/*! 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 57 In a laboratory experiment, wate... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 57

In a laboratory experiment, water flows radially outward at moderate speed through the space between circular plane parallel disks. The perimeter of the disks is open to the atmosphere. The disks have diameter \(D=150 \mathrm{mm}\) and the spacing between the disks is \(h=0.8 \mathrm{mm} .\) The measured mass flow rate of water is \(\dot{m}=305 \mathrm{g} / \mathrm{s}\). Assuming frictionless flow in the space between the disks, estimate the theoretical static pressure between the disks at radius \(r=50 \mathrm{mm} .\) In the laboratory situation, where some friction is present, would the pressure measured at the same location be above or below the theoretical value? Why?

Short Answer

Expert verified
The theoretical static pressure between the disks at radius \(r=50 \mathrm{mm}\) could be calculated using Bernoulli's Equation. If there is friction present, the pressure would be lower than the theoretical pressure because friction introduces pressure losses.

Step by step solution

01

Find velocity

First, find the radial velocity from the mass flow rate. Given that \(\dot{m}=305 \mathrm{g} / \mathrm{s}\), \(D=150 \mathrm{mm}\), and \(h=0.8 \mathrm{mm}\), radial velocity \(v=\dot{m}/(\rho \pi r h)\). Where \(\rho\) is the density of water, which is \(1000 kg/m^3\). Hence, \(v=305 \times 10^{-3} kg/s /(1000 kg/m^3 \times 3.14 \times (75 \times 10^{-3}m)^2 \times 0.8 \times 10^{-3}m)\)
02

Calculate Theoretical pressure

Now, apply Bernoulli鈥檚 equation on the streamline from the edge where pressure is atmospheric to the point where we want to locate pressure. Bernoulli鈥檚 equation is \(P_1 + 0.5 \times \rho \times v_1^2 = P_2 + 0.5 \times \rho \times v_2^2\). Assuming the fluid is at rest at the perimeter, and \(P_1\) is atmospheric pressure, \(101325 Pa = P + 0.5 \times 1000 kg/m^3 \times v^2\). This can be rearranged to \(P = 101325 - 0.5 \times 1000 \times v^2\)
03

The effect of friction

When friction is present, the pressure would be lower than the theoretical value. This is because friction introduces a pressure loss, thereby reducing the total pressure at any point within the flow field.

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

The \(x\) component of velocity in a two-dimensional, incompressible flow field is given by \(u=A x y ;\) the coordinates are measured in meters and \(A=2 \mathrm{m}^{-1} \cdot \mathrm{s}^{-1} .\) There is no velocity component or variation in the \(z\) direction. Calculate the acceleration of a fluid particle at point \((x, y)=(2,1) .\) Estimate the radius of curvature of the streamline passing through this point. Plot the streamline and show both the velocity vector and the acceleration vector on the plot. (Assume the simplest form of the \(y\) component of velocity.)

Heavy weights can be moved with relative ease on air cushions by using a load pallet as shown. Air is supplied from the plenum through porous surface \(A B .\) It enters the gap vertically at uniform speed, \(q\). Once in the gap, all air flows in the positive \(x\) direction (there is no flow across the plane at \(x=0\) ). Assume air flow in the gap is incompressible and uniform at each cross section, with speed \(u(x),\) as shown in the enlarged view. Although the gap is narrow \((h \ll L)\) neglect frictional effects as a first approximation. Use a suitably chosen control volume to show that \(u(x)=q x / h\) in the gap. Calculate the acceleration of a fluid particle in the gap. Evaluate the pressure gradient, \(\partial p / \partial x,\) and sketch the pressure distribution within the gap. Be sure to indicate the pressure at \(x=L\)

An incompressible liquid with a density of \(900 \mathrm{kg} / \mathrm{m}^{3}\) and negligible viscosity flows steadily through a horizontal pipe of constant diameter. In a porous section of length \(L=2 \mathrm{m}\) liquid is removed at a variable rate along the length so that the uniform axial velocity in the pipe is \(u(x)=U e^{-x / L},\) where \(U=20 \mathrm{m} / \mathrm{s} .\) Develop expressions for and plot the acceleration of a fluid particle along the centerline of the porous section and the pressure gradient along the centerline. Evaluate the outlet pressure if the pressure at the inlet to the porous section is \(50 \mathrm{kPa}(\text { gage })\)

A velocity field is given by \(\vec{V}=\left[A x^{3}+B x y^{2}\right] \hat{i}+\) \(\left[A y^{3}+B x^{2} y\right] \hat{j} ; A=0.2 \mathrm{m}^{-2} \cdot \mathrm{s}^{-1}, B\) is a constant, and the coordinates are measured in meters. Determine the value and units for \(B\) if this velocity field is to represent an incompressible flow. Calculate the acceleration of a fluid particle at point \((x, y)=(2,1) .\) Evaluate the component of particle acceleration normal to the velocity vector at this point.

The stream function of a flow field is \(\psi=A x^{3}+\) \(B\left(x y^{2}+x^{2}-y^{2}\right),\) where \(\psi, x, y, A,\) and \(B\) are all dimensionless. Find the relation between \(A\) and \(B\) for this to be an irrotational flow. Find the velocity potential.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Modern Physics

Read Explanation

Work Energy and Power

Read Explanation

Fundamentals of Physics

Read Explanation

Electricity and Magnetism

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.