A siphon is used to drain water from a tank, as illustrated...

A siphon is used to drain water from a tank, as illustrated in Figure P14.51. The siphon has a uniform diameter. Assume steady flow without friction. (a) If the distance $h=1.00 \mathrm{m}$ find the specd of outflow at the end of the siphon. (b) What If? What is the limitation on the height of the top of the siphon above the water surface? (For the flow of the liquid to be continuous, the pressure must not drop below the vapor pressure of the liquid.) (FIGURE CAN'T COPY)

A siphon is used to drain water from a tank, as illustrated in Figure P14.51. The siphon has a uniform diameter. Assume steady flow without friction. (a) If the distance $h=1.00 \mathrm{m}$ find the specd of outflow at the end of the siphon.
(b) What If? What is the limitation on the height of the top of the siphon above the water surface? (For the flow of the liquid to be continuous, the pressure must not drop below the vapor pressure of the liquid.)
(FIGURE CAN'T COPY)

Key Concepts

Bernoulli's Principle

Bernoulli's Principle states that for a fluid flowing without friction along a streamline, the sum of its pressure energy, kinetic energy, and potential energy per unit volume remains constant. This concept is crucial when analyzing systems like siphons, where potential energy (due to elevation differences) is converted into kinetic energy as the fluid flows from one point to another.

Conservation of Energy in Fluid Flow

The principle of energy conservation in fluid flow implies that the energy within the fluid is redistributed between pressure, velocity, and height, but the total energy remains constant in an ideal (frictionless, steady) flow. This concept underlies the calculation of the outflow speed from the siphon, linking the height difference in the fluid to the resulting velocity.

Siphon Operation

A siphon is a device that transfers liquid from a higher elevation to a lower one by creating a continuous flow driven by gravitational forces. The fluid is initially raised over an intermediate elevation, and once the siphon is primed, the fluid flows downhill, utilizing the pressure differences created by height variations. The uniform diameter and steady, frictionless flow assumptions simplify the analysis using theoretical principles like Bernoulli’s equation.

Vapor Pressure Limitation

Vapor pressure limitation refers to the condition that the pressure within the fluid must not fall below its vapor pressure, as doing so will result in the formation of vapor bubbles (cavitation), disrupting the continuous flow. In a siphon, this constraint limits the maximum allowable elevation difference between the highest point of the siphon and the free surface of the liquid, ensuring that the pressure remains above the vapor pressure for the liquid to flow continuously.

Transcript

00:01 Is about a side of the water.

00:22 And our question is to figure out the of this water.

00:32 So simply what we can do is look at the pressure of the water when it's a depth h from the surface.

00:43 Well, the difference in pressure there is just going to equal the density of water times the gravitation of acceleration, times the depth or h in this case.

01:00 This is going to equal one half row w, v squared, where v is the outgoing velocity of the water.

01:15 Since we know that, we can just figure out for v.

01:20 So you can see that the row, the density of the waters cancel out, so we're just left with gh equals 1 half v squared, which means that we'll get v equals square root of 2gh.

01:42 And now we're given that h is equal to 1 meter, we'll have 2 times 9 .8.

01:52 Meters per second squared times one meter.

01:57 And once you crunch the numbers, you get 4 .43 meters per second.

02:03 So that was part a.

02:06 In part b, we're asked to figure out the limitation of the height of this siphon.

02:12 So right now, we will note this height as y.

02:21 We're trying to find the maximum height, the limitation of this y variable.

02:30 For this, we can start out by using the bernoulli's equation, where we'll have this point as p1 and this point as p2.

02:45 So let's do that.

02:46 We have this bernoulli's equation, so we have p1 plus row density of water.

02:52 I'm just going to ignore the sub w since we're dealing with water the whole way.

02:58 Rho g, the height, plus one -half row v1 squared equals the end of this siphon is exposed to the atmosphere.

03:17 So p2 is just going to be the atmosphere plus.

03:23 And we're just going to, this is zero, since this is y starting from this point, the height of this is just zero...

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