Critical Velocity for Air pockets 3

I'm nor familiar with the formula you quote but the following thoughts may be helpful:

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

As RWF7437 said, the S within the sin function must be an angle, no matter what S is. As such, your S in the equation should be the angle determined by S. The angle in radian or degree should give you the same sin value.

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

Thank you RWF7437 and theci.
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,

Your formula apperas to be from HR Wallingford Report SR 661 April 2005.

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.

Thank you Bris, yes the formaular is from HR Wallingford. My confusion is that if I have a slope of 1/125, so do I take ATAN of this slope and then convert the result to degrees OR take the sin(1/125.

You need the angle of the slope = ATAN(1/125) in degrees

PS sorry for the errors in my response - I am having a bad tryping day.

Maybe this will help:

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."

The S in your SinS is the angle (in degree or radian) determined by slope S. For example, you have S=1/125, then the angle = ATan(1/125)=0.458 degree or 0.008 radian. Then when you sine either 0.458 degree or 0.008 radian, you will get the same reault, 0.008. Just remember to convert your calculator settings to deg or rad, accordingly.

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

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)