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Chapter 8: Problem 39

The velocity profile in fully developed laminar flow in a circular pipe of inner radius \(R=10 \mathrm{~cm}\), in \(\mathrm{m} / \mathrm{s}\), is given by \(u(r)=4\left(1-r^{2} / R^{2}\right)\). Determine the mean and maximum velocities in the pipe, and the volume flow rate.

Short Answer

Expert verified
Answer: The maximum velocity is 4 m/s, the mean velocity is 2π m/s, and the volume flow rate is 0.02π² m³/s.

Step by step solution

01

Convert Units

First, we need to convert the radius of the pipe (in cm) to meters: \(R = 10 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.1 \mathrm{~m}\).
02

Find the Maximum Velocity

Since the velocity profile contains no variable other than \(r\), we can find the maximum velocity by setting \(r=0\) (at the center of the pipe, where velocity is highest): \(u_{max} = u(0) = 4(1 - \frac{0^2}{0.1^2}) = 4 \mathrm{~m/s}\).
03

Determine the Mean Velocity

We can find the mean velocity by averaging the velocity profile over the cross-sectional area of the pipe. The mean velocity is given by: \(u_{mean} = \frac{1}{A}\int_{0}^{R} u(r)2 \pi r \mathrm{d}r\), where \(A = \pi R^2\) is the cross-sectional area of the pipe.
04

Calculate the Mean Velocity

Now we can plug in the values we know and perform the integration: \(u_{mean} = \frac{1}{\pi (0.1)^2}\int_{0}^{0.1} 4\left(1 - \frac{r^2}{0.1^2}\right)2 \pi r \mathrm{d}r\), \(u_{mean} = \frac{1}{0.01\pi}\int_{0}^{0.1} 8\pi r - \frac{8\pi r^3}{0.01} \mathrm{d}r\), \(u_{mean} = 100 \pi \left[ 4\pi r^2 - \frac{2\pi r^4}{0.01} \right]^0.1_0\), \(u_{mean} = 100 \pi \left( 4\pi (0.1)^2 - \frac{2\pi (0.1)^4}{0.01} \right) = 100 \pi (0.004\pi - 0.002\pi) = 2 \pi \mathrm{~m/s}\).
05

Calculate the Volume Flow Rate

The volume flow rate is the product of the mean velocity and the cross-sectional area of the pipe: \(Q = u_{mean} \times A = 2 \pi \times 0.01\pi = 0.02\pi^2 \mathrm{~m^3/s}\). In summary, the maximum velocity in the pipe is \(u_{max} = 4 \mathrm{~m/s}\), the mean velocity is \(u_{mean} = 2 \pi \mathrm{~m/s}\), and the volume flow rate is \(Q = 0.02\pi^2 \mathrm{~m^3/s}\).

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Most popular questions from this chapter

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

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A \(15-\mathrm{cm} \times 20\)-cm printed circuit board whose components are not allowed to come into direct contact with air for reliability reasons is to be cooled by passing cool air through a 20 -cm-long channel of rectangular cross section \(0.2 \mathrm{~cm} \times 14 \mathrm{~cm}\) drilled into the board. The heat generated by the electronic components is conducted across the thin layer of the board to the channel, where it is removed by air that enters the channel at \(15^{\circ} \mathrm{C}\). The heat flux at the top surface of the channel can be considered to be uniform, and heat transfer through other surfaces is negligible. If the velocity of the air at the inlet of the channel is not to exceed \(4 \mathrm{~m} / \mathrm{s}\) and the surface temperature of the channel is to remain under \(50^{\circ} \mathrm{C}\), determine the maximum total power of the electronic components that can safely be mounted on this circuit board. As a first approximation, assume flow is fully developed in the channel. Evaluate properties of air at a bulk mean temperature of \(25^{\circ} \mathrm{C}\). Is this a good assumption?

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: \(8 \mathrm{~m} / \mathrm{s}\)

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