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Chapter 8: Problem 44

\(\mathrm{A}\) 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 \(L_{7}=8 \mathrm{m}\).

Short Answer

Expert verified
Upon solving the system of nonlinear equations using numerical methods or software packages, we will obtain the flow rates \(Q_2, Q_3, Q_4, Q_5, Q_6, Q_7, Q_8\), and \(Q_9\). These values will allow us to compute the flow in every pipe length as per the given problem statement.

Step by step solution

01

Write down the given information and equations

We are given the following flow balance equations: \(Q_1 = Q_2 + Q_3 \\ Q_3 = Q_4 + Q_5 \\ Q_5 = Q_6 + Q_7 \) And the pressure drop equation for each pipe: \(\Delta P = \frac{16}{\pi^2} \frac{fL\rho}{2D^5} Q^2\) We are also given: \(Q_1 = 1 \, \mathrm{m^3/s}, \rho = 1.23 \, \mathrm{kg/m^3}, D = 0.5 \, \mathrm{m}, f = 0.005\) And pipe lengths: \(L_3 = L_5 = L_8 = L_9 = 2 \, \mathrm{m}, L_2 = L_4 = L_6 = 4 \, \mathrm{m}, L_7 = 8 \, \mathrm{m}\)
02

Set up the pressure drop equations for the three right-hand loops

We have three loops with pressure drops equal to zero (right-hand loops): Loop 1: \(\Delta P_3 - \Delta P_4 - \Delta P_8 = 0\) Loop 2: \(\Delta P_5 - \Delta P_6 - \Delta P_9 = 0\) Loop 3: \(\Delta P_8 + \Delta P_7 - \Delta P_9 - \Delta P_2 = 0\) Plug the pressure drop equation into the loops' equations: Loop 1: \(\frac{16}{\pi^2} \frac{fL_3\rho}{2D^5} Q_3^2 = \frac{16}{\pi^2} \frac{fL_4\rho}{2D^5} Q_4^2 + \frac{16}{\pi^2} \frac{fL_8\rho}{2D^5} Q_8^2\) Loop 2: \(\frac{16}{\pi^2} \frac{fL_5\rho}{2D^5} Q_5^2 = \frac{16}{\pi^2} \frac{fL_6\rho}{2D^5} Q_6^2 + \frac{16}{\pi^2} \frac{fL_9\rho}{2D^5} Q_9^2\) Loop 3: \(\frac{16}{\pi^2} \frac{fL_8\rho}{2D^5} Q_8^2 + \frac{16}{\pi^2} \frac{fL_7\rho}{2D^5} Q_7^2 = \frac{16}{\pi^2} \frac{fL_9\rho}{2D^5} Q_9^2 + \frac{16}{\pi^2} \frac{fL_2\rho}{2D^5} Q_2^2\)
03

Solve the system of nonlinear equations

We now have a system of five nonlinear equations (the original three flow balance relations and the two new loop pressure equations) with seven unknowns (flow rates in pipes 2 to 9). One strategy to solve this system is to use the Newton-Raphson method for simultaneous equations or an optimization algorithm to minimize the error in the equations. An alternative approach is to use a numerical solver or a symbolic computation software package (such as MATLAB, Python with the sympy library, or Wolfram Mathematica) to solve the equations. Once we obtain the solutions for the flow rates in the pipes, we can use them to compute the flow in every pipe length according to the problem statement.

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Key Concepts

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

Fluid Mechanics
Understanding fluid mechanics is essential when analyzing the flow through a system of pipes. Fluid mechanics is the branch of physics concerned with the motion of fluids (liquids and gases) and the forces acting on them. In the context of pipe flow analysis, one of the primary goals is to understand how fluids behave while moving through different pipe geometries and how variables like pressure, flow rate, and resistance due to friction affect this behavior.

Key principles in fluid mechanics, such as continuity and Bernoulli's equation, underpin the equations we use to describe flow in pipes. The continuity principle states that, for incompressible flow, the volume flowing into a pipe system must equal the volume flowing out. This conservation of mass is reflected in the flow balance equations:
  • \(Q_1 = Q_2 + Q_3\)
  • \(Q_3 = Q_4 + Q_5\)
  • \(Q_5 = Q_6 + Q_7\)
These equations ensure that the flow into each junction matches the flow out, a critical condition for analyzing steady-state behavior in a network of pipes.
Pressure Drop Calculation
The pressure drop across a pipe segment is a crucial aspect of fluid dynamics, which must be understood for the design and analysis of fluid networks. A pressure drop occurs when frictional forces, caused by the fluid's viscosity and the pipe's roughness, slow down the flow of the fluid. The formula used in the provided exercise

\[\Delta P = \frac{16}{\pi^2} \frac{fL\rho}{2D^5} Q^2\]

breaks down the relationship between the pressure drop (\(\Delta P\)) and important factors like the flow rate (\(Q\)), pipe diameter (\(D\)), fluid density (\(\rho\)), and the friction factor (\(f\)), alongside the length of the pipe (\(L\)). Notably, this equation reveals how sensitive the pressure drop is to the diameter of the pipe, with the dependence being on the fifth power of the diameter. The exercise puts this equation into practice by calculating pressure drops in a network where these variables interplay to maintain a zero-pressure drop around loops, ensuring steady and continuous flow.
Newton-Raphson Method
The Newton-Raphson method is a powerful tool to find successively better approximations to the roots of a real-valued function. It's particularly useful in systems of equations arising from fluid mechanics problems where analytical solutions might be cumbersome or impossible to find. In our pipe flow analysis, we're faced with a system of nonlinear equations that model the flow rates and pressure drops in a network.

The Newton-Raphson method allows us to iterate towards the solution to these equations by making an initial guess and then improving that guess repeatedly. It does this by considering the derivative of the function we're trying to find roots for - which, in a system of equations, translates to using a Jacobian matrix that represents how each equation changes with all the variables. Applying this method to the nonlinear system given in the exercise, you'd start with initial guesses for the flow in each pipe and iteratively solve until the changes between successive iterations are negligible, revealing the flow rates with acceptable accuracy.

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Most popular questions from this chapter

The volume \(V\) of liquid in a spherical tank of radius \(r\) is related to the depth \(h\) of the liquid by \\[V=\frac{\pi h^{2}(3 r-h)}{3}\\] Determine \(h\) given \(r=1 \mathrm{m}\) and \(V=0.75 \mathrm{m}^{3}\)

\( \mathrm{A}\) total charge \(Q\) is uniformly distributed around a ringshaped conductor with radius \(a\). A charge \(q\) is located at a distance \(x\) from the center of the ring (Fig. \(\mathrm{P} 8.31\) ). The force exerted on the charge by the ring is given by \\[F=\frac{1}{4 \pi e_{0}} \frac{q Q x}{\left(x^{2}+a^{2}\right)^{3 / 2}}\\] where \(e_{0}=8.85 \times 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \mathrm{m}^{2}\right) .\) Find the distance \(x\) where the force is \(1.25 \mathrm{N}\) if \(q\) and \(Q\) are \(2 \times 10^{-5} \mathrm{C}\) for a ring with a radius of \(0.9 \mathrm{m}\).

The operation of a constant density plug flow reactor for the production of a substance via an enzymatic reaction is described by the equation below, where \(V\) is the volume of the reactor, \(F\) is the flow rate of reactant \(C, C_{\text {in }}\) and \(C_{\text {out }}\) are the concentrations of reactant entering and leaving the reactor, respectively, and \(K\) and \(k_{\max }\) are constants. For a 500 -L reactor, with an inlet concentration of \(C_{\text {in }}=0.5 \mathrm{M},\) an inlet flow rate of \(40 \mathrm{L} / \mathrm{s}, k_{\max }=5 \times 10^{-3} \mathrm{s}^{-1},\) and \(K=0.1 \mathrm{M},\) find the concentration of \(C\) at the outlet of the reactor \\[\frac{V}{F}=-\int_{C_{i n}}^{C_{m}} \frac{K}{k_{\max } C}+\frac{1}{k_{\max }} d C\\]

Aerospace engineers sometimes compute the trajectories of projectiles like rockets. A related problem deals with the trajectory of a thrown ball. The trajectory of a ball is defined by the \((x, y)\) coordinates, as displayed in Fig. P8.36. The trajectory can be modeled as \\[y=\left(\tan \theta_{0}\right) \cdot x-\frac{g}{2 v_{0}^{2} \cos ^{2} \theta_{0}} x^{2}+y_{0}\\] Find the appropriate initial angle \(\theta_{0},\) if the initial velocity \(v_{0}=20 \mathrm{m} / \mathrm{s}\) and the distance to the catcher \(x\) is \(35 \mathrm{m}\). Note that the ball leaves the thrower's hand at an elevation of \(y_{0}=2 \mathrm{m}\) and the catcher receives it at \(1 \mathrm{m}\). Express the final result in degrees. Use a value of \(9.81 \mathrm{m} / \mathrm{s}^{2}\) for \(g\) and employ the graphical method to develop your initial guesses.

For fluid flow in pipes, friction is described by a dimensionless number, the Fanning friction factor \(f\). The Fanning friction factor is dependent on a number of parameters related to the size of the pipe and the fluid, which can all be represented by another dimensionless quantity, the Reynolds number Re. A formula that predicts \(f\) given \(\operatorname{Re}\) is the von Karman equation, \\[\frac{1}{\sqrt{f}}=4 \log _{10}(\operatorname{Re} \sqrt{f})-0.4\\] Typical values for the Reynolds number for turbulent flow are 10,000 to 500,000 and for the Fanning friction factor are 0.001 to \(0.01 .\) Develop a function that uses bisection to solve for \(f\) given a user-supplied value of Re between 2,500 and 1,000,000 . Design the function so that it ensures that the absolute error in the result is \(E_{a, d}<0.000005\).
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