Hot Water Tank Heat Loss
(for an electric water heater)
From the School of Hard Knocks. Last updated February 1, 2011
Hot Water Heat--The Big Picture
Hot Water Tank Heat Loss
Should You Shut Off Your Hot Water Tank When on Vacation?
How Quickly Does Water Cool in the Pipes?
Newton's Law of Cooling
Modern hot water tanks may feel cool to the touch, but they do lose heat. This web page investigates hot water tank standby heat loss using both a heat flow formula and also empirical measurement.
Using the formula for heat flow I show how to compute the heat loss through the tank walls in btus/hour, then convert that to watts, kwh per year, and dollars on your electric bill. Based on these calculations, our 80 gallon water tank with R-16 insulation should need an average of about 41 watts to replace heat that escapes through the tank walls. In the process of supplying that heat the water heater should consume about $36 of electricity per year (at 10 cents per kwh). The total electric bill for the water heater is of course much higher than this--the $36 is just to replace the heat that flows out the tank walls.
To check this result I measured the heat loss over time when all the hot water faucets were cut off but the tank was still on. This showed about twice the rate of heat loss as the theoretical calculation--a little over 80 watts--which implies about $72 per year worth of electricity.
The basic heat loss formula from physics is:
H = A(Thot-Tcold)/R
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.
Thot: 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.
Tcold: 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.)
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 mounted on 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. Both the pressure relief valve and the water pipe feel hot to the touch. This suggests we need an empirical measurement to get a handle on the actual loss.
By monitoring how often and for how long the heater element turns on I computed the rate of heat loss through the tank walls. The hardest part of this experiment was getting the family to do without hot water for a couple of days while I took measurements. Why does it take so long? It takes a number of hours for the water temperature inside the tank to stabilize, and then the heater element only turns on about every 5 to 7 hours, which is not surprising given that we are dealing with a thermocouple heat sensor. However, the variability meant that I needed to take several measurements to get a good average.
Sensing when the heater element turns on is not trivial. I created a current sensor by putting the wire to the heater through a torroidal core with several turns of magnet wire wrapped around it. This fed up to a full-wave bridge rectifier and a few other components in the kitchen, which in turn operated an LED, an audible alarm, and an alarm clock with a second hand. The alarm clock was powered by the current sensor, so only turned on when the heater element turned on. This provided a good way of telling how long the element ran.
Averaging the data gave a heat loss rate of a little over 80 watts, about double the theoretical calculation. This means there is quite a bit of heat loss from the pressure relief valve, water pipes, and other things protruding from the water heater.
We can reduce some of this loss through insulation. See the home page for information on insulating the hot water pipes.
You can e-mail me at support(@ sign goes here)leaningpinesoftware.com.