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Maximum Air Flow Through A Hole

 I couldn't find any reference to the gate valve setting in any of the Welte references. Didn't you measure the distance before taking the valve apart? I guessing not.... Generally speaking, a gate valve (or cut-out valve) doesn't have to open any more the 1/4" to allow maximum air flow. Look at it this way, the maximum amount of air that can possibly flow 'by' the valve is equal to the square area of the hole that the valve covers. So, if the hole is 1-1/8" in diameter, the square area = (pi x r2) or 3.14159265 x .56252 = 0.9940 sq in. or 1 sq in. Now, looking at the valve, we have to consider the point where the valve seals the hole because that is the 'limiting' factor in the equation. So, if the hole has a diameter of 1-1/8", the circumference of the hole (or the perimeter) = (2 x pi x r) or 3.5342 in. Now, you have to consider that the 'perimeter' times the valve clearance is actually a rectangular area through which the air will pass. In other words, it's like a wall. So, if the valve clearance is 1/4" (0.250") and we multiply that by 3.5342", we come up with 0.8835 sq in. Oopppps. Not enough..... Actually, the solution is to divide the square area of the hole (pi x r2) by the circumference of the hole (2 x pi x r). pi = 3.14159265 (most people use 3.14) Example: d = 1" r = .5" pi = 3.1415 pi x r2 (area) / pi x d (circumference) 0.7854 / 3.1416 = 0.25" Naturally, your answer will be dependent on the actual sizes of the parts. This page was last revised July 1, 2017 by John A. Tuttle, who Assumes No Liability
For The Accuracy or Validity of the Statements and/or Opinions
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