Question

At one point in a pipeline the water's speed is 3.00 m/s and the gauge pressure is 5.00 \(\times\) 10\(^4\) Pa. Find the gauge pressure at a second point in the line, 11.0 m lower than the first, if the pipe diameter at the second point is twice that at the first.

Short answer

The gauge pressure at the second point is 1.81 \( \times \) 10\(^5\) Pa.

Step-by-step solution

Understanding Bernoulli's Equation

We begin by noting that Bernoulli's equation is given by \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2, \]where \( P \) is the pressure, \( \rho \) is the fluid density, \( g \) is the acceleration due to gravity, and \( h \) is the height above some reference point.

Setting Known Values

We are given the water's speed \( v_1 = 3.00 \) m/s, the gauge pressure \( P_1 = 5.00 \times 10^4 \) Pa, and the height difference \( h_2 - h_1 = -11.0 \) m. The density of water \( \rho \) is approximately 1000 kg/m³ and \( g = 9.81 \) m/s².

Relating Speeds at Two Points

The area of the pipe at the second point is four times greater than at the first point (since the diameter is doubled), therefore, by the principle of continuity \( A_1v_1 = A_2v_2 \), \[ v_2 = \frac{v_1}{4} = \frac{3.00}{4} = 0.75 \text{ m/s}. \]

Applying Bernoulli's Equation

Substituting into Bernoulli's equation and solving for \( P_2 \), we have:\[ 5.00 \times 10^4 + \frac{1}{2}(1000)(3.00)^2 = P_2 + \frac{1}{2}(1000)(0.75)^2 + (1000)(9.81)(-11.0). \]Simplifying gives:\[ 5.00 \times 10^4 + 4500 = P_2 + 281.25 - 107910. \]

Solving for \( P_2 \)

Rearrange the equation from step 4 to isolate \( P_2 \): \[ P_2 = 5.00 \times 10^4 + 4500 - 281.25 + 107910 = 1.81 \times 10^5 \text{ Pa}. \]

Conclusion: Final Answer

The gauge pressure at the second point, 11.0 m lower than the first point where the pipe diameter is doubled, is \( 1.81 \times 10^5 \text{ Pa} \).

Key concepts

Fluid Dynamics

Fluid dynamics refers to the study of fluids—liquids and gases—in motion. A key aspect of this subject is to understand how these substances flow through channels like pipes. Bernoulli's equation is a cornerstone in fluid dynamics as it relates various factors like pressure, velocity, and elevation in a flowing fluid. This equation helps us analyze and solve many practical problems in engineering and physics.

When applying fluid dynamics concepts, it's important to consider parameters such as fluid speed and the cross-sectional area of the pipes. These factors directly affect how the fluid moves.

• Fluid speed: determines how quickly a fluid travels through a pipe.
• Cross-sectional area: often influences the speed due to its relationship with the fluid's continuity.

Understanding these principles aids in solving more complex fluid movement issues.

Bernoulli's equation not only helps calculate pressures but links velocity with how height changes affect flow, which makes it invaluable in practical applications like the one in the problem scenario.

Pressure Calculations

Pressure calculations are crucial in understanding how fluids behave under different conditions. In the context of the original exercise, the gauge pressure measures the water pressure against the pipeline walls and is impacted by both velocity changes and elevation differences.

The exercise leverages Bernoulli's equation to compute the gauge pressure at two points along a pipe. Several factors influence these calculations:

• The initial and final velocities of the fluid: Fluid velocity affects the kinetic energy term in Bernoulli's equation.
• The fluid density: For water, this is typically 1000 kg/m³, and it factors into both kinetic and gravitational potential energy terms.
• Elevation change: A fluid falling or rising under gravity experiences pressure changes, influenced by height differences in the system.

These elements are combined to solve for pressure at given points, as illustrated by the modifications in pressure from one pipeline point to another through Bernoulli's principles.

Continuity Equation

The continuity equation is fundamental in fluid dynamics and expresses the principle of mass conservation within a flowing fluid. It stipulates that the mass flow rate must remain constant from one cross-section of a pipeline to another, assuming no fluid is added or removed.

This principle is reflected in the equation: \[ A_1v_1 = A_2v_2 \] where:

• \( A_1 \) and \( A_2 \) are the cross-sectional areas at two different sections of the pipeline,
• \( v_1 \) and \( v_2 \) are the velocities of the fluid at these sections.

In the problem, the diameter of the second section is twice that of the first. This results in an area four times larger, thereby reducing the velocity by a quarter, highlighting the interdependence of area and speed in fluid flow. This reduction impacts the dynamics along the pipe, demonstrating how the continuity equation can simplify complex fluid movements by using straightforward relationships between area and velocity.