44A? If the appliance manufacturer provided different ratings at different voltages, then the 240V rating would be 38A. But absent the manufacturer providing this rating, are we allowed to calculate it?
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We can, but we can't use it for anything other than an academic exercise.

Unless you say get the customer hot water now, then get the right unit installed ASAP.. This won't be the first or only thing eaten to keep a customer happy
 
winnie said:
absent the manufacturer providing this rating, are we allowed to calculate it?
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No sparky can be expected to get this math conversion right, for proper conductor size, without engineering supervision.
It took me 3 screw ups after reading the right answer twice, and studying the NEC for 23 years.

The prevalence of such ridiculous calcs, and fungible-code exceptions, is why AHJ's train inspectors to check listing violations first, sometimes exclusively.

Inspectors who caught the listing violation would be correct to red tag it, since conversion attempts wont be sizing that wire correctly.
 
As I commented earlier a 22A 277V resistive heating element can be run at a voltage less than its rating just like you could use a 480V circuit breaker at 240V, the OP has 12.579 Ohm resistors in parallel that are rated for ~22 amps of current max.

Hot tubs do it all the time they can operate at 120 or 240 just takes longer to warm up at 120.
I'd be inclined to run two 10 awg circuits to the elements, two 25A breakers rather than one 50A though, as a 50A OCPD may not trip on a faulted element.
 
tortuga said:
I'd be inclined to run two 10 awg circuits to the elements, two 25A breakers rather than one 50A though, as a 50A OCPD may not trip on a faulted element.
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Well it was originally intended to be 44 amps at 277 volts actual load, so it was probably expecting to have a 60 amp breaker in the first place, a 50 amp breaker would protect it just fine.

This is no different than feeding a 240 volt water heater, oven, dryer, or air conditioner at 208 volts.
 
Unless I have it wrong, the resistance of the element would change with voltage as the temperature of the resistor is different and therefore the resistance is also different. Probably not by enough to matter.

Mark
 
Birken Vogt said:
277 elements on 240 = 75% of rated power

9.15 kW

38 amps actual

Storage WH = continuous load, so 48 amps calculated

Change to 8 Cu and 50 amp breaker, done.

Family will never run out of hot water again. Sequencing stats not necessary, performance will be even better this way.
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Believe the 75% of rated power is incorrect. I came up with 86.64% ( 44.043 amps ÷ 38.16 amps ). I know that if you halve the applied voltage to any heater you only produce 25% of wattage. A 12,000 watt heater rated at 277 volts would be 9.15 KW.
 
busman said:
Unless I have it wrong, the resistance of the element would change with voltage as the temperature of the resistor is different and therefore the resistance is also different. Probably not by enough to matter.
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Joethemechanic said:
The current being at 87% would be "I" and once that is squared 0.87 x 0.87 = 75%.
"R" is unknown, but R stays the same
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Actually " R" is known. 12,200 watts ÷ 277 volts = 44.042 ohms. That value remains constant no matter what the applied voltage is.
 
garbo said:
Believe the 75% of rated power is incorrect. I came up with 86.64% ( 44.043 amps ÷ 38.16 amps ). I know that if you halve the applied voltage to any heater you only produce 25% of wattage. A 12,000 watt heater rated at 277 volts would be 9.15 KW.
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86.6% = sqrt(75%)

voltage is 86.6% of rated
current is 86.6% of rated
power is 75% of rated
 
garbo said:
Actually " R" is known. 12,200 watts ÷ 277 volts = 44.042 ohms. That value remains constant no matter what the applied voltage is.
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But it will change with temperature. And the temperature will change with applied voltage. As Busman said, “is it enough to matter?” Without a R vs T curve of the element, we don’t know.
 
busman said:
Unless I have it wrong, the resistance of the element would change with voltage as the temperature of the resistor is different and therefore the resistance is also different. Probably not by enough to matter.

Mark
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Correct.

The temperature of the water is the same, but when you run the element at different wattage the temperature of the resistance wire itself is slightly different because of the thermal resistance of the insulation. So the resistance is slightly different. (Going from memory, the element has resistance wire, covered in mineral insulation, encased in a metal sheath, so you have a small thermal resistance between the heating wire and the water.)

But this difference is going to be _tiny_. Nichrome has a temperature coefficient of resistance of 0.0004/C, meaning you would need a 25C temperature difference to get a 1% change in resistance. The resistance difference running at full power vs 75% power sitting in water will be tiny.

-Jonathan
 
winnie said:
Correct.

The temperature of the water is the same, but when you run the element at different wattage the temperature of the resistance wire itself is slightly different because of the thermal resistance of the insulation. So the resistance is slightly different. (Going from memory, the element has resistance wire, covered in mineral insulation, encased in a metal sheath, so you have a small thermal resistance between the heating wire and the water.)

But this difference is going to be _tiny_. Nichrome has a temperature coefficient of resistance of 0.0004/C, meaning you would need a 25C temperature difference to get a 1% change in resistance. The resistance difference running at full power vs 75% power sitting in water will be tiny.

-Jonathan
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Totally agree with you - tiny or insignificant in the scheme of things compared like the temperature of the building.
 
If the heating element is run colder due to reduced voltage, the resistance will be lower, so the current will be higher. but hardly enough to measure with your amp clamp

Running on 240 instead of 277 is going to reduce your wattage to 75%, and your current down to 87%
 
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