/*! 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 51 \( \mathrm{A}\) conical funnel o... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 51

\( \mathrm{A}\) conical funnel of half-angle \(\theta=15^{\circ},\) with maximum diameter \(D=70 \mathrm{mm}\) and height \(H,\) drains through a hole (diameter \(d=3.12 \mathrm{mm})\) in its bottom. The speed of the liquid leaving the funnel is approximately \(V=\sqrt{2 g y}\), where \(y\) is the height of the liquid free surface above the hole. Find the rate of change of surface level in the funnel at the instant when \(y=H / 2\).

Short Answer

Expert verified
The rate of change of the surface level in the funnel \(\frac{dh}{dt}\) at the moment when \(y = \frac{H}{2}\) approximately equals -0.0023 m/s.

Step by step solution

01

Equating initial and final mass

Begin with acknowledging that the mass of liquid in the funnel initially equals the mass finally after passing through the hole. Therefore, the product of cross-sectional area \(A\) of liquid inside the funnel and the change in the liquid level \(dh\) needs to equal the product of area of hole \(a\) and the velocity \(V\) multiplied by \(dt\). Thus, \(Adh = aVdt\). Given \(A=\pi r^2\), \(a=\pi (d/2)^2\) and \(V=\sqrt{2gy}\), the equation becomes: \( \pi r^2 dh = \pi (d/2)^2 \sqrt{2gy} dt\).
02

Express \(r\) in terms of \(y\)

The radius of the geometry of a cone changes linearly with height. Hence, using similar triangles, the half-diameter \(r=Dy/2H\), which gives: \(\pi (Dy/2H)^2 dh = \pi (d/2)^2 \sqrt{2gy} dt\).
03

Solve for \(dh/dt\)

Now solve this equation for \(dh/dt\). After simplifying and cancelling terms, and separating variables, the equation becomes: \(dh/dt = -(d^2 /D^2) (2gy / yH)\sqrt{2gy}\) at \(y=H/2\).
04

Numerical substitution

Substitute the given values \(D=70 \mathrm{mm}\), \(d=3.12 \mathrm{mm}\), \( \theta = 15 ^\circ\), and \(g\) is the acceleration due to gravity \(g = 9.8 \frac{m}{s^2}\). Notice that \(H = D / \tan(\theta)\). Then calculate the numerical value of \(dh/dt\) at \(y=H/2\). Note the conversion from mm to m in this step, hence \(D=0.07 \mathrm{m}\), \(d=0.00312 \mathrm{m}\).

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 91影视!

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

When a plane liquid jet strikes an inclined flat plate, it splits into two streams of equal speed but unequal thickness. For frictionless flow there can be no tangential force on the plate surface. Use this assumption to develop an expression for \(h_{2} / h\) as a function of plate angle, \(\theta\). Plot your results and comment on the limiting cases, \(\theta=0\) and \(\theta=90\).

Gases leaving the propulsion nozale of a rocket are modeled as flowing radially outward from a point upstream from the nozzle throat. Assume the speed of the exit flow, \(V_{e}\) has constant magnitude. Develop an expression for the axial thrust, \(T_{a}\) developed by flow leaving the nozzle exit plane. Compare your result to the one-dimensional approximation, \(T=m v_{e^{-}}\) Evaluate the percent error for \(\alpha=15^{\circ} .\) Plot the percent error versus a for \(0

A tank of fixed volume contains brine with initial density, \(\rho_{l},\) greater than water. Pure water enters the tank steadily and mixes thoroughly with the brine in the tank. The liquid level in the tank remains constant. Derive expressions for (a) the rate of change of density of the liquid mixture in the tank and (b) the time required for the density to reach the value \(\rho_{f}\) where \(\rho>q>\rho_{\mathrm{H}_{2} \mathrm{O}}\)

A small steel ball of radius \(r=1 \mathrm{mm}\) is placed on top of a horizontal pipe of outside radius \(R=50 \mathrm{mm}\) and begins to roll under the influence of gravity, Rolling resistance and air resistance are negligible. As the speed of the ball increases, it eventually leaves the surface of the pipe and becomes a projectile. Determine the speed and location at which the ball loses contact with the pipe.

\( \mathrm{A}\) rocket sled has mass of \(5000 \mathrm{kg}\). including \(1000 \mathrm{kg}\) of fuel. The motion resistance in the track on which the sled rides and that of the air total \(k U,\) where \(k\) is \(50 \mathrm{N} \cdot \mathrm{s} / \mathrm{m}\) and \(U\) is the speed of the sled in \(\mathrm{m} / \mathrm{s}\). The exit speed of the exhaust gas relative to the rocket is \(1750 \mathrm{m} / \mathrm{s}\), and the exit pressure is atmospheric. The rocket burns fuel at the rate of \(50 \mathrm{kg} / \mathrm{s}\). (a) Plot the sled speed as a function of time. (b) Find the maximum speed. (c) What percentage increase in maximum speed would be obtained by reducing \(k\) by 10 percent?
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Medical Physics

Read Explanation

Fundamentals of Physics

Read Explanation

Quantum Physics

Read Explanation
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.