<html xmlns:v="urn:schemas-microsoft-com:vml" xmlns:o="urn:schemas-microsoft-com:office:office" xmlns:w="urn:schemas-microsoft-com:office:word" xmlns:m="http://schemas.microsoft.com/office/2004/12/omml" xmlns="http://www.w3.org/TR/REC-html40"><head><meta http-equiv=Content-Type content="text/html; charset=us-ascii"><meta name=Generator content="Microsoft Word 12 (filtered medium)"><!--[if !mso]><style>v\:* {behavior:url(#default#VML);}
o\:* {behavior:url(#default#VML);}
w\:* {behavior:url(#default#VML);}
.shape {behavior:url(#default#VML);}
</style><![endif]--><style><!--
/* Font Definitions */
@font-face
{font-family:Calibri;
panose-1:2 15 5 2 2 2 4 3 2 4;}
@font-face
{font-family:Tahoma;
panose-1:2 11 6 4 3 5 4 4 2 4;}
/* Style Definitions */
p.MsoNormal, li.MsoNormal, div.MsoNormal
{margin:0in;
margin-bottom:.0001pt;
font-size:12.0pt;
font-family:"Times New Roman","serif";}
a:link, span.MsoHyperlink
{mso-style-priority:99;
color:blue;
text-decoration:underline;}
a:visited, span.MsoHyperlinkFollowed
{mso-style-priority:99;
color:purple;
text-decoration:underline;}
p
{mso-style-priority:99;
mso-margin-top-alt:auto;
margin-right:0in;
mso-margin-bottom-alt:auto;
margin-left:0in;
font-size:12.0pt;
font-family:"Times New Roman","serif";}
span.EmailStyle18
{mso-style-type:personal-reply;
font-family:"Calibri","sans-serif";
color:#1F497D;
font-weight:normal;
font-style:normal;
text-decoration:none none;}
.MsoChpDefault
{mso-style-type:export-only;
font-size:10.0pt;}
@page WordSection1
{size:8.5in 11.0in;
margin:1.0in 1.0in 1.0in 1.0in;}
div.WordSection1
{page:WordSection1;}
--></style><!--[if gte mso 9]><xml>
<o:shapedefaults v:ext="edit" spidmax="1026" />
</xml><![endif]--><!--[if gte mso 9]><xml>
<o:shapelayout v:ext="edit">
<o:idmap v:ext="edit" data="1" />
</o:shapelayout></xml><![endif]--></head><body bgcolor=white lang=EN-US link=blue vlink=purple><div class=WordSection1><p class=MsoNormal style='margin-top:4.0pt'><span style='font-family:"Calibri","sans-serif";color:#1F497D'>Luke,<o:p></o:p></span></p><p class=MsoNormal style='margin-top:4.0pt'><span style='font-family:"Calibri","sans-serif";color:#1F497D'>Pressurizing the intake of the gasifier air seems more complex than running at sub-atmospheric pressure and the adding the pressure boost before the I.C. intake manifold, if a person is power-hungry. Keeping the whole system airtight (for safety) would be an issue and the gas pressure in the gasifier would have to account for all of the pressure drops downstream as well as the manifold boost. Just cooling the gas has a huge effect on its specific volume as we have discussed in the last week. But most onerous may be the airlock on the fuel feed. Controlling primary air with a fuel hopper pressurized at 7PSIG or so feels like a controls nightmare.<o:p></o:p></span></p><p class=MsoNormal style='margin-top:4.0pt'><span style='font-family:"Calibri","sans-serif";color:#1F497D'>The assisted flare, segueing to air being drawn through the system by a big, undersquare, I.C. engine, just fits together too well to want to mess with it much!<o:p></o:p></span></p><p class=MsoNormal style='margin-top:4.0pt'><span style='font-family:"Calibri","sans-serif";color:#1F497D'>Best, Mark<o:p></o:p></span></p><p class=MsoNormal><span style='font-family:"Calibri","sans-serif";color:#1F497D'><o:p> </o:p></span></p><div><div style='border:none;border-top:solid #B5C4DF 1.0pt;padding:3.0pt 0in 0in 0in'><p class=MsoNormal style='margin-left:.5in'><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"'> gasification-bounces@lists.bioenergylists.org [mailto:gasification-bounces@lists.bioenergylists.org] <b>On Behalf Of </b>Luke Gardner<br><b>Sent:</b> Saturday, March 05, 2011 8:35 PM<br><b>To:</b> Discussion of biomass pyrolysis and gasification<br><b>Subject:</b> Re: [Gasification] Benefits of boosting compression ratio<o:p></o:p></span></p></div></div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>Daniel,</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>the math pencils out in the real world nearly as well as on paper.</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>case in point I have an Uncle who I accompanied on many weekend drag boat racing excursions.</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>His boat has a blown alcohol 454 with a very expensive valve train.</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>short and sweet, the blower crams air into the engine at near 30 psig, thus tripling normal atmospheric-- result 1250 hp. 1/4 mile in six seconds, starting at near walking speed, ending at near 170mph.</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'> Back to topic, it seems like people want to boost the engines output via turbo or other means and I have seen much discussion toward that topic, along with all the potential problems. I agree with many that choosing a bigger engine is a simpler way to overcome the power loss of using producer gas instead of petrol. But the reality of it is this; regular people buy regular stuff. Meaning that a generator coupled to an engine is the cost effective package needed to implement in many cases, the problem is they are matched up incorrectly for woodgas use, and in many cases ownership is already established. for most the more cost effective way of utilizing a generators full output is indeed to run it at a higher intake pressure. It seems to me it would be easier to set up the system to pressurize the intake air of the gasifier, thus boosting the engine downstream as well, rather than putting the pump precariously between the reactor and engine. </span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>No condensates to worry about, </span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>No detrimental turbo suction issue to deal with,</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>one could use a portion of this boosted air as secondary air as well.</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>fuel bin would need to be able to handle the boost pressure, and safety lid would need to be different in design,but startups would be nice, and means for switching to non boosted primary air would need to be implemented for "depressurizing" while refueling, at a reduced engine output, but that's pretty harmless.</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>I would be verry surprised, if this is not standard operating procedure for someone, just haven't run across any reports. My limited scope of gasification experience leaves me wondering, any success or failure stories out there with boosting primary air to drive produced gas pressure up as normal operating conditions?</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>Luke Gardner</span><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'> <o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'> <o:p></o:p></p></div><blockquote style='border:none;border-left:solid black 1.5pt;padding:0in 0in 0in 4.0pt;margin-left:3.75pt;margin-top:5.0pt;margin-right:0in;margin-bottom:5.0pt'><div><p class=MsoNormal style='margin-left:.5in'><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>----- Original Message ----- <o:p></o:p></span></p></div><div><p class=MsoNormal style='margin-left:.5in;background:#E4E4E4'><b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>From:</span></b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'> <a href="mailto:dmc@danielchisholm.com" title="dmc@danielchisholm.com">Daniel Chisholm</a> <o:p></o:p></span></p></div><div><p class=MsoNormal style='margin-left:.5in'><b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>To:</span></b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'> <a href="mailto:gasification@lists.bioenergylists.org" title="gasification@lists.bioenergylists.org">Discussion of biomass pyrolysis and gasification</a> <o:p></o:p></span></p></div><div><p class=MsoNormal style='margin-left:.5in'><b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>Sent:</span></b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'> Thursday, March 03, 2011 4:33 AM<o:p></o:p></span></p></div><div><p class=MsoNormal style='margin-left:.5in'><b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'>Subject:</span></b><span style='font-size:10.0pt;font-family:"Arial","sans-serif"'> Re: [Gasification] Benefits of boosting compression ratio<o:p></o:p></span></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>First the answer, then a long-winded explanation if you care for it...<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'> <o:p></o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 width=618 style='width:463.2pt;margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal>Then change the pressure to 5 psig (say ~19 psia) with the same parameters and calculate the mass of oxygen. Do the masses come out with the ratio of 9.8/19 or about 1/2 for my comparison between natural aspirated and (5 psig) low pressure?<o:p></o:p></p></div></td></tr></table></blockquote><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Yes.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'> <o:p></o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 width=618 style='width:463.2pt;margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal> If so, then there is twice the oxygen mass in a 5 psig pressure going into the engine, as compared to 20 in hg (9.8 psia).<o:p></o:p></p></div></td></tr></table></blockquote><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Yes.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>And you could therefore burn twice as much fuel, and therefore produce twice as much gross horsepower. ("gross horsepower" means how much power your pistons are extracting from the gas; most of this will make it to the crankshaft output but some will be lost to sliding friction and pumping losses)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Although your net horsepower output at the crank won't be exactly twice as much (it'll probably be a little bit better than that), it's a very good starting point to make the approximation that it is.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><b>Bottom line: power output is proportional to (absolute) manifold pressure.</b><o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 style='margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal>Change the temperature to 95C for both. Is it the same?<o:p></o:p></p></div></td></tr></table></blockquote><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Yes.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>You get twice as much air mass flow at 19.7psia as you do at 9.8psia.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>You didn't ask this but since you probably know that your engine will make less horsepower with 95C inlet air temperature than with 15C air temperature I'll ask and answer a question here:<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>How much power does an engine make at 95C inlet air temperature versus at 15C inlet air temperature?<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>The first thing to do is to convert these into absolute temperatures, 15C is 288 Kelvin (15+273), and 95C is 368 Kelvin.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Your engine's displacement is unchanged by the inlet air temperature. But the density of the air that it moves is changed, in inverse proportion to the (absolute) temperature ratio:<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>288/368 = 0.78<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>So when your inlet air temp is 95C (368K), your engine will make 78% of the power it makes when your inlet air temp is 15C (288K).<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>This is because it is only flowing 78% as much air (mass-wise), therefore it is only able to burn 78% as much fuel, and therefore there is only 78% as much energy available to be extracted.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 width=618 style='width:463.2pt;margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal>I'm sorry to ask, but as I read the ideal gas law equasions, my eyes glass over and confusion sets in.<o:p></o:p></p></div></td></tr></table></blockquote><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>You'll notice that I never did plug any numbers into "PV=nRT" (or P = rho*R*T) in any of the above.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>A great deal of useful insight can be had from looking at the form of the equations, from this you can get useful relationships like<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>- If temperature is constant, density is proportional to pressure (I used that in the first part)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>- For any particular pressure, density changes inversely proportional to temperature (that's what I used in the second part)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Now to the first part of your email and longer-winded bit..<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>On Wed, Mar 2, 2011 at 14:11, Toby Seiler <<a href="mailto:seilertechco@yahoo.com">seilertechco@yahoo.com</a>> wrote:<o:p></o:p></p><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 width=624 style='width:6.5in;margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal>Mark, Daniel C., list,<o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal>OK I admit that I'm lost when I get to "mole" in the ideal gas law. (A mole is a furry creature my dog brings up on the porch to chew up.)<o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal>So please help me in another way if you understand the physics. First assume I use a 460 cubic inch engine and have a 10/1 compression ratio. In total then all cylinders compress the gas/air to an area of 46 cubic inches...easy enough. At 12/1 it is compressed to 38.3.<o:p></o:p></p></div></td></tr></table><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>There's a small error there which I'll point out for completeness.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>(firstly good for you for choosing a big-block in your example, even if I would have used 440 or 426 cubic inches in my 'fer-instance! :-)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Your pistons *displace* 460 cubic inches, that is when they move from BDC to TDC they have pushed through 460 cubic inches of volume.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>At TDC the gas will be in the combustion chamber volume. At BDC the gas will occupy this volume *PLUS* 460 cubic inches.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>So we're naturally into algebra (sorry! ;-)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>CR = Vol_BDC / Volume_TDC<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>CR= (displacement + CombChmbr_volume) / (CombChmbr_volume)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>After a bit of algebra we get:<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>CombChmbrvolume = displacement/(CR-1)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>For displacement=460 in^3 and and CR=10 we get CombChmbr_volume = 460/9 in^3 or 51.1 in3<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>What is happening is that we start with 511.1 in^3 of gas, which we then compress by 460 in^3 down to 51.1 in^3 (we change the volume by 10:1 ---> our compression ratio!)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>For displacement=460 in^3 and and CR=12 we get CombChmbr_volume = 460/11 in^3 or 41.8 in3 (we compress 501.8 in^3 of gas down to 41.8 in^3, i.e. by 12:1)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 width=618 style='width:463.2pt;margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal>Now help me find the mass of the oxygen (we will disregard the nitrogen, argon, etc. for now),<o:p></o:p></p></div><div><p class=MsoNormal>if my manifold is 20 hg (about 9.8 psi absolute) and I know oxygen weighs 32 grams per mole and I assume 23C (of course it will be higher in a hot engine)...how much oxygen mass is going into the cylinders (with oxygen at 21% of air and for now disregarding the fuel). <o:p></o:p></p></div></td></tr></table></blockquote><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>It's easier if you consider the air, and then when you're done that figure out the oxygen.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>You can calculate how much air flows in one cycle (which in a four stroke engine is two crank rotations, or you can figure how much air flows when the engine is turning at a certain RPM. I'll do the latter since it helps to have in hand some real mass flow numbers for your air, which closely relate to fuel flow rates, which also relate to power outputs - and fuel flow rates and power outputs are something that we probably have a certain intuitive grasp of.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Let's say you are turning your engine at 1800 RPM, a nice respectable synchronous genset speed. So your crank makes 1800 rotations in a minute, which is 900 engine cycles. So your pistons will have displaced 460 cubic inches times 900 cycles which is 414,000 cubic inches. Which is also a number that means very little to me intuitively. It is 414000/1728 = 239.6 cubic feet, which might be a bit easier to grasp physically. Also for anybody who has race engine experience, many carburetors are rated in cfm and here we see that we need a 240cfm carbureter. A typical bigblock might be fitted with a racing carburetor rated for 850cfm or 1050cfm (which is how much air your engine will need at 6120rpm or 7875rpm respectively, which are the sort of RPMs a race engine might be run at)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>So <b>your 460 c.i.d. engine is displacing 239 cubic feet of air per minute when it turns 1800rpm</b>. It displaces, in its cylinders, 239 cubic feet per minute no matter what the throttle setting is and no matter what the intake manifold pressure is.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>In SI units, 239 ft^3/minute is 6.77 cubic metres per minute. What Canadians call a "45 gallon drum" and what Americans call a "55 gallon drum" is about 200 litres which is 0.2 cubic metres. This airflow is about 34 drums per minute, or about half a drum of air per second.<o:p></o:p></p></div><div><div><p class=MsoNormal style='margin-left:.5in'><br>In US Customary units the density of air is .076 pounds per cubic foot (at 59F and atmospheric pressure).<o:p></o:p></p></div></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>So let's say you set your throttle so that the manifold pressure is 20"Hg absolute (on most engine vacuum gauges which go from 0"Hg at atmospheric pressure to 29.92"Hg of vacuum, this will read as 10"Hg of vacuum).<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>At this throttle setting your manifold pressure will be about 9.8psia as you said. This is 20/29.92 = 0.67 times atmospheric pressure.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>When you throttle a gas flow, its temperature remains unchanged, its pressure drops, and its density drops by the same proportion. So if you are throttling your air to 0.67 times atmospheric pressure, its density will be 0.67 times as big.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>So your air mass flow rate is 239 ft^3/minute times .076 pounds/ft^3 times 0.67 = 12.17 pounds of air per minute.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>I did promise to tell you the oxygen flow rate so this is where we do it: 0.21 (pounds of oxygen per pound of air) times 12.17 lbs/minute air = 2.56 pounds per minute of oxygen. <o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>It might be more tangible though to think in terms of air mass flow rates. If we are burning gasoline we'd likely be using an air:fuel mixture ratio somewhere in the range of 12:1 to 15:1. These mixture ratios are "pounds of air per pound of fuel".<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>Since we are flowing 12.17 pounds per minute of air, we'll need to flow about 1 pound per minute of gasoline.<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'>One pound per minute is sixty pounds per hour, which is 10 US gallons per hour. Does this seem "about right" for a 460 c.i.d engine turning 1800rpm and producing about 65% of its wide-open power output at that speed? (It passes my smell test, because it seems "close enough" to me to a 470 in^3 aircraft engine turning 2000-2300RPM, at 65% power setting, burning 12-15 USgph)<o:p></o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'><o:p> </o:p></p></div><div><p class=MsoNormal style='margin-left:.5in'> <o:p></o:p></p></div><blockquote style='border:none;border-left:solid #CCCCCC 1.0pt;padding:0in 0in 0in 6.0pt;margin-left:4.8pt;margin-right:0in'><table class=MsoNormalTable border=0 cellspacing=0 cellpadding=0 style='margin-left:.5in'><tr><td valign=top style='padding:0in 0in 0in 0in'><div><p class=MsoNormal>Best regards, <o:p></o:p></p></div><div><p class=MsoNormal> <o:p></o:p></p></div><div><p class=MsoNormal>Toby <o:p></o:p></p></div></td></tr></table><p class=MsoNormal style='mso-margin-top-alt:0in;margin-right:0in;margin-bottom:12.0pt;margin-left:.5in'><br><br>_______________________________________________<br>Gasification mailing list<br><br>to Send a Message to the list, use the email address<br><a href="mailto:Gasification@bioenergylists.org">Gasification@bioenergylists.org</a><br><br>to UNSUBSCRIBE or Change your List Settings use the web page<br><a href="http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org" target="_blank">http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org</a><br><br>for more Gasifiers, News and Information see our web site:<br><a href="http://gasifiers.bioenergylists.org/" target="_blank">http://gasifiers.bioenergylists.org/</a><o:p></o:p></p></blockquote></div><p class=MsoNormal style='margin-left:.5in'><br><br clear=all><br>-- <br>- Daniel<br>Fredericton, NB Canada<o:p></o:p></p><div class=MsoNormal align=center style='margin-left:.5in;text-align:center'><hr size=2 width="100%" align=center></div><p class=MsoNormal style='margin-left:.5in'>_______________________________________________<br>Gasification mailing list<br><br>to Send a Message to the list, use the email address<br>Gasification@bioenergylists.org<br><br>to UNSUBSCRIBE or Change your List Settings use the web page<br>http://lists.bioenergylists.org/mailman/listinfo/gasification_lists.bioenergylists.org<br><br>for more Gasifiers, News and Information see our web site:<br>http://gasifiers.bioenergylists.org/<o:p></o:p></p></blockquote></div></body></html>