[Greenbuilding] Water heater timer energy savings?

[Jeff Martin]

Nick,

Interesting set of calculations! Obviously, actual usage patterns could 
dramatically influence the energy savings effected with the timer (e.g., 
if the heater is not used at all on weekdays, but is used every weekend).

At first glance, it would appear that the heater blanket would almost 
double the R-value of the tank insulation, so nearly halve the heat 
loss, leading to a payoff on the blanket in a little more than a year. 
However, as you noted earlier, a significant part of the heat loss might 
be from conduction along the pipes, so pipe wrap insulation at those 
points might an important addition to the blanket wrap, in order to 
achieve that saving.

Jeff

On 4/9/2016 12:15 PM, Nick Pine wrote:
> >This water heater 
> http://m.lowes.com/pd/Whirlpool-50-Gallon-240-Volt-6-Year-Regular-Electric-Water-Heater/50397576 
> has an EF = 0.95 energy factor, ie it uses Qdm Btu/day, about 5% more 
> than
> the average daily useful water heating energy 
> https://en.m.wikipedia.org/wiki/Energy_factor  as tested in a 67.5 F 
> room with 6 64.3 gallon draws of 135 F water heated from 58 F in the 
> first 6 hours of the day, with an 18 hour rest period after that.
> Oops. https://en.m.wikipedia.org/wiki/Energy_factor says:
> "During the test 64.3±1.0 gallons of water are drawn from the water 
> heater in six equally spaced draws that begin one hour apart."
> After I calculated the 88 F water temp after 6 hours and realized that 
> 6x64 gallons is a lot of hot water for a family and reflected on the 
> R1.82 wall insulation calc, I realized that the _total_ draw is 64.3 
> gallons, which makes the useful water heating energy Qu = 
> 64.3galx8.33lb/gal(135-58) = 41243 Btu, approximately. And EF = 0.95 = 
> Qu/Qdm by definition, so Qdm = Qu/0.95 = 43413 Btu, and the standby 
> loss with no water draws is 43413-41243 = 2170 Btu/day, approximately, 
> ie 0.64 kWh/day, or 1628 Btu during the 18-hour rest period, when a 
> timer might help save energy.
> With no timer, 1628 Btu = 18h(135-67.5)G makes the water heater's 
> thermal conductance G = 1.34 Btu/h-F. With 50 gallons of water, 
> thermal capacitance C = 50x8.33 = 416.5 Btu/F, approximately, so the 
> cooling time constant RC = C/G = 416.5Btu/F/(1.34Btu/h-F) = 311 hours, 
> so with a timer, the water would cool from 135 F to 67.5 + 
> (135-67.5)e^(-18h/311h) = 131.2 F in 18 hours. Reheating it to 135 
> before the next draw would require 416.5(135-131.2) = 1583 Btu. So the 
> timer saves 1628-1583 = 45.3 Btu/day, or 1.9 Btu/h or 1.9/3.412 = 0.55 
> watts, on a continuous basis.
> How long would it take this $21.57 R10 water heater blanket to pay for 
> itself at 15 cents/kWh? 
> http://www.lowes.com/pd_24399-1410-SP57/11C___?Ntt=water+heater+blanket&UserSearch=water+heater+blanket&productId=3133229
> Nick
>
>
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