Critical Velocity for Air pockets
Critical Velocity for Air pockets
(OP)
I am trying to use the formular
V/sqrt(gD) = a + 0.56sqrt(sin S)
to calculate the critical velocity on a downward slope of 0.0057 and I can't quiet get IF S (slope) is in radians or degrees. Using either unit gives diffrerent answers. I would like to know if its radians or degrees.
V/sqrt(gD) = a + 0.56sqrt(sin S)
to calculate the critical velocity on a downward slope of 0.0057 and I can't quiet get IF S (slope) is in radians or degrees. Using either unit gives diffrerent answers. I would like to know if its radians or degrees.





RE: Critical Velocity for Air pockets
First, go back to wherever you found this formula and see if that source tells you what the units are for the term "S".
Second, if the term "S" is a slope then the units are usually, feet per foot or meters per meter in which case it would be "dimensionless".
Third, but it appears from the formula that "S" is not a slope but an angle otherwise you would not be able to find the sine of it. The sine of a slope is meaningless, although the sine of a slope angle has meaning.
Which leads to a fourth approach. If you have any known data you could substitute it into the formula and see if using radians or degrees gives you the correct answer for velocity. That would tell you which is the correct one.
good luck
RE: Critical Velocity for Air pockets
Go back to the reference of the equation and you will be clear on that.
The equation for velocities required to clear or scour air pockets from a pipe that slopes downward in "Pumping Station Design" by Sanks is as follows:
V = sqrt(gD)[0.825 + 0.25sqrt(sin?)]
where D is the ID of the pipe, and ? is the angle of the downward gradient of the pipe profile.
theci
RE: Critical Velocity for Air pockets
theci in your formular, if you have a slope of 1/125 what will be the ?. Is it the ATAN(1/125) or ATAN(1/125) converted to degrees.
Thanks,
RE: Critical Velocity for Air pockets
Which proposes V/(gD)^0.5 = a+ 0.56(sin S)^0.5
a is dependent on the volume of the air pocket and is in the range 0.45 to 0.61
a = 4 * Vair/(Pi * D^3)
S is slope in degrees. Formula is stated to be valid for slopes up tro 22 degrees, but therte is a huge moount of scatter in the data and i would suggest that you use with caution.
RE: Critical Velocity for Air pockets
RE: Critical Velocity for Air pockets
RE: Critical Velocity for Air pockets
RE: Critical Velocity for Air pockets
The Water Pollution Control Federation Manual of Practice No. FD-4 "Design of Wastewater and Stormwater Pumping Stations" states "A minimum velocity of 1.2 m/s (4 ft/sec) is required in the pipeline to shear the bubble and keep it moving downgrade."
RE: Critical Velocity for Air pockets
I think you are confused about why slope S = 1/125 =0.008 is the same as as the angle in radians, 0.008 (actually is 0.0079998). When a angle theta, is small, theta(in radian) ~= tan(theta) ~= sin(theta)
theci
RE: Critical Velocity for Air pockets
The recent Wallingford Report SR 661 compares previous research and presents a best fit solution for movement of air in pipes variables are slope, diameter, velocity and air pocket shape and volume. The equations presented are best fit. (The study report was available free on the Internet but it appears that they have now updated the report to a design manual which costs some $50 to download).
Using the HR proposed formula, as stated above by Balebeng ,and applying a FOS of 1.1, as recommended by HR, gives a critical velocity to move air along a horizontal pipe of about 1.13 m/sec. (c.f 1.2 m/sec stated above by BIMR)
The report uses S in degrees, but if you are working on an Excel spread sheet it is easier to work in radians and as Theci states you can use either as long as you are consistent.
For a slope of 1/125 (0.46 degrees) the slope is not significant in comparisaon to the scatter of the data and to be safe, unless you have a velocity somewhat in excess of 1.2 m/sec, I would consider that the air pockets may not move downstream and would provide for air release valves on the downward slope. ( depending on the length and frequency of pumping - obviously air will travel back upstream to high points when the pumps are off)
(PS Falvey's paper USDI - Engineering Monograph No 41 is available on the internet with a google search)