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Velocity distributions for laminar and turbulent flow in a circular pipe of radius \(R\) carrying an incompressible liquid of density \(\rho\) are given, respectively, by $$ \begin{aligned} &\mathrm{V} / \mathrm{V}_{0}=\left[1-(r / R)^{2}\right] \\ &\mathrm{V} / \mathrm{V}_{0}=[1-(r / R)]^{1 / /} \end{aligned} $$ where \(r\) is the radial distance from the pipe centerline and \(\mathbf{V}_{0}\) is the centerline velocity. For each velocity distribution (a) plot \(\mathrm{V} / \mathrm{V}_{0}\) versus \(r / R\). (b) derive expressions for the mass flow rate and the average velocity of the flow, \(\mathrm{V}_{\text {ave }}\), in terms of \(\mathrm{V}_{0}, R\), and \(\rho\), as required. (c) derive an expression for the specific kinetic energy carried through an area normal to the flow. What is the percent error if the specific kinetic energy is evaluated in terms of the average velocity as \(\left(\mathrm{V}_{\mathrm{ave}}\right)^{2} / 2\) ? Which velocity distribution adheres most closely to the idealizations of one- dimensional flow? Discuss.

Short Answer

Expert verified
a) Plot velocity profiles. b) Derive expressions for mass flow rate and average velocity. c) Calculate specific kinetic energy and percent error.

Step by step solution

01

Understanding the Problem

The problem involves velocity distributions for laminar and turbulent flow in a circular pipe. The goal is to (a) plot the distributions, (b) derive expressions for mass flow rate and average velocity, and (c) derive the expression for specific kinetic energy and find the percent error.
02

Plot the Velocity Distributions

For both laminar and turbulent flow, plot \( \mathrm{V} / \mathrm{V}_{0} \) versus \( r / R \).For laminar flow: \[ \mathrm{V} / \mathrm{V}_{0} = 1 - \left( \frac{r}{R} \right)^2 \]For turbulent flow: \[ \mathrm{V} / \mathrm{V}_{0} = 1 - \left( \frac{r}{R} \right)^{1/7} \]Use graphical software or hand-draw the plots for the range \( 0 \le r/R \le 1 \).
03

Derive the Expression for Mass Flow Rate

Use the relationship \( Q = \int \! V \, dA \) where \( dA = 2 \pi r \, dr \).For laminar flow: \[ \dot{m} = \rho \int_{0}^{R} V \cdot 2 \pi r \, dr \]Substitute \( V = V_{0} \left[ 1 - \left( \frac{r}{R} \right)^2 \right] \):\[ \dot{m} = \rho \, V_{0} \int_{0}^{R} \left[ 1 - \left( \frac{r}{R} \right)^2 \right] 2 \pi r \, dr \]Solve the integral.
04

Find the Average Velocity

For average velocity, use \( V_{\text{ave}} = \frac{Q}{A} \), where \( A = \pi R^2 \).For laminar flow:\[ V_{\text{ave}} = \frac{2V_0 \int_{0}^{R} \left[ 1- \left( \frac{r}{R} \right)^2 \right] r \, dr}{\pi R^2} \]Evaluate the integral and simplify.
05

Specific Kinetic Energy and Percent Error

Specific kinetic energy is given by: \[ E_{k} = \frac{1}{2} \frac{\dot{m} \int_{0}^{R} V^2 dA}{\dot{m}} \]For laminar flow:\[ V^2 = V_{0}^2 \left[ 1 - \left( \frac{r}{R} \right)^2 \right]^2 \]Evaluate \( E_{k} \) and compare it to \( (V_{\text{ave}})^2 / 2 \). Calculate the percent error as: \[ \frac{E_{k} - \left( V_{\text{ave}} \right)^2 / 2}{E_{k}} \times 100 \% \]
06

Evaluate One-Dimensional Flow

Assess which flow (laminar or turbulent) adheres more closely to one-dimensional flow. Based on velocity uniformity and simplifications, determine if the turbulent or laminar profile fits the one-dimensional idealization better.

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Key Concepts

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

Laminar Flow
Laminar flow is a type of fluid flow where the fluid moves in smooth and orderly layers or paths. In this type of flow, each layer of the fluid slides past the adjacent layers with minimal mixing. The velocity distribution for laminar flow in a pipe is given by the equation: \ \[ \frac{V}{V_0} = 1 - \frac{r^2}{R^2} \] Here, \(V\) is the velocity at a distance \(r\) from the center of the pipe, \(V_0\) is the maximum velocity at the center, and \(R\) is the radius of the pipe. This parabolic profile indicates that the fluid moves slower near the edges and faster in the center, leading to a smooth and predictable flow pattern. Laminar flow is typically found in situations where the fluid viscosity is high, and the flow rate is low, resulting in a Reynolds number below 2000.
Turbulent Flow
Turbulent flow is characterized by chaotic, irregular fluid motion, where there is significant mixing between the layers of fluid. Unlike laminar flow, turbulent flow does not have a smooth velocity distribution. The velocity distribution for turbulent flow in a pipe can be approximated by the equation: \ \[ \frac{V}{V_0} = 1 - \frac{r}{R}^{1/7} \] In this equation, the velocity drop-off from the center to the edges is much less steep compared to laminar flow, reflecting a flatter velocity profile. Turbulent flow occurs at higher velocities and lower viscosities, typically when the Reynolds number exceeds 4000. This type of flow is common in many natural and industrial applications due to its ability to enhance mixing and increase heat and mass transfer.
Mass Flow Rate
Mass flow rate is a measure of the amount of mass passing through a given cross-sectional area per unit time. It is denoted by \( \dot{m}\ \) and can be calculated using the velocity distribution and the density of the fluid \( \rho\ \). For laminar flow, the mass flow rate is derived by integrating the product of velocity and the cross-sectional area over the radius: \ \[ \dot{m} = \rho \int_0^R V \cdot 2 \pi r \, dr \] Substituting the velocity distribution for laminar flow: \ \[ \dot{m} = \rho V_0 \int_0^R \[ 1 - ( \frac{r}{R} \ )^2 \] 2 \pi r \, dr \] The integral yields the total mass flow rate through the pipe. A similar approach is used for turbulent flow by substituting its corresponding velocity profile.
Average Velocity
The average velocity \( V_{\text{ave}} \) is a representative velocity of the fluid passing through the pipe. It is calculated by dividing the volumetric flow rate \( Q \) by the cross-sectional area \( A \): \ \[ V_{\text{ave}} = \frac{Q}{A} = \frac{ \dot{m} }{ \rho A } \] For laminar flow, using the derived mass flow rate expression, we find: \ \[ V_{\text{ave}} = \frac{ 2 V_0 \int_0^R \[ 1 - ( \frac{r}{R} )^2 \] r \, dr }{ \pi R^2 } \] Solving this integral gives us the average velocity in terms of \( V_0 \), \( R \), and fluid density \( \rho \). The process is similar for turbulent flow, but with its respective velocity profile.
Specific Kinetic Energy
Specific kinetic energy is the energy per unit mass associated with the velocity of the fluid. It is given by: \ \[ E_{k} = \frac{1}{2} \frac{ \dot{m} \int_0^R V^2 dA }{ \dot{m} } \] For laminar flow, we use the squared velocity distribution: \ \[ V^2 = V_0^2 \[ 1 - \frac{r^2}{R^2} \]^2 \] Evaluating the integral and finding \( E_{k} \) allows us to compare with \( \frac{ (V_{\text{ave}})^2 }{2} \). The percent error is calculated as: \ \[ \% \text{error} = \frac{ E_{k} - \( V_{\text{ave}} \)^2 / 2 }{ E_{k} } \times 100 \] This comparison helps us assess the accuracy of using average velocity for kinetic energy calculations. Laminar flow typically adheres closer to one-dimensional flow due to its orderly layers, compared to the chaotic nature of turbulent flow.

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Most popular questions from this chapter

Vegetable oil for cooking is dispensed from a cylindrical can fitted with a spray nozzle. According to the label, the can is able to deliver 560 sprays, each of duration \(0.25 \mathrm{~s}\) and each having a mass of \(0.25 \mathrm{~g}\). Determine (a) the mass flow rate of each spray, in \(\mathrm{g} / \mathrm{s}\). (b) the mass remaining in the can after 560 sprays, in \(\mathrm{g}\), if the initial mass in the can is \(170 \mathrm{~g}\).

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A rigid, well-insulated tank of volume \(0.5 \mathrm{~m}^{3}\) is initially evacuated. At time \(t=0\), air from the surroundings at 1 bar, \(21^{\circ} \mathrm{C}\) begins to flow into the tank. An electric resistor transfers energy to the air in the tank at a constant rate of \(100 \mathrm{~W}\) for \(500 \mathrm{~s}\), after which time the pressure in the tank is 1 bar. What is the temperature of the air in the tank, in \({ }^{\circ} \mathrm{C}\), at the final time?

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