weight of air in 1 (cubic foot) = 0.0807 lbs ( Air_density ????? )
Quote:
Originally Posted by
Eric@HPTuners
I just read the PDFs, and I would like thank you again for doing this. Excellent job.
Engine Displacement is the dry air mass to displace the volume of one cylinder. Ford uses to different conditions, depending if the car is N/A or Boosted. The N/A cars use 60°F and 29.95 inHG BP, the Boosted cars use 100°F and 29.95 inHG BP.
Hi Eric@HPTuners,
Engine Displacement is the dry air mass to displace the volume of one cylinder.
Ford uses to different conditions, depending if the car is N/A or Boosted.
Conditions_1 - The N/A cars use 60°F and 29.95 inHG BP,
Conditions_2 - The Boosted cars use 100°F and 29.95 inHG BP.
Question_1 , What does BP stand for at the end of all the conditions ? is this a standard it reverts back to as i have tried to work what BP means and haven't worked it out.
Question_2 , What value do you have for air density at those settings that affects cylinder fill ?
My calculations are base on air density Wikipedia encyclopedia and compared to your N/A 60°[email protected] inHG BP, and Boosted 100°[email protected] inHG BP , one cylinder fill will vary depending what part of the world one lives? This is just bugging me and a issue for me as my calculations are anything between 2.5% to 3% lower &/or higher values. I tried these calculation for the Australian ford engines GT335 coyote(USA) V8 5lt bore 92.2mm and stroke 92.7mm , one cylinder volume I calculated is 0.0017394 approx 2.8% difference ?????
Ford Australia Barra engines I6- Bore: 92.26 mm (3.6 in) X Stroke: 99.31 mm (3.9 in) = Displacement: 3984 cc (243 cu in) Ford rounds of to 4lt.
The Boss 290 4-valve DOHC 5.4 L V8, 389 hp (290 kW) @ 5500 rpm, 384 lb·ft (521 N·m) @ 4500 rpm I haven't bother doing a calculation as i was ripping out what little hair i have left. Bore: 90.2 mm (3.552 in) X Stroke: 105.8 mm (4.165 in) which will also have a different result.
" The density of air, ρ (Greek: rho) (air density), is the mass per unit volume of Earth's atmosphere. Air density, like air pressure, decreases with increasing altitude. It also changes with variation in temperature or humidity. At sea level and at 15 °C, air has a density of approximately 1.225 kg/m3 (0.001225 g/cm3, 0.0023769 slug/ft3, 0.0765 lbm/ft3) according to ISA (International Standard Atmosphere)."
weight of air in 1 (cubic foot) = 0.0807 lbs ( Air_density ????? )
1 (cubic foot) = 28.3168466 litres
One cubic meter equals 35.3 cubic feet
One cubic meter also equals 1000 liters or one million cubic centimeters.)
Sorry for the mumbo jumbo question but it is bugging me LOL:banghead: :doh: :shrug:
Cheers
MumboJumbo “Wracking”or“Racking”my brain out -That Big Orange Heavy Thing By Paul Yaw
Sill “Wracking” or “Racking” my brain out? and the little percentage that i use of it, more reading and more reading " That Big Orange Heavy Thing " by Paul Yaw@InjectorDynamics great read but would of liked to have seen more BOOBIES ;) :P cause when I READ playboy magazines its the other way around. OOPS sorry its a G-rate forum, back on my delamar.
This is all cut'n'paste of Paul's works that has shed some light on me being so anal inattentive fixated on a number.
Quote:
-That Big Orange Heavy Thing By Paul Yaw - " you get no boobies! That’s right, just math and physics. Suck it up boys, and prepare to crunch some numbers.
Ford states displacement as lbs per induction event at 100% VE, under standard temperature and pressure conditions. Just not the standard temperature and pressure conditions you may be used to. Not SAE J607 (60 F, 29.921 inHg) not SAEJ1349 (77 F, 29.234 inHg) not even the more obscure J1995 (77 F, 29.53 inHg) No…that would be too much like everyone else. Ford’s standard is 100 F, and 29.921 inHg.
if you were wondering how to calculate displacement Ford style, here it is.
1. – Calculate the density of dry air at 100F and 29.921 inHg.
1.326 * (29.921 / (100 + 459.7)) = .07089 lbs/ft^3
2. – Multiply this by the single cylinder volume displacement in ft^3.
For a 330 cubic inch, 8 cylinder engine – (330 / 8) / 1728 = .02388 ft^3
3. Multiply air density by single cylinder volume displacement.
.07089 * .02388 = .001692 lb
Simplified – (Displacement in cubic inches / Number of Cylinders) / 24392
So now we get to the final steps. The steps that take us to the old familiar VE tables
Ford’s version of speed density is quite clever, ignoring volumetric efficiency and getting straight to the point, which is air mass. After all, we don’t calculate air fuel ratio based on volume, we calculate air fuel ratio based on mass. So why bother with a volume term?"
This article is worth a read, amongst many other topics Paul Yaw written articles.
Head down bum up more work for me to warp my little brain about Ford HPTuner VCM suite.
cheers
Volumetric cylinder fill VE's should change dramatically.
Hi IH8TOADS & DarrylC , thanks guys , yep read that whole thread/posts 4 times :O
Atmospheric barometric pressure, air temperature, humidity, sea level to higher altitudes all affect cylinder fill so the Volumetric cylinder fill VE's should change dramatically.
Just getting my head around it LOL
Cheers