Laminar Flow Velocity Ratio in Circular Pipes

In fluid mechanics, understanding the velocity distribution in a pipe is crucial. For laminar flow through a circular pipe, the fluid moves in smooth layers, and the velocity is not uniform across the pipe's cross-section.

Understanding Laminar Flow Velocity Profile

For fully developed laminar flow in a circular pipe, the velocity profile is parabolic. This means the velocity is maximum at the center of the pipe and decreases to zero at the pipe walls due to the no-slip condition.

The velocity distribution $u(r)$ at a radial distance $r$ from the center is given by:

$\qquad u(r) = U_{max} \left(1 - \left(\frac{r}{R}\right)^2 \right)$

where:

• $U_{max}$ is the maximum velocity at the center ($r=0$).
• $R$ is the radius of the pipe.
• $r$ is the radial distance from the center ($0 \le r \le R$).

Calculating Maximum Velocity and Average Velocity

The maximum velocity ($U_{max}$) occurs at the center of the pipe, where $r=0$. From the velocity profile equation, when $r=0$, $u(0) = U_{max}(1-0) = U_{max}$. So, the maximum velocity is simply $U_{max}$ as defined.

The average velocity ($U_{avg}$) is the average speed of the fluid across the entire cross-sectional area of the pipe. It is calculated by integrating the velocity profile over the area and dividing by the area:

$\qquad U_{avg} = \frac{1}{A} \int_A u(r) dA$

For a circular pipe, the area element $dA = 2\pi r dr$ and the total area $A = \pi R^2$. The integration is performed from the center ($r=0$) to the wall ($r=R$).

$\qquad U_{avg} = \frac{1}{\pi R^2} \int_0^R U_{max} \left(1 - \left(\frac{r}{R}\right)^2 \right) 2\pi r dr$

We can pull out the constants:

$\qquad U_{avg} = \frac{2 \pi U_{max}}{\pi R^2} \int_0^R \left(r - \frac{r^3}{R^2} \right) dr$

$\qquad U_{avg} = \frac{2 U_{max}}{R^2} \int_0^R \left(r - \frac{r^3}{R^2} \right) dr$

Now, perform the integration:

$\qquad U_{avg} = \frac{2 U_{max}}{R^2} \left[ \frac{r^2}{2} - \frac{r^4}{4R^2} \right]_0^R$

Evaluate the expression at the limits:

$\qquad U_{avg} = \frac{2 U_{max}}{R^2} \left[ \left(\frac{R^2}{2} - \frac{R^4}{4R^2} \right) - (0 - 0) \right]$

$\qquad U_{avg} = \frac{2 U_{max}}{R^2} \left( \frac{R^2}{2} - \frac{R^2}{4} \right)$

$\qquad U_{avg} = \frac{2 U_{max}}{R^2} \left( \frac{2R^2 - R^2}{4} \right)$

$\qquad U_{avg} = \frac{2 U_{max}}{R^2} \left( \frac{R^2}{4} \right)$

$\qquad U_{avg} = \frac{U_{max}}{2}$

Ratio of Maximum Velocity to Average Velocity

The question asks for the ratio of maximum velocity to average velocity, which is $\frac{U_{max}}{U_{avg}}$.

Using the result $U_{avg} = \frac{U_{max}}{2}$:

$\qquad \frac{U_{max}}{U_{avg}} = \frac{U_{max}}{U_{max}/2} = 2$

Thus, for laminar flow through a circular pipe, the ratio of maximum velocity to average velocity is 2.0.

The final answer is $\mathbf{2.0}$.