hot water heater calculation
- Thread starter GSXR600
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mgookin
Senior Member
- Location
- Fort Myers, FL
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- Retired inspector, plans examiner & building official
How are you getting 240vac out of a 208vac 3p service?
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
Technically you could get called on it but no you can't do what you are doing. You have to calculate ohms and then use ohms law to solve at 208V
- Location
- Chapel Hill, NC
- Occupation
- Retired Electrical Contractor
If you calculated the resistance of the element at 240V it is 4.43. So at 208V the wattage will be reduced to 9766 watts not 13,000.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
The calculation you use would be correct for a motor load run at full power from either voltage. But that is not what you have.
The explanation given takes into account the fact that the heater is a simple resistive load.
- Location
- Illinois
- Occupation
- retired electrician
And the 25% reduction in wattage probably means that the water will not get hot enough.
- Location
- Placerville, CA, USA
- Occupation
- Retired PV System Designer
Since this is a tankless water heater I have to agree, but the trademarked "InstaHot" (TM) and its brothers are tank type heaters with a one or two quart tank and a thermostat for kitchen or bar use. Nowhere near the 13000 watt range though.
So we correct the OP to say that he is wiring up a tankless heater and now the result will depend on the exact design of the heater. Most likely it will get hot just fine (since there is a thermostat or proportional control) but it will not deliver the rated maximum flow rate at that temperature.
The manufacturer may even give a separate rating for use on 208, or they just may not support it at all.
The old fashioned UK geysers often did not have a thermostat at all and you regulated the temperature by controlling the flow rate. That would also be a different story.
- Location
- Lockport, IL
- Occupation
- Semi-Retired Electrical Engineer
There is a quicker way. Since power is V*V/R, and since R is constant, you get the new value of power by taking (208/240) squared, and multiplying that times the original 13000 watts. There is an even quicker way, and Don hinted at it. Running a 240 volt rated resistive load at 208 volts gives you a 25% loss of power (208/240 squared is 75%).
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