The Siphon

© 1998 by F.P.Gram

This will be more than you wanted to know about the theory of the siphon, so don't read it. All you really need to know is that you fill a tube with the liquid, insert one end into the liquid, and make sure the other end is below the liquid level. The atmospheric pressure is needed to make it work, so there is a maximum height you can siphon over. For mercury, it is 76 cm, about 2.5 ft. For water it is about 34 ft. That’s it- do not read more.

There are 10 kinds of people: those who know binary and those who don’t.

If a horizontal tube of length L and inner radius r is filled with a liquid of viscosity h , and if there is a pressure difference DP between the ends, the liquid will flow from the higher P to the lower with a volume flow rate of (and let's call this "current")

D Q/D t = D P/R (equation 1)

with R = 8h L/p r⁴. (Look up Poiseuille's law to check this out. A French name, pronounced something like pwah zweez, but then I suppose you don't say Day-twah for Detroit. ) This is the R for a tube with circular cross section. One might expect pr² in the denominator, but there is drag due to the tube wall, and this makes the difference. See Ohm’s law below.

This is like Ohm's law, with DP driving the flow through resistance R. (Ohm's law: electric current through resistance R is DV/R, where DV is the voltage difference.) Electric current is charge flow, and a better analogy with liquids is mass flow, but volume/time is the convention, and we will go with the flow. The counterpart to Poiseuille's law is R = rL/A, where A is cross-sectional area.

If the outlet is distance h₂ below the inlet, the current is assisted by gravity and we add rgh₂ to the DP in the above equation (i.e., replace DP with DP + rgh₂).

If we make a siphon (by filling the tube with liquid and inserting one end in a container of the liquid as shown below) with the inlet is at depth h₁, it would seem that we could calculate the pressure at the inlet (P_o + rgh₁, where P_o is atmospheric pressure) and, noting that the pressure at the outlet is atmospheric pressure, we find that the DP + rgh₂ is rgh where h is h₁+h₂. We can then calculate the DQ/Dt.

This is not quite right, however. The pressure at depth h₁ is rgh₁ where the velocity is zero, but then fluid accelerates toward the inlet, so the P must be smaller there. See Bernoulli's equation, and show that the P at the inlet is P_o + rgh₁ - ¹/₂ rv², where v is the velocity in the tube. (Note: here and in all that follows, v is the average speed in the tube. It varies with distance from the center, with the fastest speed at the center, but we do not need to be concerned about that.) Hence DP in equation (1) should be replaced with rgh - ¹/₂ rv² where h is h₁ + h₂ in the picture below.

So we need v. Well, it is easy to show that another expression for the current is

DQ/Dt = Av (equation 2)

where A is the cross-sectional area of the tube, pr². If we combine this with equation (1), we get a quadratic in v, with a v term from equation (2) and a v² term from the DP expression. Solve for v and obtain

v = (v₁² + v₂²)^1/2 - v₂ (equation 3)

where v₁ = (2gh)^1/2, the no-drag speed, and v₂ = 8h L/rr² = 8ph L²/m where m is the mass of fluid in the tube (which can be measured with better precision than the radius). Then use v times m/rL to get the fluid current (because m/rL = pr²). Incidentally, v₂ has units of velocity, but I know of no physical significance to it.

Equation 3 at large r or small h is the v unimpeded by viscous effects, and thus it is worth noting that it agrees with Torricelli's equation, which says that in the absence of viscous drag, the fluid emerges with the same velocity as if it fell freely for distance h, v = (2gh)^1/2. Check it out.

At the other extreme, v₁ is small compared to v₂, and
v = (v₁² + v₂²)^1/2 - v₂ reduces to approximately v₁²/2v₂. (Use a Taylor series expansion of the radical. This series is the same as the binomial series, but with a non-integer exponent.) Now if you multiply this by p r² to get the current, you will find that it is the same as equation 1. This gives me confidence that it is correct.

Water at 20^o C (room temperature) has h = 1.0 x 10^-3 Pascal second (Pa s). (Note: Pa is the SI unit of pressure. Pa = N/m² . Another unit of viscosity is the Poise. 1 Pa s = 10 Poise. Use Pa s in these equations.)

Siphoning water at 20^o C with a tube of length 1 meter and radius 1 millimeter (but use 10^-3 meters) and with h = ¹/₂ m, you will find that equation 1 with D P = r gh yields a result about 3% higher than equation 3 velocity times m/r L, the "exact" equation. This gets worse rapidly as you raise r above a millimeter.

The rest of this paper involves calculus.

(You are not interested in this, so skip to the next paragraph.) If you plot the log of D Q/D t vs log of r, it will follow a straight line with slope 4 at low r and another straight line with slope 2 at high r. This shows the r⁴ dependence for small r (see equation 1) and the r² dependence for large r.

Next:

a) the initial variation in velocity when the siphon starts,

b) the variation in fluid current as a tank is emptied with the lower end of the siphon not immersed, and

c) the variation in fluid current as a tank is siphoned to another tank, and the final fluid levels are equal: i.e., h approaches zero.

For (a) write Newton's 2^nd law with drag proportional to velocity (F_drag = 8p h Lv), then we find (r gh - ¹/₂ r v²)A - 8p h Lv = r ALdv/dt. This can be rearranged so that dt is on one side and a mess of the form dv/(av² + bv + c) is on the other side. Now let SQR = (b²-4ac)^1/2. The indefinite integral is (1/SQR)ln[(2av+b-SQR)/(2av+b+SQR)]. Do the tedious work and get

v = v₁²/[v₂(1+exp(-t/t ))/(1-exp(-t/t )) + v_H ] (equation 4)

where v₁ = (2gh)^1/2, v₂ = 8h L/r r² = 8p h L²/m, where m is the mass of liquid in the tube, and v_H = (v₁+v₂)^1/2, the hypotenuse of a right triangle with no physical significance(?). The time constant, t , is L/v_H. In this derivation we have assumed that in the time it takes to reach its maximum v, h is constant. Now show that v_max = v_H - v₂ .(Yes, v₁²/(v₂+v_H) really is the same as v_H - v₂). Check it out, and note that it agrees with equation 3. Incidentally, the start-up time of an electric circuit is also on the order of the length of the circuit divided by a velocity, but it is the velocity of light, so t for this is extremely small.

In these equations, the time constant, or "mean time" is t. You can see that when t = t, e^-1 appears, and it is useful to know that e^-1 = 0.37, approximately.

b) In this derivation, assume that the surface area A of the liquid in the source tank is constant as it drains (tank has constant horizontal cross-section). Also assume that v is small, so neglect the v² term in D P. (I was able to integrate the thing using equation 3, but it did not lead to an equation that one could solve explicitly for v, so one would have to do iterations or something.) Then the current, -Adh/dt, is a function of h, and this leads to

v = v_oe^-t/t (equation 5)

where v_o = the maximum v in part a) and t = Av₂/p r²g, and substituting for v₂ and r leads to t =8p h r AL³/m²g. (Experimentally, you can avoid the measurement of r by using r = m/p r²L, so p r² = m/r L. You can get better precision by measuring mass m of liquid in the tube than by measuring r.) Another way of expressing t is to define C = A/r g, analogous to capacitance in electricity. Then t = RC, where R is the hydraulic resistance, 8h L/p r⁴ = 8p h r ²L³/m². One can also write h = h_oe^-t/t and it has the same time constant. This is the equivalent of discharging a capacitor through a resistor.

c) If you siphon from one tank to an identical tank, the result is the same as in b) but with half the time constant. This is the equivalent of charging a capacitor with a capacitor. If you siphon between non-identical tanks, the time constant is RC_s, where C_s is C₁C₂/(C₁+C₂). This is the formula for the effective total capacitance for two capacitors in series, and the same result is found for the electrical case. By the way, in this case h is the difference in liquid levels.

I have not seen this stuff beyond equation 2 worked out anywhere, so let me know if you find anything wrong with it.

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