/*! 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 109 When a plane liquid jet strikes ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \(\theta=90\).

Short Answer

Expert verified
Ultimately, the expression for the ratio of thickness of the two streams is \(h_{2}/h = 1 / cos( \theta )\). Upon plotting, this expression will yield a curve depending on the value of \(\theta\). The special cases reveal that both streams have the same thickness when \(\theta=0\) and when \(\theta=90\), the function becomes undefined as the thickness \(h_{2}\) becomes infinitely large.

Step by step solution

01

Identify the problem

From the problem, it's understood that due to the impact of the jet on the plate, the liquid splits into two streams of equal velocity but of different thickness \(h\) and \(h_{2}\). The angle of the plate \(\theta\) plays a crucial role in determining the thickness of streams.
02

Visualize Problem

Draw a sketch of the plane fluid striking the inclined plate. This helps you in understanding the problem better and working out the relationships between various entities like velocity, thickness, angle.
03

Apply Newton’s second law for flow pressure

The pressure in the fluid jet is atmospheric, thus the forces in the direction normal to the plate are balanced. So we have, \(h \cdot v^{2} = h_{2} \cdot v^{2} \cdot cos( \theta ) \). As stated in the problem, the speed of both streams are equal, so the velocities cancel out. Rearrange the terms, the ratio \(h_{2} / h = 1 / cos( \theta )\)
04

Plot the function

Plot the function \(h_{2}/h = 1 / cos(\theta)\) using a graphing tool. Alternatively, you can create a table of values by choosing some representative values of \(\theta\) and calculating \(h_{2}/h\) for each.
05

Analyze the special cases

For special cases, when \(\theta=0\), the function \(h_{2}/h = 1 / cos( \theta )\) becomes 1 i.e., the thickness of both streams is equal. And, when \(\theta=90\), the function value becomes undefined.

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!

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

Airat standard conditions enters a compressor at \(75 \mathrm{m} / \mathrm{s}\) and leaves at an absolute pressure and temperature of \(200 \mathrm{kPa}\) and \(345 \mathrm{K}\), respectively, and speed \(V=125 \mathrm{m} / \mathrm{s}\). The flow rate is \(1 \mathrm{kg} / \mathrm{s}\). The cooling water circulating around the compressor casing removes 18 kJ/kg of air. Determine the power required by the compressor.

\( \mathrm{A}\) wet cooling tower cools warm water by spraying it into a forced dry-air flow. Some of the water evaporates in this air and is carried out of the tower into the atmosphere; the evaporation cools the remaining water droplets, which are collected at the exit pipe \((6 \text { in. } \mathrm{ID})\) of the tower. Measurements indicate the warm water mass flow rate is \(250,000 \mathrm{lb} / \mathrm{hr},\) and the cool water \(\left(70^{\circ} \mathrm{F}\right)\) flows at an average speed of 5 ft/s in the exit pipe. The moist air density is 0.065 Ib/ft \(^{3}\). Find (a) the volume flow rate \(\left(\mathrm{ft}^{3} / \mathrm{s}\right)\) and mass flow rate (lb/hr) of the cool water, (b) the mass flow rate \((\mathrm{lb} / \mathrm{hr})\) of the moist air, and (c) the mass flow rate (lb/hr) of the dry air. Hint: Google "density of moist air" for information on relating moist and dry air densities!

In a laboratory experiment, the water flow rate is to be measured catching the water as it vertically exits a pipe into an empty open tank that is on a zeroed balance. The tank is \(10 \mathrm{m}\) directly below the pipe exit, and the pipe diameter is \(50 \mathrm{mm} .\) One student obtains a flow rate by noting that after 60 s the volume of water (at \(4^{\circ} \mathrm{C}\) ) in the tank was \(2 \mathrm{m}^{3}\). Another student obtains a flow rate by reading the instantaneous weight accumulated of \(3150 \mathrm{kg}\) indicated at the \(60-\mathrm{s}\) point. Find the mass flow rate each student computes. Why do they disagree? Which one is more accurate? Show that the magnitude of the discrepancy can be explained by any concept you may have.

A small rocket motor is used to power a "jet pack" device to lift a single astronaut above the Moon's surface. The rocket motor produces a uniform exhaust jet with constant specd, \(V_{e}=3000 \mathrm{m} / \mathrm{s},\) and the thrust is varied by changing the jet size. The total initial mass of the astronaut and the jet pack is \(M_{0}=200 \mathrm{kg}, 100 \mathrm{kg}\) of which is fuel and oxygen for the rocket motor. Find (a) the exhaust mass flow rate required to just lift off initially, (b) the mass flow rate just as the fuel and oxygen are used up, and (c) the maximum anticipated time of flight. Note that the Moon's gravity is about 17 percent of Earth's.

Oil flows steadily in a thin layer down an inclined plane. The velocity profile is \\[ u=\frac{\rho g \sin \theta}{\mu}\left[h y-\frac{y^{2}}{2}\right] \\] Express the mass flow rate per unit width in terms of \(\rho, \mu, g\) \(\theta,\) and \(h\)

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.