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</o:shapelayout></xml><![endif]--></head><body lang=EN-CA link=blue vlink=purple><div class=WordSection1><p class=MsoNormal><span style='color:#1F497D'>Dear Frank<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'>What a great question.<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'>></span><span lang=EN-US style='color:#1F497D'>It seems that there may be enough moisture in the air to be a factor in cooling considering all the air that is needed? </span><span lang=EN-US style='color:#1F497D'><o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Although we talk (sometimes) about the preheating needed for clean combustion, and we also mention preheated secondary air quite frequently, it is not common to hear about the amount of heat (in Watts) required to do this heating. As you have just shown indirectly, the quantum of heat required to get air hot enough for it to be a significant factor in clean combustion varied dramatically with humidity.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>What I suggest as the path to understanding this well is to compare the heat needed to increase the temperature a cubic meter of air and that required to heat a cubic metre of moist air. That should give everyone a view of the scale of the problem/issue.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Let’s take air at 1.2 kg per cubic metre which it is at a certain altitude and temperature and it is a nice round number. The heat capacity of dry air is x1x per kg (that x1x is a place holder for Kevin!)<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Add 10 g of water vapour. The heat capacity of water vapour is x2x per g.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Let’s say we want to heat the air from 20 to 420 C, an increase of 400 degrees<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>1.2 kg * x1x * 400 + 10g * x2x * 400 = Heat required (Joules)<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Flow rate in cubic metres per second = x3x<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Heat required = x3x * Heat required per second (Joules) = Watts required.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>I have found that a modest stove needs about 3-500 Watts for air preheating.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Kevin, please filling in the variables and let’s see what the power requirement is. <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Let’s burn 0.15 grams of wood per second with an excess air ratio of 200%. That means we need a total of 3 times as much air as the fuel appears to need for complete (‘stoichiometric’) combustion.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Thanks<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Crispin<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Especially when we get the noise level of fuel and un-even combustion under control.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>I was just wondering if there was a way to easily determine the amount of actual moisture in the air should we want to add that to the calculations.<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Regards<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Frank<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Frank Shields<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>42 Hangar Way<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>Watsonville, CA 95076<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>(831) 724-5244 tel<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'>(831) 724-3188 fax<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><a href="mailto:frank@bioCharlab.com">frank@bioCharlab.com</a><o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'> <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US style='color:#1F497D'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0mm 0mm 0mm'><p class=MsoNormal><b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'>From:</span></b><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'> <a href="mailto:stoves-bounces@lists.bioenergylists.org">stoves-bounces@lists.bioenergylists.org</a> <a href="mailto:[mailto:stoves-bounces@lists.bioenergylists.org]">[mailto:stoves-bounces@lists.bioenergylists.org]</a> <b>On Behalf Of </b>Crispin Pemberton-Pigott<br><b>Sent:</b> Tuesday, December 20, 2011 7:57 PM<br><b>To:</b> 'Discussion of biomass cooking stoves'<br><b>Subject:</b> Re: [Stoves] Water vapor calc.<o:p></o:p></span></p></div></div><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>Dear Frank<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'>Humidity is just a way of expressing the moisture content of the air. Instead of expressing it in % (of what it <i>could</i> hold without condensing at that temperature and pressure) one could express it as the mass of water per cubic metre of air at that temperature and pressure. For example you could say 14 g of water vapour per m<sup>3</sup>. <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'>Cooling that air causes the moisture to release its latent heat, warming the air (opposing the cooling) causing it to cool more slowly. If the humidity is really low, air will drop in temperature very rapidly at night.<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 could calculate (in absolute terms) the quantum of energy in any particular cubic metre of air if you know the temperature, pressure and mass of water vapour in it.<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'>Regards<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'>Crispin<o:p></o:p></span></p><p class=MsoNormal><span style='color:#1F497D'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0mm 0mm 0mm'><p class=MsoNormal><span style='color:#1F497D'>+++++++</span><span lang=EN-US style='font-size:10.0pt;font-family:"Tahoma","sans-serif"'><o:p></o:p></span></p></div></div><p class=MsoNormal><o:p> </o:p></p><p class=MsoNormal><span lang=EN-US>Dear Stovers,<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>A few questions if you don’t mind. <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>Enthalpy of water is 104.73 kj/kg at 25 deg. C<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>We need to add 4.186 kj/kg/deg. C (4.186 X 75) 314 to equal 419 to reach 100 deg. C<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>We need to add (2675 – 419) =2256 kj/kg to get to a vapor. And have enthalpy of 2675 kj/kg at 100 deg. C. <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>But then the water vapor goes up to 400 deg. C in a stack. <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>So we add 1.72 kj/kg / deg. C or 300 X 1.72 = 516. So 516 + 2675 = 3191 kj/kg in the water vapor at 400 deg C. (?)<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>When the vapor cools off it releases a lot of energy. Its interesting to me that water vapor at 25 deg. C has so much more energy than water in a liquid! <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>How does humidity calculate into all this? <o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>Thanks<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>Frank<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US>Frank Shields<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>42 Hangar Way<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>Watsonville, CA 95076<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>(831) 724-5244 tel<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US>(831) 724-3188 fax<o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><a href="mailto:frank@bioCharlab.com">frank@bioCharlab.com</a><o:p></o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p><p class=MsoNormal><span lang=EN-US><o:p> </o:p></span></p></div></body></html>