[murf_au]

Can any one help me please? For ease I have being creative with some numbers but it is the theory I am interested in.

What is the weight of a vertical pipe of water, given:
Pipe supported at top only.
Pipe is 100m, vertical.
Open at bottom, no restriction
Inflow is enough to fill pipe
Volume of water in pipe = 100kg
Friction in pipe = 2 m of head per 100m
Weight of pipe = 50 Kg

I figure on 2 answers though neither is probably correct.

A. 150Kg - A full column water plus the pipe

B 52 Kg - The friction head plus the pipe

I tend towards answer B as there is no restriction to the flow and forces are equal and opposite.

[ebaines]

What is the weight of a vertical pipe of water, given:
Weight of pipe = 50 Kg

You answered your own question - 50Kg! But I'm guessing you don't want the weight of the pipe but rather the force that must be carried by the support of the pipe. In this case you would add together the weight of the pipe and the downward friction caused by the water flow. But be careful of units - weight is measured in N, not Kg, and the friction force needs to also be expressed in newtons. You say the friction is 2m of head per 100 m of pipe length - that's a measure of pressure per unit length, so to convert to a force you multiply by the internal cicumference of the pipe. So in total you get:

F = mg + 2m/100m x 9804 (N/m^2)/m x L x C

where L = length of pipe and C = internal circumference.

It's possible to calculate the cross-sectional area of the pipe (and hence the circumference) given that it contains 100 Kg of water:

Mass = density x volume = 1000 Kg/m^3 x 100m x A;
A = 100 Kg/(1000 Kg/m^3 x 100m) = 0.001 m^2

and hence its circumference C is:

C = 2 x sqrt(A x pi) = 0.112m

So the friction force of water is:
2m/100m x 9804 (N/m^2)/m x L x C = 2/100 x 9804 x 100 x 0.112 = 2198N

Note that this friction force is greater than the weight of the water in the pipe!

Add that to the pipe's weight to get the total force of the support:

F = mg + 784N = 50Kg x 9.8m/s^2 + 2198N = 2688N

[murf_au]

Thanks for that. I shall now use that formula to do the calculations and yes I shall use newtons etc.
This will be a great help
Regards.
Murf_au