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Volumetric Effeciency

  1. Mar 13, 2009 #1
    I'm in an arguement with some other car nuts, about whether volumetric effeciency can exceed 100%. I think that if you're getting a number over 100%, you miscalculated something or left a variable out. They think that a value over 100% is completely normal especially in turbocharged and supercharged engines. If they are right and I am wrong, what the hells the point of measuring it's effeciency if it can go above 100%? Would this avoid the whole conservation of energy thing haha?
     
  2. jcsd
  3. Mar 14, 2009 #2
    volumetric efficiency doesn't relate to energy at all. It can be over 100%, if certain "things" outweigh others, say ram effect is soo much at high speeds that it pushes more air in.
     
  4. Mar 14, 2009 #3

    Ranger Mike

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    Internal combustion engines are basically one big air pumps
    the ratio of the actual air capacity to ideal air capacity is commonly known as " Volumetric Efficiency"

    If an engine ran slowly enough so that there was no pressure loss in the inlet system,with inlet and exhaust manifold pressures equal, valve openings and closings at top and bottom dead center, and inlet air temperature high enough so that no temperature rise would take place on the way to the cylinders then inlet air density, air capacity of the cylinder and volumetric efficiency would be 100 percent.

    Volumetric Efficiency will vary due to inertia and fluid friction if the gases in the inlet and exhaust manifolds, valve timing, inadequacies of the pressure through the system. Inlet and exhaust temperatures vary.
    it gets close when you get direct port injection. equal length headers, stainless steel to keep in the heat but air still heats up the closer to the cylinder it gets..versus a static capacity air capacity.
    you can " squeeze " the charge with a blower or turbo charger..but
    detonation is the ultimate determiner of efficency.
    don't forget , you are compressing the fuel / air charge by a factor of 8 to 10 (street compressions ratios) and maybe 14 to one..we ran Nascar one year with this squeeze...
    normally aspriated...when you add a supercharger , you got to reduce compression ratio to about 4 to 6 to one..depending on the blower.
    finally, the piston rings used today are good but even in the best race prepared mills, you have at best 3 percent leakage thru the piston, ring combo.

    food for thought..racers!!
     
  5. Mar 14, 2009 #4
    Volumetric efficiency can be over 100%, as the equation is:

    VE= 2*volume flow rate/displaced vol*N

    Supercharging and ram effects allow the cylinder to fill above what it could normally suck in. Its even possible to get a naturally aspirated engine above 100% with a well designed airbox and appropriate tuning.

    To answer your second question; The Volumetric efficiency its not really an efficiency at all really, as its not a useful output/total input. Its would be more accurate to describe it as "the ratio of volumetric air flow into the cylinder to the volume of air it could naturally draw in under static conditions".

    Of course efficiency is far more snappy to use in everyday language.
     
    Last edited: Mar 15, 2009
  6. Mar 15, 2009 #5

    Ranger Mike

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    Venik did you get your question answered?
     
  7. Mar 15, 2009 #6

    brewnog

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    Good answers here guys.
     
  8. Mar 16, 2009 #7

    Ranger Mike

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    I agree ...excellent learning site!

    In a four-stroke naturally aspirated engine, the theoretical maximum volume of air that each cylinder can take in during the intake cycle is equal to the swept volume of that cylinder (0.7854 x bore x bore x stroke).

    Since each cylinder has one intake stroke every two revolutions of the crankshaft, then the theoretical maximum volume of air it can ingest during each rotation of the crankshaft is equal to one-half its displacement. The actual amount of air the engine takes in compared to the theoretical maximum is called volumetric efficiency (VE). An engine operating at 100% VE is using its total displacement every two crankshaft revolutions.

    Its all about the air pimp as I said many times. Its how much air / fuel mixture you can stuff into the cylinder... That mass is directly proportional to (a) the air density and (b) the volumetric efficiency. There is a VERY close relationship between an engine's VE curve and its torque curve.

    In Old tech naturally-aspirated, two-valve-per-cylinder, pushrod engines a VE over 95% is excellent, and 100% is achievable, but extremely difficult to attain. Only full blown race prepped engine reach 110%, and that is by means of extremely expensive specialized intake manifolds, combustion chambers, exhaust header technology and super trick valve system components. The practical limit for normally-aspirated engines, typically DOHC layout with four or more valves per cylinder, is about 115%, which can only be achieved under the most highly-developed conditions, with precise intake and exhaust passage tuning.

    Generally, the RPM at peak VE is same as the RPM at the torque peak. Automotive engines rarely exceed 90% VE. There is a lot of good reasons for that performance, including the design requirements for automotive engines (good low-end torque, good throttle response, high mileage, low emissions, low noise, inexpensive production costs, restrictive form factors, etc.), as well as the allowable tolerances for components in high-volume production.

    For a known engine displacement and RPM, you can calculate the engine airflow at 100% VE, in sea-level-standard-day cubic feet per minute (scfm) as follows:

    100% VE AIRFLOW (scfm) = DISPLACEMENT (ci) x RPM / 3456

    Using that equation to evaluate a 540 cubic-inch engine operating 2700 RPM reveals that, at 100% VE, the engine will flow 422 SCFM.

    If we know how to calculate the fuel flow required for a given amount of power produced, we can calculate the mass airflow required for that amount of fuel, then by using that calculated airflow along with the engine displacement, the targeted operating RPM, and the achievable VE values, we can pretty much determine the resulting performance.

    Once we know the required fuel flow, we can determine the required airflow. Assumes 12.5 air fuel mixture.

    Using that best-power air-to-fuel ratio, you calculate required airflow:

    MASS AIRFLOW (pph) = 12.5 (Pounds-per-Pound) x FUEL FLOW (pph)

    But airflow is usually discussed in terms of volume flow (Standard Cubic Feet per Minute, SCFM). One cubic foot of air at standard atmospheric conditions (29.92 inches of HG absolute pressure, 59°F temperature) weighs 0.0765 pounds. So the volume airflow required is:

    AIRFLOW (scfm) = 12.5 (ppp) x FUEL FLOW (pph) / (60 min-per-hour x 0.0765 pounds per cubic foot)

    That equation reduces to:

    REQUIRED AIRFLOW (scfm) = 2.723 x FUEL FLOW (pph)

    OK, hang on. The really useful stuff is next.

    If I know my FUEL FLOW, I get:

    FUEL FLOW (pph) = HP x BSFC

    Replacing "FUEL FLOW" in with "HP x BSFC" :

    REQUIRED AIRFLOW (scfm) = 2.723 x HP x BSFC

    Now, we can estimate the airflow required for a given amount of horsepower, we can calculate the 100% VE airflow your engine can generate at a known RPM.

    Combining those equations will give us one equation which tells us how close we are to our performance goal by knowing Requited HP, RPM, CID and BSFC

    REQUIRED VE = ( 9411 x HP x BSFC ) / (DISPLACEMENT x RPM)

    Here is an example of how useful that relationship can be. Suppose you decide that a certain 2.2 liter engine would make a great Mini Stock powerplant. You decide that 300 HP is a nice number, and 5200 RPM produces an acceptable mean piston speed

    The required VE for that engine will be:

    Required VE = (9411 x 300 x .45 ) / (134 x 5200 ) = 1.82 (182 %) no way..no how...

    Clearly that's not going to happen with a normally aspirated engine. Here's another example. Suppose you want 300 HP from a 540 cubic inch engine at 2700 RPM, and assume a BSFC of 0.45. Plugging the known values into the equation =

    Required VE = (9411 x 300 x .45 ) / (540 x 2700) = 0.87 (87 %)

    hope this clears things up a little
    I'm out of beer..later
     
  9. Mar 16, 2009 #8
    Well that was almost tl;dr in this modern day of the internets, but that was a brilliant post Mike.

    It also quite nicely answers another question on here somewhere, about how designers choose power output and rpm.

    Edit: Which i've now just read and a nice post there too.
     
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