[Greenbuilding] Water heater timer energy savings?

[Nick Pine]

barbara deane-gillett <deaneg at hotmail.com> writes:

>typical loss from an electric water heater is 1 kwh per day or 40 watts not 
>4 watts

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.

The useful water heating energy would be Qu = 6x6galx8.33lb/gal(135-58) = 
247456 Btu, approximately, and EF = 0.95 = Qu/Qdm by definition, so Qdm = 
Qu/0.95 = 260480 Btu, so the standby loss with no water draws would be 
260480-247456 = 13024 Btu/day, approximately, ie 3.8 kWh/day, or 9768 Btu 
during the rest period, when a timer might help save energy.

With no timer, 9768 Btu = 18h(135-67.5)G makes the water heater's thermal 
conductance G = 8.04 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/(8.04Btu/h-F) = 51.8 hours, so with a timer, the water 
would cool from 135 F to 67.5 + (135-67.5)e^(-18h/51.8h) = 115.2 F. 
Reheating it to 135 before the next draw would require 416.5(135-115.2) = 
8252 Btu. So the timer would save 9768-8252 = 1516 Btu/day, or 63 Btu/h or 
63/3.412 = 18.5 watts, on a continuous basis.

The water heater is 24" in diameter and 50" tall, with 14.6 ft^2 of exterior 
surface, counting the top and bottom, which makes the surface R value 
14.6ft^2/8.04Btu/h-F = 1.82 h-Fft^2/Btu. Not much, since the specs say it 
has 3" of foam insulation. Maybe it loses lots of heat through the 
connecting pipes and wires. A 3" fiberglass R11 blanket might lower G to 
14.6/12.82 = 1.14 Btu/h-F, with a lower 18-hour heat loss of 
18h(135-67.5)1.14 = 1385 Btu without a timer. With RC = 416.5/1.14 = 365 
hours and a timer, the water would only cool to 67.3+(135-67.5)e^(-18/365) = 
131.75 F in 18 hours. Reheating it to 135 would require 416.5(135-131.75) = 
1353 Btu.  So the timer would save 1385-1353 = 32 Btu/day, or 1.3 Btu/h or 
0.39 watts, on a continuous basis, if I did that right.

Why doesn't everyone add thermal blankets and pipe insulation to water 
heaters?

Nick