Daniel,
the math pencils out in the real world nearly as well as on paper.
case in point I have an Uncle who I accompanied on many weekend drag boat
racing excursions.
His boat has a blown alcohol 454 with a very expensive valve train.
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.
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.
No condensates to worry about,
No detrimental turbo suction issue to deal with,
one could use a portion of this boosted air as secondary air as well.
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.
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?
Luke Gardner
----- Original Message -----
From: Daniel Chisholm
To: Discussion of biomass pyrolysis and gasification
Sent: Thursday, March 03, 2011 4:33 AM
Subject: Re: [Gasification] Benefits of boosting compression ratio
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.
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.
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 <[email protected]> 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! ;-)
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. 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.
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)
Best regards,
Toby
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--
- Daniel
Fredericton, NB Canada
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