[Greenbuilding] Hot water pipe insulation

[Nick Pine]

Antonioli Dan <solardan26 at gmail.com> wrote:

>... 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). 

Hmm. The $21.57 blanket paid for itself in 1.39 years.

http://www.engineeringtoolbox.com/copper-pipe-heat-loss-d_19.html 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.

Adding a $6.44 piece of “R3.2” pipe insulation with a 1/2” wall thickness 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 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 http://www.kflexusa.com/HomePages/Downloads.aspx) 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.

But http://www.engineeringtoolbox.com/thermal-conductivity-d_429.html says copper has a 400 W/(mK) thermal conductivity, ie 400x0.5779 = 231 Btu/(h-F-ft), and http://www.petersenproducts.com/Specifications/Pipe_Copper.aspx 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.

So the first 6’ of bare pipe looks like this, approximately:

      50.9
68 ---www------- T      I = 68/(50.9+1/3) = 1.32735 Btu/h.
            |           T = I/3 = 0.44245 F.
            |
          | >
        I | > 1/3 
          v >
            |
            |
           0 F

And the first 6’ of insulated pipe looks like this, approximately:

      50.9
68 ---www------- T      I = 68/(50.9+2) = 1.28544 Btu/h.
           |           T = I/2 = 0.64272 F.
            |
          | >
        I | > 0.5 
          v >
            |
            |
           0 F

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.

Meanwhile, http://www.engineeringtoolbox.com/water-content-steel-copper-pipes-tubes-d_1617.html 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.

I wonder which green certification systems require insulating the entire hot water line, and why.

Nick

PS: Do vertical pipe loop heat traps prevent thermosyphoning, or just slow it down?
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