How to Calculate Water Line for Floating Pipe
How to Calculate Water Line for Floating Pipe
(OP)
If I have a pipe that will be floated in water, how do I calculate how deep the pipe will sink, or the height at which the water rises along the pipe? Assume the pipe is horizontal and closed at both ends...thanks for the help!





RE: How to Calculate Water Line for Floating Pipe
suppose a pipe with radius R and length L is sunk in water by depth x(from surface of water to the bottom of the pipe)
so:in eqilibrium F=W which F is hydraulic force and=P.A, P:pressure and A:effective area.and W is weight of the pipe.
P is proportional to x[P=p(roh).g.x]and since only the vertical component of P is impportant and horizontal one is normal to weight direction,you hove to consider that vertical component of P varies again with x because of the shape of the pipe, and finally A is function of x.
i wanted to draw details in autocad but copy and paste doesn't work here.i hope you got your answer. FRED70
RE: How to Calculate Water Line for Floating Pipe
HTH,
Carl
RE: How to Calculate Water Line for Floating Pipe
If you can use Excel, by following the procedure below you'll get a solution:
1) Take the mass of the pipe in kilograms and divide it by its length in decimeters and its outer radius (also in decimeters) squared and call it Y=M/L/R2
2) If Y>π(pi or 3.14...) then the pipe will sink
3) In a new Excel sheet put the value of Y in A1, the formula =PI()+SIN(A3)*COS(A3) in A2, the formula =A2-A1 in A3 and the formula =1+COS(A3) in A4
4) Hit repeatedly (or keep pressed) F9 till you see the numbers stabilize
5) The number in A4 multiplied by R(outer radius) gives you the height of immersion.
To test put 3.14 in A1 and A4 should go close to 2, put 0 in A1 and A4 should approach zero.
prex
http://www.xcalcs.com
Online tools for structural design
RE: How to Calculate Water Line for Floating Pipe
Assuming your pipe will flow, the weight of your pipe will be equal to the weight of the fluid to be displaced. Knowing the density of the fluid, you can know out the volume of the fluid to be displaced. What remaining is the height of the fluid line on the pipe giving that volume.
Perhaps you may find these formular usfeful.
Displaced Volume = submerged cross-sectional area * length of pipe.
If the fluid height is less than 1/2 your pipe diamter (i.e. volume of fluid to be displaced is less than 1/2 of the volume of the pipe), then
A = h(3h^2 + 4s^2)/6s
s^2 = 4h(r-h)
where
A = submerged cross-section area
h = submerged height (segment depth)
r = external radius of pipe
s = segment length
If the fluid will rise more than half the pipe diamemter, then do it the other way round and consider the area of pipe which will not be submerged and use the above formulars.
RE: How to Calculate Water Line for Floating Pipe
A = Wt/rho where A=submerged Area, Wt=Total weight of pipe and rho is the density of the fluid in which the pipe will float. So now A is a known number...
Then, the Area = h/6s(3h^2 + 4s^2)
and the radius, r, equals h/2 + s^2/8h
solving for s...sqrt[8h(r-h/2)] and substituting this into A, you can now iterate for h. Once h is solved, s is easy.
However, I did not realize that this may not work for a pipe submerged more than it's radius...what about if the pipe is submerged completely under the water?
As an example, 24" pipe (125.49 #/ft) with no liquid inside it and no coating outside is floating in a fresh water lake. This is what I get:
A = 125.49#/ft / 62.4#/ft^3 = 2.011 ft^2
h = 1.201 ft (which is a little more than the radius)
s = 1.959 ft