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A 30 gal water tank, left by itself will not use any kilowatt hours - in a month or ever!

However if it is plugged into the power system and set to maintain a specific temperature it will draw power at a rate equivalent to the heat lost to the exterior environment. This depends on the size of the tank, the insulation on the tank, the interior temperture and the exterior temperature. Considering that a hot water tank is essentially a cylinder, this is tied together by the equation:

q = 2 π k (ti - to) / ln(ro / ri)

where

q = heat transferred per unit time per unit length of cylinder or pipe (W/m, Btu/hr ft)

k = thermal conductivity of the wall(W/m.K or W/m oC, Btu/(hr oF ft2/ft))

to = temperature outside pipe or cylinder (K or oC, oF)

ti = temperature inside pipe or cylinder (K or oC, oF)

ln = the natural logarithm

ro = cylinder or pipe outside radius (m, ft)

ri = cylinder or pipe inside radius(m, ft)

A tank in active use will require more power based on the troughput of water. Each unit mass will have to be heated from its input condition to at operating temperature. Each peson in a North American home uses about 250 litres a day divided between cold and hot water.

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Q: How many kilowatt hours does a 30 gal water tank use a month?
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