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Chapter 8: Problem 33

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.

Short Answer

Expert verified
The simple equation used by the person is only valid for laminar flow, not for the turbulent flow indicated by the Reynolds number 8000. A better estimate for the friction factor can be obtained using the Blasius equation. The estimation error can be computed as the difference between the actual and estimated friction factor values, divided by the actual value, and multiplied by 100%.

Step by step solution

01

Identify the problem with the approach

The equation \(f=64 / \mathrm{Re}\) is only valid for laminar flow, where the Reynolds number is less than 2000. If the Reynolds number is above that, the flow is considered turbulent and the equation is not valid. In this case, the Reynolds number is 8000, indicating turbulent flow. Therefore, using the specified equation to calculate the friction factor is not correct.
02

Calculate the actual value of the friction factor

Since the flow is turbulent, we have to use a different equation to calculate the friction factor. A commonly used approximation for turbulent flow in smooth pipes is the Blasius equation: \(f = 0.316/Re^{0.25}\). Substituting Re=8000 into this equation gives us the actual friction factor, f.
03

Calculate the estimated friction factor and the estimation error

The estimated friction factor calculated using the incorrect equation is \(f_e=64/8000=0.008\). The estimation error is calculated by taking the difference between the actual and estimated values, divided by the actual value, and multiplied by 100% to express it as a percentage. This gives us the estimation error, \((f-f_e)/f*100%\).

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Most popular questions from this chapter

A thief siphoned 15 gal of gasoline from a gas tank in the middle of the night. The gas tank is 12 in. wide, 24 in. long, and 18 in. high and was full when the thief started. The siphoning plastic tube has an inside diameter of 0.5 in. and a length of \(4.0 \mathrm{ft}\). Assume that at any instant of time, the steady-state mechanical energy equation is adequate to predict the gasoline flow rate through the tube. As 15 gal is 3465 in \(^{3}\), the gasoline level in the tank will drop 12.0 in. You may use the gasoline level after it has dropped 6.0 in. to estimate the average gasoline flow rate. Use this flow rate to estimate the time needed to siphon the 15 gal of gasoline. Compare your answer with the answer of 190 sec found in problem 3.107 using Bernoulli's equation. The siphon cischarges at the level of the bottom of the gasoline tank. You may find it useful to use the Blasius equation for smooth pipes found in problem 8.45

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.

Water flows downward through a vertical 10 -mm-diameter galvanized iron pipe with an average velocity of \(5.0 \mathrm{m} / \mathrm{s}\) and exits as a free jet. There is a small hole in the pipe \(4 \mathrm{m}\) above the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe through the hole? Repeat the problem if the average velocity is \(0.5 \mathrm{m} / \mathrm{s}\)

Water at \(10^{\circ} \mathrm{C}\) flows through a smooth 60 -mm-diameter pipe with an average velocity of \(8 \mathrm{m} / \mathrm{s}\). Would a layer of rust of height \(0.005 \mathrm{mm}\) on the pipe wall protrude through the viscous sublayer? Justify your answer with appropriate calculations.

A viscous fluid flows in a 0.10 -m-diameter pipe such that its velocity measured \(0.012 \mathrm{m}\) away from the pipe wall is \(0.8 \mathrm{m} / \mathrm{s}\) If the flow is laminar, determine the centerline velocity and the flowrate.
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