0
$\begingroup$

Given a tank with a certain height of water above the discharge pipe of know length, open to the atmosphere at the other end. I know the flow rate can be calculated if the pipe was horizontal by using Poiseuille's equation (No entry losses, no turbine or pump work, pipe cross-section is constant) by considering the pressure loss to be equal to the difference between the hydrostatic pressure at the inlet of the pipe and the atmospheric pressure.

If the pipe was not horizontal.. ie the pipe outlet is at a certain distance below the inlet to the pipe, by knowing the length of the pipe and not considering any pipe bend losses can we calculate the flow rate?....I considered using Bernoulli's equation by considering the pressure drop to be equal to the difference between the hydrostatic pressure at the inlet of the pipe and the atmospheric pressure and as the height difference between the inlet and outlet is know. I couldn't figure out how to calculate the pressure loss term in the equation. Is this data sufficient to calculate the flow rate...if not what other data would be required?

$\endgroup$
0
$\begingroup$

I considered using Bernoulli's equation by considering the pressure drop to be equal to the difference between the hydrostatic pressure at the inlet of the pipe and the atmospheric pressure and as the height difference between the inlet and outlet is known.

Assuming an open vessel and the outlet to atmospheric pressure also:

Vessel

Then the total pressure available to overcome pipe pressure losses is:

$$\Delta P=\rho g h$$

(with $\rho$ the density of the fluid)

As regards the pressure loss in the pipe, assume no losses in the bends (if any) and assume laminar flow, so you can use the Hagen Poiseulle equation:

$$\Delta P=\frac{8\mu Q}{\pi R^4}L$$

Then calculate the flow rate. Check the Reynolds number $Re$ to see if flow is indeed laminar.

Should flow prove not to be laminar ($Re>2600$), you will need the Darcy Weisbach equation for turbulent flow to calculate the pressure loss in the pipe:

$$\Delta P=f_D\frac{8Q^2}{\pi^2 g D^5}L$$

$\endgroup$
  • $\begingroup$ Thank you for taking the time to answer. Hagen Poiseulle equation would give the irreversible pressure loss in the pipe. Pressure drop in the pipe would include reversible and irreversible loss. Given a horizontal pipe pressure drop (ΔP=ρgh) would be equal to pressure loss. But i am not talking about a horizontal pipe here. [Continued] $\endgroup$ – GRANZER Nov 23 '17 at 14:25
  • $\begingroup$ .... P1/ρg +(V1)^2/2g+z1=P2/ρg+(V2)^2/2g+z2+hloss .... is the equation I think, where ΔP=ρgh=P1-P2 is the pressure drop. How to calculate hloss which is the irreversible pressure loss? $\endgroup$ – GRANZER Nov 23 '17 at 14:28
  • $\begingroup$ ALL pressure loss in the pipe is irreversible, because it is due to friction and thus not conserved. $\endgroup$ – Gert Nov 23 '17 at 14:28
  • $\begingroup$ The calculation I gave you is standard textbook, for horizontal or vertical (and in between) pipes. It's approximate because it doesn't take into account losses in the vessel (small), inlet of pipe, bends etc. $\endgroup$ – Gert Nov 23 '17 at 14:30
  • $\begingroup$ ...Not all pressure losses are irreversible. Consider the pipe going up and down....as the pipe goes up the pressure reduces due to gravitational effects...and as the pipe goes down that pressure is recovered. $\endgroup$ – GRANZER Nov 23 '17 at 14:31
0
$\begingroup$

I had the same problem solved by following method not sure if its right: Suppose a tank with cross section are of A And the outlet area is a Flow rate of pipe is I So after t seconds I.t volume of water comes out and the water is coming from top of tank So there is almost I.t/A height decreased so its energy is of this voulme comes from top of tank to inlet: U = K => g h = v v /2 V = sqrt(2 g h) V a t = volume of water out = I.t I = sqrt(2 g h) * a This assumes the tank has opening on the top

$\endgroup$
  • $\begingroup$ This is a haphazard way to answer. Difficult to understand. I think this will be proper because u have equated the water lever decrease in tank 'h' to kinetic energy increase ....so by this as water level goes on decreasing kinetic energy should go on increasing ..which would mean velocity and flow rate should go on increasing in a pipe of constant cross-section area which is wrong. $\endgroup$ – GRANZER Nov 23 '17 at 12:25
  • $\begingroup$ h is distance from surface of water to oulet $\endgroup$ – Nemexia Nov 23 '17 at 15:08
  • $\begingroup$ Ok...If h is the height of water from the outlet then according to g h = v v /2 potential head is fully converted to the dynamic head which > But I am looking for an answer when there are losses in the flow. $\endgroup$ – GRANZER Nov 23 '17 at 15:26
0
$\begingroup$

You are really asking how you calculate the term $h_{loss}$ in applying the Bernoulli equation to a non-horizontal pipe. The equations given in @Gert's response are the correct ones to use if you simply replace the $\Delta P$'s on the left hand sides of his equations by $\rho gh_{loss}$ and regard L as the length of pipe. This will give you the correct result for $\rho gh_{loss}$ both for horizontal and vertical pipes.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.