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?????

** Also noting that during the winter at least, the lost heat adds
very slightly to warming this all electric house which consumes a year
round average of about $8 per day (all costs including sales taxes
included). Again, just hypothesising, if heating is say 60% of the $8
(say $4.80?), the rest being hot water, lighting computers, TV, radio,
work shop etc. (some of which also aid heating) then the 'lost' heat
from the hot water tank is of the order of, say 0.25 / 4.8 = about 5%
of the house heating cost! i.e. Not very significant?

"terry" wrote in message
...
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.

On Sep 10, 1:05*am, "Ralph Mowery" wrote:
"terry" wrote in message

...





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.- Hide quoted text -

- Show quoted text -

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.

terry wrote in news:ff861cfc-0530-4b46-955d-
:

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/h...nk_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.

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.

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