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<DIV><FONT face=Arial size=2>Daniel,</FONT></DIV>
<DIV><FONT face=Arial size=2>the math pencils out in the real world nearly as
well as on paper.</FONT></DIV>
<DIV><FONT face=Arial size=2>case in point I have an Uncle who I accompanied on
many weekend drag boat racing excursions.</FONT></DIV>
<DIV><FONT face=Arial size=2>His boat has a blown alcohol 454 with a very
expensive valve train.</FONT></DIV>
<DIV><FONT face=Arial size=2>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.</FONT></DIV>
<DIV><FONT face=Arial size=2> 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. </FONT></DIV>
<DIV><FONT face=Arial size=2>No condensates to worry about, </FONT></DIV>
<DIV><FONT face=Arial size=2>No detrimental turbo suction issue to deal
with,</FONT></DIV>
<DIV><FONT face=Arial size=2>one could use a portion of this boosted air as
secondary air as well.</FONT></DIV>
<DIV><FONT face=Arial size=2>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.</FONT></DIV>
<DIV><FONT face=Arial size=2>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?</FONT></DIV>
<DIV><FONT face=Arial size=2>Luke Gardner</FONT></DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
<DIV><FONT face=Arial size=2></FONT> </DIV>
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<DIV style="FONT: 10pt arial">----- Original Message ----- </DIV>
<DIV
style="BACKGROUND: #e4e4e4; FONT: 10pt arial; font-color: black"><B>From:</B>
<A title=dmc@danielchisholm.com href="mailto:dmc@danielchisholm.com">Daniel
Chisholm</A> </DIV>
<DIV style="FONT: 10pt arial"><B>To:</B> <A
title=gasification@lists.bioenergylists.org
href="mailto:gasification@lists.bioenergylists.org">Discussion of biomass
pyrolysis and gasification</A> </DIV>
<DIV style="FONT: 10pt arial"><B>Sent:</B> Thursday, March 03, 2011 4:33
AM</DIV>
<DIV style="FONT: 10pt arial"><B>Subject:</B> Re: [Gasification] Benefits of
boosting compression ratio</DIV>
<DIV><BR></DIV>
<DIV>First the answer, then a long-winded explanation if you care for
it...</DIV>
<DIV> </DIV>
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<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>
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<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>
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<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>
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<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>
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<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>
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<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>
<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>
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<DIV>Best regards, </DIV>
<DIV> </DIV>
<DIV>Toby
</DIV></TD></TR></TBODY></TABLE><BR><BR>_______________________________________________<BR>Gasification
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clear=all><BR>-- <BR>- Daniel<BR>Fredericton, NB Canada<BR>
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