Chapter 11: Problem 71
A liquid is flowing through a horizontal pipe whose radius is \(0.0200 \mathrm{m}\). The pipe bends straight upward through a height of \(10.0\) \(\mathrm{m}\) and joins another horizontal pipe whose radius is \(0.0400\) \(\mathrm{m}\). What volume flow rate will keep the pressures in the two horizontal pipes the same?
Short Answer
Step by step solution
Understand the Problem
Apply Continuity Equation
Compute Cross-Sectional Areas
Use Bernoulli's Equation
Cancel Common Terms
Relate Velocities from Continuity
Solve for Velocity
Calculate Final Velocity and Flow Rate
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity Equation
Mathematically, the Continuity Equation is expressed as:
- \[ A_1 V_1 = A_2 V_2 \]
- \(A_1\) and \(A_2\) are the cross-sectional areas of two different sections of the pipe. They can be calculated using the formula for the area of a circle: \(A = \pi r^2\).
- \(V_1\) and \(V_2\) are the fluid velocities at these sections.
Bernoulli's Equation
Bernoulli's Equation is given as:
- \[ P_1 + \frac{1}{2} \rho V_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho V_2^2 + \rho g h_2 \]
- \(P_1\) and \(P_2\) are the pressures at two different points.
- \(\rho\) is the fluid density.
- \(V_1\) and \(V_2\) are the fluid speeds at these points.
- \(h_1\) and \(h_2\) are the heights relative to a datum.
- \(g\) is the acceleration due to gravity.
Volume Flow Rate
It is directly related to both the Continuity Equation and Bernoulli's Equation and is mathematically defined by the product of the cross-sectional area of the pipe and the velocity of the fluid. That is:
- \[ Q = A V \]
- \(Q\) is the volume flow rate, typically measured in cubic meters per second (\(m^3/s\)).
- \(A\) is the cross-sectional area through which the fluid flows.
- \(V\) is the velocity of the fluid.