Hot Water Tank Heat Loss

(for an electric water heater)

From the School of Hard Knocks.  Last updated Feb 11, 2010

The basic heat loss formula from physics is:

     H = A(T_hot-T_cold)/R

     where
       H = Heat loss in btu/hr
       A = Surface area in square feet
       T_hot = Hot water temperature in F
       T_cold = Air temperature in F
       R = R-value of the insulation in ft²hrF/btu

Surface Area: You can compute the hot water tank surface area from the tank dimensions. First measure the tank diameter in inches, then the tank height in inches. Take half the diameter to get the radius. The total surface area of the tank, in square inches, is then the sum of the circular area for the top and bottom of the tank, and the area of the cylinder in between.

         A = 2*3.14*(radius)² + 2*3.14*radius*height

Now divide this result by 144 to get the total area in square feet. The number should be greater than 5 and less than 50 square feet. If you get a number outside of this range you probably did something wrong in the measurement or calculation.

T_hot: You can measure the hot water temperature directly at the tap. Run hot water onto a thermometer until the temperature stabilizes. Preferably do this at the closest faucet to the tank so you don't get much heat loss in the pipes.

T_cold: The cold temperature is the air temperature that the outside of the water heater is exposed to. This could vary greatly depending on the season and on where the water tank is located. If the tank is in an uninsulated garage, for example, it could be 90F in the summer, but 30F in the winter. In this case you might want to take a rough annual average, like 60F.

R-value: This should be written somewhere on the tank or be listed in the tank literature. If you cannot find it you can always call the manufacturer. It is also possible to estimate the R-value experimentally. (Someday I will create a web page on that.)

Sample calculation:

Our water heater is 24.5 inches in diameter, 58 inches high, and has an R-value of 16. The water temperature is about 120F and the basement temperature I will assume averages 60F.

        Surface area = 37.5 square feet

        Heat Loss = H = 37.5*(120-60)/16 = 140.8 btu/hour

A btu is a British Thermal Unit, which is the amount of heat required to raise the temperature of one pound of water by one degree Fahrenheit. To make sense of this we need to convert this to kilowatt-hours, which is the unit of energy used by the electric company and that shows up on your electric bill every month.

Converting to kwh:

One kwh of electricity, when used in a resistance heater, generates 3,412 btus of heat. So to convert btu/hr to kwh/hr we divide by 3,412. But we can cancel the hours in the dimension of kwh/hr to get just kw:

        H (in kw) = H (in btu/hr) / 3412

So for our hot water heater:

        H = 140.8 btu/hr / 3,412 = 0.0413 kw

        H = 41.3 watts

So the hot water tank loses heat at the rate of about 41 watts. The reason the tank walls feel cool is that the 41 watts are spread over a large surface area.

To compute the kwh used in a day, multiply the kw by 24 hours. For our hot water heater:

        Daily kwh = 0.0413 kw x 24 hours/day = .99 kwh/day

This is about 1 kwh per day, which implies about 30 kwh/month and 365 kwh/year.

Converting to cost:

In our area electricity costs about 10 cents per kwh, so the cost per month from tank loss is about 30kwh x $0.10/kwh = $3.00/month, or about $36 per year.

The above analysis doesn't capture all the losses. There is also a loss from the brass pressure-relief valve on the top of the water heater, and a loss from hot water that rises up the hot water pipe even when the unit is not in use. These losses are not trivial, and could add up to more than the loss through the tank walls.

See the home page for information on insulating the hot water pipes.

This site is still under construction…to be continued…

You can e-mail me at support(@ sign goes here)leaningpinesoftware.com.