Problem 59 - Fundamentals of Physics

Question

Water is moving with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) through a pipe with a cross-sectional area of \(4.0 \mathrm{~cm}^{2}\). The watcr gradually descends \(10 \mathrm{~m}\) as the pipe cross-sectional area increases to \(8.0 \mathrm{~cm}^{2}\). (a) What is the spced at the lower level? (b) If the pressure at the upper level is \(1.5 \times 10^{5} \mathrm{~Pa}\), what is the pressure at the lower level?

Step-by-Step Solution

Answer

(a) 2.5 m/s; (b) 2.0825 × 10^5 Pa

Step 1: Understand Continuity Equation

The continuity equation for fluid flow states that the product of the cross-sectional area and the velocity of the fluid remains constant. For two sections in a pipe, we have: \( A_1v_1 = A_2v_2 \), where \( A_1 \) and \( A_2 \) are the areas of the cross-sections, and \( v_1 \) and \( v_2 \) are the velocities at those sections.

Step 2: Calculate New Velocity Using Continuity

Given \( A_1 = 4.0 \text{ cm}^2 \) and \( A_2 = 8.0 \text{ cm}^2 \), and \( v_1 = 5.0 \text{ m/s} \). Convert areas from cm² to m²: \( A_1 = 4.0 \times 10^{-4} \text{ m}^2 \), \( A_2 = 8.0 \times 10^{-4} \text{ m}^2 \). Using the continuity equation, \( (4.0 \times 10^{-4}) \, (5.0) = (8.0 \times 10^{-4}) \, v_2 \). Solve for \( v_2 \): \( v_2 = 2.5 \text{ m/s} \).

Step 3: Understand Bernoulli's Equation

Bernoulli's equation relates pressure, fluid speed, and height in a flowing fluid. It 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 \( \rho \) is the fluid density, and \( g \) is the acceleration due to gravity.

Step 4: Calculate Pressure Using Bernoulli's Equation

Rearrange Bernoulli's equation to find \( P_2 \): \( P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 + \rho g (h_1 - h_2) \). Given \( P_1 = 1.5 \times 10^5 \text{ Pa} \), \( \rho = 1000 \text{ kg/m}^3 \) (density of water), \( g = 9.81 \text{ m/s}^2 \), \( h_1 - h_2 = -10 \text{ m} \). Substitute \( v_1 = 5.0 \text{ m/s} \) and \( v_2 = 2.5 \text{ m/s} \) into the equation: \( P_2 = 1.5 \times 10^5 + \frac{1}{2}(1000)(5.0)^2 - \frac{1}{2}(1000)(2.5)^2 - 1000 \times 9.81 \times 10 \). Solve for \( P_2 \): \( P_2 = 2.0825 \times 10^5 \text{ Pa} \).

Key Concepts

Continuity Equation

In fluid dynamics, the continuity equation is a fundamental principle that ensures the conservation of mass within a closed system. For any fluid flowing through a pipe, the amount of fluid entering a section must equal the amount leaving it, assuming the density remains constant. This relationship is mathematically represented as:\[ A_1v_1 = A_2v_2 \]where:

• \(A_1\) and \(A_2\) represent the cross-sectional areas of two different sections of the pipe.
• \(v_1\) and \(v_2\) are the fluid velocities at those corresponding sections.

This equation implies that a fluid's velocity will increase if it passes through a narrowing section of a pipe, and decrease if it flows through a widening section. In the provided exercise, the fluid's speed was recalculated at a wider section of the pipe. The resulting decrease in velocity, from 5.0 m/s to 2.5 m/s, highlights the inverse relationship between cross-sectional area and speed in flow rate equations.

Bernoulli's Equation

Bernoulli's equation is pivotal in understanding how fluid motion relates to pressure, speed, and height. It combines principles of energy conservation in a fluid and can be expressed as:\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]where:

• \(P_1\) and \(P_2\) are the pressures at the two points within the fluid.
• \(\rho\) is the fluid density.
• \(v_1\) and \(v_2\) represent the velocities at those points.
• \(g\) is the acceleration due to gravity.
• \(h_1\) and \(h_2\) are the heights at those points.

This equation highlights the interplay between velocity, pressure, and gravity. As a fluid speeds up, kinetic energy must come from potential energy, leading to a decrease in pressure or height. In this exercise, knowing the fluid's speeds and elevation drop, we could find the pressure decrease using Bernoulli's conservation principle.

Pressure Calculation

The calculation of pressure changes in a pipe involves understanding the variations in kinetic and potential energy as a fluid moves. By applying Bernoulli's equation, we can solve for new pressures given changes in height and speed. The exercise requires finding the lower section's pressure, considering:

• Pressure loss due to the elevation drop (potential energy translates to kinetic energy).
• Changes in speed affecting the kinetic energy term.

Substituting known values into the rearranged Bernoulli's formula:\[ P_2 = P_1 + \frac{1}{2} \rho v_1^2 - \frac{1}{2} \rho v_2^2 + \rho g (h_1 - h_2) \]Given parameters allow us to resolve for \(P_2\), leading to a calculated pressure at the lower point of 208,250 Pa. This task demonstrates how fluid dynamics principles can quantitatively describe physical systems, including the transformation and conservation of energy within a fluid.