Flow calculation help

[krugan]

Hello,

I am looking for the formula(s) to use to find the flow rate of water from a tank to another tank under gravity.

Diameter of the first feed tank is 86". Height of water in feed tank will be constant @ 68". The height to the center of the outlet drain is 33.25" and the diameter is 4". There is 10' of PVC piping in between to the other tank and the center of the inlet of second tank is also 33.25". Both tanks are on the same level/grade.

The level in the second tank will be constant at 41". I'm trying to find out what the maximum flow rate will be using these conditions.

 

[francesca]

Try the bernoulli equation, with the Darcy-Weisbach equation to get head losses due to friction.

Add the friction loss to your Bernoulli equation (it's a conservation of energy equation) and the pressure term should be zero. You want to establish conditions for both ends of the system (p1 = p2 = 0 i.e. atmospheric); v1 = 0, v2 = x (i.e. you'll solve for v2); h1 = 68", h2 = 41"; etc.

You'll set (bernoulli eqn terms for the first tank) = (bernoulli eqn terms for the second tank) and solve for v2.

You can also add inlet/outlet losses and losses due to fittings, etc. There's a rule-of-thumb that if (length of pipe)/(pipe diameter) > 1000 then these "minor" losses can be ignored. If you want to be conservative, you could deduct 2' from the higher tank water elevation.

 

[Latexman]

Look at thread378-152231

Good luck,
Latexman

 

[Ashereng]

Is this a homework assignment question?

"Do not worry about your problems with mathematics, I assure you mine are far greater."
Albert Einstein
Have you read FAQ731-376 to make the best use of Eng-Tips Forums?

 

[bimr]

Use the Darcy method from Cranes Technical Paper No. 410

K Entrance Loss = .5
10' Pipe loss = fL/d = .018 x 10 x 12 / 4 = .54
K exit loss = 1.0

Total K = .5 + .54 + 1 = 2.04

Q = 19.65 d(2)sq rt (h/K) (Page 3-4)

= 19.65 (4 X 4) sq rt ((60-41)/12)/2.04) = 277 gallons per minute.

The effect of friction factor is negligible since the pipe length is short.

 

[bimr]

Sorry the was a typo. Correction is here.

Use the Darcy method from Cranes Technical Paper No. 410

K Entrance Loss = .5
10' Pipe loss = fL/d = .018 x 10 x 12 / 4 = .54
K exit loss = 1.0

(Assumes 0.18 friction factor for PVC)

Total K = .5 + .54 + 1 = 2.04

Q = 19.65 d(2)sq rt (h/K) (Page 3-4)

= 19.65 (4 X 4) sq rt ((68-41)/12)/2.04) = 330 gallons per minute.

The effect of friction factor is negligible since the pipe length is short.

 

[krugan]

Thanks for your help everyone! I know my original question is basic to most of you, but my new job requires me to use fluid formulas for design purposes which I am just learning. I will need to take some fluid mechanics courses to help. In the mean time I will purchase Cranes Technical Papers as it seems to be very popular.

Can anyone else recommend some good introductory books on fluid mechanics which show examples and solutions?

bimir,

What does 19.65 represent in the equation you gave? Thanks again for all that replied.

 

[francesca]

The Darcy-Weisbach equation is empirical, so the 19.65 is just the constant that works.

 

[Latexman]

Crane TP 410 has an entire chapter of examples.

Good luck,
Latexman

 

[bimr]

headloss = f (L/D) (v2/2g) (Darcy Formula)

f (L/d) = K

headloss = 522 K q2/d4 = 0.00259 K Q2/d4

q = cfs
Q = gpm
d = internal pipe dia in inches
D = internal pipe dia in feet

Q = 19.65 d2 sqrt(hL /K)

The 19.65 represents the acceleration of gravity as well as unit conversions (cfs >> gpm).