/*! 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 43 Figure \(P 8.43\) shows three re... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 43

Figure \(P 8.43\) shows three reservoirs connected by circular pipes. The pipes, which are made of asphalt-dipped cast iron \((\varepsilon=0.0012 \mathrm{m}),\) have the following characteristics: $$\begin{array}{lccc}\text { Pipe } & 1 & 2 & 3 \\\\\text { Length, } \mathrm{m} & 1800 & 500 & 1400 \\\\\text { Diomeler, } \mathrm{m} & 0.4 & 0.25 & 0.2 \\\\\text { Flow, } \mathrm{m}^{3} / \mathrm{s} & 2 & 0.1 & ? \\\\\hline\end{array}$$ If the water surface elevations in Reservoirs \(A\) and \(C\) are 200 and \(172.5 \mathrm{m},\) respectively, determine the elevation in Reservoir \(\mathrm{B}\) and the flows in pipes 1 and 3 . Note that the kinematic viscosity of water is \(1 \times 10^{-6} \mathrm{m}^{2} / \mathrm{s}\) and use the Colebrook equation to determine the friction factor (recall Prob. 8.12 ).

Short Answer

Expert verified
The elevation in Reservoir B is \(172.6 \ m\) and the flow in pipe 3 is \(2.1 \ m^3/s\).

Step by step solution

01

Calculate the Reynolds number

To calculate the Reynolds number, we can use the formula: \[Re = \frac{VD}{\nu}\] where \(V\) is the flow velocity, \(D\) is the pipe diameter, and \(\nu\) is the kinematic viscosity of the fluid. First, we need to calculate the flow velocities in the pipes using the known flow rates (\(Q\)), using the formula: \[V = \frac{Q}{A}\] where \(A\) is the cross-sectional area of the pipe. For Pipe 1: \(V_1 = \frac{2 \ m^3/s}{\pi (0.4 \ m / 2)^2} = 6.366 \ m/s\) For Pipe 2: \(V_2 = \frac{0.1 \ m^3/s}{\pi (0.25 \ m / 2)^2} = 2.035 \ m/s\) Now, let's calculate the Reynolds number for each pipe using the given kinematic viscosity, \(\nu = 1 \times 10^{-6}\ m^2/s\). For Pipe 1: \(Re_1 = \frac{6.366 \ m/s \times 0.4 \ m}{1 \times 10^{-6} \ m^2/s} = 2.546 \times 10^6\) For Pipe 2: \(Re_2 = \frac{2.035 \ m/s \times 0.25 \ m}{1 \times 10^{-6} \ m^2/s} = 508.75 \times 10^3\) Now, we will need to determine the friction factor for both pipes using the next step.
02

Use the Colebrook equation to find the friction factor

The Colebrook equation is as follows: \[\frac{1}{\sqrt{f}} = -2 \log_{10}\left(\frac{\varepsilon/D}{3.7} + \frac{2.51}{Re\sqrt{f}}\right)\] For Pipe 1: The given relative roughness of the pipe is \(\varepsilon = 0.0012 \ m\). So, \(\frac{\varepsilon}{D} = \frac{0.0012 \ m}{0.4 \ m} = 0.003\). Use an iterative process to solve the Colebrook equation for the friction factor and calculate \(f_1 \approx 0.026\). For Pipe 2: Again, the given relative roughness of the pipe is \(\varepsilon = 0.0012 \ m\). So, \(\frac{\varepsilon}{D} = \frac{0.0012 \ m}{0.25 \ m} = 0.0048\). Use an iterative process to solve the Colebrook equation for the friction factor and calculate \(f_2 \approx 0.029\).
03

Use the energy equation to find the elevation of Reservoir B and flow in pipe 3

To find the elevation of Reservoir B, let's first determine the head loss due to friction in pipes 1 and 2, using the Darcy-Weisbach equation: \[h_f = f \frac{L}{D} \frac{V^2}{2g}\] where \(h_f\) is the head loss due to friction, \(L\) is the pipe length, \(f\) is the friction factor, and \(g\) is the acceleration due to gravity. For Pipe 1: \(h_{f1} = 0.026 \frac{1800 \ m}{0.4 \ m}\frac{(6.366 \ m/s)^2}{2(9.81 \ m/s^2)} = 24.2 \ m\) For Pipe 2: \(h_{f2} = 0.029 \frac{500 \ m}{0.25 \ m}\frac{(2.035 \ m/s)^2}{2(9.81 \ m/s^2)} = 3.2 \ m\) Now, we can use the energy equation to determine the elevation of Reservoir B: \[z_B = z_A - h_{f1} - h_{f2}\] \[z_B = 200 \ m - 24.2 \ m - 3.2 \ m = 172.6 \ m\] To find the flow in Pipe 3, we can use the continuity equation: \[Q_3 = Q_1 + Q_2\] \[Q_3 = 2 \ m^3/s + 0.1 \ m^3/s = 2.1 \ m^3/s\] So, the elevation in Reservoir B is \(172.6 \ m\) and the flow in pipe 3 is \(2.1 \ m^3/s\).

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!

Key Concepts

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

Understanding the Reynolds Number
When studying fluid flow in pipes, one of the first steps is to calculate the Reynolds number (Re). This dimensionless number helps us understand the nature of the flow鈥攚hether it's laminar or turbulent. Laminar flow occurs at lower Reynolds numbers and is characterized by smooth, orderly motion of fluid particles in layers. Turbulent flow, on the other hand, occurs at higher Reynolds numbers and is marked by chaotic fluid motion and mixing.

Re is expressed as:\[Re = \frac{VD}{u}\]where:\[\begin{align*}V & = \text{flow velocity (m/s)} \D & = \text{pipe diameter (m)} \u & = \text{kinematic viscosity (m}^2\text{/s)}\end{align*}\]
For any given pipe, as the flow velocity increases or the viscosity of the fluid decreases, Re will rise, indicating a potential shift from laminar to turbulent flow. This shift has a direct impact on how we calculate the frictional losses in the pipe system.
The Colebrook Equation for Friction Factor
The Colebrook equation is a semi-empirical formula used to calculate the friction factor (f) for turbulent flow in pipes when the Reynolds number is known. It accounts for the effects of pipe roughness and flow kinematics:\[\frac{1}{\sqrt{f}} = -2 \log_{10}\left(\frac{\varepsilon/D}{3.7} + \frac{2.51}{Re\sqrt{f}}\right)\]
Here, \(\varepsilon\) is the absolute roughness of the pipe's inner surface, and \(D\) is the pipe's internal diameter. The term 'friction factor' is crucial for determining the resistance to flow due to the surface conditions of the pipe, which can vary based on the material鈥攍ike asphalt-dipped cast iron in our example.

Finding f requires an iterative approach or use of a Colebrook equation solver since f appears on both sides of the equation. This complexity underlines the equation's use in understanding the interplay between the physical roughness of a pipe and the flow dynamics within it.
Head Loss Due to Friction with the Darcy-Weisbach Equation
The head loss due to friction in pipe flow is a critical aspect of fluid mechanics. We can calculate it using the Darcy-Weisbach equation:\[h_f = f \frac{L}{D} \frac{V^2}{2g}\]where:\[\begin{align*}h_f & = \text{head loss due to friction (m)} \L & = \text{length of the pipe} \text{(m)} \V & = \text{flow velocity (m/s)} \g & = \text{acceleration due to gravity (m/s}^2\text{)}\end{align*}\]
This equation not only incorporates the friction factor (f) but also ties in the velocity of the fluid and the geometry of the pipe. Essentially, it allows one to predict how much energy is lost as water travels through the pipe due to resistance from the pipe walls.

By using the Darcy-Weisbach equation, engineers can design systems that ensure efficient flow, accounting for the necessary pumps or pressure to overcome these losses.
Calculating Flows and Elevations in Pipe Systems
The complexities of pipe networks, such as those connecting reservoirs, require precise calculations to ensure that water is delivered efficiently. By using the energy equation along with considerations like head loss due to friction computed from the Darcy-Weisbach equation, the elevation of water in different reservoirs can be identified given certain assumptions.

The elevation in a reservoir can significantly influence the flow through a pipe. When elevations are known, the flow in connecting pipes can additionally be determined by the continuity equation, or conservation of mass principle, stating that the amount of fluid entering a system must equal the amount of fluid leaving the system.

This continuity equation, combined with the energy equation:\[z_B = z_A - h_{f1} - h_{f2}\]enables us to systematically solve for unknown flows or elevations within the network. By following these principles, students can not only solve textbook problems but also apply these methodologies to real-world hydraulic engineering projects.

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 general form for a three-dimensional stress field is given by \\[\left[\begin{array}{lll}\sigma_{x x} & \sigma_{x y} & \sigma_{x z} \\ \sigma_{x y} & \sigma_{y y} & \sigma_{y z} \\\\\sigma_{x z} & \sigma_{y z} & \sigma_{z z}\end{array}\right]\\] where the diagonal terms represent tensile or compressive stresses and the off-diagonal terms represent shear stresses. A stress field (in MPa) is given by \\[\left[\begin{array}{ccc}10 & 14 & 25 \\\14 & 7 & 15 \\\25 & 15 & 16\end{array}\right]\\] To solve for the principal stresses, it is necessary to construct the following matrix (again in MPa): \\[\left[\begin{array}{ccc}10-\sigma & 14 & 25 \\\14 & 7-\sigma & 15 \\\25 & 15 & 16-\sigma\end{array}\right]\\] \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) can be solved from the equation \\[\sigma^{3}-I \sigma^{2}+I I \sigma-I I I=0\\] where \\[\begin{array}{l}I=\sigma_{x x}+\sigma_{y y}+\sigma_{z z} \\\I I=\sigma_{x x} \sigma_{y y}+\sigma_{x x} \sigma_{z z}+\sigma_{y y} \sigma_{zz}-\sigma_{x y}^{2}-\sigma_{x z}^{2}-\sigma_{y z}^{2} \\\I I=\sigma_{x x} \sigma_{y y} \sigma_{z z}-\sigma_{x x} \sigma_{y z}^{2}-\sigma_{y y} \sigma_{x z}^{2}-\sigma_{z z} \sigma_{x y}^{2}+2 \sigma_{x y} \sigma_{x z} \sigma_{y z}\end{array}\\] \(I, I I,\) and \(I I I\) are known as the stress invariants. Find \(\sigma_{1}, \sigma_{2},\) and \(\sigma_{3}\) using a root-finding technique.

The following chemical reactions take place in a closed system \\[\begin{array}{l}2 \mathrm{A}+\mathrm{B} \rightleftharpoons \mathrm{C} \\\\\mathrm{A}+\mathrm{D} \rightleftharpoons \mathrm{C}\end{array}\\] At equilibrium. they can be characterized by \\[K_{1}=\frac{c_{c}}{c_{d}^{2} c_{b}}\\] \\[K_{2}=\frac{c_{c}}{c_{a} c_{d}}\\] where the nomenclature \(c_{i}\) represents the concentration of constituent i. If \(x_{1}\) and \(x_{2}\) are the number of moles of \(C\) that are produced due to the first and second reactions, respectively, use an approach similar to that of Prob. 8.5 to reformulate the equilibrium relationships in terms of the initial concentrations of the constituents. Then, use the Newton-Raphson method to solve the pair of simultaneous nonlinear equations for \(x_{1}\) and \(x_{2}\) if \(K_{1}=4 \times 10^{-4}, K_{2}=3.7 \times 10^{-2}\) \(c_{u, 0}=50, c_{b, 0}=20, c_{c .0}=5,\) and \(c_{d, 0}=10 .\) Use a graphical approach to develop your initial guesses.

The Redlich-Kwong equation of state is given by \\[p=\frac{R T}{v-b}-\frac{a}{v(v+b) \sqrt{T}}\\] where \(R=\) the universal gas constant \([=0.518 \mathrm{kJ} /(\mathrm{kg} \mathrm{K})], T=\) absolute temperature (K), \(p=\) absolute pressure (kPa), and \(v=\) the volume of a kg of gas \(\left(\mathrm{m}^{3} / \mathrm{kg}\right)\). The parameters \(a\) and \(b\) are calculated by \\[a=0.427 \frac{R^{2} T_{c}^{2.3}}{p_{c}} \quad b=0.0866 R \frac{T_{c}}{p_{c}}\\] where \(p_{c}=\) critical pressure (kPa) and \(T_{c}=\) critical temperature (K). As a chemical engineer, you are asked to determine the amount of methane fuel \(\left(p_{c}=4580 \mathrm{kPa} \text { and } T_{c}=191 \mathrm{K}\right)\) that can be held in a \(3-m^{3}\) tank at a temperature of \(-50^{\circ}\) C with a pressure of 65.000 kPa. Use a root-locating method of your choice to calculate \(v\) and then determine the mass of methane contained in the tank.

\(\mathrm{A}\) fluid is pumped into the network of pipes shown in Fig. \(P 8.44 .\) At steady state, the following flow balances must hold,\\[ \begin{array}{l}Q_{1}=Q_{2}+Q_{3} \\\Q_{3}=Q_{4}+Q_{5} \\\Q_{5}=Q_{6}+Q_{7}\end{array}\\] where \(Q_{i}=\) flow in pipe \(i\left[\mathrm{m}^{3} / \mathrm{s}\right] .\) In addition, the pressure drops around the three right-hand loops must equal zero. The pressure drop in each circular pipe length can be computed with \\[\Delta P=\frac{16}{\pi^{2}} \frac{f L \rho}{2 D^{5}} Q^{2} \\]where \(\Delta P=\) the pressure drop \([\mathrm{Pa}], f=\) the friction factor [dimensionless], \(L=\) the pipe length \([\mathrm{m}],\rho=\) the fluid density \(\left[\mathrm{kg} / \mathrm{m}^{3}\right],\) and \(D=\) pipe diameter \([\mathrm{m}] .\) Write a program (or develop an algorithm in a mathematics software package) that will allow you to compute the flow in every pipe length givep that \(Q_{1}=1 \mathrm{m}^{3} / \mathrm{s}\) and \(\rho=1.23 \mathrm{kg} / \mathrm{m}^{3} .\) All the pipes have \(D=500 \mathrm{mm}\) and \(f=0.005 .\) The pipe lengths are: \(L_{3}=L_{5}=L_{8}=L_{9}=2 \mathrm{m}\) \(L_{2}=L_{4}=L_{6}=4 \mathrm{m} ;\) and \(L_{7}=8 \mathrm{m}\).

The upward velocity of a rocket can be computed by the following formula: \\[v=u \ln \frac{m_{0}}{m_{0}-q t}-g t\\] where \(v=\) upward velocity, \(u=\) the velocity at which fuel is expelled relative to the rocket, \(m_{0}=\) the initial mass of the rocket at time \(t=0, q=\) the fuel consumption rate, and \(g=\) the downward acceleration of gravity (assumed constant \(=9.81 \mathrm{m} / \mathrm{s}^{2}\) ). If \(u=2000 \mathrm{m} / \mathrm{s}, m_{0}=150,000 \mathrm{kg},\) and \(q=2700 \mathrm{kg} / \mathrm{s},\) compute the time at which \(=750 \mathrm{m} / \mathrm{s}\). (Hint: \(t\) is somewhere between 10 and 50 s.) Determine your result so that it is within \(1 \%\) of the true value. Check your answer.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Electricity and Magnetism

Read Explanation

Physics of Motion

Read Explanation

Waves Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Modern Physics

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.