/*! 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 52 Discharging Water from a Tank A ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Discharging Water from a Tank A container that has a constant cross section \(A\) is filled with water to height \(H\). The water is discharged through an opening of cross section \(B\) at the base of the container. By using Torricelli's Law, it can be shown that the height \(h\) of the water at time \(t\) satisfies the initialvalue problem $$ \frac{d h}{d t}=-\frac{B}{A} \sqrt{2 g h} \quad h(0)=H $$ a. Find an expression for \(h\). b. Find the time \(T\) it takes for the tank to empty. c. Find \(T\) if \(A=4\left(\mathrm{ft}^{2}\right), B=1\left(\mathrm{in} .^{2}\right), H=16(\mathrm{ft})\), and \(g=32\left(\mathrm{ft} / \mathrm{sec}^{2}\right)\)

Short Answer

Expert verified
a. The expression for the height $h$ of the water at time $t$ is: $$ h(t)= \frac{A^2 t^2 g + 4 H g^2 - 4\sqrt{(A^2 t^2 g + 2Hg) Hg }}{4g}. $$ b. The time $T$ it takes for the tank to empty is: $$ T=\frac{1}{2} \sqrt{\frac{8 H}{g}}\left(\frac{5-(3 \times \frac{B}{A})}{2}\right). $$ c. For the given values of $A=4\mathrm{ft}^2$, $B=1\mathrm{in}^2$, $H=16\mathrm{ft}$, and $g=32\mathrm{ft}/\mathrm{sec}^2$, the time $T$ it takes for the tank to empty is: $$ T= 3.5 \text{ sec}. $$

Step by step solution

01

Separate the variables

Rewrite the given differential equation as follows to separate variables: \[ \frac{dh}{\sqrt{2gh}} = -\frac{B}{A} dt. \]
02

Integrate both sides

Integrate both sides of the equation with respect to their respective variables: \[ \int \frac{dh}{\sqrt{2gh}} = - \frac{B}{A} \int dt. \] Notice we have h(0) = H, so we have limits on the left side from H to h, and on the right side from 0 to t.
03

Solve the integrals

The integral on the left side can be solved by substitution, setting u = 2gh, then du = 2g dh. Solving the integrals: \[ \frac{1}{g} \int \frac{du}{\sqrt{u}} = -\frac{B}{A} \int dt, \] \[ \frac{2}{g} \left[ \sqrt{u} \right]_H^h = -\frac{B}{A} \left[ t \right]_0^t. \] The above equation becomes: \[ \frac{2}{g}(\sqrt{2gh} - \sqrt{2gH}) = - \frac{B}{A} t. \]
04

Find an expression for h

Solve the equation above for h in terms of t: \[ h(t)= \frac{A^2 t^2 g + 4 H g^2 - 4\sqrt{(A^2 t^2 g + 2Hg) Hg }}{4g}. \]
05

Find the time T when the tank is empty

As the tank becomes empty, the height h becomes 0. Set h(t) = 0, and solve for t=T: \[ 0= \frac{A^2 T^2 g + 4 H g^2 - 4\sqrt{(A^2 T^2 g + 2Hg) Hg }}{4g}. \]
06

Find the specific value of T

Substitute the given values of A, B, H, and g into the formula found in step 5 to find the specific value of T: \[ 0= \frac{(4)^2 T^2 (32) + 4 (16) (32)^2 - 4\sqrt{((4)^2 T^2 (32) + 2(16)(32)) (16)(32) }}{4(32)}. \] Solve for T: \[ T=\frac{1}{2} \sqrt{\frac{8 H}{g}}\left(\frac{5-(3 \times \frac{B}{A})}{2}\right). \] Now plug in the given values for A, B, H, and g: \[ T=\frac{1}{2} \sqrt{\frac{8 (16)}{32}}\left(\frac{5-(3 \times \frac{1}{4})}{2}\right) = 3.5 \text{ sec}. \] Thus, the time it takes for the tank to become empty in the given conditions is 3.5 seconds.

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.

Differential Equations
Differential equations are mathematical tools that describe relationships involving rates of change and the quantities associated with those rates. They are used to model a vast range of physical phenomena, such as growth, decay, oscillation, and, as in our textbook example, fluid dynamics.

In the context of Torricelli's Law, we encounter a differential equation that represents the rate at which the height of water, denoted by \( h \), changes over time \( t \). Having the form \( \frac{d h}{d t} = -\frac{B}{A} \sqrt{2 g h} \), this equation expresses how the height of the water in a tank decreases as the water is discharged through an opening at the bottom.

Understanding how to manipulate and solve these equations is essential for predicting system behavior, and in this case, it allows us to determine how long it takes for a tank to drain.
Variable Separation
Separation of variables is a method used to solve differential equations, in which we rearrange the equation so that each variable and its differential are on opposite sides of the equation. This is often one of the first strategies employed when faced with a differential equation.

For Torricelli's problem, separating variables involves manipulating the original equation to isolate \( dh \) and \( dt \) on different sides, yielding \( \frac{dh}{\sqrt{2gh}} = -\frac{B}{A} dt \). The ability to separate variables is particularly useful because it converts the problem into an easier-to-handle format where we can integrate both sides independently to find a relationship between the variables.
Integrating Factors
Integrating factors are a technique used in solving differential equations that are not readily separable. An integrating factor is a function that is multiplied by both sides of the differential equation to facilitate its solution, often transforming it into an exact equation that we can integrate directly.

In the solution to Torricelli's Law, integrating factors are not explicitly needed because the separation of variables suffices to solve the initial value problem. However, the concept is invaluable when dealing with more complex differential equations, especially linear first-order equations where the standard approach is to multiply through by an integrating factor to achieve a solution.
Initial Value Problem
An initial value problem in the realm of differential equations is a problem where the equation is supplemented by specifying the value of the unknown function at a specific point. This additional information allows for the determination of a unique solution to the differential equation.

In our example with the draining tank, we have the initial condition \( h(0) = H \) specifying the height of water at time \( t = 0 \). With this initial value, we can calculate the particular solution that describes how the water level \( h \) decreases with time \( t \), eventually finding the time \( T \) it takes for the tank to empty, which is another distinct value that characterizes our system.

The process of solving the initial value problem involves integrating the separated equation with the specified boundary conditions to find an expression for \( h \), and then determining the time \( T \) at which \( h(T) = 0 \) for the complete emptying of the tank.

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

Lambert's Law of Absorption According to Lambert's Law of Absorption, the percentage of incident light \(L\), absorbed in passing through a thin layer of material \(x\), is proportional to the thickness of the material. For a certain material, if \(\frac{1}{2}\) in. of the material reduces the light to half of its intensity, how much additional material is needed to reduce the intensity to one fourth of its initial value?

A Falling Raindrop As a raindrop falls, it picks up more moisture, and as a result, its mass increases. Suppose that the rate of change of its mass is directly proportional to its current mass. a. Using Newton's Law of Motion, \(\frac{d}{d t}(m v)=F=m g\), where \(m(t)\) is the mass of the raindrop at time \(t, v\) is its velocity (positive direction is downward), and \(g\) is the acceleration due to gravity, derive the (differential) equation of motion of the raindrop. b. Solve the differential equation of part (a) to find the velocity of the raindrop at time \(t\). Assume that \(v(0)=0\). c. Find the terminal velocity of the raindrop, that is, find \(\lim _{t \rightarrow \infty} v(t)\)

Carbon-14 Dating Skeletal remains of the so-called Pittsburgh Man unearthed in Pennsylvania had lost \(82 \%\) of the carbon14 they originally contained. Determine the approximate age of the bones. (The half-life of carbon C-14 is 5730 years.)

Von Bertalanffy Growth Model The von Bertalanffy growth model is used to predict the length of commercial fish. The model is described by the differential equation $$ \frac{d x}{d t}=k(L-x) $$ where \(x(t)\) is the length of the fish at time \(t, k\) is a positive constant called the von Bertalanffy growth rate, and \(L\) is the maximum length of the fish. a. Find \(x(t)\) given that the length of the fish at \(t=0\) is \(x_{0}\). b. At the time the larvae hatch, the North Sea haddock are about \(0.4 \mathrm{~cm}\) long, and the average haddock grows to a length of \(10 \mathrm{~cm}\) after 1 year. Find an expression for the length of the North Sea haddock at time \(t\). c. Plot the graph of \(x\). Take \(L=100(\mathrm{~cm})\). d. On average, the haddock that are caught today are between \(40 \mathrm{~cm}\) and \(60 \mathrm{~cm}\) long. What are the ages of the haddock that are caught?

Mixture Problem A tank initially holds 40 gal of pure water. Brine that contains \(2 \mathrm{lb}\) of salt per gallon enters the tank at the rate of \(1.5 \mathrm{gal} / \mathrm{min}\), and the well-stirred mixture leaves at the rate of \(2 \mathrm{gal} / \mathrm{min}\). a. Find the amount of salt in the tank at time \(t\). b. Find the amount of salt in the tank after \(20 \mathrm{~min}\). c. Find the amount of salt when the tank holds 20 gal of brine. d. Find the maximum amount of salt present.

See all solutions

Recommended explanations on Math 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.