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Phil Sherrod
 
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Default Power cost of idle electric water heater


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.