# Calculating time constant for storage-tank water heater

I was curious what the R-C time constant is, for a conventional storage-tank style electric water heater; I came up with an answer that seems very counter-intuitive.

("RC time constant" is the product of a resistance and capacitance, with the dimensions of time. In that time, a given capacitance C is charged or discharged, through a given resistance R, about 63%. For thermal systems, R can have the dimensions: degrees-F per btu per hour and C the dimensions: btu per degree-F).

For my 10 gal kitchen water heater, the surface area of the tank is about 8 sq-ft, and the insulation is about 8 sq-ft * degree-F per btu/hr (aka. R-8). Thus the overall thermal resistance of the tank's envelope is about 1 degree-F per btu/hr.

The thermal "capacitance" is about 80 btu per degree-F (10 gals of water at 8 lbs per gallon, with a btu defined as the energy to raise one lb of water by a degree-F).

Thus the thermal R-C time constant is (1 degree-F per btu/hr) * (80 btu per degree-F), or 80 hours. That means that in 3+ days, after losing power, the tank would still retain 37% of its heat (or cool only 63% of the way from its set temperature to ambient temperature). Maybe that's right, but it seems pretty counterintuitive. Am I calculating wrong somehow ?

Best I can tell, your math is right. I've never done thermal R-C calculations before, but what I got agrees with your numbers. Here's a plot of the temperature over 10 days if it started at 140°F and ambient temperature is 70°F: Basically what this is saying is there is a lot of energy in water. It takes about 8 BTU to heat a gallon of water only one degree F. It only takes about 0.2 BTU to heat the same volume of air one degree F. It takes about 400 times more energy to heat water than air per unit volume. Since the heat is so concentrated in the water's volume, it's easier to insulate and keep warm because there is less surface area.

Here's the thing though, this model assumes a sphere of liquid completely separated by insulation from the environment. In reality, you're going to have pipes, other components and the water itself which will also dissipate heat into the environment affecting the final temperature.

Here's a related discussion on a Thermos bottle

• Interesting but I think the calculation also neglects thermal migration. Although newer water heaters have check valves to reduce this effect. – ArchonOSX Nov 13 '17 at 17:10
• I have insulated the pipes in/out of my unit very well. And (I didn't mention) it's electric, so none of the large loss that gas units have out the flue. – RustyShackleford Nov 13 '17 at 19:06
• Thanks, @Jonathan727 for checking my calculation. Maybe the result is correct (notwithstanding the 2nd-order effects that you and ArchonOSX mention). I guess I could power the unit down and see what happens ! – RustyShackleford Nov 13 '17 at 19:12
• I think the only significant thing other than the ideal cylinder of water (the shape doesn't matter, as long as I get the surface area right) is the water in the pipes. There's the heating element, but even it is fairly well insulated (except for the wires themselves). The T&P valve might be significant. – RustyShackleford Nov 13 '17 at 19:13
• However, there's another factor which should make the cooling slower than the ideal: the boundary effect where the water nearest the edge of the tank cools off and effectively insulates the water farther from the edge; so for the ideal model, there would have to be something keeping the water mixed so its temperature is uniform. – RustyShackleford Nov 13 '17 at 19:14