There are two ways to solve this problem in my opinion.
First if we treat air as an ideal gas, then PV=nRT and if we assume that the temperature does not vary significantly (air is in pipe so it should remain close to the same temperature), then P1V1 = P2V2. The pipe, as shown is 58 ft long and...
here is a sketch of the pipe...thanks for all your input..for some reason is is sideways sorry for that...more later thankshttp://files.engineering.com/getfile.aspx?folder=9b2d6bf4-e4d1-4989-99e0-f4646c63da07&file=pipe_sketch.pdf
A sketch would help but I do not have one I can post at this time...perhaps tonight
In my mind D is the height of the water inside the pipe, the pipe has 58 ft of water over it and the top of the pipe is 58-41 = 17 ft below the surface. Another way to look at this is that the pressure in the...
I think you guys are close
BIG INCH I am not sure your equation is completely correct if D in your equation is the Depth of the water in the pipe, then the air deptch would be 41-D, Vw = D*1639/41 and Va = (41-D)* 1639/41
LITTLE INCH I am not sure this is a valid equation
If P1 = 1bara and SG...
I have a 6 ft diameter pipe that is sealed at the top (effectively air tight) that drains into a empty pond. The elevation change is 41 ft. The pipe is at a 45 degree angle and is about 58 ft long, the volume is 1639 cubic ft. When the pond fills with water, to a height of 58 ft (~25 psig)...