91Ó°ÊÓ

To find the radius of the pipe, we have to find the point where the velocity profile becomes zero (at the pipe wall). So, we will solve the equation \(u(r) = 0\) for \(r\). \(u(r)=6\left(1-100 r^{2}\right) = 0\) Now, solve for \(r\): \(1-100r^2 = 0\) \(r^2 = \frac{1}{100}\) \(r=\sqrt{\frac{1}{100}}\) \(r=0.1 \thinspace\text{m}\) So, the radius of the pipe is \(0.1 \thinspace\text{m}\).

To find the mean velocity, we will integrate the velocity profile over the pipe's cross-sectional area and divide by the area. The cross-sectional area of a circular pipe is given by \(A = \pi R^2\), where \(R\) is the radius of the pipe. The mean velocity, \(U\), is given by: \(U = \frac{1}{A} \int_{0}^{R} u(r) \cdot 2\pi r \thinspace dr\) Using the given velocity profile, \(u(r) = 6(1 - 100r^{2})\), and radius \(R = 0.1\thinspace\text{m}\), we get: \(U = \frac{1}{\pi (0.1)^2} \int_{0}^{0.1} 6(1-100r^2) \cdot 2\pi r \thinspace dr\) \(U = \frac{1}{0.01\pi} \int_{0}^{0.1} 12\pi r(1-100r^2) \thinspace dr\) Now, integrate with respect to r: \(U = \frac{1}{0.01\pi} \left[ 6\pi r^2 - 2\pi r^4\right]_0^{0.1}\) \(U = \frac{1}{0.01\pi}\left[ 6\pi (0.1)^2 - 2\pi (0.1)^4\right]\) \(U = \frac{1}{0.01\pi} \left[0.06\pi - 0.0002\pi \right]\) \(U = 6-0.02 \thinspace\text{m/s}\) Thus, the mean velocity through the pipe is \(5.98\thinspace\text{m/s}\).

The maximum velocity is at the centerline of the pipe, where the derivative of the velocity profile is zero. To find the maximum velocity, we need to calculate the derivative of the velocity profile with respect to \(r\) and find its value at \(r = 0\). Take the derivative of \(u(r) = 6\left(1-100 r^{2}\right)\) with respect to \(r\): \(\frac{du}{dr} = 6 \cdot (-200r)\) Now, evaluate the velocity profile at \(r = 0\): \(u_\text{max} = u(0) = 6\left(1-100(0)^2\right)\) \(u_\text{max} = 6 \thinspace\text{m/s}\) So, the maximum velocity in the pipe is \(6 \thinspace\text{m/s}\).