Goodday . i wonder if anyone can help with this .

I looking for the formula to the below problem inorder to use same to transpose in case i shoild need the pipe size in another case

Two tanks , tank A and tank B.

How much time in hours to fill tank B. Where Tank A is a full water tank( always being filled up from the top)of 100m^3 with a pipe on the bottom filling into tank B , an empty water tank of 50m^3 . The pipe is 100mm in diametre.

Both tanks are 10 m high

And where 1m^3 of water = 1 ton

The problem I.m having is that the flow rate slows down as the tank fills'

## decanting one tank into another

### Re: decanting one tank into another

This is non-steady flow problem for which differential equation should be written.

I will try to provide one later today.

I will try to provide one later today.

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### Re: decanting one tank into another

Here is how you can start:

Pressure drop in pipe due to friction is:

Dp=rho * f * L * V^2 / (2 * D), or

(A) Dp = 8 * rho * f *L * Q^2 / (D^5 * pi^2);

where is:

(B) Dp = rho * g * (H - Dh); - available energy due to level difference if both reservoirs are on atmospheric pressures

(Dh - is the change in elevation due to tank B being filled in time)

Dp - pressure drop due to friction [Pa]

rho - fluid density [kg/m3]

f - friction factor [ - ]

L - pipe length [m]

V - flow velocity [m/s]

D - pipe diameter [m]

pi = 3.1416

Q - flow rate - [m3/s]

Flow rate in tank B is: Q = Dh * D^2 * pi / 4 * Dt,

where is

Dt - change in time

Dh - change in level

As (A) has to be equal to (B) (if local resistances are negligible) than we get:

rho * g * (H - Dh) = 8 * rho * f *L * Q^2 / (D^5 * pi^2)

g * (H - Dh) = (f * L / (2 * D)) * (Dh / Dt)^2

Dt^2 = (f * L / (2 * D * g)) * ((Dh^2)/(H - Dh))

where is once again:

Dh - change in elevation - rise of level in tank B

Dt - time change

f - friction factor - for turbulent flow it is constant value

L - pipe length

D - pipe diameter

H - elevation difference on the start of filling

As result of this equation you can get how the elevation h will change in respect to time change t with start conditions - when t=0, elevation h - equals some height.

This is now mathematical problem - differential equation for which I (or you) have to take a look some where for solution.

Hope this gives some clue...

Pressure drop in pipe due to friction is:

Dp=rho * f * L * V^2 / (2 * D), or

(A) Dp = 8 * rho * f *L * Q^2 / (D^5 * pi^2);

where is:

(B) Dp = rho * g * (H - Dh); - available energy due to level difference if both reservoirs are on atmospheric pressures

(Dh - is the change in elevation due to tank B being filled in time)

Dp - pressure drop due to friction [Pa]

rho - fluid density [kg/m3]

f - friction factor [ - ]

L - pipe length [m]

V - flow velocity [m/s]

D - pipe diameter [m]

pi = 3.1416

Q - flow rate - [m3/s]

Flow rate in tank B is: Q = Dh * D^2 * pi / 4 * Dt,

where is

Dt - change in time

Dh - change in level

As (A) has to be equal to (B) (if local resistances are negligible) than we get:

rho * g * (H - Dh) = 8 * rho * f *L * Q^2 / (D^5 * pi^2)

g * (H - Dh) = (f * L / (2 * D)) * (Dh / Dt)^2

Dt^2 = (f * L / (2 * D * g)) * ((Dh^2)/(H - Dh))

where is once again:

Dh - change in elevation - rise of level in tank B

Dt - time change

f - friction factor - for turbulent flow it is constant value

L - pipe length

D - pipe diameter

H - elevation difference on the start of filling

As result of this equation you can get how the elevation h will change in respect to time change t with start conditions - when t=0, elevation h - equals some height.

This is now mathematical problem - differential equation for which I (or you) have to take a look some where for solution.

Hope this gives some clue...

Pipe flow calculations - free fluid flow calculators

http://www.pipeflowcalculations.com/

Find pipe flow calculations on the Facebook

http://www.facebook.com/PipeFlowCalcs

http://www.pipeflowcalculations.com/

Find pipe flow calculations on the Facebook

http://www.facebook.com/PipeFlowCalcs