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Chapter 8: Problem 45

Repeat Prob. 8.44 , but incorporate the fact that the friction factor can be computed with the von Karman equation, $$\frac{1}{\sqrt{f}}=4 \log _{10}(\operatorname{Re} \sqrt{f})-0.4$$ where \(\operatorname{Re}=\) the Reynolds number $$\operatorname{Re}=\frac{\rho V D}{\mu}$$ where \(V=\) the velocity of the fluid in the pipe \((\mathrm{m} / \mathrm{s})\) and \(\mu=\) dynamic viscosity \(\left(\mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\right)\). Note that for a circular pipe \(V=4 Q / \pi D^{2} .\) Also, assume that the fluid has a viscosity of \(1.79 \times 10^{-5} \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^{2}\).

Short Answer

Expert verified
To find the friction factor using the von Karman equation, first obtain the Reynolds number using the given viscosity and velocity equation: \[\operatorname{Re} = \frac{4 \rho Q}{\pi D \mu}\] Then, solve the von Karman equation iteratively with the obtained Reynolds number: \[\frac{1}{\sqrt{f}} = 4 \log_{10}(\operatorname{Re}\sqrt{f}) - 0.4\] Repeat the iterations until the friction factor converges. The resulting friction factor is the solution for the given problem.

Step by step solution

01

Write the von Karman equation and Reynolds number equation

The von Karman equation is: \[\frac{1}{\sqrt{f}} = 4 \log_{10}(\operatorname{Re}\sqrt{f}) - 0.4\] The Reynolds number equation is: \[\operatorname{Re} = \frac{\rho VD}{\mu}\] #Step 2: Write down the equation for fluid velocity in a circular pipe#
02

Write down the equation for fluid velocity in a circular pipe

The equation for fluid velocity in a circular pipe is: \[V = \frac{4Q}{\pi D^2}\] #Step 3: Obtain the Reynolds number using the given viscosity and velocity equation#
03

Obtain the Reynolds number

From the Reynolds number equation, we can replace the velocity V using the equation for velocity in a circular pipe: \[\operatorname{Re} = \frac{\rho (4Q / \pi D^2) D}{\mu}\] Now, we can simplify the equation: \[\operatorname{Re} = \frac{4 \rho Q}{\pi D \mu}\] where \(\mu = 1.79 \times 10^{-5}\, \mathrm{N} \cdot \mathrm{s} / \mathrm{m}^2\) #Step 4: Solve the von Karman equation iteratively to find the friction factor#
04

Solve the von Karman equation iteratively

To find the friction factor \(f\), we need to solve the von Karman equation iteratively. Start with an initial guess for \(f\) and use the Reynolds number obtained in Step 3: 1. Calculate \(\operatorname{Re}\sqrt{f}\). 2. Compute the expression \(4 \log_{10}(\operatorname{Re}\sqrt{f}) - 0.4\). 3. Update the value of \(f\) using the obtained result: \(f = \left(\frac{1}{4\log_{10}(\operatorname{Re}\sqrt{f}) - 0.4}\right)^2\). 4. Repeat steps 1-3 until the value of \(f\) converges (i.e., the difference between two consecutive iterations is less than a desired small value). The converged value of \(f\) is the friction factor for the given problem.

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Key Concepts

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

Reynolds number
Understanding the Reynolds number is crucial when dealing with fluid dynamics in systems like pipes. It's a dimensionless quantity that helps predict flow patterns in different fluid flow situations. The Reynolds number is calculated by the formula:
\[\begin{equation}\text{Re} = \frac{\rho VD}{\mu}\end{equation}\]
where \( \rho \) is the fluid density, \( V \) is the fluid velocity, \( D \) is the characteristic diameter of the pipe, and \( \mu \) is the dynamic viscosity of the fluid. When \( \text{Re} < 2000 \), flow is typically laminar (smooth), and when \( \text{Re} > 4000 \), flow tends to be turbulent (chaotic). Understanding where the flow regime stands in relation to these values helps in predicting the type of flow and accurately calculating the friction factor using the von Karman equation.
Fluid velocity in a pipe
The fluid velocity in a pipe is a key factor in many calculations within fluid mechanics. For a circular pipe, the velocity \( V \) can be calculated based on the volumetric flow rate \( Q \), using the formula:
\[\begin{equation}V = \frac{4Q}{\pi D^2}\end{equation}\]
This relationship implies that for a given flow rate, the velocity is inversely proportional to the square of the pipe's diameter. Larger pipes can carry the same flow rate at a slower velocity, which generally reduces the occurrence of turbulent flow, potentially minimizing energy loss due to friction.
Friction factor
The friction factor is a dimensionless number indicative of the resistance or friction that occurs when a fluid flows through a pipe or duct. It's influenced by both the nature of the fluid and the characteristics of the pipe, like roughness and diameter. The von Karman equation allows us to compute the friction factor when dealing with turbulent flow, which can be more complex to assess compared to laminar flow. An accurate determination of the friction factor is essential for various calculations, such as pressure loss in a pipe, which is critical for the design and analysis of fluid systems.
To calculate this using the von Karman equation, we can't do it directly because it relies on the Reynolds number, which in turn depends on the velocity of the fluid and consequently the friction factor itself. This complex relationship necessitates an iterative solution to identify the accurate friction factor.
Iterative solution
Solving for the friction factor using the von Karman equation requires an iterative solution. This computational technique involves making an initial guess and then refining that guess through repeated application of the equation. Each iteration provides a new value, which is then used in the next loop until the answer changes very little between iterations, indicating that a solution has been reached. The process requires patience and attention to detail, ensuring that each iteration brings us closer to the accurate value. In the context of finding the friction factor, the iterations involve applying the Reynolds number and the initial guess of the friction factor to the von Karman equation repeatedly until the value stabilizes.

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Most popular questions from this chapter

A fluid is pumped into the network of pipes shown in Fig. P8.44. 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 (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 given 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 \(L_{7}=8 \mathrm{m}\).

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

A total charge \(Q\) is uniformly distributed around a ring-shaped conductor with radius a. A charge \(q\) is located at a distance \(x\) from the center of the ring (Fig. \(P 8.31\) ). 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 \(1 \mathrm{N}\) if \(q\) and \(Q\) are \(2 \times 10^{-5} \mathrm{C}\) for a ring with a radius of \(0.9 \mathrm{m}\).

The Manning equation can be written for a rectangular open channel as $$Q=\frac{\sqrt{S}(B H)^{5 / 3}}{n(B+2 H)^{2 / 3}}$$ where \(Q=\) flow \(\left[\mathrm{m}^{3} / \mathrm{s}\right], S=\) slope \([\mathrm{m} / \mathrm{m}], H=\) depth \([\mathrm{m}],\) and \(n=\) the Manning roughness coefficient. Develop a fixed-point iteration scheme to solve this equation for \(H\) given \(Q=5, S=0.0002\) \(B=20,\) and \(n=0.03 .\) Prove that your scheme converges for all initial guesses greater than or equal to zero.

Many fields of engineering require accurate population estimates. For example, transportation engineers might find it necessary to determine separately the population growth trends of a city and adjacent suburb. The population of the urban area is declining with time according to $$P_{u}(t)=P_{u, \max } e^{-k_{a t} t}+P_{u, \min }$$ while the suburban population is growing, as in $$P_{s}(t)=\frac{P_{s, \max }}{1+\left[P_{s, \max } / P_{0}-1\right] e^{-k_{s} t}}$$ where \(P_{u, \max }, k_{u}, P_{s, \max }, P_{0},\) and \(k_{s}=\) empirically derived parameters. Determine the time and corresponding values of \(P_{u}(t)\) and \(P_{s}(t)\) when the suburbs are \(20 \%\) larger than the city. The parameter values are \(P_{u, \max }=75,000, \quad k_{u}=0.045 / \mathrm{yr}, \quad P_{u, \min }=100,000\) people, \(P_{s, \max }=300,000\) people, \(P_{0}=10,000\) people, \(k_{s}=\) \(0.08 /\) yr. To obtain your solutions, use (a) graphical, (b) falseposition, and (c) modified secant methods.
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