Chapter 13: Problem 53
Water discharges from a horizontal cylindrical pipe at the rate of \(465 \mathrm{~cm}^{3} / \mathrm{s}\). At a point in the pipe where the radius is \(2.05 \mathrm{~cm},\) the absolute pressure is \(1.60 \times 10^{5} \mathrm{~Pa}\). What is the pipe's radius at a constriction if the pressure there is reduced to \(1.20 \times 10^{5} \mathrm{~Pa}\) ?
Short Answer
Step by step solution
Identify Known Values
Continuity Equation
Substitute Flow Rate to Find Velocity
Apply Bernoulli's Equation
Solve for Constriction Radius
Final Calculation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bernoulli's Equation
- \( P \) is the pressure energy per unit volume,
- \( \frac{1}{2} \rho v^2 \) is the kinetic energy per unit volume,
- \( \rho gh \) is the potential energy per unit volume.
Continuity Equation
- \( A_1 \) and \( A_2 \) are the cross-sectional areas of the pipe at two different points,
- \( v_1 \) and \( v_2 \) are the fluid velocities at these points.
Volumetric Flow Rate
- \( A \) is the cross-sectional area of the pipe,
- \( v \) is the velocity of the fluid.