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If water flow out of a vertical pipe at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\), how far below the outlet will the radius be half the original valuc of radius of suream? (a) \(110 \mathrm{~m}\) (b) \(11 \mathrm{~m}\) (c) \(1.1 \mathrm{~m}\) (d) \(2 \mathrm{~m}\)

Short Answer

Expert verified
The radius is half the original at \(1.1 \mathrm{~m}\) below the outlet.

Step by step solution

01

Identify the Known Variables

The problem states that water flows out of a pipe at a speed of \(1.2\, \text{m/s}\). We need to find the distance below the outlet where the radius of the stream is half its original size.
02

Apply the Continuity Equation

The Continuity Equation for fluid flow states that \(A_1 v_1 = A_2 v_2\), where \(A\) is the cross-sectional area and \(v\) is the velocity. The area \(A\) is related to the radius \(r\) by \(A = \pi r^2\). If \(r_2 = \frac{1}{2}r_1\), then \(A_2 = \pi \left(\frac{1}{2}r_1\right)^2 = \frac{1}{4}\pi r_1^2\). This implies \(v_2 = 4v_1\).
03

Apply the Bernoulli's Equation

According to Bernoulli's principle: \(v_1^2 + 2gh = \left(4v_1\right)^2\), where \(h\) is the height below the outlet. Rewrite this as \(1.2^2 + 2\times 9.81 \times h = \left(4\times 1.2\right)^2\).
04

Solve for h

Substitute and rearrange the equation: The initial kinetic energy is \(1.44\), and the final kinetic energy is \(23.04\), which gives us \(1.44 + 19.62h = 23.04\). Solving for \(h\) gives \(19.62h = 21.6\), thus \(h = \frac{21.6}{19.62} \approx 1.1\) meters.

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Key Concepts

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

Continuity Equation
Fluid dynamics includes understanding the continuity equation, a fundamental principle used to analyze fluid flow. It states that the product of cross-sectional area and velocity of fluid remains constant along a streamline. This can be expressed by the equation:
  • \( A_1 \cdot v_1 = A_2 \cdot v_2 \)
where:
  • \( A \) is the cross-sectional area
  • \( v \) is the fluid velocity
If a pipe narrows, the velocity of the fluid must increase to satisfy the equation, assuming the fluid is incompressible and there is steady flow. Keeping the area and velocity paired helps us understand situations like water flowing out of a pipe and changing in shape as it falls. A reduction in radius, for example, leads to an increase in velocity downstream, illustrating conservation of flow dynamics and helping predict flow behavior.
Bernoulli's Principle
Bernoulli's Principle is another cornerstone in fluid dynamics that describes how the speed of a fluid relates to its pressure and potential energy. Essentially, as the speed of a fluid increases, its pressure decreases, which can be stated with the Bernoulli's equation:\[ v_1^2 + 2gh = \left(4v_1\right)^2 \]This equation is crucial as it explains the interchange between kinetic and potential energy along a streamline, factoring in gravity and acceleration of the fluid. In practical terms, understanding this principle is key to analyzing situations where fluid speeds change, such as with water exiting a pipe. The height below where a stream radius changes involves evaluating energy conservation where initial potential and kinetic energies balance out with final energies.
Streamflow
Streamflow examines the behavior of fluids as they move in a path, often focusing on interactions with surfaces or other elements like air. In studying streamflow in a vertical setting, such as our exercise, it's important to consider how external forces impact the fluid. A key observation is how gravity helps accelerate fluid as it falls, affecting both speed and pressure. As stream diameter changes—like halving in our exercise—the velocity must change to maintain the continuity equation. Observing free-falling streams, you'll notice the stream narrows as it speeds up, showcasing dynamic equilibrium among forces involved. With accurately understanding streamflows, predictions about fluid paths and behaviors can be made, aiding engineering, meteorology, and environmental science by indicating how real-world systems react under various conditions.

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Most popular questions from this chapter

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