/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 A large storage tank, open at th... [FREE SOLUTION] | 91Ó°ÊÓ

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A large storage tank, open at the top and filled with water, develops a small hole in its side at a point \(16.0 \mathrm{m}\) below the water level. If the rate of flow from the leak is equal to \(2.50 \times 10^{-3} \mathrm{m}^{3} / \mathrm{min},\) determine (a) the specd at which the water leaves the hole and (b) the diameter of the hole.

Short Answer

Expert verified
To solve the problem, the first step is to reach the speed at which the liquid leaves the hole by calculating \(v = \sqrt{(2)(9.8 \mathrm{m/s}^2)(16.0 \mathrm{m})}\). The next step is to find the diameter of the hole. The diameter is reached using the formula \(d = 2 * \sqrt{A/\pi}\) where \(A\) is the area, computed as \(Q/v\).

Step by step solution

01

Apply Torricelli's theorem

The formula based off Torricelli's Theorem calculates the speed \(v\) at which the fluid leaves the hole. \(v = \sqrt{2gh}\) where \(g\) is the acceleration due to gravity (9.8 m/s²) and \(h\) is the height of fluid above the hole (16m).
02

Calculate the speed

Plug the values into the formula: \(v = \sqrt{(2)(9.8 \mathrm{m/s}^2)(16.0 \mathrm{m})}\). This will give the speed \(v\) at which the fluid leaves the hole.
03

Apply formula for Volume Flow Rate

Next, calculate the diameter of the hole, using the formula for volume flow rate: \[Q = A*v\] where \(Q\) is the flow rate \((2.50 x 10^{-3} m^3/min)\) and \(v\) is the speed at which fluid leaves the hole, calculated above.
04

Calculate hole diameter

First, calculate the area \(A\) of the hole: \[ A = Q/v \]. The hole is circular, so the area of the hole is given by:\[ A = \pi*(d/2)^2 \]. Isolate \(d\) in the above formula to get the diameter \(d\) of the hole: \[ d = 2 * \sqrt{A/\pi} \].Insert the previously calculated area into this formula to find the diameter of the hole.

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is a subdiscipline of fluid mechanics dealing with the flow of fluids — liquids and gases. It involves the study of the conditions under which fluids move, the forces involved, and the energy associated with them. Torricelli's theorem, utilized in the textbook exercise, is a powerful concept in fluid dynamics that describes the speed of a fluid flowing out of an orifice under the influence of gravity. The theorem is applicable to the scenario where a fluid is exiting a hole in a tank. The speed at which the fluid leaves is directly related to the height of the fluid column above the exit point because the potential energy due to the fluid's height gets converted into kinetic energy.

Understanding the principles of fluid dynamics, such as continuity and the Bernoulli equation, aids in comprehending Torricelli's theorem. The continuity equation states that the mass flow rate must remain constant from one cross-section of a pipe to another, which lays the groundwork for understanding the relationship between the speed of the fluid and the cross-sectional area of the hole, represented by the formula for volume flow rate.
Volume Flow Rate
The volume flow rate is a measure of how much volume of fluid passes through a cross-section of a pipe or an opening per unit time. It is a crucial concept in both theoretical and practical applications of fluid mechanics, and is given by the equation \( Q = A \times v \), where \(Q\) is the volume flow rate, \(A\) is the cross-sectional area, and \(v\) is the velocity of the fluid.

In the context of the textbook exercise, the volume flow rate is given, and the task is to deduce information about the hole through which the fluid escapes. Once the speed \(v\) is calculated using Torricelli's theorem, the cross-sectional area \(A\) can be determined. Since the hole is circular, the area is related to the diameter \(d\) of the hole using the formula \( A = \frac{\text{\textpi} d^2}{4} \). By understanding how the area relates to diameter, students can apply algebra to solve for \(d\), providing an excellent example of how volume flow rate connects fluid dynamics with geometry.
Acceleration Due to Gravity
Acceleration due to gravity is a constant force that imparts on all objects on Earth due to the planet's mass. It is denoted as \( g \) and has a value of approximately \(9.8 \text{ m/s}^2\). This force is crucial to understand various phenomena in physics and plays a pivotal role in fluid dynamics, specifically in scenarios involving fluid flow affected by gravity.

In our exercise, acceleration due to gravity is central to Torricelli's theorem, where it is used to calculate the speed at which water leaves the hole in the storage tank. The formula for this speed is based on the conversion of potential energy to kinetic energy, and gravity's acceleration is a defining factor in this energy conversion. Essentially, it provides a link between the energy state of the fluid in the tank and its velocity upon exiting, where a higher height of the fluid column, or a greater gravitational force, would lead to a faster exit speed according to \( v = \sqrt{2gh} \). This highlights gravity's fundamental role in the characteristics of fluid flow and dynamics.

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