Just to stoke up the discussion about heat loss from an electric (or gas) hot water tank/heater, versus those instant-on type, herewith some current numbers.

Left the house at around 2.00 PM Thursday Sept 4th. turning off both the water supply and the circuit breaker for the electric hot water tank. Returned today, Sept 9th. at around 2.00 PM and noting that water was still quite warm, after an absence from a completely vacant house except for neighbour checking, of 5 days, measured the water temperature.

It was 80 deg F.

The foam insulated 40 US gallon tank is located in the basement which has an air temperature at this time of year of about 60 deg F. The water is normally heated to about 150 deg F.

So in 5 days the 40 gallons lost heat to the 60 deg. basement dropping in temp by 150 - 80 = 70 degrees.

Leaving aside for the moment that the 'rate' of heat loss was probably directly proportional to the difference in temperatures; i.e. highest at the start when the tank was at full temperature, what are members opinions on the following.

40 gallons x 8.33 = 333 pounds of water. 333 x 70 = 23,324 BTUs of heat lost during the 5 days. 23,324 / 3.41 = 6,840 watts of electricity. Previously used to heat water. 6,840 = 6.84 kilowatts of electrcity Which at 0.1 $ per k.watt hr. is/was a heat cost of 68.4 cents. 68.4 / 5 = 13.6 cents, per day. So at a rough approximation 14 cents per day (from a tank that was cooling down gradually over the 5 days). When the same tank is being maintained at 'normal' full temperature the heat loss; could we assume no more than twice that? Perhaps of the order of say 25 cents per day?

Maybe someone with better mathematical skills could to do a more informed calculation? Using dy/dt etc. Also btw anything wrong with the numbers used?????

Not sure how much it would change things, but the greater the differance in the tank temperature and the basement temperature the greater the loss. That is you loose more heat when the tank is 150 deg and the basement is 60 deg than you do when the tank is 120 deg and the basement is still at 60 deg in the same length of time.

Exactly; while this time we do have the start and finish readings for our hot water heater (tank), instead of a bunch of assumptions, to do acompletely accurate calculation would require a some form of exponential calculation of the RATE of heat loss.

For example the amount of heat lost during the first of the five days, while the water is hottest, would be much greater than the amount of heat lost during say the fifth day by which time the water would cooled quite a bit?

However we do know that with no one in the house, electrcity shut off and no one using water at all, the 40 US gallons cooled down in 60 degree basement from 150 degrees to 80 degrees.

Also when it comes to cost of the electricity that had been previously used to raise the temperature of the water to 150 deg. we could prevaricate a bit. The overall cost of electricity here averages out to just over 10 cents per kilowatt hour. But the 'incremental' cost (that is to say for each additional single kilowatt hr. used it costs about 9.1 cents) the other costs being a monthly account charge, taxes etc. Overall ten cents per kw.hr is close enough.

So in summary our 40 gallons of water lost 70 degrees of temperature over a period of five days. A few more days, a week etc. probably would have cooled to exactly the 60 deg. temperature of the basement air; correct.

Using the idea that the hottest coolest fastest.

It might be that of the 23,324 BTUs of heat lost during the five days, an average of about 4,700 BTUs per day (for a cooling off of an un- reheated tank) might have varied from say 10,000 BTU per day (first day as the tanks starts to cool) to say less that 3,000 BTU fifth day etc.

If so; one could then argue that the heat loss-cost for a continually being reheated hot water tank could be as high as 12,000 / 341 = 35.19 kw.hrs which at 0.1$ per is closer to 35 cents per day due to heat losses, compared to `the originally postulated 25 cents per day for a cooling tank.

So the heat loss could be as much as 365 x 0.35 = $128 per year; which heat being lost warms our basement.

A neigbour has a 75 watt mercury vapour yard lamp. On for an average of say 10 hours per night, again at 0.1$ per kw.hr that's 75/1000 x 10 x 365 x 0.1 = $27 yr. About $2 per month.

You can't get from heat lost in a week to the steady heat loss of a tank
being maintained at operating temperature. The heat lost every hour is what
needs to be replaced to maintain an even temperature. You can get that from
this web site;

http://www.leaningpinesoftware.com/hot_water_heater_tank_insul.shtml

The formula from that web site is below.

The basic heat loss formula from physics is:

H = A(Thot-Tcold)/R

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

For an 80 gallon tank they show an example calculation for about 140 watts per hour.

http://www.leaningpinesoftware.com/hot_water_heater_tank_insul.shtml

The formula from that web site is below.

The basic heat loss formula from physics is:

H = A(Thot-Tcold)/R

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

For an 80 gallon tank they show an example calculation for about 140 watts per hour.

Yes you can!

It's not linear so you need to do exponential calculations but YES you can get it...

For the numbers the OP gave,

150 deg F initial temperature cooling down to 80 deg F in 5 days in 60 deg F air

I got 110 Watts steady state heat loss...

For that particular example, it takes 110 Watts to maintain the temp at 150 deg F.

Mark

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