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Chapter 3: Problem 32

The velocity distribution in a pipe with a circular cross section under laminar flow conditions can be approximated by the equation $$ v(r)=V_{0}\left[1-\left(\frac{r}{R}\right)^{2}\right] $$ where \(v(r)\) is the velocity at a distance \(r\) from the centerline of the pipe, \(V_{0}\) is the centerline velocity, and \(R\) is the radius of the pipe. (a) Calculate the average velocity and volume flow rate in the pipe in terms of \(V_{0}\). (b) Based on the result in part (a), assess the extent to which the velocity can be assumed to be constant across the cross section.

Short Answer

Expert verified
Average velocity is \( \frac{V_0}{2} \), volume flow rate is \( \frac{V_0 \pi R^2}{2} \). Velocity is not constant across the pipe cross-section.

Step by step solution

01

Understand the Problem

We need to find the average velocity and volume flow rate of a fluid in a pipe under laminar flow conditions. The velocity is given by a function of the radial distance from the center of the pipe. We also need to assess how constant the velocity is across the pipe's cross-section.
02

Define the Average Velocity Formula

The average velocity \( \bar{v} \) for flow through a pipe is defined by integrating the velocity across the entire cross-sectional area and dividing by the area:\[\bar{v} = \frac{1}{\pi R^2} \int_0^R 2\pi r v(r) \; dr\]where \( 2\pi r \) is used to convert the integral into polar coordinates to account for the circular cross-section area.
03

Plug in Velocity Function

Insert the given velocity function into the average velocity formula:\[\bar{v} = \frac{1}{\pi R^2} \int_0^R 2\pi r V_{0}\left(1 - \left(\frac{r}{R}\right)^2\right) \; dr\]This simplifies to\[\bar{v} = \frac{2V_{0}}{R^2} \int_0^R \left(r - \frac{r^3}{R^2}\right) dr\]
04

Integrate the Function

Perform the integration:\[\int_0^R \left(r - \frac{r^3}{R^2}\right) dr = \left[\frac{r^2}{2} - \frac{r^4}{4R^2}\right]_0^R\]Substitute the limits:\[= \frac{R^2}{2} - \frac{R^4}{4R^2}\]\[= \frac{R^2}{2} - \frac{R^2}{4}\]\[= \frac{R^2}{4}\]
05

Calculate Average Velocity

Insert the integrated value back into the average velocity equation:\[\bar{v} = \frac{2V_{0}}{R^2} \times \frac{R^2}{4} = \frac{V_{0}}{2} \]Thus, the average velocity \( \bar{v} \) is \( \frac{V_{0}}{2} \).
06

Calculate Volume Flow Rate

The volume flow rate \( Q \) is given by multiplying the average velocity by the cross-sectional area:\[Q = \bar{v} \cdot A = \left(\frac{V_{0}}{2}\right) \cdot \pi R^2 = \frac{V_{0} \pi R^2}{2}\]Thus, \( Q = \frac{V_{0} \pi R^2}{2}. \)
07

Assess Constant Velocity Across Cross Section

The velocity distribution is parabolic, meaning it is highest at the center and decreases towards the edges. The average velocity being half of the maximum (\( V_0 \)) suggests significant variation across the cross section.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Velocity Distribution
In a laminar flow within a circular pipe, the velocity distribution is not uniform across the pipe's diameter. Instead, it follows a specific pattern determined by the equation:\[v(r)=V_{0}\left[1-\left(\frac{r}{R}\right)^{2}\right]\]where the velocity, \(v(r)\), decreases as you move away from the centerline towards the pipe walls. This occurs because, near the wall, friction slows down the fluid.
  • At \(r = 0\), the velocity is maximum, \(V_{0}\).
  • At \(r = R\), the velocity is zero.

This profile is known as a parabolic velocity distribution. Understanding this distribution helps in several flow analyses and is essential in determining the fluid flow's average behavior and other characteristics.
Average Velocity
The average velocity of fluid flow in a pipe under laminar conditions is a significant metric. It's found by considering the velocity distribution across the pipe's cross-section.To compute the average velocity \(\bar{v}\), we integrate the velocity profile over the pipe's cross-sectional area using the formula:\[\bar{v} = \frac{1}{\pi R^2} \int_0^R 2\pi r v(r) \, dr\]Substituting the velocity equation, the integral evaluates to yield the result:\[\bar{v} = \frac{V_{0}}{2}\]This means that the average velocity is half of the maximum velocity at the centerline. This average provides insight into the fluid's overall movement through the pipe.
Volume Flow Rate
The volume flow rate, \(Q\), indicates the volume of fluid passing through the pipe per unit time. It is directly connected to the average velocity and the cross-sectional area of the pipe.To find the volume flow rate, we use the relationship:\[Q = \bar{v} \cdot A = \frac{V_{0}}{2} \cdot \pi R^2\]Calculating this, we get:\[Q = \frac{V_{0} \pi R^2}{2}\]This showcases how much fluid the pipe can move, which is crucial for designing and analyzing fluid transport systems in engineering applications.
Parabolic Velocity Profile
A parabolic velocity profile is characteristic of laminar flow in pipes. It depicts that the fluid velocity is highest at the center and diminishes towards the walls.Key aspects include:
  • Maximum velocity occurs at the centerline \(V_{0}\).
  • The velocity profile is symmetric concerning the pipe's centerline.
  • Velocity decreases in a quadratic manner towards the edges.
  • This results in zero velocity at the pipe walls.

This distribution signifies the non-uniform nature of flow in the pipe, influencing factors like friction and pressure drop. Understanding this profile is pivotal for predicting flow behavior under laminar conditions.

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Most popular questions from this chapter

An orifice meter consists of an arrangement in which a plate containing a circular orifice at its center is placed normal to the flow in a conduit, and the flow rate in the conduit is related to the measured difference in pressure across the orifice. Consider the orifice meter shown in Figure \(3.52,\) where the orifice has a diameter of \(50 \mathrm{~mm}\) and a discharge coefficient of 0.62 and the conduit has a diameter of \(100 \mathrm{~mm} .\) The fluid is water at \(20^{\circ} \mathrm{C},\) and the differential pressure gauge shows a pressure difference of \(30 \mathrm{kPa}\). What is the volume flow rate in the conduit?

A closed \(15-\mathrm{cm}\) -diameter cylindrical tank is filled with water and is rotated at \(600 \mathrm{rpm} .\) What is the maximum pressure difference between the center and the wall of the \(\operatorname{tank} ?\)

An airplane operates at a low elevation where the pressure is \(101 \mathrm{kPa}\) and the temperature is \(20^{\circ} \mathrm{C}\). (a) Neglecting compressibility effects, estimate the flight speed at which the stagnation pressure is \(15 \%\) higher than the static pressure. (b) Estimate by how much the result in part (a) would differ if compressibility effects were taken into account.

Airflow within the eye of a hurricane can be approximated as a spiral free vortex. The velocity at the center of the eye can be approximated as being equal to zero, and the velocity at the eye wall can be associated with the maximum wind velocity in the hurricane. Consider Hurricane Wilma, which attained pressure of \(88.2 \mathrm{kPa}\) at the center of the eye and a wind velocity of \(295 \mathrm{~km} / \mathrm{h}\) in the eye wall. (a) Estimate the difference in pressure between the center of the eye and the eye wall. What is the difference in sea-surface elevation between the center of the eye and eye wall caused by this difference in pressure? (b) If the pressure at the outer limit of the hurricane is \(101 \mathrm{kPa}\) with negligible wind velocities, estimate the difference in sea-surface elevation between the wall of the hurricane and the outer limit of the hurricane. What are the implications of this sea-surface difference to coastal communities? Assume a uniform atmospheric temperature of \(25^{\circ} \mathrm{C},\) assume a seawater temperature of \(20^{\circ} \mathrm{C}\) and neglect air density variations in the eye of the hurricane.

A velocity field, \(\mathbf{v}\), is spatially uniform and varies with time according to the following relation: $$ \mathbf{v}=\left\\{\begin{array}{ll} 3 \mathbf{i}+\mathbf{j} \mathrm{m} / \mathrm{s}, & t \in[0 \mathrm{~s}, 8 \mathrm{~s}] \\ 5 \mathbf{i}-4 \mathbf{j} \mathrm{m} / \mathrm{s}, & t \in(8 \mathrm{~s}, 15 \mathrm{~s}] \end{array}\right. $$ If an injection point is located at the origin of a Cartesian coordinate system at ( \(0 \mathrm{~m}\), \(0 \mathrm{~m}\) ), sketch to scale the following at \(t=15 \mathrm{~s}\) along with their key coordinates: (a) the pathline of a particle released at the injection point at \(t=0 \mathrm{~s},(\mathrm{~b})\) the streakline of dye continuously released at the injection point starting at \(t=0 \mathrm{~s},\) and (c) the streamlines in the flow field at \(t=15 \mathrm{~s}\).
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