<html><head><meta http-equiv="Content-Type" content="text/html charset=windows-1252"></head><body style="word-wrap: break-word; -webkit-nbsp-mode: space; -webkit-line-break: after-white-space;"><div>Most modern hot water heaters do not need blankets as they are redundant, and redundant insulation doesn’t do much. </div><div><br></div><div>Insulating both hot and cold lines 2’ from the heater makes a difference, and in places like Berkeley, CA you have to do this when you sell/buy a home as part of the city’s energy conservation program. Cold lines emit heat, and if you don’t believe it and have a tank hot water heater then go check it out….many do!</div><div><br></div><div>That aside, I posted this to see if there are any integrated timers and watt meters. This particular application will be for a solar thermal electric tank heater, where 80%+ of the heat will come from solar thermal, and the remaining 20% will come from grid-tie pv. We want to measure every watt of actual usage and generation. During the summer we will probably turn off the electric back-up altogether if we can get the usage pattern down…..but it’s always a question with the kids!</div><div><br></div><div>Dan Antonioli</div><div><br></div><div><br></div><div><br></div><br><div><div>On Apr 9, 2016, at 4:18 PM, Jeff Martin <<a href="mailto:jeff@open2learn.ca">jeff@open2learn.ca</a>> wrote:</div><br class="Apple-interchange-newline"><blockquote type="cite">
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Nick,<br>
<br>
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).<br>
<br>
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.<br>
<br>
Jeff<br>
<br>
<div class="moz-cite-prefix">On 4/9/2016 12:15 PM, Nick Pine wrote:<br>
</div>
<blockquote cite="mid:5F3835D6E10F479A81F3256A1B225090@Fred" type="cite">
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<div>>This water heater <a moz-do-not-send="true" href="http://m.lowes.com/pd/Whirlpool-50-Gallon-240-Volt-6-Year-Regular-Electric-Water-Heater/50397576">http://m.lowes.com/pd/Whirlpool-50-Gallon-240-Volt-6-Year-Regular-Electric-Water-Heater/50397576</a>
has an EF = 0.95 energy factor, ie it uses Qdm Btu/day,
about 5% more than </div>
<div>the average daily useful water heating energy
<a class="moz-txt-link-freetext" href="https://en.m.wikipedia.org/wiki/Energy_factor">https://en.m.wikipedia.org/wiki/Energy_factor</a> 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.</div>
<div> </div>
<div>Oops. <a moz-do-not-send="true" href="https://en.m.wikipedia.org/wiki/Energy_factor">https://en.m.wikipedia.org/wiki/Energy_factor</a>
says:</div>
<div> </div>
<div>"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."</div>
<div> </div>
<div>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 <u>total</u> 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.</div>
<div> </div>
<div>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.</div>
<div> </div>
<div>How long would it take this $21.57 R10 water heater
blanket to pay for itself at 15 cents/kWh? <a moz-do-not-send="true" title="http://www.lowes.com/pd_24399-1410-SP57/11C___?Ntt=water+heater+blanket&UserSearch=water+heater+blanket&productId=3133229" href="http://www.lowes.com/pd_24399-1410-SP57/11C___?Ntt=water+heater+blanket&UserSearch=water+heater+blanket&productId=3133229"></a><a class="moz-txt-link-freetext" href="http://www.lowes.com/pd_24399-1410-SP57/11C___?Ntt=water+heater+blanket&UserSearch=water+heater+blanket&productId=3133229">http://www.lowes.com/pd_24399-1410-SP57/11C___?Ntt=water+heater+blanket&UserSearch=water+heater+blanket&productId=3133229</a></div>
<div> </div>
<div>Nick</div>
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