Dear all
Not quite a mining history topic but here goes.
Can anyone provide a formula for the area below a partially closed gate
valve.
On a direct ratio valve I have two turns per inch plus two for the
"slogger". So, for a six inch main, 6x2 + 2 = 14 turns to close the valve
from the fully open position.
The question is, if a six inch gate valve has been closed 8 turns what is
the cross sectional area of free space below the valve?
Regards
brian
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