/*! 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 13 For the general case of laminar ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For the general case of laminar flow in a pipe of diameter \(D,\) at what distance from the pipe centerline is the velocity equal to the average velocity? Give your answer as a fraction of the pipe diameter.

Short Answer

Expert verified
The velocity is equal to the average velocity at a distance of \( \frac{1}{2\sqrt{3}} \) times the pipe diameter from the centerline.

Step by step solution

01

Understanding the Problem

We need to find the radial position in a pipe at which the velocity equals the average velocity, for laminar flow. This requires using the parabolic velocity profile typical for laminar flow between fixed boundaries.
02

Setting up the Velocity Profile Equation

In laminar flow through a cylindrical pipe, the velocity profile is parabolic. The velocity at a radial distance \( r \) from the centerline is given by \[ v(r) = v_{max} imes \left(1 - \left( \frac{r}{R} \right)^2 \right) \]where \( R = \frac{D}{2} \) is the pipe radius.
03

Calculating the Average Velocity

The average velocity \( v_{avg} \) for laminar flow in a pipe is given by \[ v_{avg} = \frac{1}{A} \int_{0}^{R} v(r) imes 2\pi r \, dr \]where \( A = \pi R^2 \) is the cross-sectional area of the pipe. This simplifies to \[ v_{avg} = \frac{2}{3} v_{max} \].
04

Equating Velocity to Average Velocity

We set the velocity profile equal to the average velocity to find the radial position:\[ v_{max} \left(1 - \left( \frac{r}{R} \right)^2 \right) = \frac{2}{3} v_{max} \].
05

Solving for Radial Position

Cancelling \( v_{max} \) and rearranging terms, we have:\[ 1 - \left( \frac{r}{R} \right)^2 = \frac{2}{3} \]\[ \left( \frac{r}{R} \right)^2 = \frac{1}{3} \]\[ \frac{r}{R} = \frac{1}{\sqrt{3}} \].
06

Expressing as a Fraction of the Pipe Diameter

Since \( R = \frac{D}{2} \), multiplying both sides by \( \frac{D}{2} \) gives \[ r = \frac{D}{2\sqrt{3}} \], which simplifies to \( \frac{r}{D} = \frac{1}{2\sqrt{3}} \).

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.

Understanding Velocity Profile in Pipes
In laminar flow within a pipe, the velocity profile describes how the fluid velocity changes across the pipe's cross-section.
This velocity distribution is parabolic in nature, meaning that the fluid moves fastest at the center and slows down as it approaches the pipe wall.
This steady variation is crucial for understanding other flow characteristics like average velocity. A parabolic profile is represented mathematically as: - The speed at any point in the pipe depends on its distance from the centerline.
Near the center, the speed approaches the maximum velocity, while near the edges, it drops to zero. Understanding this profile helps in:
  • Predicting the behavior of fluids in pipes
  • Calculating various flow properties such as the average velocity
  • Designing pipelines with specific flow characteristics
Unraveling Average Velocity
Average velocity in pipe flow represents the mean speed of the fluid elements passing through the entire cross-section of the pipe.
Unlike the velocity at a specific point, it averages out the speed of all the fluid particles. To find the average velocity, the area under the velocity profile curve is divided by the total cross-sectional area of the pipe.
For laminar flow, it is calculated using the integral formula:\[ v_{avg} = \frac{1}{A} \int_{0}^{R} v(r) \times 2\pi r \, dr \]Simplified, this results in the formula:- Average velocity is generally two-thirds of the maximum velocity: \[ v_{avg} = \frac{2}{3} v_{max} \]Knowing the average velocity helps:
  • Estimate flow rates in engineering applications
  • Ensure that systems are designed to handle expected flow conditions
  • Analyze differences in speed throughout the pipe cross-section
The Dynamics of Pipe Flow
Pipe flow is the movement of fluid through a cylindrical channel, guided by pressure differences.
In laminar flow, this motion occurs in parallel layers, reducing turbulence.
Understanding this helps explain various phenomena within a piping system. Key characteristics include:
  • Smooth layers of fluid that slide over one another
  • Stability in the flow pattern over time
  • Velocity distribution that follows a predictable pattern (like the parabolic profile)
This regularity in flow means: - Calculations about velocity and pressure drop can be highly accurate.
Engineers can design efficient and safe pipeline systems, selecting materials and diameters that meet specific demands. Challenges, such as pressure loss and frictional resistance, must also be considered in designing pipe systems.
Examining Parabolic Velocity Distribution
Parabolic velocity distribution is a hallmark of laminar flow in pipes.
It describes the way velocity changes from the pipe's center to its edges, following a parabolic curve. The velocity is highest at the center and decreases symmetrically toward the walls due to viscous effects.
Mathematically, this is expressed as:\[ v(r) = v_{max} \times \left(1 - \left( \frac{r}{R} \right)^2 \right) \]Where:
  • \( v(r) \) is the velocity at radial position \( r \)
  • \( v_{max} \) is the maximum centerline velocity
  • \( R \) is the pipe radius
The parabolic distribution indicates the pressure-driven nature of laminar flow, where the:- Velocity diminishes due to frictional forces at the pipe wall.
Understanding this distribution aids in calculating the average velocity and designing pipe systems that optimize flow efficiency. This profile is utilized in many engineering disciplines to predict the fluid movement in conduits and optimize flow-related processes.

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 is to be delivered from a public water supply line to a two-story building. Under design conditions, each floor of the building is to be simultaneously supplied with water at a rate of \(200 \mathrm{~L} / \mathrm{min}\). The pipes in the building plumbing system are to be made of galvanized iron. The length of pipe from the public water supply line to the delivery point on the first floor is \(20 \mathrm{~m},\) the length of pipe from the delivery point on the first floor to the delivery point on the second floor is \(5 \mathrm{~m},\) the water delivery point on the first floor is \(2 \mathrm{~m}\) above the water main connection, and the delivery point on the second floor is \(3 \mathrm{~m}\) above the delivery point on the first floor. If the water pressure at the water main is \(380 \mathrm{kPa}\), what is the minimum diameter pipe in the building plumbing system to ensure that the pressure is at least \(240 \mathrm{kPa}\) on the second floor? Neglect minor losses and consider pipe diameters in increments of \(\frac{1}{4} \mathrm{~cm},\) with the smallest allowable diameter being \(\frac{1}{2} \mathrm{~cm} .\) For the selected diameter under design conditions, what is the water pressure on the first floor?

A fluid flows at a rate of \(6 \mathrm{~L} / \mathrm{min}\) in a straight 20 -mm-diameter pipe. The fluid has a density of \(918 \mathrm{~kg} / \mathrm{m}^{3}\) and a viscosity of \(440 \mathrm{mPa}\).s. Pressure measurements at two locations \(12 \mathrm{~m}\) apart show an upstream pressure of \(400 \mathrm{kPa}\) and a downstream pressure of \(340 \mathrm{kPa}\). The downstream section is \(1.75 \mathrm{~m}\) higher than the upstream section. (a) Estimate the average shear stress on the surface of the pipe. (b) Estimate the friction factor for the flow. (c) Estimate the distance from the pipe entrance required to establish fully developed flow. (d) Would your answers to part (a) and part (b) be different if the flow between the upstream and downstream sections was not fully developed?

Water at \(20^{\circ} \mathrm{C}\) is to be delivered at a rate of \(100 \mathrm{~L} / \mathrm{s}\) through a \(200-\mathrm{m}\) -long pipe with an estimated roughness height of \(0.1 \mathrm{~mm}\) and an estimated local loss coefficient of 3.4 . The downstream end of the pipe has an elevation that is \(1.8 \mathrm{~m}\) higher than the upstream end of the pipe. The pressure at the upstream end of the pipe can be maintained at \(450 \mathrm{kPa}\) using a pump, and the minimum allowable pressure at the downstream end of the pipe is \(200 \mathrm{kPa}\). What is the minimum diameter pipe that can be used?

Water leaves a treatment plant in a 500 -mm-diameter ductile iron pipeline at a pressure of \(600 \mathrm{kPa}\) and at a flow rate of \(0.50 \mathrm{~m}^{3} / \mathrm{s}\). If the elevation of the pipeline at the treatment plant is \(120 \mathrm{~m}\), estimate the pressure in the pipeline \(1 \mathrm{~km}\) downstream where the elevation is \(100 \mathrm{~m}\). Assess whether the pressure in the pipeline would be sufficient to serve the top floor of a ten- story building (approximately \(30 \mathrm{~m}\) high).

Water at \(20^{\circ} \mathrm{C}\) flows at \(6 \mathrm{~L} / \mathrm{s}\) through a 75 -mm-diameter pipe, and a fixture is installed such that the diameter is suddenly reduced to \(50 \mathrm{~mm}\). (a) Estimate the head loss due to the sudden contraction. (b) What pressure change is expected at the contraction? (c) What would be the pressure change if the local head loss was not taken into account?

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.