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How to Calculate Water Line for Floating Pipe
- Thread starter SuperG
- Start date
i hope my answer will be correct and useful:
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
ressure 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
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
ressure 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
You do not need any complicated force evaluations, since as Archimedes would tell you when the pipe is floating the weight of the pipe is equal to the weight of the displaced fluid. So you just have a geometry problem; at what depth does the the weight of the volume of water displaced equal the pipe weight. All the math you need is the equation for a circle (the pipe cross-section) and a sector of a circle (the portion of the pipe above the water level).
HTH,
Carl
HTH,
Carl
Your problem involves trigonometric functions and has no direct compact solution.
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>&[ignore]pi[/ignore];(pi or 3.14...) then the pipe will sink
3) In a new Excel sheet put the value of Y in [tt]A1[/tt], the formula [tt]=PI()+SIN(A3)*COS(A3)[/tt] in [tt]A2[/tt], the formula [tt]=A2-A1[/tt] in [tt]A3[/tt] and the formula [tt]=1+COS(A3)[/tt] in [tt]A4[/tt]
4) Hit repeatedly (or keep pressed) [tt]F9[/tt] till you see the numbers stabilize
5) The number in [tt]A4[/tt] multiplied by R(outer radius) gives you the height of immersion.
To test put 3.14 in [tt]A1[/tt] and [tt]A4[/tt] should go close to 2, put 0 in [tt]A1[/tt] and [tt]A4[/tt] should approach zero.
prex
Online tools for structural design
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>&[ignore]pi[/ignore];(pi or 3.14...) then the pipe will sink
3) In a new Excel sheet put the value of Y in [tt]A1[/tt], the formula [tt]=PI()+SIN(A3)*COS(A3)[/tt] in [tt]A2[/tt], the formula [tt]=A2-A1[/tt] in [tt]A3[/tt] and the formula [tt]=1+COS(A3)[/tt] in [tt]A4[/tt]
4) Hit repeatedly (or keep pressed) [tt]F9[/tt] till you see the numbers stabilize
5) The number in [tt]A4[/tt] multiplied by R(outer radius) gives you the height of immersion.
To test put 3.14 in [tt]A1[/tt] and [tt]A4[/tt] should go close to 2, put 0 in [tt]A1[/tt] and [tt]A4[/tt] should approach zero.
prex
Online tools for structural design
I agree with CarlB.
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.
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.
- Thread starter
- #6
Thanks for the replies...I have grabbed a trig book and looked up the equations for segments of circles and did the following, which looks like what "bhwtam" said:
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
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
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