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Chapter 8: Problem 33

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 \(E_{a, d}<0.000005\).

Short Answer

Expert verified
In summary, to solve for Fanning friction factor (f) given a user-supplied Reynolds number (Re) between 2,500 and 1,000,000, using the bisection method with an absolute error tolerance Ea,d < 0.000005: 1. Define the given von Karman equation as a continuous function g(f), \( g(f) = \frac{1}{\sqrt{f}} - (4 \log_{10}(\operatorname{Re} \sqrt{f}) - 0.4) \). 2. Define the function 'bisection_method' to find the root of g(f) within a given interval [a, b] and absolute error tolerance Ea,d < 0.000005. 3. Define and implement a primary function 'compute_f' that takes user-supplied Reynolds number Re, and initial guess values of f (f_lower and f_upper) as input. Call the 'bisection_method' function to compute the Fanning friction factor (f). The 'compute_f' function returns the value of Fanning friction factor (f) calculated using the bisection method, meeting the absolute error tolerance criterion.

Step by step solution

01

Understand the bisection method

The bisection method is a root-finding numerical algorithm, which finds the root of a continuous function in a given closed interval, suppose [a,b]. Given that the function is continuous, and there's a root between a and b, then we can find the midpoint (c) of the interval [a,b]. The bisection method then checks the midpoint (c) value against the given tolerance value, and if it's not within the tolerance value, it selects which half of the interval should continue searching, depending on the sign of the function.
02

Define the given equation

We have the given Von Karman equation: \( \frac{1}{\sqrt{f}} = 4 \log_{10}(\operatorname{Re} \sqrt{f}) - 0.4 \). To make it suitable for solving using the bisection method, we can re-write the equation as: \( g(f) = \frac{1}{\sqrt{f}} - (4 \log_{10}(\operatorname{Re} \sqrt{f}) - 0.4) \) Now, we can find the root of this continuous function g(f) using the bisection method.
03

Define the bisection method function

We can define a function named bisection_method to find the root of the given equation g(f) within a given interval [a, b] and absolute error tolerance Ea,d < 0.000005. The function should have the following parameters: g(f), Interpolation values a and b, Reynolds number Re as input, and the stop criterion |g(c)|< Ea,d.
04

Implement the bisection method

Iterate the bisection method until the absolute error tolerance is met: 1. Calculate the function value at a and b, g(a) and g(b). Ensure that the product g(a)g(b) < 0, meaning that there's a root between the values a and b. 2. Compute the midpoint of the interval: \( c = \frac{a + b}{2} \) 3. Calculate the function value at the midpoint: g(c) 4. Check if the absolute error tolerance is met: \( |g(c)| < Ea,d \). If true, return c as the root value for f. 5. If the absolute error tolerance is not met, update the interval [a, b] to [a, c] or [c, b] based on the signs of g(a)g(c) and g(b)g(c). 6. Repeat steps 2-5 until the absolute error tolerance is met.
05

Define the main function to compute f

Define a primary function called 'compute_f', which takes user-supplied Reynolds number Re, and initial guess values of f (f_lower and f_upper) as input. Call the 'bisection_method' function within 'compute_f' using these input values, to compute the Fanning friction factor (f). Return the value of f.

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Key Concepts

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

Fanning friction factor
The Fanning friction factor, denoted as \( f \), is a dimensionless number that quantifies the frictional resistance experienced by fluid flowing through a pipe. It plays an essential role in fluid dynamics as it indicates the effect of viscous forces in a pipe's pressure drop.

Understanding the Fanning friction factor can help engineers and scientists analyze and optimize piping systems, ensuring that energy losses due to friction are minimized. The Fanning friction factor is dependent on factors such as:
  • The pipe's material and surface roughness
  • The fluid's velocity and viscosity
  • The pipe's diameter
This factor is especially critical in turbulent flow conditions, where the flow patterns are chaotic and irregular. Typical values range from 0.001 to 0.01. A lower friction factor indicates a smoother and less resistant flow, while higher values suggest more resistance.

The Fanning friction factor is determined empirically or calculated using equations like the von Karman equation based on the Reynolds number, which we will cover next.
Reynolds number
The Reynolds number, often denoted as \( ext{Re} \), is a fundamental dimensionless parameter in fluid mechanics. It helps in predicting the flow regime of a fluid in a pipe or over a surface. The Reynolds number provides insights into whether the flow will be laminar or turbulent.

The Reynolds number is calculated using the formula: \[ ext{Re} = \frac{{ ho v D}}{{u}} = \frac{{v D}}{{u/\rho}} = \frac{{v D}}{{u_{ ext{kin}}}}\]where:
  • \( \rho \) is the fluid density
  • \( v \) is the velocity of the fluid
  • \( D \) is the characteristic length (such as diameter for pipe flow)
  • \( u \) is the dynamic viscosity of the fluid
  • \( u_{kin} \) is the kinematic viscosity (i.e., \( u/\rho \))
Different ranges of Reynolds numbers indicate different flow regimes:
  • Re < 2000: Laminar flow
  • 2000 < Re < 4000: Transitional flow
  • Re > 4000: Turbulent flow
Understanding these ranges is important for engineers to design efficient systems and predict behavior in different scenarios.

For the bisection method exercise, where we solve for \( f \), it is essential to know the Reynolds number because it is a significant input for the von Karman equation.
von Karman equation
The von Karman equation is a specialized equation used in fluid dynamics to relate the Fanning friction factor to the Reynolds number. It is especially useful for solving complex flow problems through empirical correlations. The equation is given by: \[\frac{1}{\sqrt{f}}=4 \log_{10}(\operatorname{Re} \sqrt{f})-0.4\]
This formula is utilized specifically for turbulent flow, and it allows the calculation of the friction factor \( f \) when the Reynolds number is known.

Solving the von Karman equation is not straightforward because it is implicit in nature; that is, \( f \) appears on both sides of the equation. Thus, numerical methods like the bisection method become essential in finding the friction factor accurately.

The equation shows the logarithmic relationship between the two variables, indicating that small changes in the Reynolds number can lead to significant impacts on the friction factor. Therefore, accurate estimation and calculation are crucial in engineering applications.

When using this equation in calculations, it's vital to ensure that the inputs (like Reynolds number) fall within realistic ranges to achieve practical and meaningful results. This understanding helps in optimizing systems and effectively controlling costs and efficiency.

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Most popular questions from this chapter

The pressure drop in a section of pipe can be calculated as \\[\Delta p=f \frac{L \rho V^{2}}{2 D}\\] where \(\Delta p=\) the pressure drop (Pa), \(f=\) the friction factor, \(L=\) the length of pipe \([\mathrm{m}], \rho=\) density \(\left(\mathrm{kg} / \mathrm{m}^{3}\right), V=\) velocity \((\mathrm{m} / \mathrm{s})\) and \(D=\) diameter (m). For turbulent flow, the Colebrook equation provides a means to calculate the friction factor, \\[\frac{1}{\sqrt{f}}=-2.0 \log \left(\frac{\varepsilon}{3.7 D}+\frac{2.51}{\operatorname{Re} \sqrt{f}}\right)\\] where \(\varepsilon=\) the roughness \((\mathrm{m}),\) and \(\mathrm{Re}=\) the Reynolds number \\[\mathrm{Re}=\frac{\rho V D}{\mu}\\] where \(\mu=\) dynamic viscosity \(\left(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\right)\) (a) Determine \(\Delta p\) for a 0.2 -m-long horizontal stretch of smooth drawn tubing given \(\rho=1.23 \mathrm{kg} / \mathrm{m}^{3}, \mu=1.79 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\) \(D=0.005 \mathrm{m}, V=40 \mathrm{m} / \mathrm{s},\) and \(\varepsilon=0.0015 \mathrm{mm} .\) Use a numer- ical method to determine the friction factor. Note that smooth pipes with \(\mathrm{Re}<10^{5},\) a good initial guess can be obtained using the Blasitus formula, \(f=0.316 / \mathrm{Re}^{0.25}\) (b) Repeat the computation but for a rougher commercial steel pipe \((\varepsilon=0.045 \mathrm{mm})\)

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.

The space shuttle, at lift-off from the launch pad, has four forces acting on it, which are shown on the free-body diagram (Fig. P8.46). The combined weight of the two solid rocket boosters and external fuel tank is \(W_{B}=1.663 \times 10^{6} \mathrm{lb}\). The weight of the orbiter with a full payload is \(W_{s}=0.23 \times 10^{6} \mathrm{Jb}\). The combined thrust of the two solid roctet boosters is \(T_{B}=5.30 \times 10^{6}\) lb. The combined thrust of the three liquid fuel orbiter engines is \(T_{S}=1.125 \times 10^{6} \mathrm{lb}\) At liftoff, the orbiter engine thrust is directed at angle \(\theta\) to make the resultant moment acting on the entire craft assembly (external tank, solid rocket boosters, and orbiter) equal to zero. With the resultant moment equal to zero, the craft will not rotate about its mass center \(G\) at liftoff. With these forces, the craft will have a resultant force with components in both the vertical and horizontal direction. The vertical resultant force component is what allows the craft to lift off from the launch pad and fly vertically. The horizontal resultant force component causes the craft to fly horizontally. The resultant mornent acting an the craft will be zero when \(\theta\) is adjusted to the proper value. If this angle is not adjusted properly, and there is some resultant moment acting on the craft, the craft will tend to rotate about it mass center. (a) Resolve the orbiter thrust \(T_{S}\) into horizontal and vertical components, and then sum moments about point \(G,\) the craft mass center. Set the resulting moment equation equal to zero. This equation can now be solived for the value of \(\theta\) required for liftoff. (b) Derive an equation for the resutent moment acting on the craft in terms of the angle \(\theta\). Plot the resultant moment as a function of the angle \(\theta\) over a range of -5 radians to +5 radians. (c) Write a computer program to solve for the angle \(\theta\) using Newton's method to find the root of the resultant moment equation. Make an initial first guess at the root of interest using the plot. Terminate your iterations when the value of \(\theta\) has better than five significant figures. (d) Repeat the program for the minimum payload weight of the orbiter of \(W_{S}=195,000 \mathrm{lb}\).

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 ).
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