Velocity Profiles in a Pipe and a Free Surface (Open Channel)

In the context of pipe flow, the speed difference across the pipe's cross-section is often described by the velocity profile. For a fully developed laminar flow in a circular pipe, the velocity profile is parabolic. This means that the fluid velocity is highest at the center of the pipe and decreases towards the pipe walls.

Figure 1: Velocity Profiles of a Laminar Flow in Pipe

In this illustration, we explore three distinct cases of fluid flow:

Flow Inside a Pipe: This scenario involves laminar flow within a circular pipe, where the fluid moves in smooth, orderly layers. The velocity profile is parabolic, with the highest velocity at the center of the pipe and decreasing towards the walls.

Flow Inside a Non-Circular Closed Channel: Here, the fluid flows within a closed channel that is not circular, such as between two parallel plates. The fluid is in contact with the walls of the channel, and the velocity profile is also parabolic but follows a different equation compared to pipe flow.

Open Channel Flow: This case represents flow in open channels like rivers or streams, where the fluid is exposed to the atmosphere. The flow characteristics can vary significantly depending on the channel shape, slope, and roughness, often leading to complex velocity profiles.

Laminar Flow in a Pipe Sample Problem

Problem Statement:

Water flows through a circular pipe with a radius of 0.05 meters. The mean velocity of the water is 0.8 meters per second. Calculate the velocity of the water at a radial distance of 0.03 meters from the center of the pipe.

Given Data:

• Radius of the pipe, $R = 0.05$ meters
• Mean velocity, $u_{\text{mean}} = 0.8$ meters per second
• Radial distance from the center, $r = 0.03$ meters

Solution

1. Calculate the Maximum Velocity:

The maximum velocity $u_{\text{max}}$ is related to the mean velocity $u_{\text{mean}}$ by the following relation:

$u_{\text{max}} = 2 \times u_{\text{mean}}$

Substituting the given mean velocity:

$u_{\text{max}} = 2 \times 0.8 \, \text{m/s} = 1.6 \, \text{m/s}$

2. Calculate the Velocity at $r = 0.03$ meters:

Using the velocity profile equation:

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

Substituting the values:

$u(0.03) = 1.6 \, \text{m/s} \left( 1 - \frac{(0.03 \, \text{m})^2}{(0.05 \, \text{m})^2} \right)$

$u(0.03) = 1.6 \, \text{m/s} \left( 1 - \frac{0.0009 \, \text{m}^2}{0.0025 \, \text{m}^2} \right)$

$u(0.03) = 1.6 \, \text{m/s} \left( 1 - 0.36 \right)$

$u(0.03) = 1.6 \, \text{m/s} \times 0.64$

$u(0.03) = 1.024 \, \text{m/s}$

Answer:

The velocity of the water at a radial distance of 0.03 meters from the center of the pipe is 1.024 meters per second.

Open Channel Flow Sample Problem

Problem Statement:

A river has a rectangular cross-section with a width of 10 meters and a depth of 2 meters. The average velocity of the water flow is 1.2 meters per second. Calculate the flow rate of the river in cubic meters per second.

Given Data:

• Width of the river, $W = 10$ meters
• Depth of the river, $D = 2$ meters
• Average velocity, $u_{\text{mean}} = 1.2$ meters per second

Solution

1. Calculate the Cross-Sectional Area:

The cross-sectional area $A$ of the river is given by:

$A = W \times D$

Substituting the given values:

$A = 10 \, \text{m} \times 2 \, \text{m} = 20 \, \text{m}^2$

2. Calculate the Flow Rate:

The flow rate $Q$ is given by the product of the cross-sectional area $A$ and the average velocity $u_{\text{mean}}$:

$Q = A \times u_{\text{mean}}$

Substituting the values:

$Q = 20 \, \text{m}^2 \times 1.2 \, \text{m/s}$

$Q = 24 \, \text{m}^3/\text{s}$

Answer:

The flow rate of the river is 24 cubic meters per second.

Laminar Flow Between Two Parallel Plates Sample Problem

Problem Statement:

Water flows between two parallel plates that are 0.02 meters apart. The mean velocity of the water is 0.5 meters per second. Calculate the velocity of the water at a distance of 0.005 meters from the bottom plate.

Given Data:

• Distance between the plates, $h = 0.02$ meters
• Mean velocity, $u_{\text{mean}} = 0.5$ meters per second
• Distance from the bottom plate, $y = 0.005$ meters

Solution

1. Calculate the Maximum Velocity:

The maximum velocity $u_{\text{max}}$ is related to the mean velocity $u_{\text{mean}}$ by the following relation:

$u_{\text{max}} = \frac{3}{2} u_{\text{mean}}$

Substituting the given mean velocity:

$u_{\text{max}} = \frac{3}{2} \times 0.5 \, \text{m/s} = 0.75 \, \text{m/s}$

2. Velocity Profile Equation:

The velocity $u(y)$ at a distance $y$ from the bottom plate is given by the equation:

$u(y) = 6u_{\text{max}} \left( \frac{y}{h} - \left( \frac{y}{h} \right)^2 \right)$

3. Substitute the Given Values:

$u(0.005) = 6 \times 0.75 \, \text{m/s} \left( \frac{0.005 \, \text{m}}{0.02 \, \text{m}} - \left( \frac{0.005 \, \text{m}}{0.02 \, \text{m}} \right)^2 \right)$

4. Calculate the Terms:

$ \frac{0.005 \, \text{m}}{0.02 \, \text{m}} = 0.25 $

$ \left( \frac{0.005 \, \text{m}}{0.02 \, \text{m}} \right)^2 = 0.25^2 = 0.0625 $

5. Substitute and Simplify:

$ u(0.005) = 6 \times 0.75 \, \text{m/s} \left( 0.25 - 0.0625 \right) $

$ u(0.005) = 6 \times 0.75 \, \text{m/s} \times 0.1875 $

$ u(0.005) = 4.5 \, \text{m/s} \times 0.1875 $