/*! 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 132 The nozzle has a diameter of \(4... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The nozzle has a diameter of \(40 \mathrm{~mm}\). If it discharges water uniformly with a downward velocity of \(20 \mathrm{~m} / \mathrm{s}\) against the fixed blade, determine the vertical force exerted by the water on the blade. \(\rho_{w}=1 \mathrm{Mg} / \mathrm{m}^{3}\).

Short Answer

Expert verified
The calculation of the vertical force exerted by the water on the blade involves determining the cross-sectional area of the nozzle, calculating the mass flow rate of the water and then applying the momentum equation to find the force. The actual numerical value would be obtained by replacing the Area \(A\) and mass flow rate \(\dot{m}\) with the calculated values in step 1 and step 2 respectively.

Step by step solution

01

Calculation of the Area of the Nozzle

Begin by calculating the cross-sectional area of the nozzle using the given diameter. The formula for the area of a circle is \(A=\pi(D/2)^2\), where \(D\) is the diameter of the nozzle. So, the Area \(A\) of the nozzle in square meters is equal to \(\pi (0.04/2)^2\).
02

Calculation of the Mass Flow Rate

Next, calculate the mass flow rate of the water. The formula for the mass flow rate is \(\dot{m}=\rho V A\), where \(V\) is the velocity of the water, \(\rho\) is the density of the water, and \(A\) is the cross-sectional area of the circle you calculated in the previous step. Therefore, \(\dot{m} = 1 Mg/m^3 \times 20 m/s \times A\).
03

Calculation of the Vertical Force on the Blade

Now, determine the vertical force exerted by the water on the blade, making use of the momentum equation. In the vertical direction, the force exerted by the fluid is equal to the time rate of change of momentum of the fluid. As the fluid is brought to rest by the blade, the force can be expressed as \( F = \dot{m} V\) where \(\dot{m}\) is the mass flow rate and \(V\) is the velocity of the fluid. Substituting the values from the previous steps gives \(F = \dot{m} \times 20 m/s\)

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.

Mass Flow Rate
Understanding the concept of mass flow rate is crucial when dealing with fluid dynamics and various applications such as the scenario presented by the nozzle and blade exercise. In layman's terms, mass flow rate represents the amount of mass passing through a given surface per unit of time. This is often measured in kilograms per second (\text{kg/s}) in the metric system.

When calculating the mass flow rate, one must multiply the fluid's velocity by its density and the cross-sectional area through which it flows. The formula looks like this: \[ \dot{m}=\rho VA \] where \(\rho\) stands for the fluid's density, \(V\) for the velocity, and \(A\) for the area. In practical terms, if you have a garden hose discharging water, the mass flow rate would tell you how much water (in terms of mass) is coming out of it each second.

For the exercise on hand, with the known velocity, density, and area, we can determine the mass flow rate of the water. This value serves as a cornerstone to further calculations such as the determination of forces exerted by the flowing fluid, showcasing mass flow rate's implication in engineering and physics problems.
Momentum Equation
The momentum equation is a fundamental principle within fluid mechanics, particularly when analyzing forces exerted by moving fluids. Momentum is a measure of the motion of an object and is calculated as the product of mass and velocity. In fluid dynamics, the momentum equation based on Newton's second law shows the relationship between the forces applied on a fluid and the change in momentum of the fluid.

In its simplest form, the momentum equation for a steady flow of fluid can be expressed as: \[ F = \dot{m} \Delta V \] In this expression, \( F \) represents the force exerted by the fluid, \( \dot{m} \) denotes the mass flow rate, and \( \Delta V \) is the change in velocity of the fluid.

In our example, the water strikes the blade and its vertical component of velocity is brought to rest, which implies a change in velocity from its initial value to zero. Applying the momentum equation enables us to quantify the vertical force exerted by the water on the blade, which has practical relevance in designing turbines and understanding fluid impact on surfaces.
Fluid Mechanics
At its core, fluid mechanics is the branch of physics concerned with the behavior of fluids—liquids and gases—and the forces on them. Fluid mechanics can be broken down into fluid statics, which deals with fluids at rest, and fluid dynamics, which concerns the movement of fluids.

Fluid dynamics studies various properties such as flow velocity, pressure, density, and temperature as functions of space and time. Understanding these properties allows us to solve complex problems like calculating the force on a blade exerted by a jet of water in the problem presented.

The application of fluid mechanics is extensive and includes designing hydraulic systems, predicting weather patterns, understanding blood flow in the human body, and analyzing pollutant dispersion in the environment. The principles form the foundation for aerospace and marine engineering, amongst many other fields.

In the context of our exercise, fluid mechanics principles let us calculate the vertical force exerted by the fluid by applying the concepts of mass flow rate and the momentum equation to understand the fluid's interaction with the blade.

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 \(15-\mathrm{kg}\) block \(A\) slides on the surface for which \(\mu_{k}=0.3 .\) The block has a velocity \(v=10 \mathrm{~m} / \mathrm{s}\) when it is \(s=4 \mathrm{~m}\) from the \(10-\mathrm{kg}\) block \(B\). If the unstretched spring has a stiffness \(k=1000 \mathrm{~N} / \mathrm{m},\) determine the maximum compression of the spring due to the collision. Take \(e=0.6\).

Each ball has a negligible size and a mass of \(10 \mathrm{~kg}\) and is attached to the end of a rod whose mass may be neglected. If the rod is subjected to a torque \(M=\left(t^{2}+2\right) \mathrm{N} \cdot \mathrm{m},\) where \(t\) is in seconds, determine the speed of each ball when \(t=3 \mathrm{~s}\). Each ball has a speed \(v=2\) \(\mathrm{m} / \mathrm{s}\) when \(t=0\)

The 50 -kg crate is pulled by the constant force \(\mathbf{P}\). If the crate starts from rest and achieves a speed of \(10 \mathrm{~m} / \mathrm{s}\) in \(5 \mathrm{~s}\), determine the magnitude of \(\mathbf{P}\). The coefficient of kinetic friction between the crate and the ground is \(\mu_{k}=0.2\).

An earth satellite of mass \(700 \mathrm{~kg}\) is launched into a free-flight trajectory about the earth with an initial speed of \(v_{A}=10 \mathrm{~km} / \mathrm{s}\) when the distance from the center of the earth is \(r_{A}=15 \mathrm{Mm}\). If the launch angle at this position is \(\phi_{A}=70^{\circ},\) determine the speed \(v_{B}\) of the satellite and its closest distance \(r_{B}\) from the center of the earth. The earth has a mass \(M_{e}=5.976\left(10^{24}\right) \mathrm{kg} .\) Hint: Under these conditions, the satellite is subjected only to the earth's gravitational force, \(F=G M_{e} m_{s} / r^{2},\) Eq. \(13-1\). For part of the solution, use the conservation of energy.

A \(50-\mathrm{kg}\) crate rests against a stop block \(s\), which prevents the crate from moving down the plane. If the coefficients of static and kinetic friction between the plane and the crate are \(\mu_{s}=0.3\) and \(\mu_{k}=0.2,\) respectively, determine the time needed for the force \(\mathbf{F}\) to give the crate a speed of \(2 \mathrm{~m} / \mathrm{s}\) up the plane. The force always acts parallel to the plane and has a magnitude of \(F=(300 t) \mathrm{N},\) where \(t\) is in seconds. Hint: First determine the time needed to overcome static friction and start the crate moving.

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.