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<p class=MsoNormal><span lang=EN-US style='font-size:10.0pt;font-family:"Arial","sans-serif";
color:black'>Cheney, can you post the chapter in the ASHRAE Handbook of
Fundamentals where I can find the equation<o:p></o:p></span></p>
<p class=MsoNormal><span lang=EN-US style='font-size:10.0pt;font-family:"Arial","sans-serif";
color:black'><o:p> </o:p></span></p>
<p class=MsoNormal><span lang=EN-US style='font-size:10.0pt;font-family:"Arial","sans-serif";
color:black'>Johannes</span><span lang=EN-US style='font-size:8.0pt;font-family:
"Arial","sans-serif";color:#29457B'><o:p></o:p></span></p>
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<br>
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<p class=MsoNormal style='margin-left:35.4pt'><b><span style='font-size:10.0pt;
font-family:"Tahoma","sans-serif"'>Von:</span></b><span style='font-size:10.0pt;
font-family:"Tahoma","sans-serif"'> bldg-sim-bounces@lists.onebuilding.org
[mailto:bldg-sim-bounces@lists.onebuilding.org] <b>Im Auftrag von </b>Cheney<br>
<b>Gesendet:</b> Mittwoch, 6. Oktober 2010 22:43<br>
<b>An:</b> creations.deepa@gmail.com<br>
<b>Cc:</b> Bldg-Sim<br>
<b>Betreff:</b> Re: [Bldg-sim] Negative pressure for EQ cr 5<o:p></o:p></span></p>
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<p class=MsoNormal style='margin-left:35.4pt'><o:p> </o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>Hi Deepa, <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>My two cents. <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>For your question, you need to
know: <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>1. the air volume in room
and its pressure<o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>2. To achieve 5Pa, how much more
air should be introduced - <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>Higher pressure x air volume in
room/atmospheric pressure <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>Now you get the number for the
static condition... ;)<o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>to maintain the pressure in
a leaky room you have to pump in more air, right? <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>3. from the ASHRAE Handbook
of Fundamentals:<br>
<br>
Q = 2610 x A x (DP)1/2<br>
<br>
• Q is the room leakage flow rate in cfm,<br>
• 2610 is a conversion factor,<br>
• A is the net open crack area of the room (generally not measurable), and<br>
• DP is the differential pressure.<o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>4. If you'd like to maintain
certain pressure in the room, you need to supply 2+3 first. When the
pressure is established in dynamic condition, you need to supply only 3 to
maintain the pressure you want. <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'> <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>This is only the theoretical
calculation for your reference. <o:p></o:p></p>
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<p class=MsoNormal style='margin-left:35.4pt'>Cheers, <br>
Cheney <o:p></o:p></p>
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