/*! 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 54 A large storage tank, open to th... [FREE SOLUTION] | 91Ó°ÊÓ

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A large storage tank, open to the atmosphere 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 \(2.50 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{min}\), determine (a) the speed at which the water leaves the hole and (b) the diameter of the hole.

Short Answer

Expert verified
The speed at which the water leaves the hole is 17.9 m/s and the diameter of the hole is roughly 1.13 mm.

Step by step solution

01

Apply Torricelli's theorem

Torricelli's theorem states that the speed of outflow (v) from the hole is given by: \(v = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity and \(h\) is the height of the water level above the hole. Here, \(g = 9.81 m/s^2\) and \(h = 16 m\). Therefore, by substituting these values into the equation, we get the speed of the water as it leaves the hole.
02

Compute The Speed Of The Water

By substituting \(g = 9.81 m/s^2\) and \(h = 16 m\) into \(v = \sqrt{2gh}\), we get: \(v = \sqrt{(2)(9.81 m/s^2)(16 m)} = 17.9 m/s\).
03

Apply The Continuity Equation

The continuity equation is given by \(\text{Rate of flow} = \text{Cross-sectional area} \times \text{speed}\). The rate of flow given is \(2.50 \times 10^{-3} m^3/min\) or \(\frac{2.50 \times 10^{-3} m^3}{60 s} = 4.167 \times 10^{-5} m^3/s\). We can express the cross-sectional area of the hole as \(\pi(\frac{d}{2})^2\), where \(d\) is the diameter of the hole. Rearranging for \(d\) gives \(d = \sqrt{\frac{4 \times \text{Rate of flow}}{\pi \times v}}\).
04

Compute The Diameter Of The Hole

Substituting \(4.167 \times 10^{-5} m^3/s\) for Rate of flow and \(17.9 m/s\) for \(v\) into \(d = \sqrt{\frac{4 \times \text{Rate of flow}}{\pi \times v}}\), we get: \(d = \sqrt{\frac{4 \times 4.167 \times 10^{-5} m^3/s}{3.14 \times 17.9 m/s}} = 0.00113 m\) or \(1.13 mm\).

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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 the study of the motion of fluids, including liquids and gases. It is a sub-discipline of fluid mechanics that deals with fluid flow—the science of the forces that affect fluids in motion. In our everyday experiences, we see examples of fluid dynamics in the form of water flowing from a tap, air streaming around a moving car, or the behavior of a large storage tank when a hole appears on its side, causing water to gush out. For such a scenario, principles that govern the fluid's behavior are crucial to understanding the situation comprehensively.

Specifically, the notion that the pressure within a fluid and the fluid's velocity are inversely related is an essential aspect of fluid dynamics. This principle is observed when water jets out at a high speed from a small hole at the tank's side, caused by the high water pressure at the depth where the hole is located. The study of dynamics of the leaking tank involves complex interactions between fluid statics, which is the study of fluids at rest, and kinematics, which describes the motion of objects without reference to the forces that cause the motion.
Continuity Equation
The continuity equation is a mathematical expression of the principle of conservation of mass in fluid dynamics. It essentially states that within a closed system, the mass flowing into a system must be equal to the mass flowing out, assuming there is no accumulation of mass within the system. When applied to a steady, incompressible flow, the continuity equation simplifies to the statement that the product of the cross-sectional area of the fluid flow (A) and its velocity (v) is a constant.

This can be expressed mathematically as: \[ A_1v_1 = A_2v_2 \] where \( A_1 \) and \( A_2 \) are the cross-sectional areas of the fluid at two different points and \( v_1 \) and \( v_2 \) are the respective velocities at those points. In the context of our tank with a hole, the continuity equation enables us to determine the speed of water flow or the size of the hole when provided with the rate of flow through the hole, as seen in the step-by-step solution.
Rate of Flow
The rate of flow, often referred to as volumetric flow rate, is one of the imperative characteristics in the study of fluid dynamics. It represents the volume of fluid that passes through a given surface area per unit time. Mathematically, it is described as: \[ Q = A \times v \] where \( Q \) stands for the rate of flow, \( A \) for the cross-sectional area of the passage, and \( v \) for the velocity of the fluid. Importantly, in practical situations, the rate of flow will determine how rapidly a tank can be emptied or filled and is directly used in the design of efficient systems involving fluid transfer.

Furthermore, when a leakage happens as in the exercise, knowing the rate of flow aids in evaluating the severity of the leakage and the potential consequences. Once provided with the rate of flow and the speed of the fluid outflow, one can deduce characteristics of the outflow such as the area of the hole through which the fluid is escaping as seen in the provided step-by-step solution.

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