0
$\begingroup$

I currently have a ditch that fills with water draining out into an area lower than the water (via a hose), but the hose for most of the travel extend above the hight of the water.

As long as the hose stays filled with water, which it should assuming the water doesn't evaporate enough, this system works. Unfortunately, the water doesn't drain as quickly as I'd like after a rain storm.

Does the piping at all effect (e.g. the volume it holds) at what level the the basin will drain to?

If I increase the diameter of the piping will the volume of water flowing out per minute increase?

Bonus points for providing the physics principles being applied here. I'm not familiar with the names.

basin drainage diagram

$\endgroup$
  • $\begingroup$ Is your system stationary? Is the basin continuously refilled? $\endgroup$ – MrYouMath Apr 26 '17 at 11:38
  • $\begingroup$ The system is stationary and will be be intermittently​ refilled at different reasonable rates $\endgroup$ – TruthOf42 Apr 26 '17 at 11:55
  • $\begingroup$ Google "siphon" $\endgroup$ – Carl Witthoft Apr 26 '17 at 12:57
5
$\begingroup$

Does the piping at all effect (e.g. the volume it holds) at what level the basin will drain to?

The outlet level of the pipe work determines the minimum level the basin will drain to. At this point, the water is static and reaches its natural level, which will be to just fill the discharge (any excess will overflow).

If I increase the diameter of the piping will the volume of water flowing out per minute increase?

Yes. The higher the velocity, the greater the head loss through the pipe. The head loss is fixed by the difference between the water levels in the basin and the discharge pipe. So for a fixed, head loss, you want to reduce the velocity to increase the flow rate. To do this, you need to increase the diameter of the pipe.

| improve this answer | |
$\endgroup$
1
$\begingroup$

While the other answers give a relationship, there are factors that need to be included such as friction factor, inlet & outlet coefficients and length, so this link gives a more practical approach : http://files.engineering.com/getfile.aspx?folder=23da7aca-8762-4e14-b029-09

| improve this answer | |
$\endgroup$
0
$\begingroup$

This is answering the second part of your question to get a mathematical relationship for the parameters that are involved.

We write down the continuity equation in integral form.

$$\int_{V}\dfrac{\partial \rho}{\partial t}dV+\oint_S\rho \mathbf{u}\cdot\mathbf{n}dS=0$$

The Greek letter $\rho$ is the density of your fluid. $\mathbf{u}\cdot\mathbf{n}$ is the dot product of the velocity at the surface $\mathbf{u}$ and the outward normal unit vector $\mathbf{n}$.

For stationary flow, the volume integral vanishes. At the inlet side, we assume a uniform velocity $u_{\text{i}}$, which is parallel with the unit vector of the surface, and a surface area of $A_{\text{i}}$. Same procedure at the outlet with $u_{\text{o}}$ and $A_{\text{o}}$. Evaluation of the surface integral is only necessary at the inlet and outlet. We get:

$$-\rho u_{\text{i}}A_{\text{i}}+\rho u_{\text{o}}A_{\text{o}}=0.$$

Note, that $u_{\text{i}}A_{\text{i}}$ is the volumetric flowrate at the inlet and $\rho u_{\text{i}}A_{\text{i}}$ is representing the massflowrate at the inlet. Assuming circular inlet and outlet geometries we obtain:

$$u_{\text{i}}\frac{\pi}{4} d^2_{\text{i}}=u_{\text{o}}\frac{\pi}{4} d^2_{\text{o}} \implies u_{\text{i}} d^2_{\text{i}}=u_{\text{o}}d^2_{\text{o}}.$$

So if you know the flow rate at the inlet $Q_{\text{i}}=u_{\text{i}}A_{\text{i}}$ (can be determined by measuring the volume of output in a certain time). You can easily infer the velocity at the outlet for a given outlet $d_{\text{o}}$ and vice versa.

In order to get more information about the system behavior it is possible to apply Bernoulli's equation with empirical pressure loss $\Delta p_{\text{loss}}$ due to dissipative effects (can be looked up in tables for different pipings):

$$p_{\text{i}}+\frac 12\rho u^2_{\text{i}}+\rho g h_{\text{i}}=p_{\text{o}}+\frac 12 \rho u^2_{\text{o}}+\rho g h_{\text{o}}+\Delta p_{\text{loss}}.$$

We now assume that the pressure at the outlet and the inlet is the same, which is a good assumption for slowly moving air as surrounding "fluid". Hence, we can drop the pressure terms. Subtract $\frac 12 \rho u^2_{\text{i}}$ and use the continuity equation to replace $u_{\text{i}}=\frac {A_{\text{o}}}{A_{\text{i}}}u_{\text{o}}$. Additionally, we subtract $\rho g h_{\text{o}}$ and $\Delta p_{\text{loss}}$. We get the following equation:

$$\rho g(h_{\text{i}}-h_{\text{o}}) -\Delta p_{\text{loss}}=\frac 12 \rho u^2_{\text{o}}\left(1-\frac{A_{\text{o}}}{A_{\text{i}}} \right).$$

Solving this equation for $u_{\text{o}}$:

$$u_{\text{o}}=\sqrt{\frac{2g(h_{\text{i}}-h_{\text{o}}) -2\frac{\Delta p_{\text{loss}}}{\rho}}{1-\frac{A_{\text{o}}}{A_{\text{i}}}}}.$$

If you additionally assume the following simplifications

  • Assume that there is no pressure loss: $\Delta p_{\text{loss}}=0$
  • Assume that the inlet area is much larger than the outlet area: $\frac{A_{\text{o}}}{A_{\text{i}}} \ll 1$

you will obtain Torricelli's law as a crude estimate:

$$u_{\text{o}} \approx \sqrt{2g(h_{\text{i}}-h_{\text{o}})}.$$

| improve this answer | |
$\endgroup$
  • $\begingroup$ How would he know the flow rate at the inlet? He's asking whether this flow rate is affected by the diameter of the tubing. $\endgroup$ – Max Apr 28 '17 at 13:42
  • 1
    $\begingroup$ Why assume that the pressure drop is zero? One expression is Δploss=4 Cf L/D 0.5 ρ v^2. Cf depends on the pipe material and surface roughness. $\endgroup$ – Solar Mike Apr 29 '17 at 21:49
  • $\begingroup$ You are totally right. I just wanted to simplify matters for the OP to get a crude estimate. Added this in the post. $\endgroup$ – MrYouMath Apr 29 '17 at 21:52

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.