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Chapter 11: Problem 66

A small crack occurs at the base of a \(15.0-\mathrm{m}\) -high dam. The effective crack area through which water leaves is \(1.30 \times 10^{-3} \mathrm{m}^{2}\) (a) Ignoring viscous losses, what is the speed of water flowing through the crack? (b) How many cubic meters of water per second leave the dam?

Short Answer

Expert verified
(a) The speed of water flow is approximately 17.15 m/s. (b) The flow rate is approximately 0.0223 m鲁/s.

Step by step solution

01

Understand Torricelli's Theorem

Torricelli's Theorem states that the speed of efflux of a fluid under gravity from a hole at height \( h \) is given by the formula \( v = \sqrt{2gh} \), where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
02

Apply the Formula for Speed

Use Torricelli鈥檚 theorem to find the speed \( v \): \( v = \sqrt{2gh} \). Substitute \( g = 9.81 \, \text{m/s}^2 \) and \( h = 15.0 \, \text{m} \). Calculate:\[ v = \sqrt{2 \times 9.81 \, \text{m/s}^2 \times 15.0 \, \text{m}} \approx \sqrt{294.3} \approx 17.15 \, \text{m/s}. \]
03

Calculate Volume Flow Rate

The volume flow rate \( Q \) is given by \( Q = A \times v \), where \( A \) is the area of the crack. Given \( A = 1.30 \times 10^{-3} \, \text{m}^2 \) and \( v = 17.15 \, \text{m/s} \), calculate \( Q \):\[ Q = 1.30 \times 10^{-3} \, \text{m}^2 \times 17.15 \, \text{m/s} \approx 2.23 \times 10^{-2} \, \text{m}^3/ ext{s}. \]

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Key Concepts

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

Torricelli's Theorem
Torricelli's Theorem offers a fascinating insight into fluid dynamics. Developed by Evangelista Torricelli, it allows us to calculate the speed at which a fluid will exit a hole due to gravity. The theorem is summarized by the formula \( v = \sqrt{2gh} \), where \( v \) represents the efflux speed, \( g \) is the acceleration due to gravity (about \( 9.81 \, \text{m/s}^2 \)), and \( h \) denotes the height of the fluid above the opening.
This theorem is particularly useful for determining how fast a liquid will flow from a reservoir, such as our exercise involving a dam.
  • It assumes the liquid is non-viscous, meaning it neglects internal friction and other losses.
  • The theorem applies under ideal fluid conditions, which means the only force is gravity and the liquid surface is open to the atmosphere.
On solving, Torricelli's Theorem helps find the speed at which water exits through a crack in a dam.
Efflux Speed
Efflux speed refers to the velocity at which a fluid exits an opening due to the influence of gravity. This concept is crucial in scenarios where a fluid flows freely out of a container, such as a dam or a tank.
The efflux speed can be easily determined using Torricelli鈥檚 Theorem. In the problem given, the speed of the water flowing from the base of a dam is computed using the equation \( v = \sqrt{2gh} \).
  • Substitute \( g = 9.81 \, \text{m/s}^2 \) as the value for gravity.
  • Use \( h = 15.0 \, \text{m} \), the height of the water above the crack.
Calculating gives us \( v \approx 17.15 \, \text{m/s} \), demonstrating how quickly water escapes under these conditions.
Volume Flow Rate
The volume flow rate is the quantity of fluid that flows through a cross-section per unit time, often measured in cubic meters per second \((\text{m}^3/\text{s})\). This is an essential concept when analyzing systems involving fluid transfer.
To compute the volume flow rate, you multiply the efflux speed by the area of the opening. In our scenario, this is \( Q = A \times v \).
  • Given \( A = 1.30 \times 10^{-3} \, \text{m}^2 \), which is the area of the crack.
  • The speed \( v \) is calculated as \( 17.15 \, \text{m/s} \).
The resulting volume flow rate is \( Q \approx 2.23 \times 10^{-2} \, \text{m}^3/\text{s} \), indicating the rate at which water exits the dam through the crack.
Hydrodynamics
Hydrodynamics is the branch of physics that deals with the motion of fluids, particularly liquids. It encompasses various principles and laws that explain how fluids behave when in motion, like the water exiting a crack in a dam.
Key factors influencing hydrodynamics include:
  • The fluid鈥檚 speed and direction of flow.
  • The geometrical constraints of the system, such as the area of an opening.
  • Environmental interactions like gravity, shown in Torricelli鈥檚 Theorem.
In this context, hydrodynamics helps understand how factors such as pressure differences, gravity, and flow areas impact the movement of water through a structure. Mastery of these concepts allows engineers and scientists to design better hydraulic structures and systems for efficient water management.

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Most popular questions from this chapter

Three fire hoses are connected to a fire hydrant. Each hose has a radius of 0.020 m. Water enters the hydrant through an underground pipe of radius 0.080 m. In this pipe the water has a speed of 3.0 m/s. (a) How many kilograms of water are poured onto a fire in one hour by all three hoses? (b) Find the water speed in each hose.

A mercury barometer reads 747.0 mm on the roof of a building and 760.0 \(\mathrm{mm}\) on the ground. Assuming a constant value of 1.29 \(\mathrm{kg} / \mathrm{m}^{3}\) for the density of air, determine the height of the building.

The drawing shows a hydraulic system used with disc brakes. The force \(\overrightarrow{\mathbf{F}}\) is applied perpendicularly to the brake pedal. The pedal rotates about the axis shown in the drawing and causes a force to be applied perpendicularly to the input piston (radius \(=9.50 \times 10^{-3} \mathrm{m} )\) in the master cylinder. The resulting pressure is transmitted by the brake fluid to the output plungers (radii \(=1.90 \times 10^{-2} \mathrm{m}\) ), which are covered with the brake linings. The linings are pressed against both sides of a disc attached to the rotating wheel. Suppose that the magnitude of \(\overrightarrow{\mathbf{F}}\) is 9.00 \(\mathrm{N}\) . Assume that the input piston and the output plungers are at the same vertical level, and find the force applied to each side of the rotating disc.

A water bed for sale has dimensions of 1.83 \(\mathrm{m} \times 2.13 \mathrm{m} \times 0.229 \mathrm{m} .\) The floor of the bedrroom will tolerate an additional weight of no more than 66660 \(\mathrm{N}\) . Find the weight of the water in the bed and determine whether the bed should be purchased.

A hydrometer is a device used to measure the density of a liquid. It is a cylindrical tube weighted at one end, so that it floats with the heavier end downward. The tube is contained inside a large 鈥渕edicine dropper,鈥� into which the liquid is drawn using the squeeze bulb (see the drawing). For use with your car, marks are put on the tube so that the level at which it floats indicates whether the liquid is battery acid (more dense) or antifreeze (less dense). The hydrometer has a weight of \(W=5.88 \times 10^{-2} \mathrm{N}\) and a cross-sectional area of tw \(A=7.85 \times 10^{-5} \mathrm{m}^{2} .\) How far from the bottom of the tube should the mark be put that denotes (a) battery acid \(\left(\rho=1280 \mathrm{kg} / \mathrm{m}^{3}\right)\) and (b) antifreeze \(\left(\rho=1073 \mathrm{kg} / \mathrm{m}^{3}\right) ?\)
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