Posted by Clearfield Consumer on November 29, 2006, 3:57 am
Forgive me if this is the wrong group but I know there are some smart
people here regarding energy use.
I recently moved into a home that has a new Whirlpool Energy Smart
water heater that uses the little computer control box and is claimed
to save electricty.
This water heater is also hooked up to a timer. The previous owner had
the control set to "Smart" in which case the water heater operates like
a conventional unit and also had it on a timer.
I am under the impression that a timer on a water heater this new is of
little use and that it will also interfear (sp?) with the operation of
the Energy Smart circuit. I am thinking that setting the control to
"Energy Smart " mode and not using the timer would be a more
Should I disconnect the timer and let the energy smart circuit do it's
job or can they both be used together?
Posted by Astro on November 29, 2006, 10:52 am
On Nov 28, 10:57 pm, "Clearfield Consumer"
I just installed one of these and have been doing energy measurements
on it. When idling, the unit uses less than .05kwh/day, so turning it
off at night is of minimal use. Further, while the technical details
are lacking, they describe an adaptive control system that allows the
water to cool during idle periods and ramp up during learned high use
periods. They don't indicate whether the internal clock is battery
backed, but I doubt it. If that's the case, then every time you power
the unit down then back up, it's likely to reset itself to it's
original condition and lose all its smarts.
If it were in my house, I'd remove the timer to avoid the temptation
for someone to use it. The water heater already has an energy factor of
0.95, the best in the business as far as I can tell, so turn it on and
let it do its stuff.
Posted by nicksanspam on November 29, 2006, 12:26 pm
Given good heater insulation, it might use more energy than it saves.
I'd disconnect it.
That's only 2.1 watts, vs 4 for an alarm clock motor. How much power does
the timer use? That's a very small heat loss. If a 115 F 50 gallon heater
with 25 ft^2 of surface in a 70 F room only used 0.05 kWh/day, ie 7.1 Btu/h,
the US R-value of the surface would be (115-70)25/7.1 = 158, like 46 inches
of fiberglass, at R19/5.5". Something seems wrong here. Maybe your meter is
less accurate at low power levels.
Lower the setpoint by 0.3 F (minimum 115 F) whenever the lower element
turns on and raise it by 3 F whenever the upper element turns on, up to
a max user setpoint. We could save more energy with a lower min temp or
a much cleverer scheme that leaves the heater filled with cold water
overnight or a "vacation detector."
I doubt there's an internal clock.
An Oregon web site defines the DOE energy factor as "the ratio of useful
hot water output to energy used by the heater, over a year's time." If
an electric heater with an energy factor of 0.95 loses 0.05 kWh/day, it
would supply 0.05/(1-0.95) = 1 kWh or 3412 Btu/day of hot water, enough
to heat G gallons per day from 60 to 120 F, where 8.33G(120-60) = 3412,
so G = 68, which seems reasonable.
Posted by nicksanspam on November 30, 2006, 1:22 pm
Oops. If Ein = Eout/0.95 and Ein-Eout = 0.05 kWh, Eout = 0.95 kWh, ie
3241 Btu/day = 8.33G(135-58), which makes G = 5.05 gallons per day.
G = 64 gallons per day (Eout = 41,050 Btu/day) would make Ein-Eout
= 2161 Btu/day, ie 0.63 kWh/day.
For an electric storage water heater, this may boil down to standby losses.
Posted by Clearfield Consumer on November 29, 2006, 7:30 pm
Thanks for the advice, I disconnected it this morning.