[Gasification] Benefits of boosting compression ratio

[Arnt Karlsen]

On Thu, 3 Mar 2011 12:33:41 +0000, Daniel wrote in message 
<AANLkTikYU886i8dOZAex8yKCmOUTHSX4Kg_Knj7Aj=4U at mail.gmail.com>:

> First the answer, then a long-winded explanation if you care for it...
> 
> 
> > 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?
> >
> 
> Yes.
> 
> 
> >   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).
> >
> 
> Yes.
> 
> 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)
> 
> 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.
> 
> *Bottom line: power output is proportional to (absolute) manifold
> pressure.*
> 
> 
> 
> > Change the temperature to 95C for both.  Is it the same?
> >
> 
> Yes.
> 
> You get twice as much air mass flow at 19.7psia as you do at 9.8psia.

..to be pedantic; your gas flow losses will more than double, 
which again means you'll get a bit less than twice the gas 
mass flow, on doubling the pressure.  Easily fixed, just crank 
up the boost pressure 'till you get what you want. ;o)

> 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:
> 
> How much power does an engine make at 95C inlet air temperature
> versus at 15C inlet air temperature?
> 
> The first thing to do is to convert these into absolute
> temperatures,  15C is 288 Kelvin (15+273), and 95C is 368 Kelvin.
> 
> 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:
> 
> 288/368 = 0.78
> 
> 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).
> 
> 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.

..educational point: is why gas cooling is useful. ;o)
WWII wood gas Swedes liked ~50C (~120F) in the winter, to 
avoid gas piping and filter icing, according to Gengas.
 
> > I'm sorry to ask, but as I read the ideal gas law equasions, my
> > eyes glass over and confusion sets in.
> >
> 
> You'll notice that I never did plug any numbers into "PV=nRT" (or P =
> rho*R*T) in any of the above.
> 
> 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
> 
> - If temperature is constant, density is proportional to pressure (I
> used that in the first part)
> - For any particular pressure, density changes inversely proportional
> to temperature (that's what I used in the second part)
> 
> 
> Now to the first part of your email and longer-winded bit..
> 
> On Wed, Mar 2, 2011 at 14:11, Toby Seiler <seilertechco at yahoo.com>
> wrote:
> 
> > Mark, Daniel C., list,
> >
> > 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.)
> >
> > 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.
> >
> 
> There's a small error there which I'll point out for completeness.
> 
> (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! :-)
> 
> Your pistons *displace* 460 cubic inches, that is when they move from
> BDC to TDC they have pushed through 460 cubic inches of volume.
> 
> At TDC the gas will be in the combustion chamber volume.  At BDC the
> gas will occupy this volume *PLUS* 460 cubic inches.
> 
> So we're naturally into algebra (sorry! ;-)

..bull, you're enjoying it. ;oD

> CR = Vol_BDC / Volume_TDC
> CR= (displacement + CombChmbr_volume) / (CombChmbr_volume)
> 
> After a bit of algebra we get:
> 
> CombChmbrvolume = displacement/(CR-1)
> 
> For displacement=460 in^3 and and CR=10 we get CombChmbr_volume =
> 460/9 in^3 or 51.1 in3
> 
> 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!)
> 
> 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)     
> 
> 
> 
> 
> 
> 
> > Now help me find the mass of the oxygen (we will disregard the
> > nitrogen, argon, etc. for now),
> > 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).
> >
> 
> It's easier if you consider the air, and then when you're done that
> figure out the oxygen.
> 
> 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.
> 
> Let's say you are turning your engine at 1800 RPM, a nice respectable
> synchronous genset speed.

..which is nice for 4-pole synchronous gensets.  Most people will find
asynchronous gensets much easier and cheaper to make asynchronous
squirrel cage motors, and to profit from on hooking up to the grid.  

..running an asynchronous genset made from an asynchronous squirrel
cage motor at plate speed, makes it an expensive smoky noisy gas pump.
Adding speed up to synchronous grid speed, makes it cheaper, because 
it now only draws magnetizing power from grid.   

..adding more speed, e.g. till the meter reads zero, means you feed
power back into the grid, as the meter reads zero, you match what you
draw from the grid, once you match the plate voltage and amperage
figures, you're finding out what and how much spank it will take 
from your shaft and the grid.

..a good starting point can be had looking at the plate rpm figure 
and add the sag to the grid speed, a typical 4 pole motor may have
a plate speed of say 1750rpm, 60Hz grid is 1800rpm, and you add the
"negative sag" to get 1850rpm, where you will find you are a bit 
below the plate amperage numbers; magnetizing power and losses.

..once you find your rpms, e.g. 1811 to 1877.12, you will want 
to tune it and put your peak at your most profitable high end 
output rpm, that could be say 1875.666rpm. 

..at idle, we only need it to easily tick over and to easily 
speed up past grid speed to only there pick up load.

..another way is "heat it up, spin it up to e.g. 1777rpm, 
turn on everything it needs to spool up past the grid 
and let it spool up and pick up load."  
It does not need to be capable of running under plate speed, 
to be useful and profitable feeding the grid, it _can_ be 
started on grid power. ;o)

>  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)
> 
> So *your 460 c.i.d. engine is displacing 239 cubic feet of air per
> minute when it turns 1800rpm*.  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.
> 
> 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.
> 
> In US Customary units the density of air is .076 pounds per cubic
> foot (at 59F and atmospheric pressure).
> 
> 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).
> 
> 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.
> 
> 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.
> 
> 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.

..more fun can be had here: this gas mix flows thru manifold channels,
sized to fit the valve port holes at worst, still dictating flow rates
that would land you in jail if you try match them on roads other than
German Autobahns, and sometimes, there too.

..similar speeds can also be had hoisting your ride way up high and 
cutting it loose.  Similar deformation zone displacements can be had
impacting some hard vertical road side surface, and right there you
are ready to solidarically share deformation zone displacement space
with the next idiot rear ending you to help you build up the pile-up
pressure.

..such pile-up pressures is what gas molecules may endure, when 
some wound up valve spring dives down the cam lobe to slam shut 
its valve.  
Skilful strategic gas chamber tuners puts such pile-ups inside 
the cylinders to maximize gas etc flow efficiency in normal utility
service settings, and upstream in the manifold piping, to maximize 
fuel etc economy and minimize wear on idle power settings. 

..now, do we have skilful tacticians volunteering the math? ;o)

> 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.
> 
> 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".
> 
> Since we are flowing 12.17 pounds per minute of air, we'll need to
> flow about 1 pound per minute of gasoline.
> 
> 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)


-- 
..med vennlig hilsen = with Kind Regards from Arnt Karlsen
...with a number of polar bear hunters in his ancestry...
  Scenarios always come in sets of three: 
  best case, worst case, and just in case.