I recently installed an electric water heater to service a guest bedroom
located far from the central water heater. Since water will be drawn from this
heater only when guests are visiting, I plan to leave it turned off to save
power.

But before shutting it down, I decided to take some measurements and calculate
how much it costs to run an idle water heater.

The water heater is an electric GE Smar****er 40 gallon, “lowboy” (squat) unit.
The plate on the unit says it draws 4500 watts, but my measurements show that
it actually draws about 4320 watts (18 amps at 240 volts). The EPA estimated
annual cost of operation is $401.

I used a Supco model DLAC recording clamp-on ammeter to record power (amperage)
over a 3 day interval. During the same period, I used a Supco model DLT
recording thermometer to record the ambient air temperature in the crawl space
where the water heater is located.

Here is a summary of my measurements:
Monitored interval: 3 days
Power draw when heating element is on: 4320 watts (18 amps at 240 volts)
Duty cycle when heater is running: 0.0161 (1.61%)
Average power used (heating watts times duty cycle): 69.55 watts
Temperature of hot water delivered: 114 degrees F.
Average temperature in crawl space during measurement period: 61 degrees F.
Temperature rise for water: 53 degrees F (114 - 61)

When the heater is on, it draws 4320 watts. However, the duty cycle
(proportion of time heating) is only 0.0161 (1.61%), so the average power drawn
is 4320*0.0161=69.55 watts. (On average, the heating element is on 23
minutes/day.)

An average power usage of 69.55 watts over 24 hours works out to 1.669 KWH
(kilo-watt hours) per day.

The EPA average national power rate is 8 cents per KWH. So, using the EPA
power rate, the cost of keeping the idle water heater hot is 13.35 cents/day or
$4.00/month or $48.73/year.

Here in Tennessee, we enjoy relatively cheap TVA power which costs 5.6
cents/KWH. Using that rate, the energy cost is 9.35 cents/day, $2.80/month or
$34.13/year.

The EPA estimated annual cost of operation is $401 (assuming 8 cents/KWH). So
the idle heat-loss cost of $48.73/year is about 12% of the total cost.

If you adapt these figures for another location, remember that the cost is
directly proportional to the temperature difference between the hot water and
the surrounding room temperature, and you must adjust for your KWH power cost.

Phil Sherrod
(phil.sherrod 'at' sandh.com)


Index: power, energy, cost, water heater, waterheater, KWH, energy use, cost of
hot water, hot water cost, efficiency, power rate, electric water heater,

Phil Sherrod wrote:

Temperature of hot water delivered: 114 degrees F.

Your numbers are nice and thanks for posting the info,
however it seems to me 114 is a little on the low end,
with 120-125 being more common and I run my domestic
hot water at 140F.

To complete your experiment you really ought to put an
insulating blanket over the unit to see how your numbers
improve. Ideally you should be able to cut "idle"
consumption by half without too much effort by
reducing thermal losses.

On 29-Mar-2004, Bill Vajk wrote:

Your numbers are nice and thanks for posting the info,
however it seems to me 114 is a little on the low end,
with 120-125 being more common and I run my domestic
hot water at 140F.

I agree. If this was my primary water heater, I would probably bump the
temperature up to 125. I may change it later.

To complete your experiment you really ought to put an
insulating blanket over the unit to see how your numbers
improve. Ideally you should be able to cut "idle"
consumption by half without too much effort by
reducing thermal losses.

That would be a good experiment. However, that has the potential of saving
$2/month if I leave the water heater running. With the expected usage, I would
be lucky to save $5/year.

On 29-Mar-2004, Bill Vajk wrote:

Your numbers are nice and thanks for posting the info,
however it seems to me 114 is a little on the low end,
with 120-125 being more common and I run my domestic
hot water at 140F.

Since the energy loss is directly proportional to the temperature differential
of the stored water and the air temperature around the tank, it's easy to
calculate how much more it would cost to maintain the water at 140F.

Assuming a room temperature of 61 degrees F around the water heater:

Cost to maintain water at 114F with 8 cents/KWH cost = $4.00/month (53 degree
differential)

Cost to maintain water at 140F with 8 cents/KWH cost = $5.96/month (79 degree
differential)

Phil Sherrod wrote:
On 29-Mar-2004, Bill Vajk wrote:

Your numbers are nice and thanks for posting the info,
however it seems to me 114 is a little on the low end,
with 120-125 being more common and I run my domestic
hot water at 140F.

Since the energy loss is directly proportional to the temperature differential
of the stored water and the air temperature around the tank, it's easy to
calculate how much more it would cost to maintain the water at 140F.

Assuming a room temperature of 61 degrees F around the water heater:

Cost to maintain water at 114F with 8 cents/KWH cost = $4.00/month (53 degree
differential)

Cost to maintain water at 140F with 8 cents/KWH cost = $5.96/month (79 degree
differential)

Ideally, yes. I do wonder, however, what the startup cost of
the electric heating element(s) is. A higher cycling rate
isn't going to present a linear extension based on delta T
alone since the current consumed during the period the
heating element takes while coming up to operating temperature
is higher than at near steady state operation.

The water temperature, as it rises, results in successively
higher operating temperature of the heating element with a
corresponding (albeit small) decrease in the current drawn.
I don't know what the typical slop in the thermostat
temperature is for an electric hot water heater.

Yes, I realize I'm nitpicking, but that's part of the fun
in discussions like this one. :-)

Your numbers are plenty close enough for the average guy.

On 29-Mar-2004, Bill Vajk wrote:

Ideally, yes. I do wonder, however, what the startup cost of
the electric heating element(s) is. A higher cycling rate
isn't going to present a linear extension based on delta T
alone since the current consumed during the period the
heating element takes while coming up to operating temperature
is higher than at near steady state operation.

That doesn't matter at all to the total energy usage.

If the average water temperature is stable over a long period (small
fluctuations don't matter), then the total energy going into the tank during
this time must exactly equal the total heat energy lost from the tank. This is
a consequence of the law of the conservation of energy. If we put more energy
in, the water temperature will rise over time; if we put in less, the
temperature will fall. If it is stable of an extended period, then the total
energy put in during that period must match the energy taken out. (If you find
a tank that violates this law, explain why and immediately apply for a Nobel
prize.)

So regardless of the voltage, amps, wattage, size or shape of the heating
element, as long as the heater is able to supply enough energy to match the
loss (76 watts in my case), the total energy used over a long period will be
the same; but the duty cycle will change. You could put a 76 watt heater
inside the tank, and it would use the same long-term energy as a 4500 watt
heater. It would just have a longer duty cycle -- 100% rather than 1.6%.

The water temperature, as it rises, results in successively
higher operating temperature of the heating element with a
corresponding (albeit small) decrease in the current drawn.
I don't know what the typical slop in the thermostat
temperature is for an electric hot water heater.

Irrelevant, it just changes the duty cycle -- the long term energy use is
exactly the same.

Conservation of energy -- Not just a good idea, it's the law.

In misc.industry.utilities.electric Phil Sherrod wrote:

| So regardless of the voltage, amps, wattage, size or shape of the heating
| element, as long as the heater is able to supply enough energy to match the
| loss (76 watts in my case), the total energy used over a long period will be
| the same; but the duty cycle will change. You could put a 76 watt heater
| inside the tank, and it would use the same long-term energy as a 4500 watt
| heater. It would just have a longer duty cycle -- 100% rather than 1.6%.

But you're still losing 76 watts of energy. The question is, is there a
way to recover that cheaply. In cold weather, if you could recover 100%
thay would be 76 watts less (or equivalent) energy used for other purposes.

Also, how much would these figures change if you put the water heater on a
timer to ensure that it only heated during night?

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

"Phil Sherrod" concluded with

Irrelevant, it just changes the duty cycle -- the long term energy use is
exactly the same.

Conservation of energy -- Not just a good idea, it's the law.

I wonder if there is any problem with the life span of the water
heater, being cycled from room temperature to operating temperature on
a frequent basis, vs. leaving it at a fixed temperature. I know when
a gas water heater is first installed, the first heating cycle causes
enough sweating that there is sometimes dripping water into the burner
area, causing many do-it-yourselfers to think they bought a leaker.
Do the electric water heaters sweat when first turned on?

Lena
lenagainsterathotmaildotcom

On 30-Mar-2004, wrote:

| So regardless of the voltage, amps, wattage, size or shape of the heating
| element, as long as the heater is able to supply enough energy to match the
| loss (76 watts in my case), the total energy used over a long period will
be
| the same; but the duty cycle will change. You could put a 76 watt heater
| inside the tank, and it would use the same long-term energy as a 4500 watt
| heater. It would just have a longer duty cycle -- 100% rather than 1.6%.

But you're still losing 76 watts of energy. The question is, is there a
way to recover that cheaply. In cold weather, if you could recover 100%
thay would be 76 watts less (or equivalent) energy used for other purposes.

Yes, that is correct. I suggest extra insulating blankets around the heater.
That might cut the heat loss in half and reduce the average energy usage to 37
watts which would save about $2/month.

Also, how much would these figures change if you put the water heater on a
timer to ensure that it only heated during night?

Timers that turn off the water heater for a few hours are virtually useless
because nearly all of that energy is put back in to reheat the water when the
timer turns back on. Now, if you're going to be away for a week or month, you
might save a little money because the tank will have time to cool down and stop
losing energy through the insulation. But the temperature drop over a 6 hour
period is only a few degrees, so the reduction in heat transfer through the
tank due to that small drop is practically insignificant. When the timer turns
on, the heating element runs continuously for many minutes to bring the entire
mass of water back up to the set temperature, and that energy will nearly equal
the energy "saved" during the time that the heater was turned off.

If you are going to try to increase the efficiency of a hot water heater, it is
better to use an insulating jacket than a timer. But regardless of what you
do, you won't save more than $4/month unless you either (1) reduce your hot
water usage or (2) change to a cheaper source of energy (gas, heat pump, solar,
etc.).

On 30-Mar-2004, (Lena) wrote:

I wonder if there is any problem with the life span of the water
heater, being cycled from room temperature to operating temperature on
a frequent basis, vs. leaving it at a fixed temperature.

I don't know. Heating elements tend to have pretty long lives.

I do know that dropping the temperature increases the lifespan of the annode
rod and dip tube, so you save a little there by powering down during periods of
non-use.

Do the electric water heaters sweat when first turned on?

I've never seen that.

wrote:

But you're still losing 76 watts of energy.

That's power, not energy.

In cold weather, if you could recover 100% thay would be 76 watts less
(or equivalent) energy used for other purposes.

That's power, not energy.

Phil Sherrod wrote:

If you are going to try to increase the efficiency of a hot water heater,
it is better to use an insulating jacket than a timer. But regardless of
what you do, you won't save more than $4/month unless you either (1) reduce
your hot water usage or (2) change to a cheaper source of energy (gas,
heat pump, solar, etc.).

Or

3) use cheaper off-peak electrical energy, or

4) use a greywater heat exchanger.

Nick

In misc.industry.utilities.electric Phil Sherrod wrote:

| Timers that turn off the water heater for a few hours are virtually useless
| because nearly all of that energy is put back in to reheat the water when the
| timer turns back on. Now, if you're going to be away for a week or month, you
| might save a little money because the tank will have time to cool down and stop
| losing energy through the insulation. But the temperature drop over a 6 hour
| period is only a few degrees, so the reduction in heat transfer through the
| tank due to that small drop is practically insignificant. When the timer turns
| on, the heating element runs continuously for many minutes to bring the entire
| mass of water back up to the set temperature, and that energy will nearly equal
| the energy "saved" during the time that the heater was turned off.

If your utility charges less at night, then you can see financial benefits
from such a timer. I believe more and more utilities will be going thois
route as they deploy more smart digital meters.


| If you are going to try to increase the efficiency of a hot water heater, it is
| better to use an insulating jacket than a timer. But regardless of what you
| do, you won't save more than $4/month unless you either (1) reduce your hot
| water usage or (2) change to a cheaper source of energy (gas, heat pump, solar,
| etc.).

If you are actually using water (e.g. it's flowing and regular gets heated)
then pre-heating the water by some other means (like discard heat from the
air conditioner in summer, or solar heating, etc) before it enters the tank
could save some money. Rather than merely changing the heat source, this is
more of a diversity. For example it works in winter when the A/C is not in
use, and in cloudy weather when the solar won't do so well, by drawing more
electricity only at those times, but you still get hot water at the desired
temperature.

A larger tank should help, given a smaller surface to lose heat from. I am
wondering if it is worthwhile to heavily insulate the closet the tank is in.
Or might that just end up creating too much heat rise in the closet? The
best would be very well insulated tanks.

I once was favoring tankless instant heat. But now I'm looking at that as
only a backup, if at all.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

In misc.industry.utilities.electric Phil Sherrod wrote:

| On 30-Mar-2004, (Lena) wrote:
|
| I wonder if there is any problem with the life span of the water
| heater, being cycled from room temperature to operating temperature on
| a frequent basis, vs. leaving it at a fixed temperature.
|
| I don't know. Heating elements tend to have pretty long lives.
|
| I do know that dropping the temperature increases the lifespan of the annode
| rod and dip tube, so you save a little there by powering down during periods of
| non-use.

What about heating elements that can be run at lower wattage (such as two
elements switched on in series instead of parallel) during idle times?
Do any tanks even have this capability?

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

wrote:

A larger tank should help, given a smaller surface to lose heat from...

Larger tanks have less surface?

Nick

In misc.industry.utilities.electric wrote:
| wrote:
|
|A larger tank should help, given a smaller surface to lose heat from...
|
| Larger tanks have less surface?

As the volume goes up proportionally to the cube of a dimension, the surface
area through which heat can escape goes up only proportionally to the square
of a dimension, assuming a constant shape (e.g. ratio between diameter and
length for a cylinder). A tank which doubles in size for all dimensions
will have 4 times the surface area (where the heat escapes), and 8 times the
volume (where heat is held). The heating surface may be similarly limited,
so such a tank could also be slower to heat up (because the heating elements
would probably be limited to at most 4 times the wattage, and for some other
reasons, not even that much). So much of this depends on the designs.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

wrote:

wrote:


A larger tank should help, given a smaller surface to lose heat from...


Larger tanks have less surface?

Relative to volume, yes.

wrote:

wrote:


A larger tank should help, given a smaller surface to lose heat from...


Larger tanks have less surface?

Imagine you have a cube tank that's one cubic foot.
Its surface area is six square feet. Now you take
another identical tank, cut the bottom off it, cut the
top off the first, and weld them together. You've
doubled your volume to two cubic feet, but you've only
increased your surface area by four square feet: two
1-foot cubes stacked on top of one another have an
*external* surface area of 10 square feet, not 12.

If you have a cylindrical tank, the volume is given by

h * (? * r * r)

where r * r = 'r squared' (no exponents in plain text
font) and h is the height.

The surface area is two times the surface area of the
circle on the end, plus the surface area of the sides:

(2 * (? * r * r)) + (2 * ? * r * h)

where the surface area of the sides is given by the
circumference of the cylinder's base, 2 * ? * r, times
the height.

If you now double the height of your cylinder, which
obviously doubles the volume, you can see that only the
expression to the right of the '+' sign in the formula
for the surface area is increased; 'h' doesn't appear
in the term to the left of the '+', meaning that as
height increases, that term is constant.

Jonathan Ball wrote:

wrote:

wrote:


A larger tank should help, given a smaller surface to lose heat from...



Larger tanks have less surface?


Imagine you have a cube tank that's one cubic foot. Its surface area is
six square feet.

I meant to specify it's 1 x 1 x 1.

Now you take another identical tank, cut the bottom
off it, cut the top off the first, and weld them together. You've
doubled your volume to two cubic feet, but you've only increased your
surface area by four square feet: two 1-foot cubes stacked on top of
one another have an *external* surface area of 10 square feet, not 12.

If you have a cylindrical tank, the volume is given by

h * (? * r * r)

where r * r = 'r squared' (no exponents in plain text font) and h is the
height.

The surface area is two times the surface area of the circle on the end,
plus the surface area of the sides:

(2 * (? * r * r)) + (2 * ? * r * h)

where the surface area of the sides is given by the circumference of the
cylinder's base, 2 * ? * r, times the height.

If you now double the height of your cylinder, which obviously doubles
the volume, you can see that only the expression to the right of the '+'
sign in the formula for the surface area is increased; 'h' doesn't
appear in the term to the left of the '+', meaning that as height
increases, that term is constant.

| So regardless of the voltage, amps, wattage, size or shape of the heating
| element, as long as the heater is able to supply enough energy to match the
| loss (76 watts in my case), the total energy used over a long period will
be
| the same; but the duty cycle will change. You could put a 76 watt heater
| inside the tank, and it would use the same long-term energy as a 4500 watt
| heater. It would just have a longer duty cycle -- 100% rather than 1.6%.

But you're still losing 76 watts of energy. The question is, is there a
way to recover that cheaply. In cold weather, if you could recover 100%
thay would be 76 watts less (or equivalent) energy used for other purposes.

Yes, that is correct. I suggest extra insulating blankets around the heater.
That might cut the heat loss in half and reduce the average energy usage to 37
watts which would save about $2/month.

Also, how much would these figures change if you put the water heater on a
timer to ensure that it only heated during night?

Timers that turn off the water heater for a few hours are virtually useless
because nearly all of that energy is put back in to reheat the water when the
timer turns back on. Now, if you're going to be away for a week or month, you
might save a little money because the tank will have time to cool down and stop
losing energy through the insulation. But the temperature drop over a 6 hour
period is only a few degrees, so the reduction in heat transfer through the
tank due to that small drop is practically insignificant. When the timer turns
on, the heating element runs continuously for many minutes to bring the entire
mass of water back up to the set temperature, and that energy will nearly equal
the energy "saved" during the time that the heater was turned off.

If you are going to try to increase the efficiency of a hot water heater, it is
better to use an insulating jacket than a timer. But regardless of what you
do, you won't save more than $4/month unless you either (1) reduce your hot
water usage or (2) change to a cheaper source of energy (gas, heat pump, solar,
etc.).

Bill & Phil,

I read your posts with interest and would like to comment on using a
timer to reduce the overall power consumption. I think what would be
of interest is the time to recover the energy used to heat the tank
from ambient to the target temperature. To calculate the time it
takes to recover the power used to heat the tank from ambient you
would solve E1=E2 for t2. This is where E1=P1*t1 and is the power
with the heating element on at a 100% duty cycle times the time it has
to stay on to heat the tank, and where E2=P2*t2 where P2 is the
average power to maintain the tank at the heated temperature. Solving
for t2 would answer the question: what is the minimum time the hot
water heater has to remain turned off to realize an overall power
savings. If the timer is set longer than this value, you will save
power overall. This doesn't hold up as well when people actually use
the hot water.

Let me do an example with your numbers. I will just take a guess that
it takes an hour to heat the tank to the target temperature, if you
know the real number it should be easy to plug in to get a new E1.

So we have:

E1 = 4500W * 3600 sec = 16.2M Joules (energy to heat the tank)
E2 = (4500W * .016) * t2 (power to maintain the tank and unknown t2)

so solving for t2
t2 = 16.2M Joules / 72W = 225000 seconds or 62.5 hours

It looks like that if you know the duty cycle you can just divide the
time it takes to heat the tank by the duty cycle and get the same
answer, I just wanted to go through the math. Keep in mind that this
assumes the tank has completely cooled off. Still it doesn't make a
good case for using a timer.

- James B

On 30-Mar-2004, wrote:

If your utility charges less at night, then you can see financial benefits
from such a timer. I believe more and more utilities will be going thois
route as they deploy more smart digital meters.

Most people have their timer set to turn their heaters OFF at night and back on
in the morning when they are ready to shower. Most of the energy consumption
will come during the morning reheat and during the period after their show when
the water heater is heating the cold water that was drawn in. So I don't see
how lower night power rates will apply.

If you are actually using water (e.g. it's flowing and regular gets heated)
then pre-heating the water by some other means (like discard heat from the
air conditioner in summer, or solar heating, etc) before it enters the tank
could save some money. Rather than merely changing the heat source, this is
more of a diversity. For example it works in winter when the A/C is not in
use, and in cloudy weather when the solar won't do so well, by drawing more
electricity only at those times, but you still get hot water at the desired
temperature.

Yes, I agree with that. Using a secondary source is definitely a good idea.
Waste heat from air conditioning is a good candidate.

A larger tank should help, given a smaller surface to lose heat from.

The closer the tank shape is to spherical, the better it is. You want to
minimize surface area for a given volume of water. However, you don't really
have a lot of choice here; I've never seen a spherical water heater.

I once was favoring tankless instant heat. But now I'm looking at that as
only a backup, if at all.

I don't think a tankless heater makes sense from an energy point of view. The
only advantage is that you have an unlimited quantity of hot water.

On 30-Mar-2004, (james b) wrote:

I read your posts with interest and would like to comment on using a
timer to reduce the overall power consumption. I think what would be
of interest is the time to recover the energy used to heat the tank
from ambient to the target temperature. To calculate the time it
takes to recover the power used to heat the tank from ambient you
would solve E1=E2 for t2. This is where E1=P1*t1 and is the power
with the heating element on at a 100% duty cycle times the time it has
to stay on to heat the tank, and where E2=P2*t2 where P2 is the
average power to maintain the tank at the heated temperature. Solving
for t2 would answer the question: what is the minimum time the hot
water heater has to remain turned off to realize an overall power
savings. If the timer is set longer than this value, you will save
power overall. This doesn't hold up as well when people actually use
the hot water.

I don't know how long it takes the tank to cool, but I believe it is quite a
long time. Although there was some variation in the length of the on and off
periods, typically the heater was on for about 7 minutes and off for about 2.5
hours. I don't know how many degrees variation there is between the hot and
cold setpoints on the thermostat, but I'm guessing it might be in the range of
5 degrees. If that's the case, then the water temperature drops about 2
degrees per hour. So if the timer turned the heater off for 8 hours, you would
get a 16 degree drop.

Since the rate of heat loss through the tank shell is directly proportional to
the temperature difference between the water and the air outside the tank,
there is a small savings from allowing the temperature to drop. However, the
temperature drop occurs slowly, so after 4 hours you have only about an 8
degree drop. Assuming a normal temperature differential of 60 degrees, with a
16 degree drop, the rate of heat loss would be (60-16)/60 = 0.73. But you get
to that point only at the end of the 8 hour period. The average would be about
half that or (60-8)/60 = 0.87. If we figure the energy cost is $4/month the
savings would be about $0.52. Note that you could get the same savings by
turning down the heater temperature a few degrees. And you could probably save
more by putting on an insulating jacket.

On 30-Mar-2004, wrote:

What about heating elements that can be run at lower wattage (such as two
elements switched on in series instead of parallel) during idle times?
Do any tanks even have this capability?

It makes no difference in terms of overall energy use. If the wattage is
lower, they will have to be on longer to heat the water, so the duty cycle will
be longer. But the total energy going in has to equal the total energy lost
through the shell or the water will continue to get hotter or colder over time.
You could put in a 10,000 watt heater or a 100 watt heater and the energy use
over a period of time will be exactly the same. The only difference is how
long it takes to heat the water initially and how long it would take to reheat
the water after you use some.

"Phil Sherrod" wrote in message
...
I recently installed an electric water heater to service a guest bedroom
located far from the central water heater. Since water will be drawn from
this
heater only when guests are visiting, I plan to leave it turned off to
save
power.

But before shutting it down, I decided to take some measurements and
calculate
how much it costs to run an idle water heater.

The water heater is an electric GE Smar****er 40 gallon, “lowboy” (squat)
unit.
The plate on the unit says it draws 4500 watts, but my measurements show
that
it actually draws about 4320 watts (18 amps at 240 volts). The EPA
estimated
annual cost of operation is $401.

I used a Supco model DLAC recording clamp-on ammeter to record power
(amperage)
over a 3 day interval. During the same period, I used a Supco model DLT
recording thermometer to record the ambient air temperature in the crawl
space
where the water heater is located.

Here is a summary of my measurements:
Monitored interval: 3 days
Power draw when heating element is on: 4320 watts (18 amps at 240 volts)
Duty cycle when heater is running: 0.0161 (1.61%)
Average power used (heating watts times duty cycle): 69.55 watts
Temperature of hot water delivered: 114 degrees F.
Average temperature in crawl space during measurement period: 61 degrees
F.
Temperature rise for water: 53 degrees F (114 - 61)

Um, not quite. The temperature differential your heater is maintaining is
53 degrees. The temperature rise for the water is the difference between
temperature of the cold water coming into the tank and the hot water coming
out. The air temperature in the vicinity of the tank doesn't tell you what
the inlet temperature is - it could be warmer or cooler - unless there's a
long run of pipe before it enters the tank, or possibly your water supply is
from a well and there's a pressure tank in the same vicinity - either would
allow the water to pre-warm (or cool) to the ambient temperature.

When the heater is on, it draws 4320 watts. However, the duty cycle
(proportion of time heating) is only 0.0161 (1.61%), so the average power
drawn
is 4320*0.0161=69.55 watts. (On average, the heating element is on 23
minutes/day.)

An average power usage of 69.55 watts over 24 hours works out to 1.669 KWH
(kilo-watt hours) per day.

The EPA average national power rate is 8 cents per KWH. So, using the EPA
power rate, the cost of keeping the idle water heater hot is 13.35
cents/day or
$4.00/month or $48.73/year.

Here in Tennessee, we enjoy relatively cheap TVA power which costs 5.6
cents/KWH. Using that rate, the energy cost is 9.35 cents/day,
$2.80/month or
$34.13/year.

The EPA estimated annual cost of operation is $401 (assuming 8 cents/KWH).
So
the idle heat-loss cost of $48.73/year is about 12% of the total cost.

And all these years, I thought the energy guide labels were from the
Department of Energy, not the EPA. Either way, your measurements pretty
much agree with the DOE's estimate that standby losses generally run from
10% to 20% of the total water heating bill.

In your case, it appears that you didn't draw hot water from the tank during
your test period. The standby loss figure, as well as the annual bill
estimate, assumes "normal" use of hot water.

Interesting post - it's nice to see that lab-derived government figures
agree with real world installations.

wrote in message
...
In misc.industry.utilities.electric Phil Sherrod
wrote:

I once was favoring tankless instant heat. But now I'm looking at that as
only a backup, if at all.

I love the idea behind tankless heaters. But when you can buy a tanked
heater for a couple hundred bucks (Sears) and a "whole house" tankless
heater that delivers 5 gpm can run a grand (Lehman's) and the most that
$700-$800 difference can save you is the standby losses of maybe $5/month,
it just doesn't seem worthwhile, especially if you throw financing costs for
the difference into consideration. Don't know how much installation costs
differ, if at all.

On 30-Mar-2004, "Lou" wrote:

Temperature rise for water: 53 degrees F (114 - 61)

Um, not quite. The temperature differential your heater is maintaining is
53 degrees. The temperature rise for the water is the difference between
temperature of the cold water coming into the tank and the hot water coming
out. The air temperature in the vicinity of the tank doesn't tell you what
the inlet temperature is - it could be warmer or cooler

My analysis was to measure the energy loss for an IDLE water heater. No water
was drawn from the heater during the test, so the incoming water temperature is
irrelevant -- there was no incoming water. The 53 degree figure is the
difference in the temperature between the water in the tank and the ambient air
temperature. That figure is significant because the heat loss is directly
proportional to the temperature differential.

And all these years, I thought the energy guide labels were from the
Department of Energy, not the EPA

You may be right about the department. I'll have to go under the house to
check the label again.

wrote:

|A larger tank should help, given a smaller surface to lose heat from...
|
| Larger tanks have less surface?

As the volume goes up proportionally to the cube of a dimension, the surface
area through which heat can escape goes up only proportionally to the square
of a dimension, assuming a constant shape (e.g. ratio between diameter and
length for a cylinder). A tank which doubles in size for all dimensions
will have 4 times the surface area (where the heat escapes), and 8 times the
volume (where heat is held)...

A lovely explanation, but as you write above, larger tanks
have MORE surface, so larger tanks have MORE heat loss...

Nick

In misc.industry.utilities.electric wrote:
| wrote:
|
||A larger tank should help, given a smaller surface to lose heat from...
||
|| Larger tanks have less surface?
|
|As the volume goes up proportionally to the cube of a dimension, the surface
|area through which heat can escape goes up only proportionally to the square
|of a dimension, assuming a constant shape (e.g. ratio between diameter and
|length for a cylinder). A tank which doubles in size for all dimensions
|will have 4 times the surface area (where the heat escapes), and 8 times the
|volume (where heat is held)...
|
| A lovely explanation, but as you write above, larger tanks
| have MORE surface, so larger tanks have MORE heat loss...

Compare storing 800 liters of hot water in one big 800 liter tank, versus
8 separate 100 liter tanks.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

In misc.industry.utilities.electric Phil Sherrod wrote:

| On 30-Mar-2004, wrote:
|
| If your utility charges less at night, then you can see financial benefits
| from such a timer. I believe more and more utilities will be going thois
| route as they deploy more smart digital meters.
|
| Most people have their timer set to turn their heaters OFF at night and back on
| in the morning when they are ready to shower. Most of the energy consumption
| will come during the morning reheat and during the period after their show when
| the water heater is heating the cold water that was drawn in. So I don't see
| how lower night power rates will apply.

Heat all of a day's hot water needs overnight. That does require a larger tank.


| A larger tank should help, given a smaller surface to lose heat from.
|
| The closer the tank shape is to spherical, the better it is. You want to
| minimize surface area for a given volume of water. However, you don't really
| have a lot of choice here; I've never seen a spherical water heater.

I have seen a hot water heater with a spherical tank. But it was not
very large. One has to consider space utilization issues, construction
cost issues, etc. Cylindrical does seem to wokr out best in most cases.


| I once was favoring tankless instant heat. But now I'm looking at that as
| only a backup, if at all.
|
| I don't think a tankless heater makes sense from an energy point of view. The
| only advantage is that you have an unlimited quantity of hot water.

A tankless water heater uses less energy overall. But it uses that
energy when energy costs (demand) is higher, so it is, in the end, a
disadvantage. It's higher current drawn, and lack of wattage step-up,
means it causes more significant voltage sags when it kicks in.
Electric utilities don't like them for that reason (because neighbors
complain that lights blink). If, OTOH, you have a large solar power
system, it might make sense to use them.

Energy storage is a key, here, and a hot water tank is one form of storage.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

In misc.industry.utilities.electric james b wrote:

| Bill & Phil,
|
| I read your posts with interest and would like to comment on using a
| timer to reduce the overall power consumption. I think what would be

The timer isn't to reduce energy consumption; it's to reduce cost by
deferring energy acquisition to a time when cost per energy is less.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

In misc.industry.utilities.electric Phil Sherrod wrote:

| On 30-Mar-2004, wrote:
|
| What about heating elements that can be run at lower wattage (such as two
| elements switched on in series instead of parallel) during idle times?
| Do any tanks even have this capability?
|
| It makes no difference in terms of overall energy use. If the wattage is
| lower, they will have to be on longer to heat the water, so the duty cycle will
| be longer. But the total energy going in has to equal the total energy lost
| through the shell or the water will continue to get hotter or colder over time.
| You could put in a 10,000 watt heater or a 100 watt heater and the energy use
| over a period of time will be exactly the same. The only difference is how
| long it takes to heat the water initially and how long it would take to reheat
| the water after you use some.

Or looking at it another way, at what rate you can use hot water on a
continuous basis.

--
-----------------------------------------------------------------------------
| Phil Howard KA9WGN | http://linuxhomepage.com/ http://ham.org/ |
| (first name) at ipal.net | http://phil.ipal.org/ http://ka9wgn.ham.org/ |
-----------------------------------------------------------------------------

"Phil Sherrod" wrote in message
...

On 30-Mar-2004, wrote:

If your utility charges less at night, then you can see financial
benefits
from such a timer. I believe more and more utilities will be going
thois
route as they deploy more smart digital meters.

Most people have their timer set to turn their heaters OFF at night and
back on
in the morning when they are ready to shower. Most of the energy
consumption
will come during the morning reheat and during the period after their show
when
the water heater is heating the cold water that was drawn in. So I don't
see
how lower night power rates will apply.

you have to define night time. in my area, that's 9pm to 9am M-F, and 9pm F
to 9am M. my bill is substantially less if i use electric intensive
activities to that period.

wrote:
wrote:


|A larger tank should help, given a smaller surface to lose heat from...
|
| Larger tanks have less surface?



As the volume goes up proportionally to the cube of a dimension, the surface
area through which heat can escape goes up only proportionally to the square
of a dimension, assuming a constant shape (e.g. ratio between diameter and
length for a cylinder). A tank which doubles in size for all dimensions
will have 4 times the surface area (where the heat escapes), and 8 times the
volume (where heat is held)...


A lovely explanation, but as you write above, larger tanks
have MORE surface, so larger tanks have MORE heat loss...

Larger tanks have LESS increase in surface than the
increase in volume.

Take a 1-foot cube, 1 x 1 x 1. The volume is 1 cubic
foot, and the surface area is 6 square feet. An n x n
x n cube of 2 cubic feet will have n = (approx) 1.26.
The surface area is 6 x 1.25^2 = 9.52 sqare feet. The
volume has doubled, but the surface area has gone up by
LESS than twice.

Phil Sherrod wrote:

On 30-Mar-2004, wrote:

If your utility charges less at night, then you can see financial benefits
from such a timer. I believe more and more utilities will be going thois
route as they deploy more smart digital meters.

Most people have their timer set to turn their heaters OFF at night and back on
in the morning when they are ready to shower. Most of the energy consumption
will come during the morning reheat and during the period after their show when
the water heater is heating the cold water that was drawn in. So I don't see
how lower night power rates will apply.


If your utility co. is charging less at night you would set up your
timer
to only heat the water during the times of lower charges.

For instance my utility co. offers time of day rates. When I was
participating
the electric rate was about half the normal rate from 7pm to 7am and
all day weekends and holidays, and about twice the normal rate the rest
of the time. I had a timer that would turn on the water heater from
7pm to 7am. I never remembered running out of hot water in the evenings
even though the water heater did not have power all day. Both of us
in the house worked during the day so no one was at home to use any
hot water. We waiting until after 7pm before using large amounts of
hot water such as washing clothes.

Mark

Jonathan Ball wrote:

...Take a 1-foot cube, 1 x 1 x 1. The volume is 1 cubic
foot, and the surface area is 6 square feet. An n x n
x n cube of 2 cubic feet will have n = (approx) 1.26.
The surface area is 6 x 1.25^2 = 9.52 sqare feet. The
volume has doubled, but the surface area has gone up by
LESS than twice.

A lovely explanation, but larger tanks lose more heat.

Nick

wrote:

Jonathan Ball wrote:


...Take a 1-foot cube, 1 x 1 x 1. The volume is 1 cubic
foot, and the surface area is 6 square feet. An n x n
x n cube of 2 cubic feet will have n = (approx) 1.26.
The surface area is 6 x 1.25^2 = 9.52 sqare feet. The
volume has doubled, but the surface area has gone up by
LESS than twice.


A lovely explanation, but larger tanks lose more heat.

Not relative to volume.

You're just not getting it. Here's a little experiment
even you, with your limited intelligence, can conduct
and perhaps even appreciate.

On a cool night, fill the largest kitchen pot you have
with water, almost to the brim, and bring it to a boil.
When it reaches boiling, take the pot of water and an
empty 1-cup measuring cup outside. Set the pot down,
and dip the measuring cup into the hot water and fill
it. Set it down next to the pot. Place the pot's
cover on the pot, and some type of covering on the
measuring cup; maybe a saucer. Go back into the house
and pour yourself some more of your cheap vodka.

45 minutes later, go back outside, and remove the
covers from the pot and the measuring cup. Stick your
fingers in the measuring cup, and decide if the water
feels cold, cool, tepid, warm or hot. Now do the same
with the large pot of water. You will find the water
in the large pot is substantially warmer than the water
in the cup.

The RATE of heat loss is based on the ratio of the
surface area to the volume, and because the larger
vessel has a SMALLER ratio of surface area to volume,
it will lose heat at a slower rate.

You are a bonehead.

Jonathan Ball wrote:

...Take a 1-foot cube, 1 x 1 x 1. The volume is 1 cubic
foot, and the surface area is 6 square feet. An n x n
x n cube of 2 cubic feet will have n = (approx) 1.26.
The surface area is 6 x 1.25^2 = 9.52 sqare feet. The
volume has doubled, but the surface area has gone up by
LESS than twice.

A lovely explanation, but larger tanks lose more heat.

Not relative to volume.

Who cares about volume?

The RATE of heat loss is based on the ratio of the
surface area to the volume, and because the larger
vessel has a SMALLER ratio of surface area to volume,
it will lose heat at a slower rate.

I'm afraid you are wrong, my good man. The rate of heat loss
(vs temperature change) is directly proportional to the amount
of surface. This is known as "Newton's law of cooling."

Nick

wrote:
Jonathan Ball wrote:


...Take a 1-foot cube, 1 x 1 x 1. The volume is 1 cubic
foot, and the surface area is 6 square feet. An n x n
x n cube of 2 cubic feet will have n = (approx) 1.26.
The surface area is 6 x 1.25^2 = 9.52 sqare feet. The
volume has doubled, but the surface area has gone up by
LESS than twice.

A lovely explanation, but larger tanks lose more heat.

Not relative to volume.


Who cares about volume?

Anyone who is concerned with temperature change.

Why did you snip out my experiment, bozo, without
noting your snip? Here, try it:

On a cool night, fill the largest kitchen pot you have
with water, almost to the brim, and bring it to a boil.
When it reaches boiling, take the pot of water and an
empty 1-cup measuring cup outside. Set the pot down,
and dip the measuring cup into the hot water and fill
it. Set it down next to the pot. Place the pot's
cover on the pot, and some type of covering on the
measuring cup; maybe a saucer. Go back into the house
and pour yourself some more of your cheap vodka.

45 minutes later, go back outside, and remove the
covers from the pot and the measuring cup. Stick your
fingers in the measuring cup, and decide if the water
feels cold, cool, tepid, warm or hot. Now do the same
with the large pot of water. You will find the water
in the large pot is substantially warmer than the water
in the cup.

Here's another experiment for you, bozo. Get a small
individual-serving plastic yogurt container, and a
larger 32 oz. plastic container. Fill them both with
tap water, then put them in your freezer. Check on
them every 45 minutes. Tell us which one freezes
solidly first.



The RATE of heat loss is based on the ratio of the
surface area to the volume, and because the larger
vessel has a SMALLER ratio of surface area to volume,
it will lose heat at a slower rate.


I'm afraid you are wrong, my good man. The rate of heat loss
(vs temperature change) is directly proportional to the amount
of surface. This is known as "Newton's law of cooling."

You're wrong, bozo. Newton's Law of Cooling states
that the rate of temperature change in an object is
proportional to the DIFFERENCE in temperature between
the object and the ambient air temperature. It doesn't
include a variable for surface.

Note also, bozo, that the law applies to the
temperature *of the surface* of the object.

"Phil Sherrod" wrote in message
...

On 30-Mar-2004, "Lou" wrote:

Temperature rise for water: 53 degrees F (114 - 61)

Um, not quite. The temperature differential your heater is maintaining
is
53 degrees. The temperature rise for the water is the difference
between
temperature of the cold water coming into the tank and the hot water
coming
out. The air temperature in the vicinity of the tank doesn't tell you
what
the inlet temperature is - it could be warmer or cooler

My analysis was to measure the energy loss for an IDLE water heater.

Right. So you're not measuring the cost of the water temperature rise.

"Jonathan Ball" wrote in message
ink.net...
wrote:

Jonathan Ball wrote:


...Take a 1-foot cube, 1 x 1 x 1. The volume is 1 cubic
foot, and the surface area is 6 square feet. An n x n
x n cube of 2 cubic feet will have n = (approx) 1.26.
The surface area is 6 x 1.25^2 = 9.52 sqare feet. The
volume has doubled, but the surface area has gone up by
LESS than twice.


A lovely explanation, but larger tanks lose more heat.

Not relative to volume.

For the same temperature differential, a larger tank has more surface area
than a smaller tank, and will therefore lose more BTU's per hour than will
the smaller tank.

The larger the tank, the fewer BTU's lost per unit volume in a given period
of time, but the absolute amount of heat lost varies with the radiating
surface.

It is not the case, as originally stated, that larger tanks have a smaller
surface area. It is true that larger tanks have a smaller surface area per
unit volume. The orignal statement, as posted, was incomplete and therefore
incorrect.

"Mark Sauder" wrote in message
...
For instance my utility co. offers time of day rates. When I was
participating
the electric rate was about half the normal rate from 7pm to 7am and
all day weekends and holidays, and about twice the normal rate the rest
of the time.

That sounds somewhat askew - what you've just said is that the electric rate
was _never_ the normal rate. You seem to be telling us that if, for
example, the normal rate was ten cents per kilowatt-hour, participants in
this time of day pricing scheme could buy electricity for either five cents
or for twenty cents a kilowatt-hour, but never for ten cents.

Is that right?