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Chapter 8: Problem 38

In fully developed laminar flow in a circular pipe, the velocity at \(R / 2\) (midway between the wall surface and the centerline) is measured to be \(6 \mathrm{~m} / \mathrm{s}\). Determine the velocity at the center of the pipe. Answer: \(8 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
Based on the given information, the axial velocity at the center of the pipe, \(u_c\), is 8 m/s.

Step by step solution

01

Recall the velocity profile for laminar flow in a circular pipe

The axial velocity in a laminar flow inside a circular pipe varies parabolically with the radial distance r from the center of the pipe. The velocity profile can be expressed as: \(u(r) = u_c \cdot (1 - (\frac{r}{R})^2)\) where \(u(r)\) is the axial velocity at radial distance r, \(u_c\) is the axial velocity at the center of the pipe (r = 0), and R is the radius of the pipe.
02

Use the given information and substitute into the velocity profile equation

We are given that the velocity at the point \(R / 2\) is 6 m/s. Therefore, we can write: \(6 = u_c \cdot (1 - (\frac{R / 2}{R})^2)\) Simplify the equation: \(6 = u_c \cdot (1 - (\frac{1}{2})^2)\)
03

Solve the equation for \(u_c\)

Now we need to solve the equation for \(u_c\) (velocity at the center of the pipe): \(6 = u_c \cdot (1 - \frac{1}{4})\) So, \(u_c = \frac{6}{(1 - \frac{1}{4})}\) Calculate \(u_c\): \(u_c = \frac{6}{\frac{3}{4}}\) or \(u_c = 6 \cdot \frac{4}{3}\)
04

Calculate the velocity at the center of the pipe

Finally, we can calculate the velocity at the center of the pipe: \(u_c = 8 \mathrm{~m} / \mathrm{s}\) Answer: The velocity at the center of the pipe is \(8 \mathrm{~m} / \mathrm{s}\).

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Most popular questions from this chapter

Air enters a 25-cm-diameter 12 -m-long underwater duct at \(50^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\) at a mean velocity of \(7 \mathrm{~m} / \mathrm{s}\), and is cooled by the water outside. If the average heat transfer coefficient is \(85 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and the tube temperature is nearly equal to the water temperature of \(10^{\circ} \mathrm{C}\), determine the exit temperature of air and the rate of heat transfer. Evaluate air properties at a bulk mean temperature of \(30^{\circ} \mathrm{C}\). Is this a good assumption?

How is the friction factor for flow in a tube related to the pressure drop? How is the pressure drop related to the pumping power requirement for a given mass flow rate?

Water enters a \(5-\mathrm{mm}\)-diameter and \(13-\mathrm{m}\)-long tube at \(15^{\circ} \mathrm{C}\) with a velocity of \(0.3 \mathrm{~m} / \mathrm{s}\), and leaves at \(45^{\circ} \mathrm{C}\). The tube is subjected to a uniform heat flux of \(2000 \mathrm{~W} / \mathrm{m}^{2}\) on its surface. The temperature of the tube surface at the exit is (a) \(48.7^{\circ} \mathrm{C}\) (b) \(49.4^{\circ} \mathrm{C}\) (c) \(51.1^{\circ} \mathrm{C}\) (d) \(53.7^{\circ} \mathrm{C}\) (e) \(55.2^{\circ} \mathrm{C}\) (For water, nse \(k=0.615 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, \operatorname{Pr}=5.42, v=0.801 \times\) \(\left.10^{-6} \mathrm{~m}^{2} / \mathrm{s} .\right)\)

Consider fully developed flow in a circular pipe with negligible entrance effects. If the length of the pipe is doubled, the pressure drop will \((a\) ) double, \((b)\) more than double, \((c)\) less than double, \((d)\) reduce by half, or \((e)\) remain constant.

What is hydraulic diameter? How is it defined? What is it equal to for a circular tube of diameter \(D\) ?
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