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<DIV><STRONG><FONT size=2 face="Courier New">"thorthesailor"
<sailorthor@...> wrote:</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> Psychrometric charts are
pretty important to those cargo ships<BR>transiting from one climate to
another...</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Charts are nice for understanding,
but formulas let simple computers<BR>control ventilation automatically...
</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">> ... the mates and captains
aboard ship always had to know<BR>how and when to ventilate the cargo holds to
prevent condensation</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">To COOL a cargo or basement,
ventilate whenever the outdoor air<BR>temp Ta is less than the air near the
cargo Tc. Condensation<BR>can only occur if the dew point Td of the outdoor air
is higher<BR>than Tc. If Ta < Tc and Tc < Td, then Ta < Td, but Td is
always<BR>less than Ta by definition, so we only need to measure the Ta<BR>and
Td temperatures. </FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">To WARM cargo with outdoor air
requires more control. To avoid<BR>condensation, we can only ventilate if Td
< Tc, which requires<BR>measuring the RH fraction of the outdoor air and
calculating its<BR>dewpoint. In Rankine units, Tdr = Tar/(1-Tarln(RH)/9621),
where<BR>ln() is the natural log button on a $9 fx-260 Casio
calculator<BR>(After 1619, mates and captains might have known natural logs if
<BR>not calculators </FONT></STRONG><A
href="http://en.wikipedia.org/wiki/Natural_logarithm"><STRONG><FONT size=2
face="Courier New">http://en.wikipedia.org/wiki/Natural_logarithm</FONT></STRONG></A><STRONG><FONT
size=2 face="Courier New">.) </FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">If Tc = 40 F (Tcr = 500 R) and Ta =
60 F (Tar = 520 R) and<BR>RH = 0.49 (ie 49%), Tdr = 520/(1-520ln(0.49)/9621) =
500.7 R,<BR>so ventilation would cause condensation, but RH = 0.47 would<BR>make
Tdr = 499.6 R < 500 R = Tcr, so ventilation would be
OK.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">To DRY a basement or cargo,
ventilate with outdoor air if its<BR>absolute moisture content (aka humidity
ratio) is less than<BR>the moisture content of the air near the cargo. This
requires<BR>measuring the RH of the air near the cargo as well as the RH<BR>of
the outdoor air. Air at T (R) and RH fraction R has vapor<BR>pressure P =
Re^(17.863-9621/T) "Hg, so we can dry with<BR>outdoor air if Ra <
Rce^(9621(1/Tcr-1/Tar)).</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">For instance, if Tc = 80 F (Tcr =
540) and Ta = 70 F (Tar = 530)<BR>and Rc = 0.6 (RH = 60%), we can dry by
ventilation if <BR>Ra < 0.6e^(9621(1/530-1/540) = 0.84, ie 84%. If Tc = 40
F<BR>and Ta = 100 F, we can dry if Ra < 0.6e^(9621(1/560-1/500))<BR>= 0.08,
ie 8%.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">Air with vapor pressure P "Hg
contains w = 0.62198/(29.921/P-1)<BR>pounds of water per pound of dry air... 60
F (520 R) basement<BR>air at 60% RH has Pc = 0.6e^(17.863-9621/520) = 0.317 "Hg
and<BR>humidity ratio wc = 0.62198/(29.921/Pc-1) = 0.00665. Outdoor<BR>air at 70
F and 30% RH has Pa = 0.225 "Hg and wa = 0.0047.</FONT></STRONG></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2 face="Courier New">With air weighing 0.075 lb/ft^3, a
15 watt 100 cfm fan can remove<BR>100cfmx60m/hx0.075(wc-wa) = 0.88 pounds of
water per hour (21 pints<BR>per day) from the basement with an energy factor of
0.88/2.2/0.015<BR>= 27 liters per kWh, 18 times more efficiently than a 1.5
EF<BR>Energy Star dehumidifier </FONT></STRONG><A
href="http://www.energystar.gov/index.cfm?c=dehumid.pr_basics_dehumidifiers"><STRONG><FONT
size=2
face="Courier New">http://www.energystar.gov/index.cfm?c=dehumid.pr_basics_dehumidifiers</FONT></STRONG></A></DIV>
<DIV><STRONG><FONT size=2 face="Courier New"></FONT></STRONG> </DIV>
<DIV><STRONG><FONT size=2
face="Courier New">Nick</FONT></STRONG></DIV></BODY></HTML>