/*! 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 70 A vertical cylindrical tank of c... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A vertical cylindrical tank of cross-sectional area \(A_{1}\) is open to the air at the top and contains water to a depth \(h_{0}\) . A worker accidentally pokes a hole of area \(A_{2}\) in the bottom of the tank. (a) Derive an equation for the depth \(h\) of the water as a function of time \(t\) after the hole is poked. (b) How long after the hole is made does it take for the tank to empty out?

Short Answer

Expert verified
The depth as a function of time is \(h(t) = \left(\sqrt{h_0} - \frac{A_2\sqrt{2g}}{2A_1} t\right)^2\), and the tank empties in \(t = \frac{2A_1\sqrt{h_0}}{A_2\sqrt{2g}}\)."

Step by step solution

01

Understanding the Setup

We have a cylindrical tank with cross-sectional area \(A_1\) filled with water to a depth \(h_0\). A hole of area \(A_2\) is poked at the bottom, allowing water to escape. We need to derive how the depth \(h\) changes over time \(t\) and find the time it takes to empty the tank.
02

Applying Torricelli’s Law

Using Torricelli’s Law, the speed \(v\) of water exiting the hole can be described by the equation \(v = \sqrt{2gh}\), where \(g\) is the acceleration due to gravity. This is derived from the principle of conservation of energy.
03

Relating Flow Rate to Change in Water Level

The flow rate through the hole is given by \(Q = A_2v = A_2\sqrt{2gh}\). This flow rate also equals the rate of change of volume in the tank: \(Q = -A_1\frac{dh}{dt}\).
04

Setting up the Differential Equation

Equating the two expressions for \(Q\), we have \(-A_1\frac{dh}{dt} = A_2\sqrt{2gh}\). This simplifies to \(\frac{dh}{dt} = -\frac{A_2}{A_1}\sqrt{2gh}\).
05

Separating Variables

Separate variables to solve the differential equation: \(\frac{dh}{\sqrt{h}} = -\frac{A_2\sqrt{2g}}{A_1} dt\).
06

Integrating Both Sides

Integrate both sides, with \(h = h_0\) at \(t = 0\):\[\int_{h_0}^{h} \frac{1}{\sqrt{h'}} \, dh' = -\frac{A_2\sqrt{2g}}{A_1} \int_{0}^{t} dt'\].This results in \(2\sqrt{h} - 2\sqrt{h_0} = -\frac{A_2\sqrt{2g}}{A_1} t\).
07

Solving for h(t)

Rearrange to solve for \(h\) as a function of \(t\):\[\sqrt{h(t)} = \sqrt{h_0} - \frac{A_2\sqrt{2g}}{2A_1} t\],thus, \[h(t) = \left(\sqrt{h_0} - \frac{A_2\sqrt{2g}}{2A_1} t\right)^2\].
08

Determining Time to Empty the Tank

The tank is empty when \(h(t) = 0\). Set \(\sqrt{h_0} - \frac{A_2\sqrt{2g}}{2A_1} t = 0\) and solve for \(t\):\[t = \frac{2A_1\sqrt{h_0}}{A_2\sqrt{2g}}\].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Torricelli's Law
Torricelli's Law is a fascinating concept from fluid dynamics that helps us understand how fluids exit a container. It's named after Evangelista Torricelli, who discovered that the speed of water flowing out of an opening is related to the height of the liquid above the opening. This is because the potential energy of the liquid at height is converted into kinetic energy as it exits. The formula derived from this principle is \( v = \sqrt{2gh} \), where:
  • \( v \) is the velocity of the water exiting the hole.
  • \( g \) is the acceleration due to gravity, approximately \( 9.81 \text{ m/s}^2 \).
  • \( h \) is the height of the water above the opening.
This velocity formula is crucial in solving problems involving fluid flow, such as calculating the rate at which a tank empties.
Conservation of Energy
The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In fluid dynamics, this principle explains how potential energy in a fluid converts to kinetic energy as it flows. In the scenario of a tank with water, potential energy is due to the height of the water column above the hole. When the water falls, this stored energy transforms into kinetic energy, increasing the water's speed as it exits. Therefore, the velocity formula in Torricelli's Law is a direct consequence of energy conservation. Understanding this energy transformation allows us to manipulate and predict fluid motion, making it an essential concept in many physics problems.
Differential Equations
Differential equations are vital mathematical tools in physics used to describe how quantities change. They help us model the dynamics of systems, like how the water level in a tank changes over time.In our exercise, we encounter the differential equation \( \frac{dh}{dt} = -\frac{A_2}{A_1}\sqrt{2gh} \). This equation:
  • Relates the rate of change of the water height \( h \) to the flow of water out of the tank.
  • Incorporates the areas of the tank and the hole \( A_1 \) and \( A_2 \).
By solving this differential equation, we can find \( h(t) \), the water height at any time \( t \), providing a continuous understanding of the tank's draining process.
Flow Rate
Flow rate signifies how much fluid passes through a point in a given time period, usually expressed in cubic meters per second (\( \text{m}^3/\text{s} \)). In the context of a leaking tank, flow rate is determined by the size of the hole and the speed of the exiting liquid.For the hole at the bottom of the tank, flow rate \( Q \) can be calculated using the formula \( Q = A_2v \), where:
  • \( Q \) is the volumetric flow rate.
  • \( A_2 \) is the area of the hole.
  • \( v \) is the velocity of the water, provided by Torricelli's Law.
Understanding this helps in calculating how long it takes for the tank to empty, as it's directly related to the rate at which water leaves the tank.
Physics Problem Solving
Solving physics problems involves a clear, step-by-step approach. In our tank emptying scenario, we start by understanding the physical setup and the principles at play. We:
  • Utilize relevant physical laws like Torricelli's Law and conservation of energy.
  • Set up equations based on these principles, relating given rates and changes.
  • Apply mathematical techniques, such as solving differential equations, to derive function expressions like \( h(t) \).
  • Analyze our results to answer the posed question, such as finding the time when the tank is fully drained.
By breaking down the problem into manageable parts, we harness formulae and mathematics to arrive at a solution efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Great Molasses Flood. On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a \(27.4-\mathrm{m}\) -high 27.4 - diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 9 -m-deep stream, killing pedestrians and horses, and knocking down buildings. The molasses had a density of \(1600 \mathrm{kg} / \mathrm{m}^{3} .\) If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

A plastic ball has radius 12.0 \(\mathrm{cm}\) and floats in water with 16.0\(\%\) of its volume submerged. (a) What force must you apply to the ball to hold it at rest totally below the surface of the water? (b) If you let go of the ball, what is its acceleration the instant you release it?

A piece of wood is 0.600 \(\mathrm{m}\) long, 0.250 \(\mathrm{m}\) wide, and 0.080 \(\mathrm{m}\) thick. Its density is 600 \(\mathrm{kg} / \mathrm{m}^{3}\) . What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

An ore sample weighs 17.50 \(\mathrm{N}\) in air. When the sample is suspended by a light cord and totally immersed in water, the tension in the cord is 11.20 \(\mathrm{N}\) . Find the total volume and the density of the sample.

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.