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Chapter 8: Problem 49

The Churchill formula for the friction factor is $$f=8\left[\left(\frac{8}{\mathrm{Re}}\right)^{12}+\frac{1}{(A+B)^{15}}\right]^{1 / 12}$$ where $$\begin{array}{l} A=\left\\{-2.457 \ln \left[\left(\frac{7}{\mathrm{Re}}\right)^{0.9}+\frac{\varepsilon}{3.7 D}\right]\right\\}^{16} \\ B=\left(\frac{37.530}{\mathrm{Re}}\right)^{16} \end{array}$$ Compare this equation for \(f\) for both the laminar and turbulent regimes for \(\varepsilon / D=0.00001,0.0001,0.001,\) and 0.01 and Reynolds numbers of \(10,10^{2}, 10^{3}, 10^{4}, 10^{5}, 10^{6},\) and \(10^{7}\) with the Moody chart and decide whether it is an acceptable replacement for the Colebrook formula.

Short Answer

Expert verified
The answer to whether the Churchill formula might replace the Colebrook formula highly depends on the comparability of the data acquired from the Churchill calculations with the Moody chart. Run those calculations and make that comparison to obtain a decision.

Step by step solution

01

Understand the Churchill Formula

The Churchill equation is a mathematical model to describe the friction factor involved in a fluid system. This formula varies for different ranges of Reynolds numbers and roughness levels. Here, \(f\) is the friction factor, \(Re\) is the Reynolds number, and \(\varepsilon / D\) is a roughness factor.
02

Calculate the Friction Factor

Next, you need to plug in the provided values of \(\varepsilon / D = 0.00001, 0.0001, 0.001, and 0.01\) and Reynolds number \(Re = 10, 10^{2}, 10^{3}, 10^{4}, 10^{5}, 10^{6},\) and \(10^{7}\) into the Churchill formula to calculate the corresponding friction factors \(f\). Make sure to follow the order of operations and apply the correct mathematical functions (power, square root, logarithm) where needed.
03

Compare with Moody Chart

Next, use a Moody chart to find the appropriate friction factors for the given Reynolds numbers and roughness levels and compare these factors with the calculated values from the Churchill equation. The Moody chart is a graphical representation widely used in fluid dynamics and can provide a good approximation for the friction factor.
04

Decision on Replacement for Colebrook Formula

Finally, based on the comparability of the results from the Churchill equation and Moody chart, make a judgement if the Churchill formula could be an acceptable replacement for the Colebrook formula. If the results closely match across a broad range of Reynolds numbers and roughness factors, then it could possibly replace the Colebrook formula.

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Most popular questions from this chapter

A person with no experience in fluid mechanics wants to estimate the friction factor for 1 -in.-diameter galvanized iron pipe at a Reynolds number of 8,000 . The person stumbles across the simple equation of \(f=64 / \mathrm{Re}\) and uses this to calculate the friction factor. Explain the problem with this approach and estimate the error.

{ Blood iassume } \mu=4.5 \times 10^{-5} \mathrm{lb} \cdot \mathrm{s} / \mathrm{ft}^{2}, S G=1.0\right)\( flows through an artery in the neck of a giraffe from its heart to its head at a rate of \)2.5 \times 10^{-4} \mathrm{ft}^{3} / \mathrm{s}\(. Assume the length is \)10 \mathrm{ft}\( and the diameter is 0.20 in. If the pressure at the beginning of the artery (outlet of the heart) is equivalent to \)0.70 \mathrm{ft}$ Hg, determine the pressure at the end of the artery when the head is (a) 8 ft above the heart, or (b) 6 ft below the heart. Assume steady flow. How much of this pressure difference is due to elevation effects, and how much is due to frictional effects?

A fluid flows through a horizontal 0.1-in.-diameter pipe. When the Reynolds number is \(1500,\) the head loss over a \(20-f t\) lenzth of the pipe is \(6.4 \mathrm{ft}\). Determine the fluid velocity.

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