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A small circular hole \(6.00 \mathrm{~mm}\) in diameter is cut in the side of a large water tank, \(14.0 \mathrm{~m}\) below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

Short Answer

Expert verified
The speed of efflux of the water is 16.68 m/s and the volume discharged per second is \(4.72 * 10^-4 m^3/s.\)

Step by step solution

01

Calculate Speed of Efflux

By applying the Torricelli's theorem, which is a special case of Bernoulli's theorem, the speed \(\(v\)\) of efflux can be calculated using the equation \(v = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity (in this case, \(9.81 m/s^2\)) and \(h\) is the height of the water level above the hole (in this case, \(14.0 m\)).
02

Derive the value of Speed of Efflux

By substituting the given values, the speed of efflux (\(v\)) = \sqrt{2*9.81*14.0 m} = 16.68 m/s.
03

Calculate Volume Discharge

To find out the volume discharge per second, multiply the cross-sectional area of the hole (\(A\)) with the speed of efflux (\(v\)). Area of the hole can be calculated using the formula \(A = \pi r^2\), where \(r\) is the radius of the hole. Here, \(r = diameter/2 = 6.0 mm/2 = 3.0 mm = 0.003 m.\)
04

Derive the value of Volume Discharge

By substituting the given values, the volume discharge (\(Q\)) per second = Area of hole * speed of efflux = \( \pi * (0.003 m)^2 * 16.68 m/s = 4.72 * 10^-4 m^3/s.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Speed of Efflux
Understanding the speed of efflux plays a crucial role when analyzing fluid dynamics, particularly in scenarios involving the exit velocity of fluid through an orifice. Torricelli's theorem is instrumental in determining this speed, which is the velocity at which fluid leaves a hole in a container.

The formula derived from Torricelli's theorem is given by \( v = \sqrt{2gh} \), where \( v \) is the speed of efflux, \( g \) represents the acceleration due to gravity (approximately \( 9.81 m/s^2 \) on the surface of the Earth), and \( h \) is the height of the fluid above the hole. Think of this speed as the final velocity that a freely falling object would have after descending the height \( h \) under the influence of gravity.

Applying this knowledge to the exercise, with a water level 14.0 meters above the hole, we can use Torricelli's theorem to calculate the exit velocity or speed of efflux of the water.
Volume Discharge
Moving beyond the speed at which a fluid exits, we encounter the concept of volume discharge. This term refers to the quantity of fluid that flows out of an orifice per unit of time. To find the volume discharge, \( Q \), we multiply the speed of efflux, \( v \), by the cross-sectional area, \( A \), of the hole through which the fluid is discharged.

In mathematical terms, \( Q = A \times v \). The area \( A \) can be determined for circular holes with the formula \( A = \pi r^2 \), where \( r \) is the radius of the hole. The unit of volume discharge commonly used is cubic meters per second \( m^3/s \). Hence, for our exercise, once the speed of efflux is known, calculating the volume discharged per second becomes a straightforward multiplication.
Bernoulli's Theorem
Diving deeper, Bernoulli's theorem, also known as Bernoulli's principle, forms the foundation for understanding fluid behavior in various situations. It states that within a flowing fluid, there is a constant sum of potential energy, kinetic energy, and pressure energy.

The principle can be expressed as \( P + \frac{1}{2}\rho v^2 + \rho gh = \text{constant} \), where \( P \) stands for the fluid's pressure, \( \rho \) its density, \( v \) its velocity, and \( h \) the height relative to a reference point. Bernoulli's theorem illustrates that the sum of the fluid's pressure head, velocity head, and height head remains unchanged along a streamline.

Torricelli's theorem is actually a specific case of Bernoulli’s theorem. It's applied when a fluid exits into the air, which implies that the pressure at the exit point is atmospheric pressure and can be cancelled from the equation.
Acceleration due to Gravity
The acceleration due to gravity, denoted by \( g \), is the rate at which objects accelerate towards Earth without resistance, such as air resistance, being involved. Its standard value is \( 9.81 m/s^2 \). This acceleration is a key variable in equations for projectile motion, freefall, and, as we've seen, fluid dynamics with the Torricelli's theorem.

In our problem, the effect of gravity defines the speed of efflux as it is the force causing the water to accelerate as it exits the hole. It's important to note that while \( g \) is constant in many calculations, it can vary slightly depending on altitude and location on Earth, factors that are usually negligible in most high school or introductory college physics problems.

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Most popular questions from this chapter

A firehose must be able to shoot water to the top of a building \(28.0 \mathrm{~m}\) tall when aimed straight up. Water enters this hose at a steady rate of \(0.500 \mathrm{~m}^{3} / \mathrm{s}\) and shoots out of a round nozzle. Neglect air resistance. (a) What is the maximum diameter this nozzle can have? (b) If the only nozzle available has a diameter twice as great, what is the highest point the water can reach?

A liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed \(v_{0}\) and the radius of the stream of liquid is \(r_{0}\). (a) Find an equation for the speed of the liquid as a function of the distance \(y\) it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of \(y\). (b) If water flows out of a vertical pipe at a speed of \(1.20 \mathrm{~m} / \mathrm{s},\) how far below the outlet will the radius be one-half the original radius of the stream?

A \(950 \mathrm{~kg}\) cylindrical buoy floats vertically in seawater. The diameter of the buoy is \(0.900 \mathrm{~m} .\) Calculate the additional distance the buoy will sink when an \(80.0 \mathrm{~kg}\) man stands on top of it.

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What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of \(15.0 \mathrm{~m} ?\) (Assume that the mains have a much larger diameter than the fire hose.

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