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A liquid flowing from a vertical pipe has a definite shape as it flows from the pipe. To get the equation for this shape, assume that the liquid is in free fall once it leaves the pipe. Just as it leaves the pipe, the liquid has speed \(v_{0}\) and the radius of the stream of liquid is \(r_{0 .}\) (a) Find an equation for the speed of the liquid as a function of the distance \(y\) it has fallen. Combining this with the equation of continuity, find an expression for the radius of the stream as a function of \(y\) . (b) If water flows out of a vertical pipe at a speed of 1.20 \(\mathrm{m} / \mathrm{s}\) , how far below the outlet will the radius be one-half the original radius of the stream?

Short Answer

Expert verified
At about 0.220 meters below the outlet, the radius will be half of the original radius.

Step by step solution

01

Analyze the Motion of Liquid

As the liquid leaves the pipe, it undergoes free fall due to gravity. Its initial vertical speed is already given as \( v_0 \). The speed of the liquid as a function of distance \( y \) it has fallen is determined by the equations of motion. Since only gravity acts after leaving the pipe, the increase in speed can be written as \ v(y) = \sqrt{v_0^2 + 2gy} \ where \( g \) is the acceleration due to gravity (\( 9.81 \ \text{m/s}^2 \)).
02

Apply the Equation of Continuity

The equation of continuity states that the product of the cross-sectional area and flow speed at any point is constant. For a circular cross-section, the area is \( A = \pi r^2 \). Thus, \ \pi r_0^2 v_0 = \pi r(y)^2 v(y), \ and solving for \( r(y) \), we have \ r(y) = r_0 \sqrt{\frac{v_0}{v(y)}}. \
03

Substitute the Expression for Speed

From step 1, substitute the expression for \( v(y) \) into the radius equation: \ r(y) = r_0 \sqrt{\frac{v_0}{\sqrt{v_0^2 + 2gy}}}, \ simplifying to \ r(y) = r_0 \sqrt{\frac{v_0^2}{v_0^2 + 2gy}}. \
04

Solving for a Specific Case

We need to find the distance \( y \) such that the radius \( r(y) = \frac{1}{2} r_0 \). Substitute this into the equation: \ \frac{1}{2} r_0 = r_0 \sqrt{\frac{v_0^2}{v_0^2 + 2gy}}, \ simplifying to \ \frac{1}{4} v_0^2 = \frac{v_0^2}{v_0^2 + 2gy}. \
05

Solve for y

Rearrange the equation from Step 4 to solve for \( y \): \ v_0^2 + 2gy = 4v_0^2, \ \ 2gy = 3v_0^2, \ and solving for \( y \): \ y = \frac{3v_0^2}{2g}. \
06

Calculate the Specific Distance

Substitute \( v_0 = 1.20 \ \text{m/s} \) and \( g = 9.81 \ \text{m/s}^2 \) into the equation for \( y \): \ y = \frac{3(1.20)^2}{2 \cdot 9.81}, \ which simplifies to \ y \approx 0.220 \ \text{meters}. \

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Key Concepts

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

Free Fall Equations
In fluid dynamics, understanding how a liquid behaves once it leaves a container, such as a pipe, is crucial. When a liquid exits a pipe, it experiences free fall due to gravity. This means that the only force acting on it is gravity, pulling it downward. The free fall equations help us describe this motion.

The initial speed of the liquid is denoted as \(v_0\), which is the speed as it leaves the pipe. As the liquid falls, its speed increases due to gravitational acceleration. The equation for the speed of an object in free fall, as it reaches a distance \(y\) below its starting point, is given by:
  • \(v(y) = \sqrt{v_0^2 + 2gy}\)
Here, \(g\) is the acceleration due to gravity, approximately \(9.81 \, \text{m/s}^2\).

This formula calculates the increased speed resulting from gravitational acceleration over the distance \(y\) the liquid has fallen.
Continuity Equation
The continuity equation is a fundamental principle in fluid dynamics which indicates that, for a fluid in steady flow, the mass flow rate must remain constant. When it comes to liquids flowing through pipes, the equation can be expressed in terms of the liquid's speed and the cross-sectional area it traverses.

For liquid flowing out with a circular cross-section, the area is given by \(A = \pi r^2\). The equation of continuity can then be written as:
  • \(\pi r_0^2 v_0 = \pi r(y)^2 v(y)\)
This relationship shows that as the liquid accelerates and its speed \(v(y)\) increases, its radius \(r(y)\) must adjust to ensure the product of area and velocity remains unchanged.

Solving for \(r(y)\), we utilize the continuity concept to find the radius at any given point:
  • \(r(y) = r_0 \sqrt{\frac{v_0}{v(y)}}\)
This equation ensures the conservation of mass, maintaining a steady flow regardless of the fluid speed or pipe diameter changes.
Circular Cross-Section Area
Pipes and other conduits commonly have a circular cross-section, and understanding this geometry is essential in calculating the flow characteristics of a liquid.

The area of a circular section with radius \(r\) is given as:
  • \(A = \pi r^2\)
This fundamental geometric principle helps us derive many important aspects of fluid flow, such as using it in the continuity equation to balance the flow rates.

Changes in the radius of the fluid stream, such as those due to varying speeds in free fall, affect the cross-sectional area. Therefore, by understanding how the circular area interacts with velocity, we can predict behaviors like stream contraction when fluid acceleration occurs.
Acceleration Due to Gravity
Acceleration due to gravity, commonly denoted as \(g\), is a constant that describes the acceleration all objects experience under Earth's gravitational pull. Its standard value is \(9.81 \, \text{m/s}^2\).

In free fall scenarios, such as when liquid exits a pipe and is subject to gravity alone, \(g\) plays a critical role in determining how quickly the velocity of the fluid increases. As referenced in the free fall equations, the increase in velocity over a fallen distance \(y\) is given by the term \(2gy\).

Thus, understanding gravity's influence is pivotal not just for predicting speed increases but also for applying solutions such as determining how far a fluid must fall to reach half its original radius by employing formulas like those derived from continuity.

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