A woman is draining her fish tank by siphoning the water...

A woman is draining her fish tank by siphoning the water into an outdoor drain as shown in Figure $\mathrm{P} 14.42$. The rectangular tank has footprint area $A$ and depth $h .$ The drain is located a distance $d$ below the surface of the water in the tank, where $d>>h$. The crosssectional area of the siphon tube is $A^{\prime}$. Model the water as flowing without friction. Show that the time interval required to empty the tank is given by $\Delta t=\frac{A h}{A^{\prime} \sqrt{2 g d}}$

A woman is draining her fish tank by siphoning the water into an outdoor drain as shown in Figure $\mathrm{P} 14.42$. The rectangular tank has footprint area $A$ and depth $h .$ The drain is located a distance $d$ below the surface of the water
in the tank, where $d>>h$. The crosssectional area of the siphon tube is $A^{\prime}$. Model the water as flowing without friction. Show that the time interval required to empty the tank is given by $\Delta t=\frac{A h}{A^{\prime} \sqrt{2 g d}}$

Key Concepts

Differential Equation Integration

To compute the time required to empty the tank, one must set up and integrate a differential equation that relates the rate of change of the water’s volume (derived from the tank’s footprint area A and depth h) to the outflow rate through the siphon tube (involving A' and the velocity given by sqrt(2gd)). Solving this differential equation yields the final expression for the emptying time, which highlights the interplay between geometric and gravitational parameters.

Continuity Equation

The continuity equation in fluid dynamics expresses the conservation of mass by ensuring the volume flow rate is constant along a streamline. This means that the product of the cross-sectional area and the velocity of the flow (A' and sqrt(2gd) in the siphon) must equate to the rate at which the volume of water in the tank decreases. This relationship is essential for setting up the differential equation whose integration gives the total drainage time.

Bernoulli's Principle

Bernoulli's principle states that for an incompressible, non-viscous fluid in steady flow, the sum of the pressure energy, kinetic energy, and potential energy per unit volume remains constant along a streamline. In this context, it is used to relate the gravitational potential difference (due to the vertical drop d) to the kinetic energy of the water as it exits the siphon, leading to the derivation of the outflow velocity.

Torricelli's Law

Torricelli's law is a specific application of Bernoulli’s principle that provides the speed of efflux of a fluid under gravity from an opening. It typically states that the exit speed is sqrt(2gh) when the height of the fluid above the hole is h. In this problem, because the siphon’s exit is located a distance d below the water surface (with d much larger than the tank depth), the speed becomes approximately sqrt(2gd), a key step in simplifying the overall time-to-drain formula.

Transcript

00:01 Hi there.

00:02 So for this problem, we have the situation that is shown in here.

00:06 And what we need to show is that the timing turbo required to empty the tan is given by the expression that we are given there.

00:14 So we're going to take the point one at here.

00:23 That is the point at the free water surface in the tank.

00:27 And the point two at the bottom end of the tube.

00:31 Just right here.

00:32 That is the point two.

00:33 So once we set thought, we can use the bernoules equation.

00:42 So we will have that the pressure at the point one.

00:45 Let me just move this a little bit.

00:47 Okay, the pressure or the point one, plus the density times the acceleration due to gravity times the height at the point one.

00:56 And this plus 1 divided by 2 times the density and the speed.

01:03 Oh, sorry, the density and the speed at the point 1 to the square is equal to the pressure at the point 2.

01:11 This plus the density times the acceleration due to the gravity times the height at the point 2, and this plus 1 divided by 2 times the density times the speed 2 to the square.

01:27 Now we can simplify this because we know that the speed at the point 1, we can assume that that is 0.

01:32 The speed at the point of altitude y2 is just 0...

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