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 the volume flow rate.

Short answer

Answer: The maximum velocity is 4 m/s, the mean velocity is 2π m/s, and the volume flow rate is 0.02π² m³/s.

Step-by-step solution

Convert Units

First, we need to convert the radius of the pipe (in cm) to meters: \(R = 10 \mathrm{~cm} \times \frac{1 \mathrm{~m}}{100 \mathrm{~cm}} = 0.1 \mathrm{~m}\).

Find the Maximum Velocity

Since the velocity profile contains no variable other than \(r\), we can find the maximum velocity by setting \(r=0\) (at the center of the pipe, where velocity is highest): \(u_{max} = u(0) = 4(1 - \frac{0^2}{0.1^2}) = 4 \mathrm{~m/s}\).

Determine the Mean Velocity

We can find the mean velocity by averaging the velocity profile over the cross-sectional area of the pipe. The mean velocity is given by: \(u_{mean} = \frac{1}{A}\int_{0}^{R} u(r)2 \pi r \mathrm{d}r\), where \(A = \pi R^2\) is the cross-sectional area of the pipe.

Calculate the Mean Velocity

Now we can plug in the values we know and perform the integration: \(u_{mean} = \frac{1}{\pi (0.1)^2}\int_{0}^{0.1} 4\left(1 - \frac{r^2}{0.1^2}\right)2 \pi r \mathrm{d}r\), \(u_{mean} = \frac{1}{0.01\pi}\int_{0}^{0.1} 8\pi r - \frac{8\pi r^3}{0.01} \mathrm{d}r\), \(u_{mean} = 100 \pi \left[ 4\pi r^2 - \frac{2\pi r^4}{0.01} \right]^0.1_0\), \(u_{mean} = 100 \pi \left( 4\pi (0.1)^2 - \frac{2\pi (0.1)^4}{0.01} \right) = 100 \pi (0.004\pi - 0.002\pi) = 2 \pi \mathrm{~m/s}\).

Calculate the Volume Flow Rate

The volume flow rate is the product of the mean velocity and the cross-sectional area of the pipe: \(Q = u_{mean} \times A = 2 \pi \times 0.01\pi = 0.02\pi^2 \mathrm{~m^3/s}\). In summary, the maximum velocity in the pipe is \(u_{max} = 4 \mathrm{~m/s}\), the mean velocity is \(u_{mean} = 2 \pi \mathrm{~m/s}\), and the volume flow rate is \(Q = 0.02\pi^2 \mathrm{~m^3/s}\).

Key concepts

Velocity Profile

The concept of a velocity profile is fundamental when analyzing fluid flow, especially in the context of laminar flow inside a circular pipe.
The velocity profile describes how the fluid velocity varies across the cross-section of the pipe, from the center to the edge.
In the exercise given, the velocity profile is represented by the equation \(u(r) = 4 \left(1 - \frac{r^2}{R^2}\right)\).

• Here, \(u(r)\) is the velocity at a distance \(r\) from the center of the pipe.
• \(R\) is the radius of the pipe, and in the given problem, it is 0.1 meters.
• The profile illustrates that fluid velocity is highest at the center \((u(0) = u_{max})\) and decreases towards the pipe walls, reaching zero at the surface.

This type of parabolic velocity profile is characteristic of laminar flow, where the streamlined flow causes layers of fluid to slide past one another without mixing.

Mean Velocity

Mean velocity is an average flow velocity across a pipe’s cross-section, providing a single-speed equivalent for varying velocities throughout the pipe.
It considers the velocity variations at different radial positions from the center.
The mean velocity \(u_{mean}\) can be determined using the formula:\[u_{mean} = \frac{1}{A} \int_{0}^{R} u(r) \, 2 \pi r \, dr\]

• \(A = \pi R^2\) is the cross-sectional area.
• The integration runs from the center \(r=0\) to the edge of the pipe \(r=R\).

Substituting the problem values, we calculated the mean velocity as \(2 \pi \mathrm{~m/s}\).
It's essential in understanding fluid mechanics because knowing \(u_{mean}\) helps quantify overall flow characteristics and helps in engineering applications such as pipeline design.

Volume Flow Rate

The volume flow rate, often represented by \(Q\), measures how much fluid volume passes a given cross-section of the pipe per unit time.
It's crucial for determining the efficiency and effectiveness of fluid transport systems.
To find the flow rate, multiply the mean velocity by the pipe's cross-sectional area:\[Q = u_{mean} \times A\]

• In the problem, this calculates to \(Q = 2 \pi \times 0.01 \pi = 0.02 \pi^2 \mathrm{~m^3/s}\).
• The unit \(\mathrm{m^3/s}\) implies this is a volumetric measure, suitable for various practical applications, from water pipelines to oil shipment.

Understanding volume flow rates helps engineers and designers ensure that pipelines convey the right amount of fluid, balancing efficiency and cost.

Circular Pipe

A circular pipe is a common conduit used in transporting fluids due to its symmetrical shape, which facilitates uniform flow characteristics.
A key focus in fluid dynamics, especially when considering laminar flow, is how this symmetry affects the velocity distribution and hence the flow efficiency.

• The radius of a circular pipe \(R\) is vital for calculating other parameters, such as flow rate and mean velocity.
• In our problem, the radius was 0.1 meters after conversion, impacting various calculations.
• Circular pipes are preferred because they have fewer materials and installation costs and are naturally strong against pressure due to their geometry.

Understanding the role of a circular pipe helps in predicting how fluids behave under different conditions, ensuring optimal design and function in numerous engineering applications.