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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 water graduall descends \(10 \mathrm{~m}\) as the pipe cross-sectional area increases to \(8.0 \mathrm{~cm}^{2} .\) (a) What is the speed 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?

Short Answer

Expert verified
(a) Speed at lower level: 2.5 m/s. (b) Pressure at lower level: 260,500 Pa.

Step by step solution

01

Use the Continuity Equation for Velocity

The continuity equation tells us that the product of cross-sectional area and velocity remains constant in incompressible flow. Thus, we write: \[ A_1 \times v_1 = A_2 \times v_2 \]where \( A_1 \) and \( v_1 \) are the cross-sectional area and velocity at the upper level, and \( A_2 \) and \( v_2 \) are at the lower level. We have: \[ A_1 = 4.0\, \text{cm}^2 = 4.0 \times 10^{-4} \text{m}^2 \]\[ v_1 = 5.0\, \text{m/s} \]\[ A_2 = 8.0\, \text{cm}^2 = 8.0 \times 10^{-4} \text{m}^2 \]Substituting these values in:\[ 4.0 \times 10^{-4} \times 5.0 = 8.0 \times 10^{-4} \times v_2 \]Solve for \( v_2 \):\[ v_2 = \frac{4.0 \times 10^{-4} \times 5.0}{8.0 \times 10^{-4}} \]\[ v_2 = 2.5\, \text{m/s} \]
02

Apply Bernoulli’s Equation for Pressure Calculation

Bernoulli's equation relates pressure, velocity, and height between two points in fluid flow:\[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]Given:\( \rho \) (density of water) = \( 1000 \text{ kg/m}^3 \)\( P_1 = 1.5 \times 10^5 \text{ Pa} \)\( v_1 = 5.0 \text{ m/s} \)\( v_2 = 2.5 \text{ m/s} \)\( h_1 - h_2 = 10 \text{ m} \)We need to find \( P_2 \). Rearranging Bernoulli’s equation:\[ P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) + \rho g (h_1 - h_2) \]Substitute:\[ P_2 = 1.5 \times 10^5 + \frac{1}{2} \times 1000 \times (5^2 - 2.5^2) + 1000 \times 9.8 \times 10 \]\[ P_2 = 1.5 \times 10^5 + 12500 + 98000 \]\[ P_2 = 260500 \text{ Pa} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Continuity Equation
In fluid dynamics, the Continuity Equation is a fundamental principle describing the conservation of mass in a flowing fluid. It ensures that, in an incompressible flow, the mass flow rate remains constant from one cross-section of a pipe to another. This means the product of the cross-sectional area (\( A \)) and the velocity of the fluid (\( v \)) must equal at different sections:
  • Upper section: \( A_1 \cdot v_1 \)
  • Lower section: \( A_2 \cdot v_2 \)
By knowing the initial speed and area, we can calculate the speed of the fluid at another point where the cross-sectional area changes.
In the provided problem, we're given an initial area of \( 4.0 \ \text{cm}^2 \) and speed \( 5.0 \, \text{m/s} \). The pipe then widens, doubling the area to \( 8.0 \, \text{cm}^2 \). According to the Continuity Equation:\[A_1 \times v_1 = A_2 \times v_2\]Using this equation, you can solve for the new velocity \( v_2 \), which is crucial in determining how the fluid behaves further along the pipe.
Bernoulli's Equation
Bernoulli's Equation links the pressure, velocity, and height in a flowing fluid, illustrating the principle of energy conservation. For an incompressible, steady flow, the total mechanical energy remains constant. Thus, it is expressed as:\[P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}\]In this equation:
  • \( P \) is the pressure energy.
  • \( \frac{1}{2} \rho v^2 \) captures the kinetic energy per unit volume.
  • \( \rho gh \) represents the potential energy due to elevation.
In the context of our problem, Bernoulli’s Equation helps us solve for pressure (\( P_2 \)) at the lower level of the pipe. As the fluid descends by \( 10 \, \text{m} \), the potential energy decreases, which affects the pressure. By rearranging Bernoulli’s Equation, we consider changes in kinetic and potential energies to find the new pressure value.
Understanding this equation is essential when dealing with complications involving different points in varying elevations and flow speeds.
Incompressible Flow
In fluid dynamics, the term "Incompressible Flow" refers to a flow where the density of the fluid remains constant over time and space. For most liquids, such as water, this is a valid assumption as they do not compress significantly under typical conditions.
When dealing with incompressible flow, the Continuity Equation becomes very useful, as it allows us to assume that the mass flow rate (density \( \times \) area \( \times \) velocity) is conserved across any cross-section of a streamline. It eliminates complexities involved in accounting for changes in density, simplifying the analysis.
This assumption simplifies our example problem significantly. It enables us to rely on algebraic calculations with constant density, ensuring straightforward calculations of varied speed and pressure as the pipe expands and descends.
Pressure Calculation
Pressure calculation in fluid dynamics involves determining changes in pressure across different sections of a fluid flow. Using Bernoulli's Equation, we focus on
  • Initial and resultant kinetic and potential energies.
  • Determining the effects of speed changes, cross-sectional area, and elevation.
In the given problem, you start with an initial pressure, velocity, and height. The pressure at the lower section is calculated considering the increase in pipe area and height difference:\[P_2 = P_1 + \frac{1}{2} \rho (v_1^2 - v_2^2) + \rho g (h_1 - h_2)\]The computation here factors in reduced kinetic energy as velocity decreases due to a larger area, along with the gained potential energy from descending. This detailed approach ensures a more comprehensive understanding for students tackling similar pressure-related queries.

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