Sorry I haven't been able to get back on this topic as fast as I would have liked to have but I had some personal issues that needed to be handled.
First I think "tc29" did an excellent job of summing up the questions. From the beginning of my post I've tried to communicate that calculations are validated through testing. If the results in testing are different then there's either problems in the testing method or problems with the calculations. It's pretty simple.
So in response to "prime-VTEC" and "4g62bt2c30a" as you'll see it's quite possible to acheive the targeted power using standard engineering equations. As we all know doing so in real world is much more difficult. I think the question about the injectors is also answered through my work and will be validated by Mark's testing.
As I mentioned I had concerns with two factors being used by "4g62bt2c30a", the BSFC and the duty cycle. As you'll see I'm not sure the duty cycle is an issue as Mark has communicated. The BSFC that "4g62bt2c30a" was using I believe is incorrect for the NSX. I calculated back the BSCF for the NSX engine using know published numbers (or at least the ones I could remember). If I've used the wrong numbers (someone please check as I'm in a rush to get this done) then please update/post them with the correct numbers and results. By know means do I think I couldn't have made a mistake.
So here's my analyst.
Definitions
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n = effieciency
w = useful work dine by the process
J = Joule's law coefficent = 778 ft lbf per btu
Q' = heat which flows into the system during the process
P = power
Mf = mass of fuel supplied per unit time
Ma = mass of air supplied per unit time
Qc = heat of the combustion of a unit mass of fuel
F = mass ratio of fuel to air
sfc = specific fuel consumption = fuel consumed per unit of work
sac = specific air consumption = mass of air consumed per unit of work
hp = horse power = J per Kp
Kp = value of 1 hp expressed in (force x length/time) units = 33,000 lb ft min
Kp/J = 42.41645244 Btu / min ( 33,000 / 778)
Qc = heat of combustion per unit mass ~ for gasoline between (18900 & 20,460) ~ 19000
Equations
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n = w / J Q'
n = P / (J * Mf * Qc)
n = (hp * 42.4) / (19000 * Mf)
P = J * Ma * F * Qc * n = J * Mf * Qc * n
sfc = Mf / P = 1 / (J * Qc * n)
sfc = 1 / (J * Qc * n)
sfc = (42.4 *60) / (Qc * n)
sac = Ma / P
sac = 1 / (J * F * Qc * n)
sac = sfc /F
hp = P/Kp
hp = n * Ma * F * Qc * (J / Kp)
hp = ( n * Mf * 19000) / (42.4 *60)
Start by solving for sfc and n for a stock 3.0 L NSX engine using what's known
Stock 3.0L NSX engine
Given Power = 270 hp
Injector size = 240cc
Tested Flow pressure = 43.5 psi
Operating pressure = 50 psi
Effective Flow rate = 257.3070084 cc [Square root of (Operating Pressure / Tested Pressure) * Injector size]
Max Flow rate = 24.50542937 lb/hr [Convert cc to lb/hr - divide cc by 10.5]
Desired Duty Cycle = 80%
Designed Flow rate/injector = 19.60434349 lb/hr [Injector duty cycle * max flow rate]
# of Injectors = 6
Then Mf = 117.626061 lbs/hr [# of Injectors * Design Flow rate]
sfc = 0.435652078 lbm/hp-hr [Mf / Power]
n = 30.75% [(hp * 42.4 * 60) / (19000 * Mf)]
If 3.0L NSX engine is SuperCharged
Given Injector size = 240 cc
Tested Flow pressure = 43.5 psi
Operating pressure = 96psi
Effective Flow rate = 356.535049 cc [Square root of (Operating Pressure / Tested Pressure) * Injector size]
Max Flow rate = 33.95571898 lb/hr [Convert cc to lb/hr - divide cc by 10.5]
Desired Duty Cycle = 80%
Designed Flow rate/injector = 27.16457518 lb/hr [Injector duty cycle * max flow rate]
# of Injectors = 6
Mf = 162.9874511 lbs/hr [# of Injectors * Design Flow rate]
And from stock 3.0L analysis these factors should not change dramatically
sfc = 0.435652078 lbm/hp-hr
n = 31%
then
Power = 374.1229744 hp [(effeciency * Mf * 19000) / (42.4 * 60)]
BTW I the reason why the BSFC is better is exactly because of VTEC and VVIS. It allowed the engineers to optimize the engine design across a wider range of RPM.
The reference material I used came from my old textbook used in my engine design class, ah 20 years ago! You can purchase the latest revision of this book from Amazon.com
http://www.amazon.com/exec/obidos/ASIN/0262700271/002-0965198-5489657
All the above equations are covered in Chapter 1. The last chapter gets into supercharging.
Ak - In response to you question about horse power equal torque and rpm. Yes but when working with combustion engines it's easier and a more common approach to use efficiency from the thermodynamics of a cyclic process i.e burning of the fuel.
It get's a little more difficult from here as I feel like I need to teach Thermodynamics and it's been 20 years since I done much. But I'll try and give you the basics and hopefully wont' screw up too much.
We start with the first equation, n = w / J Q'. Go back and look at the diffinitions. Except we're computing the heat released versus added , w is work done by the system minus the work done on the system and where Q is the heat added to the system minus the heat released by the system.
J has to appear in any equation to maintain deminsional homogeneity when relating heat and work. J is units of work divided by units of heat and since we can express work as force times length we can use this to figure power if we can figure the heat released.
I don't know if that makes any sense but hey I took 4 years of school to understand this stuff and it's been 20 since I've looked at.
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hejo
Lake Oswego, Oregon
95T Blk\Blk SportShift
[This message has been edited by hejo (edited 02 December 2001).]