/*! 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 2 Consider incompressible flow in ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 2

Consider incompressible flow in a circular channel. Derive general expressions for Reynolds number in terms of (a) volume flow rate and tube diameter and (b) mass flow rate and tube diameter. The Reynolds number is 1800 in a section where the tube diameter is \(10 \mathrm{mm}\). Find the Reynolds number for the same flow rate in a section where the tube diameter is \(6 \mathrm{mm}\)

Short Answer

Expert verified
The Reynolds number for the same flow rate in a section where the tube diameter is 6 mm is 3000.

Step by step solution

01

Derive a general expression for Reynolds number in terms of volume flow rate and tube diameter

The general formula for the Reynolds number (Re) is given by: \[Re = \frac{\rho u D}{\mu}\] where \(\rho\) is the fluid density, \(u\) is the flow velocity, \(D\) is the tube diameter and \(\mu\) is the fluid viscosity. Volume flow rate, \(Q\), is described by the formula \(Q = uA\), where \(A\) is the cross-sectional area of the tube. For a circular tube, \(A= \frac{\pi D^2}{4}\). Substituting this into the flow rate equation gives \(u = \frac{4Q}{\pi D^2}\).Substituting this velocity into the Reynolds number formula provides:\[Re = \frac{4 \rho Q}{\mu D}\]
02

Derive a general expression for Reynolds number in terms of mass flow rate and tube diameter

Mass flow rate, \(m\), is related to the volume flow rate through the formula \(m = \rho Q\). Substituting this into the Reynolds number expression derived in the previous step gives: \[Re = \frac{4m}{\mu D}\]
03

Calculate the Reynolds number for the given flow rates and tube diameters

The problem states that the Reynolds number is 1800 when the tube diameter is 10 mm. Assuming the same flow properties (i.e., same fluid with the same mass flow rate or volume flow rate), when reducing the diameter to 6 mm, the Reynolds number can be calculated using the inversely proportional relationship between Re and D in the derived formulas in steps one and two. If the Reynolds number is 1800 for a 10 mm tube, the Reynolds number for the same flow rate in a 6 mm tube diameter section would be:\[Re' = Re \times \left(\frac{D}{D'}\right) = 1800 \times \left(\frac{10}{6}\right) = 3000\]

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

Water discharges to atmosphere from a large reservoir through a moderately rounded horizontal nozzle of \(25 \mathrm{mm}\) diameter. The free surface is \(2.5 \mathrm{m}\) above the nozzle exit plane. Calculate the change in thow rate when a short section of 50 -mm-diameter pipe is attached to the end of the nozzle to form a sudden expansion. Determine the location and estimate the magnitude of the minimum pressure with the sudden expansion in place. If the flow were frictionless (with the sudden expansion in place), would the minimum pressure be higher, lower, or the same? Would the flow rate be higher, lower, or the same?

Water flows from a horizontal tube into a large tank. The tube is located \(2.5 \mathrm{m}\) below the free surface of water in the tank. The head loss is \(2 \mathrm{J} / \mathrm{kg}\). Compute the average flow speed in the tube.

Gasoline flows in a long. underground pipeline at a constant temperature of \(15^{\circ} \mathrm{C}\). Two pumping stations at the same elevation are located \(13 \mathrm{km}\) apart. The pressure drop between the stations is \(1.4 \mathrm{MPa}\). The pipeline is made from 0.6-m-diameter pipe. Although the pipe is made from commercial steel, age and corrosion have raised the pipe roughness to approximately that for galvanized iron. Compute the volume flow rate.

A venturi meter with a 3 -in. -diameter throat is placed in a 6 -in.-diameter line carrying water at \(75^{\circ} \mathrm{F}\), The pressure drop between the upstream tap and the venturi throat is 12 in. of mercury. Compute the rate of flow.

A continuous belt, passing upward through a chemical bath at speed \(U_{0},\) picks up a liquid film of thickness \(h,\) density \(\rho,\) and viscosity \(\mu .\) Gravity tends to make the liquid drain down, but the movement of the belt keeps the liquid from running off completely. Assume that the flow is fully developed and laminar with zero pressure gradicnt, and that the atmosphere produces no shear stress at the outer surface of the film. State clearly the boundary conditions to be satisfied by the velocity at \(y=0\) and \(y=h\). Obtain an expression for the velocity profile.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Scientific Method Physics

Read Explanation

Quantum Physics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Atoms and Radioactivity

Read Explanation

Fluids

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.