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<DIV><FONT face="Courier New">Antonioli Dan <solardan26@gmail.com>
wrote:</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">>... The biggest heat loss for a modern hot
water heater is through the uninsulated pipes, and you’ll get a much bigger bang
for the buck if you insulated the entire hot water lines (as required by many
green certification systems). </FONT></DIV>
<DIV> </DIV>
<DIV><FONT face="Courier New">Hmm. The $21.57 blanket paid for itself in 1.39
years.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><A title=http://www.engineeringtoolbox.com/copper-pipe-heat-loss-d_19.html
href="http://www.engineeringtoolbox.com/copper-pipe-heat-loss-d_19.html"><FONT
face="Courier New">http://www.engineeringtoolbox.com/copper-pipe-heat-loss-d_19.html</FONT></A><FONT
face="Courier New"> says 6” of 1/2” copper pipe that’s 68 F warmer than the
surrounding air loses 6’x34Btu/h-ft = 204 Btu/h, with Gbare = 204Btu/h/68F = 3
Btu/(h-F) and Rbare = 1/Gbare = 1/3 F-h/Btu.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">Adding a $6.44 piece of “R3.2” pipe insulation
with a 1/2” wall thickness </FONT><A
title=https://www.zoro.com/k-flex-usa-pipe-ins-elastomeric-58-in-id-6-ft-6rxl048058/i/G1474785/?gdffi=047ada998cf641fa93e55ae8579df863&gdfms=60C8D0F0830E4C0AABE51DDE36F763C1&gclid=Cj0KEQjwosK4BRCYhsngx4_SybcBEiQAowaCJZDB7jGPJOb2M9BpF2HWIgXlFM6BGAQ3-kkkNk3e6IwaAmMa8P8HAQ&gclsrc=aw.ds
href="https://www.zoro.com/k-flex-usa-pipe-ins-elastomeric-58-in-id-6-ft-6rxl048058/i/G1474785/?gdffi=047ada998cf641fa93e55ae8579df863&gdfms=60C8D0F0830E4C0AABE51DDE36F763C1&gclid=Cj0KEQjwosK4BRCYhsngx4_SybcBEiQAowaCJZDB7jGPJOb2M9BpF2HWIgXlFM6BGAQ3-kkkNk3e6IwaAmMa8P8HAQ&gclsrc=aw.ds"><FONT
face="Courier New">https://www.zoro.com/k-flex-usa-pipe-ins-elastomeric-58-in-id-6-ft-6rxl048058/i/G1474785/?gdffi=047ada998cf641fa93e55ae8579df863&gdfms=60C8D0F0830E4C0AABE51DDE36F763C1&gclid=Cj0KEQjwosK4BRCYhsngx4_SybcBEiQAowaCJZDB7jGPJOb2M9BpF2HWIgXlFM6BGAQ3-kkkNk3e6IwaAmMa8P8HAQ&gclsrc=aw.ds</FONT></A><FONT
face="Courier New"> to the first 6’ of pipe, with r1 = 0.3125” and r2 = 0.8125”
and k = 0.26 and equivalent thickness t = r2ln(r2/r1) = 0.776” and Rvalue = t/k
= 3 ft-F-h/Btu, or R0.5 for 6’ (see tech bulletin TS13 at </FONT><A
title=http://www.kflexusa.com/HomePages/Downloads.aspx
href="http://www.kflexusa.com/HomePages/Downloads.aspx"><FONT
face="Courier New">http://www.kflexusa.com/HomePages/Downloads.aspx</FONT></A><FONT
face="Courier New">) would reduce the heat loss to 68F/0.5 = 136 Btu/h, saving
204-136 = 68 Btu/h or 595680 Btu (175 kWh) per year worth $26.18 at 15
cents/kWh, for a simple payback of about $6.44/$26.18 = 0.25 years, ie 3 months.
We could save lots of energy if the pipe were always hot, with continuous hot
water flow or infinite pipe conductivity.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">But </FONT><A
title=http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html
href="http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html"><FONT
face="Courier New">http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html</FONT></A><FONT
face="Courier New"> says copper has a 400 W/(mK) thermal conductivity, ie
400x0.5779 = 231 Btu/(h-F-ft), and </FONT><A
title=http://www.petersenproducts.com/Specifications/Pipe_Copper.aspx
href="http://www.petersenproducts.com/Specifications/Pipe_Copper.aspx"><FONT
face="Courier New">http://www.petersenproducts.com/Specifications/Pipe_Copper.aspx</FONT></A><FONT
face="Courier New"> says type L copper pipe has a 0.625” OD and a 0.545” ID,
with a Pi/4(0.625^2-0.545^2) = 0.0735 in^2 (5.1E-4 ft^2) cross-sectional wall
area, so the lengthwise thermal conductance of 6’ of pipe Gpipe = 5.1E-4x231/6’
= 0.0197 Btu/(h-F), with Ripe = 1/Gpipe = 50.9 F-h/Btu, if I did that
right.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">So the first 6’ of bare pipe looks like this,
approximately:</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New"> 50.9</FONT></DIV>
<DIV><FONT face="Courier New">68 ---www------- T I
= 68/(50.9+1/3) = 1.32735 Btu/h.</FONT></DIV>
<DIV><FONT
face="Courier New">
| T = I/3 = 0.44245
F.</FONT></DIV>
<DIV><FONT
face="Courier New">
|</FONT></DIV>
<DIV><FONT
face="Courier New"> |
></FONT></DIV>
<DIV><FONT face="Courier New">
<EM>I</EM> | > 1/3 </FONT></DIV>
<DIV><FONT
face="Courier New"> v
></FONT></DIV>
<DIV><FONT
face="Courier New">
|</FONT></DIV>
<DIV><FONT
face="Courier New">
|</FONT></DIV>
<DIV><FONT
face="Courier New">
0 F<BR></FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New" new?>And the first 6’ of
insulated pipe looks like this, approximately:</FONT></DIV>
<DIV><FONT style='face: "Courier' new?></FONT><FONT
face="Courier New"></FONT> </DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?> 50.9</FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New" new?>68 ---www-------
T I = 68/(50.9+2) = 1.28544 Btu/h.</FONT></DIV>
<DIV><FONT style='face: "Courier' new?><FONT
face="Courier New">
<FONT
style="size: +0">| T
= I/2 = 0.64272 F.</FONT></FONT></FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?>
|</FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?> |<FONT
style="size: +0"> </FONT>></FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?> I | > 0.5 </FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?> v ></FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?>
|</FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?>
|</FONT></DIV>
<DIV><FONT style='face: "Courier' face="Courier New"
new?> 0
F</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">Not much savings, 1.32735-1.28544 = 0.0419 Btu/h
or 367 Btu (0.108 kWh) per year worth $0.0161/year, for a simple payback of
$6.44/$0.0161 = 400 years.</FONT></DIV>
<DIV><FONT style='face: "Courier' new?></FONT> </DIV>
<DIV><FONT style='face: "Courier' new?><FONT face="Courier New">Meanwhile,
</FONT><A
title=http://www.engineeringtoolbox.com/water-content-steel-copper-pipes-tubes-d_1617.html
href="http://www.engineeringtoolbox.com/water-content-steel-copper-pipes-tubes-d_1617.html"><FONT
face="Courier New">http://www.engineeringtoolbox.com/water-content-steel-copper-pipes-tubes-d_1617.html</FONT></A><FONT
face="Courier New"> says 6’ of 1/2” type L copper pipe contains 0.0121 gallons
of water, with C = 6x0.121x8.33 = 0.604 Btu/F, so a 6’ length of insulated pipe
would have a cooling time constant RC = 0.5Btu/(h-F)x0.604Btu/F = 0.302 hours,
ie 18.1 minutes. After a 3.6 minute 10.7 gallon hot water draw at 3 gpm, water
that’s 68 F warmer than the surrounding air would cool to 68e^(-56.4/18.1) =
3.01 F warmer than the surrounding air, just before the next water draw. After
the 6th water draw, ie after 17 hours and 56.4 minutes, ie 1076.4 minutes, it
would cool to 68e^(-1076.4/18.1) = 1.01 x 10^-24 F warmer than the surrounding
air.</FONT><FONT face="Courier New"><BR></FONT></DIV></FONT>
<DIV><FONT face="Courier New">I wonder which green certification systems require
insulating the entire hot water line, and why.</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">Nick</FONT></DIV>
<DIV><FONT face="Courier New"></FONT> </DIV>
<DIV><FONT face="Courier New">PS: Do vertical pipe loop heat traps prevent
thermosyphoning, or just slow it down?</FONT></DIV></DIV></DIV></BODY></HTML>