/*! 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 74 Water flows steadily through a f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 74

Water flows steadily through a fire hose and nozzle. The hose is \(75 \mathrm{mm}\) inside diameter, and the nozzle tip is \(25 \mathrm{mm}\) ID; water gage pressure in the hose is \(510 \mathrm{kPa}\), and the stream leaving the nozzle is uniform. The exit speed and pressure are \(32 \mathrm{m} / \mathrm{s}\) and atmospheric, respectively. Find the force transmitted by the coupling between the nozzle and hose. Indicate whether the coupling is in tension or compression.

Short Answer

Expert verified
Use the short answer field to input the calculated force and the nature of the force i.e., tension or compression.

Step by step solution

01

Determine the mass flow rate

The mass flow rate can be calculated using the formula for flow speed, cross-sectional area of the pipe, and water density. Because the flow is steady, the mass flow rate is the same at every cross-section of the hose and nozzle. The formula for flow rate is \( \dot{m} = 蟻 * A * v \), where 蟻 is the density of the water (1000 kg/m鲁), A is the cross-sectional area of the hose which is \( 蟺*(d^2)/4 \) and v is the velocity of water. Here, the diameter d of the hose is 75mm or 0.075m, and the velocity v is not given for this section of the hose but will be the same as at the exit (32 m/s). By substituting these values, compute for mass flow rate.
02

Compute for exit pressure

The exit pressure is provided as being atmospheric, which is equal to zero gauge pressure.
03

Apply momentum equation

The linear momentum equation is given by \( 危F = \dot{m}*(V_{out} - V_{in}) \), which is the sum of all forces equal to the change in momentum. Since fluid enters the control volume with negligible velocity as compared to the exit velocity, the equation simplifies to \( 危F = \dot{m}*V_{out} \). Here, 危F is the vector sum of external forces acting on the control volume, \(\dot{m}\) is the mass flow rate, \(V_{out}\) is the exit speed, and \(V_{in}\) is the inlet speed. By plugging in the known values, compute for 危F.
04

Determine the nature of the force

If the force computed is positive, the coupling is in tension. If the force is negative, it signifies that the force is acting in the opposite direction, thereby implying that the coupling is in compression.

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

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.

Ventilation air specifications for classrooms require that at least \(8.0 \mathrm{L} / \mathrm{s}\) of fresh air be supplied for each person in the room (students and instructor). A system needs to be designed that will supply ventilation air to 6 classrooms, each with a capacity of 20 students. Air enters through a central duct, with short branches successively leaving for each classroom. Branch registers are \(200 \mathrm{mm}\) high and \(500 \mathrm{mm}\) wide. Calculate the volume flow rate and air velocity entering each room. Ventilation noise increases with air velocity. Given a supply duct \(500 \mathrm{mm}\) high, find the narrowest supply duct that will limit air velocity to a maximum of \(1.75 \mathrm{m} / \mathrm{s}\).

Air enters a tank through an area of \(0.2 \mathrm{ft}^{2}\) with a velocity of 15 fl/s and a density of 0.03 slug \(f\) ft \(^{3}\). Air leaves with a velocity of \(5 \mathrm{ft} / \mathrm{s}\) and a density equal to that in the tank. The initial density of the air in the tank is 0.02 slug/ft' \(^{3}\). The total tank volume is \(20 \mathrm{ft}^{3}\) and the exit area is \(0.4 \mathrm{ft}^{2}\). Find the initial rate of change of density in the tank.

\( \mathrm{A}\) disk, of mass \(M,\) is constrained horizontally but is free to move vertically. A jet of water strikes the disk from below. The jet leaves the nozzle at initial speed \(V_{0}\), Obtain a differential equation for the disk height, \(h(t),\) above the jet exit plane if the disk is released from large height, \(H .(\mathrm{You}\) will not be able to solve this ODE, as it is highly nonlinear!) Assume that when the disk reaches equilibrium, its height above the jet exit plane is \(h_{0}\) (a) Sketch \(h(t)\) for the disk released at \(t=0\) from \(H>h_{0}\) (b) Explain why the sketch is as you show it.

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

Solid State Physics

Read Explanation

Wave Optics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Electricity and Magnetism

Read Explanation

Rotational Dynamics

Read Explanation

Magnetism

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.