Calculating Flow Rate and Loss Coefficients in a Complex Fluid Tank System

[kris]

I'm stuck with this problem, can anyone 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 receiving 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...

 

[ekrim]

I'm stuck with this problem, can anyone 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 receiving 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 don't 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 coefficient 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

 

[kris]

Thank you very much

 

[kris]

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]

I haven't had a chance to look over the numbers, but your Reynolds number looks way too low. I would double check your units on the numbers that went into that calculation.

The value for the delta P doesn't look too bad. That equates to about 27 psi for water.

 

[kris]

I re-worked it out and found that i had used of the length of the pipe in Reynolds EQn rather than the diameter. So i now get Reynolds to be about 2.8 x 10 to the 4. Which gave me a more realisistic head of 4.95m. So thanks again for your help.

PS i like you quote, its so true!