Engine Ventilation

[WestVanHan]

Originally Posted by rosbullterrier

Having just read 'High Speed Motor Boats' (John Teale), I am informed my twin 229 cu.in V6's require 398 cu. ft. per minute intake air at only 1500 rpm!
Well, my inlet vent is a plastic gunwale manifold with just two 3" diameter inlet holes!
According to Mr Teales calculation - assuming 100 lineal feet per minute through natural ventilation - my boat will need 398/100 = 3.98 sq.ft. cross section ventilator !!!!!!!!

To simplify- think of your 2 engines as one, so a 458 CID V12

Being a 4 stroke,each revolution uses half the displacement of air= 229 CI of air per revolution.

At 1500 rpm it would be 1500 x 229= 343,500 CI of air/min.

One cu.ft of air is (12x12x12) 1728 CI
-343,500/ 1728 = 199 cu. ft per minute at 1500 rpm used by both engines.

At 3000 rpm- 400 cu.ft. a minute.

At 4000 rpm- 533 cu ft/min

At 5000 rpm- 666 cu ft/min.

At any rate,that 3.98 sq ft cross section (144 x 4) would be 576 square inches or a 27" round vent.

So to support 4000 rpm using his calculations.
-(398/1500) x 4000 =1061 square inch intake

Which would need about a 36" diameter round intake.

If you were running 2 or 3 Griffons,maybe.

http://www.youtube.com/watch?v=QdfO657zHNY

[powerabout]

I dont know where the lack of knowledge comes from, every manufacturer I know will supply you air volumes and btu ratings ( heat rejection)for their engines so you can work out exactly how much air you need to keep x temperature in an engine room with x ambient.