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<p class=MsoNormal><span style='color:#1F497D'>Hmm… I read this quickly and it
seemed off so I double checked:<o:p></o:p></span></p>
<p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><b><span
style='font-size:10.0pt;font-family:"Times New Roman","serif"'>TANK-UA<o:p></o:p></span></b></p>
<p class=MsoNormal style='mso-margin-top-alt:auto;mso-margin-bottom-alt:auto'><span
style='font-size:12.0pt;font-family:"Times New Roman","serif"'>The overall heat
loss coefficient (UA) of the heater. The heat loss is calculated as this value
times the temperature difference between the water in the heater and the
environmental temperature. <o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'>You got the 90.1 calc right: SL
= 20 + 35*sqrt(240) = 562 <o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'>Rephrasing the DOE2 help entry
copied above: SL = UA * delta-T<o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'>Set: 562 = UA * 70<o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'>Then: UA = 562/70 = 8.03<o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'>As I understand it, TANK-UA rolls
together the properties of both the insulation and surface area, so given
TANK-UA you shouldn’t need to work in the tank volume or area to find the
standby loss for a given delta-T. That’s my take – hope it helps =)!<o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'>~Nick<o:p></o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
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<p class=MsoNormal><b><span style='font-size:12.0pt;font-family:"Stylus BT","sans-serif";
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color:#CC9900'><o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:10.0pt;color:#2D4D5E'>25501 west
valley parkway, suite 200<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:10.0pt;color:#2D4D5E'>olathe, ks
66061<o:p></o:p></span></p>
<p class=MsoNormal><span style='font-size:10.0pt;color:#2D4D5E'>direct
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<p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p>
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<div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0in 0in 0in'>
<p class=MsoNormal><b><span style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>From:</span></b><span
style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> equest-users-bounces@lists.onebuilding.org
[mailto:equest-users-bounces@lists.onebuilding.org] <b>On Behalf Of </b>James
Hansen<br>
<b>Sent:</b> Wednesday, July 06, 2011 10:31 AM<br>
<b>To:</b> equest-users@lists.onebuilding.org<br>
<b>Subject:</b> [Equest-users] my math on DWH efficiency?<o:p></o:p></span></p>
</div>
</div>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>I had a question about calculating baseline 90.1 efficiency
for electric DWHs (tank type).<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>Based on Table 7.8, my efficiency needs to be SL = 20 + 35*sqrt(V)
at a 70 degree delta T between water and ambient, where SL is the standby loss
in btuh, and V is the volume of the tank.<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>If, in my proposed model, I have a 240 gallon tank (meaning
my baseline has the same volume), it would appear that the maximum standby loss
at a 70 degree delta should be 562 btuh [20 + 35 * sqrt(240)]<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>Since eQuest calculates the heat loss by multiplying the UA
(thermal conductivity of storage tank) by the tank, it would appear that my UA
for this particular example would be 562 / (70*240), or 0.0334.<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>That way, at a 70 degree delta, the heat loss (SL) would be
0.0334 Btu/hr-gal-F * 240 gallons * 70 F = 562 btuh<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>Does that sound about right? <o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal>Thanks to anyone that has a quick chance to review!<o:p></o:p></p>
<p class=MsoNormal><o:p> </o:p></p>
<p class=MsoNormal><b><span style='font-family:"Arial","sans-serif";color:teal'>GHT
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</span></b><b><span style='font-family:"Arial","sans-serif"'><ns0:PersonName><span
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