| 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