Effect of initial suction power on siphon performance

Question

Background

I have a flooded crawl space and I'm using a long hose to siphon the water out.

I do this as follows:

• One side of the hose is in the crawl space, and the other side down a hill in the garden (lower than the crawl space).
• I connect a sump pump to the hose in the crawlspace, immerge the pump and start it.
• While it's underwater, I unscrew the hose from the pump. This results in getting splashed a bit, and if things go well successfully starting the siphon.

Question

Does the initial pumping power (or suction power if done from the other end) matter?

i.e. everything else being equal, will I get a faster siphon if I start the process with a powerful pump than with a weaker pump? Or does that become irrelevant once I disconnect the pump and the siphon starts flowing?

Answer 1

i.e. everything else being equal, will I get a faster siphon if I start the process with a powerful pump than with a weaker pump? Or does that become irrelevant once I disconnect the pump and the siphon starts flowing?

Simply put, only three factors significantly affect flow rate:

1. Difference in height between inlet and outlet of your siphon. Roughly, the flow speed $v$ ($\mathrm{m/s}$) through the hose is proportional to the square root of the height difference $\Delta h$:

$$v \propto \sqrt{\Delta h}$$

2. Pipe diameter: that relation is a little more complicated but volumetric throughput benefits greatly from using a smooth, large internal diameter hose or pipe.

3. Pipe length: volumetric throughput $Q$ is inversely proportional to pipe length $L$:

$$Q \propto \frac1L$$

Initial throughput caused by the pump has no effect once flow has been established and the pump has been switched off: then other laws take over that determine flow rate/throughput.

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Edit: with regards to point 3.

Darcy-Weisbach equation for laminar flow:

$$\frac{\Delta p}{L}=f_D \frac {\rho}{2}\frac{v^2}{D}$$ With $f_D=\frac{64}{Re}$ and $Re=\frac{vD}{\eta}$, and $Q=\frac{\pi D^2v}{4}$ then with reworking:

$$\frac{\Delta p}{L}=\frac{128\eta Q}{\pi D^4}$$

Bear in mind that $\Delta p$ is the pressure difference between inlet and outlet of the hose, then all other things being equal:

$$\large{Q \propto \frac{1}{L}}$$

Answer 2

The initial velocity of the water through the hose doesn’t affect the velocity of the water that the siphon quickly settles down to. After only a brief moment, if water were a perfect fluid, the steady-state velocity of the water in a siphon would be $$v=\sqrt{2 g h} ,$$ where $g$ is the acceleration due to gravity at the Earth's surface, and $h$ is the height of the surface of the source water above the siphon's drain point. This velocity can be derived from Bernoulli's equation, as in Wikipedia's Siphon article. However, it's easy to understand that velocity using much simpler physics, if you view the above equation as being a simple derivation from $$\frac{1}{2}m v^{2}=m g h ,$$ where $m$ is the mass of a small bit of water that goes through the siphon. The left hand side of that equation is the expression for the bit of water's kinetic energy as it comes out of the hose, and the right hand side of that equation is the expression for the change in gravitational potential energy that the bit of water has between the top of the source water, and the drain point. Basically, gravitational potential energy that the water has initially gets converted into kinetic energy.

For a fixed $h$, the main thing that will affect the volumetric flow rate is the hose diameter. If water were a perfect fluid, the volumetric flow rate would be $$Q=\pi r^{2} v=\pi r^{2} \sqrt{2 g h}, $$ where $r$ is the interior radius of the hose.

In the real world, water isn't precisely a perfect fluid, so some of the gravitational potential energy actually gets converted into heat due to drag, rather than being entirely converted into kinetic energy, so the real steady-state velocity of the water will be somewhat lower than the value expressed above. For a fixed $h$, the amount of energy lost to drag decreases as the hose diameter increases, and increases as the hose length increases.

The relationship between $Q$ and the hose length $L$ can be determined by once again using a statement of the conservation of energy.

At low velocities (such that there is a laminar flow), the drag force is proportional to $v$. Thus, for a given hose diameter, choice of fluid, etc., it's possible to express the drag on a bit of water with mass $m$ in the form $$F_{d}=mbv$$ for some value of $b$ which does not depend on $v$. ($m$ is included in this definition of $b$ in order to simplify later calculations.) The energy lost due to drag over the hose length $L$ is thus $$E_{d}=F_{d}L=mbvL .$$ Conservation of energy, with drag no longer neglected this time, then gives $$mgh=\frac{1}{2}mv^{2}+mbvL ,$$ which can be solved for $v$ to give $$v=\sqrt{2gh+b^{2}L^{2}}-bL$$ or $$Q=\pi r^{2} ( \sqrt{2gh+b^{2}L^{2}}-bL ) .$$ If drag is small, more precisely if $bL << \sqrt{2gh}$, a Taylor expansion of $Q$ around $bL=0$ gives the approximation $$Q \approx \pi r^{2}\sqrt{2gh}-\pi r^{2}bL .$$ I.e., in the absence of drag, the throughput is independent of $L$, but in the presence of drag, the ideal throughput is reduced (for a small amount of drag) by an amount proportional to $L$.

Answer 3

Full derivation of volumetric flow for a siphon:

Bernoulli’s equation in the absence of friction:

$p_1+\frac12 \rho v_1^2+\rho gz_1=p_2+\frac12 \rho v_2^2+\rho gz_2$

Head loss in the case of friction (Darcy – Weisbach):

$\Delta p=f \frac{L}{D} \frac12 \rho v_2^2$, where $f$ is the Moody friction coefficient.

So:

$p_1+\frac12 \rho v_1^2+\rho gz_1=p_2+\frac12 \rho v_2^2+\rho gz_2+f \frac{L}{D} \frac12 \rho v_2^2$

With speed at the water surface $v_1\approx 0$ and $p_1=p_2$:

$gz_1=gz_2+\frac12(f\frac{L}{D}+1)v^2$

$\Delta z=z_1-z_2$

$\large{v^2=\frac{2g\Delta z}{f\frac{L}{D}+1}}$

For laminar flow through a smooth hose:

$f=\frac{64}{Re}$, where $Re$ is the Reynolds number:

$Re=\frac{vD}{\eta}$

$f=\frac{64\eta}{vD}$

$v^2(\frac{64\eta L}{vD^2}+1)=2g\Delta z$

$v(\frac{64\eta L+vD^2}{D^2})=2g\Delta z$

$64\eta Lv+D^2v^2=2gD^2\Delta z$

$D^2v^2+64\eta Lv-2gD^2\Delta z=0$

Which has a positive root for:

$\large{v=\frac{-64\eta L+\sqrt{64^2 \eta^2L^2+8gD^4\Delta z}}{2D^2}}$

$Q=\frac{\pi D^2}{4}v$

Function analysis of $Q(L,D,\Delta z)$ with partial derivatives shows:

1. $Q$ continuously decreases with increasing $L$.
2. $Q$ continuously increases with increasing $D$.
3. $Q$ continuously increases with increasing $\Delta z$.

For turbulent flow the $(f,Re,v)$ relationship is much more complicated.

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Adding a pump to get the siphon flowing amounts to adding a term $\Delta p_{pump}$ to the left hand side of the Bernoulli equation. This then increases $Q$, until the pump is switched off again.

Answer 4

Simply fill a siphon hose of sufficient length that has a shutoff valve at its inlet and a check valve at its outlet and place the inlet into the deepest pool of the water you want drained and open the valve. The water will flow as long as the outlet is below the inlet elevation and when the most of the water is drained, if you close the shutoff valve before air enters the siphon line, it will remain primed for further use. This controllable siphon works by not allowing air to enter and travel up the siphon hose from the one way check valve at the long leg outlet.