# Fluid Tank Mechanics Problem

I'm stuck with this problem, can any one help me please?

Two large tanks are connected by a cast iron pipe 50mm in diameter and 50m long. There are eight 60 degree and four 90 degree bends along the length of the pipe. The inlet to and the exit from the pipe are abrupt and the rate of flow of water through the pipe is 6 litres/second. Determine the difference between the supplying and recieving tank.

The following values may be assumed for the loss coefficients k:

abrupt entrance k=0.5 abrupt exit k=1.0
60 degree bend k=0.45 90 degree bend k=1.15

I would appreciate any help anyone can give me. Thanks

FredGarvin
Basically you are working backwards to determine the difference in height of fluid in the two tanks. That difference creates the flow from one tank to the other. You have to take into account the head losses due to the minor losses mentioned, i.e. entrance and exit, bends, etc...

I'm stuck with this problem, can any one help me please?

Two large tanks are connected by a cast iron pipe 50mm in diameter and 50m long. There are eight 60 degree and four 90 degree bends along the length of the pipe. The inlet to and the exit from the pipe are abrupt and the rate of flow of water through the pipe is 6 litres/second. Determine the difference between the supplying and recieving tank.

The following values may be assumed for the loss coefficients k:

abrupt entrance k=0.5 abrupt exit k=1.0
60 degree bend k=0.45 90 degree bend k=1.15

I would appreciate any help anyone can give me. Thanks

use the energy equation. Take point one to be the surface of tank one and take point two to be the surface of tank two.
I dont know how to type all the symbols, but basically, the pressure at both surfaces is the same and that term cancels out. The same goes for the velocity term because the tanks are sufficiently large such that the surface is not descending.

What's left is the difference in height, which is equal to the head loss in the system. The head loss is equal to:

(Ke + 8Kb + 4KB + KE + fL/D)*(V^2/2g)

Ke, Kb, KB, and KE are the loss coefficients for the given bends, entrances, and expansions.

fL/D is the frictive loss coefficient, where
f is the resistance coefficent obtained from moody's diagram, L is the length of the pipe, and D is the diameter

V is the velocity in the pipe, which is constant because the pipe is of constant diameter and g is the gravitational acceleration

use the discharge value you were given to find the velocity and reynolds number for the moodys diagram

Thank you very much

Just checking

So I wondered if you could tell me what I worked out now is correct...

I got that Reynolds number is 1.43 x 10-4, and that therefore f equals 0.0275

This then gave me a value of 18.79m approx. for the difference in height.

Does this look rounghly right?

FredGarvin