## The velocity of an object falling through tubing filled with sea water

## The velocity of an object falling through tubing filled with sea water

(OP)

Can anyone help me here, I am at my wits end trying to work with this calculation and seems to have all my engineering buddies scratching their heads or throwing themselves off bridges. My problem starts with a vertical piece of tubing, approximately 100' in length. The tubing is filled with sea water and is stagnant. An object ( of flat bottom for arguments sake) is dropped into the tubing and falls freely under the force of gravity. There is very little space between the outside diameter of the object and the inside diameter of tubing. My question is, how do I work out the final velocity of the object as it hits the bottom of the 100' fall. I know i should be looking at some sort of shear force calculation as the object is basically shearing the water, but i know i should also be taking into account the drag effect the water has on the object. GGGGGGGrrrrrrrr.......

I would be happy working with this, neglecting friction of the tubing but still cant get anywhere near........Can anybody out there help?????

Please....

Kindest Regards

M.

I would be happy working with this, neglecting friction of the tubing but still cant get anywhere near........Can anybody out there help?????

Please....

Kindest Regards

M.

## RE: The velocity of an object falling through tubing filled with sea water

Is this a real life problem or a school problem?

If it is a school problem please keep away from here.

Otherwise you should give us some more figures (tube diameter, object dimensions and shape, gap around the object, expected falling velocity minimum and maximum value) to help figure out a method for the calculation.

prex

motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

## RE: The velocity of an object falling through tubing filled with sea water

This is a real life problem, a crown plug was dropped from 100' through tubing onto one of our closed valves. I am trying to determine the resultant force the plug would invoke onto the valve to prove to a client that it would be sufficient enough to cause noticeable damage.

The ID of the tubing is 6.094" and we are taking the OD of the crown plug to be 5.90". The estimated weight of the plug is 40kg. We are taking the plug as a cylindrical shape, even though this is not the case, this would give us an approximate value which is what we are looking for.

M.

## RE: The velocity of an object falling through tubing filled with sea water

d = dia of plug

area of plug = A1 = pi *d^2 /4

D = dia of pipe

area of pipe = A2 = pi * D^2 /4

Water flow area past the plug = Aa = pi * [D^2 - d^2] /4

Wetted area = C = pi * [D + d]

hydraulic dia of annulus = Dh = 4 * A / C

B = flow area ratio = Aa/A2

length of the plug = L

rho = density of the water

to fall dL, the water displaced is dQ =[pi * (D^2 - d^2)] *dL

speed of fall = dL/dt

the displacing pressure is Pdisp = {40 kg / [pi * d^2]}

this equals the pressure loss across the plug

(initially I'm ignoring friction losses because they're too difficult to estimate quickly and probably are insignificant compared to the end losses)

The velocity of flow in the annulus = V = dQ /A

the velocity head in the annulus = hv = rho * V^2/2g (g = gravitational constant if you're using engineering units)

Consider that the inlet loss into the annulus is deltaP1 = 0.5 *(1-B)* hv

and the outlet loss from the annulus is deltaP2 = (1-B)^2 * hv

deltaP = deltaP1 + deltaP2 = Pdisp

equate these and produce the equation representing the speed of fall

Find the speed of fall at the bottom of the pipe = v

and just dL before the bottom of the pipe = u

Then from Mr Newton we get v^2 = u^2 + (2 * f * dL)

and the force is then Force = mass * f

Hope this points you in a good direction

Let me know if it works

## RE: The velocity of an object falling through tubing filled with sea water

http://egweb.mines.edu/eggn350/spring02/flow.htm

and go about halfway down the page to see if some parts of the rotameter equation can apply.

## RE: The velocity of an object falling through tubing filled with sea water

Assuming turbulent flow, consider a head loss coefficient of 0.5 for the sudden contraction at object front and a coefficient of 1 for the sudden enlargement at object rear.

The pressure drop across the object is

delta

p=1.5rho*v^{2}/2At the constant speed that will develop after the initial acceleration, the delta

pwill counterbalance the driving force (weight minus lift) divided by the object front area.Now introducing the numbers and assuming your object is made out of steel:

Weight: 40x9.8=400 N

Lift: 50 N

Object area: 17640 mm

^{2}Pressure drop: (400-50)/17640 = 0.02 MPa = 20000 Pa

rho=1000 kg/m^{3}v=sqrt(20000*2/1000/1.5)=5 m/sThis is the answer to your problem.

However this is valid under certain conditions:

1) The flow must be turbulent. To check this, one may calculate

D_{h}=10 mm and, assuming 1 cp for the viscosity of water, andRe=50000. The flow is indeed turbulent.2) The shape of the object must be purely cylindrical (not shaped as a missile), otherwise the enlargement and contraction losses will display coefficients lower than 1.5 as a total. This also explains why my solution does not depend on the pipe diameter: this solution holds as far as the area around the object is much smaller than the pipe section area (but not too small, otherwise a laminar flow will arise).

prex

motori@xcalcsREMOVE.com

http://www.xcalcs.com

Online tools for structural design

## RE: The velocity of an object falling through tubing filled with sea water

## RE: The velocity of an object falling through tubing filled with sea water