Recent content by ed shaw

My drawing of this funnel shows a 12" deep flat collar. When R=600" the volume of the first 12" of the funnel is 13,564,000 in3 or 290 cu yds. Weight of air in that volume is 600 lbs of air. At about 15 mph (2 feet per second) that mass of air generates 23 HP as it moves through the funnel, at...

Part one: Supposed: A wind funnel. Circular Input radius = 600" Windspeed = 24 ft/sec Air weight = 2 lbs/yd3 Circular Output radius = 60" Disregard = backpressure. 1 HP = 560 ft lb/ sec Solve for = (1) Theoretical Wind speed at output (2) HP at output. Part two: Backpressure will...