Laminar flow takes place in a circular tube. At what distance from the boundary does the local velocity equal the average velocity?
0.29R
Laminar Flow Velocity Profile in a Circular Tube
In fully developed laminar flow through a circular tube, the velocity profile is parabolic. The velocity is maximum at the center of the tube and zero at the wall (boundary).
The local velocity, \(u(r)\), at a radial distance \(r\) from the center of the tube is given by the formula:
\[u(r) = u_{max} \left(1 - \left(\frac{r}{R}\right)^2\right)\]where:
- \(u_{max}\) is the maximum velocity (at the center, \(r=0\))
- \(R\) is the radius of the tube
- \(r\) is the radial distance from the center (\(0 \le r \le R\))
The average velocity, \(u_{avg}\), for fully developed laminar flow in a circular tube is related to the maximum velocity by:
\[u_{avg} = \frac{u_{max}}{2}\]We can express the local velocity in terms of the average velocity:
\[u(r) = 2u_{avg} \left(1 - \left(\frac{r}{R}\right)^2\right)\]The question asks for the distance from the boundary where the local velocity is equal to the average velocity. Let \(y\) be the distance from the boundary. The radial distance from the center \(r\) is related to the distance from the boundary \(y\) by \(r = R - y\). We want to find \(y\) such that \(u(r) = u_{avg}\).
Setting the local velocity equal to the average velocity:
\[u(r) = u_{avg}\] \[2u_{avg} \left(1 - \left(\frac{r}{R}\right)^2\right) = u_{avg}\]Since \(u_{avg}\) is not zero (for flow to occur), we can divide both sides by \(u_{avg}\):
\[2 \left(1 - \left(\frac{r}{R}\right)^2\right) = 1\]Now, let's solve for \(r/R\):
\[1 - \left(\frac{r}{R}\right)^2 = \frac{1}{2}\] \[\left(\frac{r}{R}\right)^2 = 1 - \frac{1}{2}\] \[\left(\frac{r}{R}\right)^2 = \frac{1}{2}\] \[\frac{r}{R} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}\]So, the radial distance from the center where the local velocity equals the average velocity is \(r = \frac{R}{\sqrt{2}}\).
The question asks for the distance from the boundary, which is \(y = R - r\). Substituting the value of \(r\):
\[y = R - \frac{R}{\sqrt{2}}\] \[y = R \left(1 - \frac{1}{\sqrt{2}}\right)\]Calculating the numerical value:
\[\frac{1}{\sqrt{2}} \approx \frac{1}{1.4142} \approx 0.7071\] \[y \approx R (1 - 0.7071)\] \[y \approx R (0.2929)\]Rounding to two decimal places, the distance from the boundary is approximately \(0.29R\).
Therefore, the local velocity equals the average velocity at a distance of approximately \(0.29R\) from the boundary of the circular tube.
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