using a 2.00 inch by 14 inch round element in a drop base AC with flat top. 28 square inch of surface area AC. top of AC clears inj pod by 1/4 inch. top plate of AC and **** clears the hood by 1/4 inch. base of AC clears linkage by 1/4 inch!. perfect. KN free flow top hits hood. cant use! will this small AC choke engine?. car not yet running. tommorrow maybe? i am aware ALDL/VE/MAP will tell me. any better suggestions? only other option is that low profile Holley i see with two sidemount KN's. expensive. sawzall is not an option.

 

Datalog the MAP at WOT and see what values you get. Id say a pressure drop over 3" of mercury with the cleaner on and it should go.

 

back at ya. think 90 map indicates a restriction? 95 map is adequate?

 

that would depend, you would of course need to be in a high gear, were the engine is under full load.

i pull about 93kpa in 1st and 2nd, 98-99 in 3rd.

you might give it a run with no AC, and then with AC and compare the diffrence.

 

good point. planning on a cold air kit. need to confiqure a round dome on top of the 5.175 aluminum tube(that is easy part) that sits on 454 TB. dont want a flat top inverted coffee can. theni will then run two outlets of say 2.5 inch each(like dual snorkel) to a Y to a vette cold air tpi (slp) kit in front of radiator. ebay $140.

 

93 kPa under worst-case load, with assumed 101.3 kPa for 1 atm. So pressure delta is 8 kPa. I'll assume the air cleaner filter element is 12-inch-diam round by 4-inch-tall, or 150 sq-inches or 1.04 sq-foot, (or 0.1 sq-meters).

Pressure*area = force so 8 kPa * 0.1 sq-meters is 800 Newtons force.

Assuming some more things: a 260 fwhp engine needs approx 388 cfm airflow, so the avg airspeed is 372 fpm when going through 1.04 sq-ft of air filter. 372 fpm is 6.2 ft/sec or 1.89 m/sec.

Using 1.89 meters/sec as the speed and 800 Newtons as the force, allows the calculation of power (force*dist = work, then take rate of work so it becomes force*velocity = power).

800 Newtons * 1.88 m/sec = 1511.8 Nm/sec. 1 Nm/sec is 1 Watt, so the peak power loss through the air filter using the above assumptions & math is 1512 Watts, or 2.03 fwhp.

2 hp loss in an air filter is pretty small, and that was one of the points of jumping into this thread --- to show that 8 kPa of pressure differential across an air filter works out to roughly a 2 hp loss for an engine that's pulling enough air to make around 260 fwhp.

IF any of the above assumptions are off, then it's fairly easy to change the numbers and go through the same math process. I'm just posting the framework on how to do it. HTH.