Question

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 determine the volume flow rate.

Short answer

Answer: The maximum velocity is 4 m/s, the mean velocity is 2/30 m/s, and the volume flow rate is π/300 m³/s.

Step-by-step solution

Identifying given values

We are given the following information and formulas: - The inner radius of the pipe, \(R = 10 \mathrm{~cm} = 0.1 \mathrm{~m}\) - The velocity profile, \(u(r)=4\left(1-\frac{r^2}{R^2}\right)\)

Find the maximum velocity

To find the max velocity, we need to find the point where the derivative of the velocity profile with respect to the radius, \(r\), is zero. So, $$ \frac{du(r)}{dr} = \frac{d}{dr}\left[4\left(1-\frac{r^2}{R^2}\right)\right] $$ Now let's find the derivative: $$ \frac{du(r)}{dr} = -\frac{8r}{R^2} $$ We can see that the derivative will be zero when \(r = 0\). So, the maximum velocity occurs at the center of the pipe (i.e., \(r = 0\)). Now let's find the maximum velocity \(u_{max}\) by inserting \(r = 0\) in the velocity profile equation: $$ u_{max} = u(0) = 4\left(1-\frac{0^2}{R^2}\right) = 4 \mathrm{~m/s} $$

Find the mean velocity

To find the mean velocity, we need to first calculate the integral of the velocity profile over the entire cross-sectional area of the pipe, and then divide by the total area. The cross-sectional area of the pipe is given by \(A = \pi R^2\). So, the mean velocity \(u_{mean}\) can be calculated as follows: $$ u_{mean} = \frac{1}{A}\int_{0}^{R} u(r) 2\pi r dr $$ Now let's insert the given velocity profile and solve the integral: $$ u_{mean} = \frac{1}{\pi R^2}\int_{0}^{R} 4\left(1-\frac{r^2}{R^2}\right) 2\pi r dr $$ $$ u_{mean} = \frac{8}{R^2}\int_{0}^{R} (R^2 - r^2) r dr $$ Now calculate the integral: $$ u_{mean} = \frac{8}{R^2}\left[\frac{1}{4}R^4 - \frac{1}{12}R^4\right] = \frac{8}{R^2}\left[\frac{1}{12}R^4\right] = \frac{2}{3}R^2 $$ Now substitute the radius \(R=0.1\,\mathrm{m}\) and find the mean velocity: $$ u_{mean} = \frac{2}{3}(0.1^2) = \frac{2}{30} \mathrm{~m/s} $$

Find the volume flow rate

The volume flow rate, \(Q\), can be calculated as the product of the mean velocity and the pipe's cross-sectional area: $$ Q = u_{mean} A $$ Now substitute the mean velocity and the area, and calculate \(Q\): $$ Q = \left(\frac{2}{30} \mathrm{~m/s}\right)\left( \pi(0.1^2)\right) = \frac{\pi}{300} \mathrm{~m^3/s} $$ In conclusion, the maximum velocity in the pipe is \(4 \mathrm{~m/s}\), the mean velocity is \(\frac{2}{30} \mathrm{~m/s}\), and the volume flow rate is \(\frac{\pi}{300} \mathrm{~m^3/s}\).