Question

The velocity profile in fully developed laminar flow in a circular pipe, in \(\mathrm{m} / \mathrm{s}\), is given by \(u(r)=4\left(1-100 r^{2}\right)\) where \(r\) is the radial distance from the centerline of the pipe in \(\mathrm{m}\). Determine \((a)\) the radius of the pipe, \((b)\) the mean velocity through the pipe, and \((c)\) the maximum velocity in the pipe.

Short answer

Based on the given velocity profile of fully developed laminar flow in a circular pipe, we determined the following values: 1. The radius of the pipe is 0.1 m (10 cm). 2. The mean velocity through the pipe is 0.4 m/s. 3. The maximum velocity in the pipe is 4 m/s at the centerline (r=0).

Step-by-step solution

Determine the Radius of the Pipe

The velocity at the pipe wall should be zero. Set the velocity profile equation to zero and solve for r: \[u(r) = 4 (1 - 100r^2) = 0\] \[100r^2 = 1\] \[r^2 = \frac{1}{100}\] \[r = \sqrt{\frac{1}{100}}\] \[r = \frac{1}{10} \mathrm{m}\] The radius of the pipe is 0.1 m (10 cm).

Calculate the Mean Velocity

To calculate the mean velocity, we need to first find the average velocity across the pipe cross-sectional profile. This can be done by integrating the velocity profile equation with respect to the radial distance r and dividing by the pipe's cross-sectional area: \[v_{mean} = \frac{1}{A} \int_0^R u(r) \cdot 2\pi r dr\] Since we know the radius R is 0.1 m, the cross-sectional area of the pipe A is: \[A = \pi R^2 = \pi \left(\frac{1}{10}\right)^2 = \frac{\pi}{100} \mathrm{m^2}\] Now, we can calculate the mean velocity: \[v_{mean} = \frac{1}{\frac{\pi}{100}} \int_0^{0.1} 4(1-100r^2) \cdot 2\pi r dr\] \[v_{mean} = \frac{100}{\pi} \int_0^{0.1} 8\pi r(1-100r^2) dr\] Now, integrate and evaluate the integral: \[v_{mean} = \frac{100}{\pi} \left[4\pi r^2 - \frac{8}{3}\pi r^4 \right]_0^{0.1}\] \[v_{mean} = \frac{100}{\pi} \left[4\pi \left(\frac{1}{10}\right)^2 - \frac{8}{3}\pi \left(\frac{1}{10}\right)^4 \right]\] \[v_{mean} = \frac{4}{10} \mathrm{m/s}\] The mean velocity through the pipe is 0.4 m/s.

Find the Maximum Velocity

The maximum velocity occurs when the first derivative of the velocity profile equation with respect to r is zero: \[\frac{d}{dr} \left[4(1 - 100r^2)\right] = 0\] \[\frac{d}{dr}\left[-400r^2 + 4\right] = 0\] Now solve for r: \[-800r = 0\] \[r = 0\] The maximum velocity occurs at r = 0 (centerline of the pipe), and it can be found by plugging this value back into the velocity profile equation: \[u_{max} = 4(1 - 100(0)^2) = 4 \mathrm{m/s}\] The maximum velocity in the pipe is 4 m/s.