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Chapter 8: Problem 60

Given two rectangular ducts with equal cross-sectional area but different aspect ratios (width/height) of 2 and \(4,\) which will have the greater frictional losses? Explain your answer

Short Answer

Expert verified
The rectangular duct with an aspect ratio of 4 will have greater frictional losses. This is because the hydraulic diameter is smaller for a duct with a higher aspect ratio, leading to increased frictional losses as they are inversely proportional to the hydraulic diameter.

Step by step solution

01

Defining Hydraulic Diameter

The hydraulic diameter (D) of a rectangular duct is given by \(\[D = \frac{4A}{P}\]\) where A is the cross-sectional area of the duct and P is the wetted perimeter. Here, the cross-sectional area (A) is constant but the aspect ratio (width/height) varies.
02

Calculating Hydraulic Diameter for Different Aspect Ratios

For a rectangular duct with width 'w' and height 'h', the wetted perimeter P is given by \(P = 2(w + h)\). Since aspect ratio is given as width/height, the height can be written as w/aspect ratio. Therefore, for the two ducts given, the hydraulic diameters should be calculated accordingly by substituting the aspect ratios in the formula provided in step 1.
03

Analyzing the Relation between Hydraulic Diameter and Frictional Losses

The frictional losses in a duct are inversely proportional to the hydraulic diameter therefore the duct with the smaller hydraulic diameter (which can be calculated from step 2) will have a greater frictional loss.

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Most popular questions from this chapter

When soup is stirred in a bowl, there is considerable turbulence in the resulting motion (see Video \(\mathrm{V} 8.7\) ). From a very simplistic standpoint, this turbulerce consists of numerous intertwined swirls, each involving a characteristic diameter and velocity. As time goes by, the smaller swirls (the fine scale structure) die out relatively quickly, leaving the large swirls that continue for quite some time. Explain why this is to be expected.

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 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.

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.

A certain process requires 2.3 cfs of water to be delivered at a pressure of 30 psi. This water comes from a large-diameter supply main in which the pressure remains at 60 psi. If the galvanized iron pipe connecting the two locations is 200 ft long and contains six threaded \(90^{\circ}\) elbows, determine the pipe diameter. Elevation differences are negligible.
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