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Chapter 8: Problem 33

Beyond the Colebrook equation, other relationships, such as the Fanning friction factor \(f,\) are available to estimate friction in pipes. 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 \(E_{a, d}<0.000005\).

Short Answer

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#tag_title#Step 2: Define the bisection method algorithm#tag_content# To implement the bisection method, we need to follow these steps: 1. Define an interval \([f_{lower}, f_{upper}]\) where \(g(f_{lower})\) and \(g(f_{upper})\) have different signs, meaning there is a root within the interval. 2. Calculate the midpoint (\(f_{mid}\)) of the interval. 3. Evaluate \(g(f_{mid})\). 4. Check the absolute error: \(E_a = |f_{mid} - f_{prev}|\), where \(f_{prev}\) is the midpoint from the previous iteration. If \(E_a < 0.000005\), we have achieved the desired accuracy, and the root is \(f_{mid}\). Otherwise, continue with step 5. 5. Determine the new interval: If \(g(f_{mid})\) and \(g(f_{lower})\) have different signs, the root is within the interval \([f_{lower}, f_{mid}]\). Otherwise, if \(g(f_{mid})\) and \(g(f_{upper})\) have different signs, the root is within \([f_{mid}, f_{upper}]\). Set the new interval accordingly, and return to step 2. We will now implement the algorithm and test it with given ranges of Re and \(f\) to solve for the Fanning friction factor.

Step by step solution

01

Convert the von Karman equation to a bisection-ready function

Let's convert the von Karman equation into a function that we can use in the bisection method. To do that, we need to isolate \(f\) on one side of the equation: \[\frac{1}{\sqrt{f}} - 4 \log_{10}(\operatorname{Re} \sqrt{f}) + 0.4 = 0\] Now, let's denote this function as \(g(f)\): \[g(f) = \frac{1}{\sqrt{f}} - 4 \log_{10}(\operatorname{Re} \sqrt{f}) + 0.4\]

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Key Concepts

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

Reynolds Number
The Reynolds number (\texttt{Re}) is a dimensionless quantity used in fluid mechanics to characterize the flow regime of fluid through a pipe or over a surface. It is defined as the ratio of inertial forces to viscous forces and provides insight into whether the flow will be laminar or turbulent.

For flow within a pipe, the Reynolds number is given by: \[ \texttt{Re} = \frac{\text{velocity} \times \text{diameter}}{\text{kinematic viscosity}} \]
A low Reynolds number, typically below about 2000, indicates laminar flow, where fluid particles move in parallel layers without disruption. Conversely, high Reynolds numbers signify turbulent flow, where the flow becomes chaotic and vortices and eddies are prevalent.
Colebrook Equation
The Colebrook equation is an empirical formula used in fluid mechanics for calculating the friction factor in turbulent, fully rough flow in circular pipes. It takes into account the effect of pipe roughness as well as the flow's Reynolds number. The equation is commonly written as:

\[ \frac{1}{\sqrt{f}} = -2 \log_{10} \left(\frac{e}{3.7D} + \frac{2.51}{\texttt{Re}\sqrt{f}}\right) \]
where \(e\) is the roughness height, \(D\) is the diameter of the pipe, \(f\) is the Darcy-Weisbach friction factor, and \(\texttt{Re}\) is the Reynolds number. Due to its implicit nature, solving the Colebrook equation typically requires iterative methods, such as the Newton-Raphson or bisection methods.
Von Karman Equation
The von Karman equation is another form used to approximate the friction factor \(f\) for turbulent flow conditions. It's derived from the Moody chart, which relates the Darcy-Weisbach friction factor to the Reynolds number. The von Karman equation is sometimes preferred because it tends to be simpler to use in computations:

\[ \frac{1}{\sqrt{f}} = 4 \log_{10}(\operatorname{Re} \sqrt{f}) - 0.4 \]
This somewhat simplified version of the full equation provides a practical way to estimate the Fanning friction factor, \(f\), given a known Reynolds number. However, just like the Colebrook equation, it is implicit in \(f\), which requires iterative methods to solve.
Bisection Method
The bisection method is a type of root-finding algorithm used in mathematics and computing to find zero points of continuous functions. It works by repeatedly bisecting an interval and then selecting a subinterval in which a root must lie, based on the sign of the function's values at the end points.

When the function changes sign across an interval, there must be a root within that interval due to the intermediate value theorem. By choosing the subinterval where the sign change occurs, the bisection method narrows down the location of the root. The process is continued until the desired level of precision is achieved, such as the absolute error condition outlined in the original exercise.
Turbulent Flow
Turbulent flow in fluid dynamics is characterized by chaotic and irregular motion, which stands in contrast to the orderly layers of laminar flow. Turbulence is associated with higher Reynolds numbers, typically greater than 4000, and involves complex interactions between the fluid's velocity components, resulting in mixing, eddies, and vortices.

In terms of pipe flow, turbulence affects the frictional resistance of the fluid moving inside the pipe, increasing the energy required to maintain the flow. This is why accurate estimation of the Fanning friction factor, which accounts for these losses, is crucial in the design and analysis of systems involving turbulent flow.

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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 \(V=0.5 \mathrm{m}^{3}\).

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The Ergun equation, shown below, is used to describe the flow of a fluid through a packed bed. \(\Delta P\) is the pressure drop, \(\rho\) is the density of the fluid, \(G_{o}\) is the mass velocity (mass flow rate divided by cross- sectional area), \(D_{p}\) is the diameter of the particles within the bed, \(\mu\) is the fluid viscosity, \(L\) is the length of the bed, and \(\varepsilon\) is the void fraction of the bed. $$\frac{\Delta P \rho}{G_{o}^{2}} \frac{D_{p}}{L} \frac{\varepsilon^{3}}{1-\varepsilon}=150 \frac{1-\varepsilon}{\left(D_{p} G_{o} / \mu\right)}+1.75$$ Given the parameter values listed below, find the void fraction \(\varepsilon\) of the bed. $$\begin{aligned}&\frac{D_{p} G_{o}}{\mu}=1000\\\&\frac{\Delta P \rho D_{p}}{G_{o}^{2} L}=10\end{aligned}$$

In chemical engineering, plug flow reactors (that is, those in which fluid flows from one end to the other with minimal mixing along the longitudinal axis) are often used to convert reactants into products. It has been determined that the efficiency of the conversion can sometimes be improved by recycling a portion of the product stream so that it returns to the entrance for an additional pass through the reactor (Fig. \(P 8.2\) ). The recycle rate is defined as \(R=\frac{\text { volume of fluid returned to entrance }}{\text { volume leaving the system }}\) Suppose that we are processing a chemical A to generate a product B. For the case where A forms B according to an autocatalytic reaction (that is, in which one of the products acts as a catalyst or stimulus for the reaction \(),\) it can be shown that an optimal recycle rate must satisfy $$\ln \frac{1+R\left(1-X_{A f}\right)}{R\left(1-X_{A f}\right)}=\frac{R+1}{R\left[1+R\left(1-X_{A f}\right)\right]}$$ where \(X_{A f}=\) the fraction of reactant A that is converted to product B. The optimal recycle rate corresponds to the minimum-sized reactor needed to attain the desired level of conversion. Use a numerical method to determine the recycle ratios needed to minimize reactor size for a fractional conversion of \(X_{A f}=0.96\).

The following equation pertains to the concentration of a chemical in a completely mixed reactor: $$c=c_{\mathrm{in}}\left(1-e^{-0.04 t}\right)+c_{0} e^{-0.04 t}$$ If the initial concentration \(c_{0}=4\) and the inflow concentration \(c_{\mathrm{in}}=10,\) compute the time required for \(c\) to be 93 percent of \(c_{\mathrm{in}}\).
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