<meta http-equiv="content-type" content="text/html; charset=utf-8"><div>First the answer, then a long-winded explanation if you care for it...</div><div> </div><blockquote class="gmail_quote" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; padding-left: 1ex; ">
<table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font: inherit; "><div>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?</div>
</td></tr></tbody></table></blockquote><div><br></div><div>Yes.</div><div> </div><blockquote class="gmail_quote" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; padding-left: 1ex; ">
<table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font: inherit; "><div> 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).</div>
</td></tr></tbody></table></blockquote><div><br></div><div>Yes.</div><div><br></div><div>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)</div>
<div><br></div><div>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.</div>
<div><br></div><div><b>Bottom line: power output is proportional to (absolute) manifold pressure.</b></div><div><br></div><div><br></div><blockquote class="gmail_quote" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; padding-left: 1ex; ">
<table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font: inherit; "><div> </div><div>Change the temperature to 95C for both. Is it the same?</div></td></tr></tbody></table></blockquote>
<div>
<br></div><div>Yes.</div><div><br></div><div>You get twice as much air mass flow at 19.7psia as you do at 9.8psia.</div><div><br></div><div>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:</div>
<div><br></div><div>How much power does an engine make at 95C inlet air temperature versus at 15C inlet air temperature?</div><div><br></div><div>The first thing to do is to convert these into absolute temperatures, 15C is 288 Kelvin (15+273), and 95C is 368 Kelvin.</div>
<div><br></div><div>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:</div><div><br>
</div><div>288/368 = 0.78</div><div><br></div><div>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).</div><div><br></div><div>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.</div>
<div><br></div><div><br></div><blockquote class="gmail_quote" style="margin-top: 0px; margin-right: 0px; margin-bottom: 0px; margin-left: 0.8ex; border-left-width: 1px; border-left-color: rgb(204, 204, 204); border-left-style: solid; padding-left: 1ex; ">
<table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font: inherit; "><div> </div><div>I'm sorry to ask, but as I read the ideal gas law equasions, my eyes glass over and confusion sets in.</div>
</td></tr></tbody></table></blockquote><div><br></div><div>You'll notice that I never did plug any numbers into "PV=nRT" (or P = rho*R*T) in any of the above.</div><div><br></div><div>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</div>
<div><br></div><div>- If temperature is constant, density is proportional to pressure (I used that in the first part)</div><div>- For any particular pressure, density changes inversely proportional to temperature (that's what I used in the second part)</div>
<div><br></div><div><br></div><div>Now to the first part of your email and longer-winded bit..</div><div><br></div><div class="gmail_quote">On Wed, Mar 2, 2011 at 14:11, Toby Seiler <span dir="ltr"><<a href="mailto:seilertechco@yahoo.com">seilertechco@yahoo.com</a>></span> wrote:<br>
<blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font:inherit"><div>Mark, Daniel C., list,</div>
<div> </div>
<div>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.)</div>
<div> </div>
<div>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.</div>
</td></tr></tbody></table></blockquote><div><br></div><div>There's a small error there which I'll point out for completeness.</div><div><br></div><div>(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! :-)</div>
<div><br></div><div>Your pistons *displace* 460 cubic inches, that is when they move from BDC to TDC they have pushed through 460 cubic inches of volume.</div><div><br></div><div>At TDC the gas will be in the combustion chamber volume. At BDC the gas will occupy this volume *PLUS* 460 cubic inches.</div>
<div><br></div><div>So we're naturally into algebra (sorry! ;-)</div><div><br></div><div>CR = Vol_BDC / Volume_TDC</div><div>CR= (displacement + CombChmbr_volume) / (CombChmbr_volume)</div><div><br></div><div>After a bit of algebra we get:</div>
<div><br></div><div>CombChmbrvolume = displacement/(CR-1)</div><div><br></div><div>For displacement=460 in^3 and and CR=10 we get CombChmbr_volume = 460/9 in^3 or 51.1 in3</div><div><br></div><div>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!)</div>
<div><br></div><div>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)</div><div><br></div><div><br></div><div><br></div>
<div><br></div><div><br></div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font:inherit">
<div> </div>
<div>Now help me find the mass of the oxygen (we will disregard the nitrogen, argon, etc. for now),</div>
<div>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). </div>
</td></tr></tbody></table></blockquote><div><br></div><div>It's easier if you consider the air, and then when you're done that figure out the oxygen.</div><div><br></div><div>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.</div>
<div><br></div><div>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)</div>
<div><br></div><div>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.</div>
<div><br></div><div>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.</div>
<div><meta http-equiv="content-type" content="text/html; charset=utf-8"><div><br class="Apple-interchange-newline">In US Customary units the density of air is .076 pounds per cubic foot (at 59F and atmospheric pressure).</div>
</div><div><br></div><div>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).</div>
<div><br></div><div>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.</div><div><br></div><div>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.</div>
<div><br></div><div>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.</div><div><br></div><div>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. </div>
<div><br></div><div>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".</div>
<div><br></div><div>Since we are flowing 12.17 pounds per minute of air, we'll need to flow about 1 pound per minute of gasoline.</div><div><br></div><div>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)</div>
<div><br></div><div> </div><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex;"><table cellspacing="0" cellpadding="0" border="0"><tbody><tr><td valign="top" style="font:inherit">
<div>Best regards, </div>
<div> </div>
<div>Toby </div></td></tr></tbody></table><br>
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