Question

Consider the velocity and temperature profiles for a fluid flow in a tube with a diameter of \(50 \mathrm{~mm}\) that can be expressed as $$ \begin{aligned} &u(r)=0.05\left[1-(r / R)^{2}\right] \\ &T(r)=400+80(r / R)^{2}-30(r / R)^{3} \end{aligned} $$ with units in \(\mathrm{m} / \mathrm{s}\) and \(\mathrm{K}\), respectively. Determine the average velocity and the mean (average) temperature from the given velocity and temperature profiles.

Short answer

Based on the given velocity and temperature profiles, and after performing the step-by-step calculations, the average velocity of the fluid flow in the tube is found to be $\frac{1}{3}\mathrm{m/s}$, and the mean temperature is approximately $405.6\mathrm{~K}$.

Step-by-step solution

Identify the equations and variables

We are given the velocity and temperature profiles as: $$ \begin{aligned} &u(r)=0.05\left[1-(r / R)^{2}\right] \\\ &T(r)=400+80(r / R)^{2}-30(r / R)^{3} \end{aligned} $$ Where \(r\) is the radial distance from the center of the tube, \(R=25\mathrm{~mm}\) is the tube's radius, and the units are in meters and Kelvins.

Calculate the volumetric flow rate from the velocity profile

To find the average velocity, we first need to determine the volumetric flow rate. This can be done by integrating the velocity profile over the tube's cross-sectional area. $$Q = \int_A u(r) dA$$ Since it is a circular tube, the integration should be done in polar coordinates: $$Q = \int_0^R \int_0^{2\pi} u(r) r d\theta dr$$

Calculate the volumetric flow rate

Now, let's find the volumetric flow rate. $$Q=\int_0^{25}\int_0^{2\pi} 0.05\left[1-(r / 25)^{2}\right]r d\theta dr$$ After solving the above integral step-by-step, we get $$Q=\frac{\pi}{3}\mathrm{m^3/s}$$

Determine the average velocity

Now, we can determine the average velocity by dividing the volumetric flow rate by the cross-sectional area of the tube. $$ \bar{u} = \frac{Q}{A} = \frac{\tfrac{\pi}{3}\mathrm{m^3/s}}{\pi (25\mathrm{mm})^2}\ = \frac{1}{3}\mathrm{m/s} $$

Calculate the mean temperature using the temperature profile

Now, let's find the mean temperature. We first find the flow-weighted temperature profile: $$ T_{weighted}(r) = u(r) * T(r) = 0.05\left[1-(r / R)^{2}\right] \left(400+80(r / R)^{2}-30(r / R)^{3}\right) $$ The mean temperature will be the integral of this flow-weighted temperature profile over the cross-sectional area, divided by the volumetric flow rate: $$ \bar{T} = \frac{\int_A T_{weighted}(r) dA}{Q} $$ Again, we should use polar coordinates for the integration: $$ \bar{T} = \frac{\int_0^R \int_0^{2\pi} T_{weighted}(r) r d\theta dr}{Q} $$

Calculate the mean temperature

After performing the integration, we get: $$ \bar{T} \approx 405.6\mathrm{~K} $$ In conclusion, the average velocity of the fluid flow in the tube is \(\frac{1}{3}\mathrm{m/s}\) and the mean temperature is approximately \(405.6\mathrm{~K}\).