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 P14.50. Assume steady flow without friction. (a) If $h=1.00 \mathrm{m},$ find the speed 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 end of the siphon? Note: For the flow of the liquid to be continuous, its pressure must not drop below its vapor pressure. Assume the water is at $20.0^{\circ} \mathrm{C},$ at which the vapor pressure is $2.3 \mathrm{kPa}$. (FIGURE CANT COPY)

Key Concepts

Bernoulli's Principle

Bernoulli's Principle states that in a streamline flow, the sum of the pressure energy, kinetic energy per unit volume, and gravitational potential energy per unit volume remains constant. This principle is used to relate the speed, elevation, and pressure at different points in the flow of an ideal, incompressible, and frictionless fluid, which is essential in analyzing the flow through a siphon.

Hydrostatic Pressure

Hydrostatic pressure refers to the pressure exerted by a fluid at rest due to the force of gravity. The pressure increases linearly with depth, and this concept is crucial for determining the pressure difference between two points in the tank and within the siphon, which in turn allows the calculation of the fluid velocity at the siphon’s outlet.

Siphon Operation

The siphon is a device that uses gravity and atmospheric pressure differences to transfer liquid from one level to a lower level, even when the source is above the exit point. Understanding how the siphon works involves applying energy conservation and recognizing that fluid will continue to flow as long as the pressure at the highest point does not fall below the vapor pressure.

Vapor Pressure and Cavitation

Vapor pressure is the pressure at which a liquid begins to vaporize. In the context of a siphon, the fluid must maintain a pressure above its vapor pressure to remain in liquid form and avoid cavitation, which is the formation of vapor bubbles that can interrupt smooth and continuous flow. This concept sets an upper limit on how high the top of the siphon can be relative to the exit point.

Transcript

00:01 In this problem, we're going to talk about bernoulli's equation.

00:04 So what we need to remember is that for a streamline of an incompressible fluid, this constant is concerned.

00:15 We have p, that's the pressure plus one -half, of row, which is the density times the velocity squared, and this is the kinetic term, plus row g times y, y is the height of the fluid, is constant.

00:30 So what we have in our problem is a fluid, actually this density here is wrong.

00:40 We have a fluid, actually it's water inside a siphon, as shown here in the picture.

00:47 And the value of h is equal to one meter.

00:54 In question a, we have to calculate what is the velocity v with which the fluid exits the siphon.

01:06 So we're going to use bernoula's equation.

01:08 And we're going to consider that at this point here, which i'm going to call a, this one here, the surface of the water first, the pressure is equal to the atmospheric pressure, v0, and v is equal to zero.

01:31 So we're going to assume that the water almost doesn't move at this point.

01:37 Okay, and at this point, i'm going to set y equals zero.

01:46 Okay, because actually, instead of y equals zero, i'm going to set y equals 1, such that y equals zero is at the point where the water exits the siphon.

02:02 Okay, this is in point a.

02:05 Now in point b, in this point, the pressure will also be.

02:15 The atmospheric pressure.

02:16 The speed is exactly what we want to find.

02:20 And y is zero.

02:21 So now we can apply bernouil's equation to our problem...

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