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Chapter 13: Problem 95

The stream of water from a faucet decreases in diameter as it falls (Fig. \(13-58\) ). Derive an equation for the diameter of the stream as a function of the distance \(y\) below the faucet, given that the water has speed \(v_{0}\) when it leaves the faucet, whose diameter is \(d\).

Short Answer

Expert verified
The diameter of the stream as a function of distance is \( D = d \sqrt{\frac{v_0}{\sqrt{v_0^2 + 2gy}}} \).

Step by step solution

01

Understand the Problem

The problem involves deriving an equation for the diameter of the water stream as it falls under gravity. The stream's diameter changes because the speed of the falling water increases due to gravity.
02

Use the Continuity Equation

The principle of conservation of mass requires the flow rate to be constant. If the cross-sectional area of the stream at the faucet is \( A_0 = \frac{\pi d^2}{4} \) and the speed there is \( v_0 \), then:\[A_0 v_0 = A v \]where \( A \) is the cross-sectional area at distance \( y \) and \( v \) is the speed at that point.
03

Calculate the Area as a Function of Diameter

Since the cross-sectional area of the stream changes with diameter, express the area at distance \( y \) in terms of diameter:\[A = \frac{\pi D^2}{4}\]where \( D \) is the unknown diameter at distance \( y \).
04

Apply Bernoulli's Equation

Using energy conservation between the faucet and the point at depth \( y \):\[v^2 = v_0^2 + 2gy\]where \( g \) is the acceleration due to gravity. This relates the velocities at different heights.
05

Substitute for Velocity

Solve the previous step for \( v \):\[v = \sqrt{v_0^2 + 2gy}\]
06

Relate Area and Diameter

Combine the continuity equation and the expression for velocity:\[A_0 v_0 = \frac{\pi D^2}{4} \sqrt{v_0^2 + 2gy} \]Solve for \( D \):\[D^2 = \frac{4 A_0 v_0}{\pi \sqrt{v_0^2 + 2gy}}\]
07

Solve for Final Diameter Expression

Substituting \( A_0 = \frac{\pi d^2}{4} \) into the equation for \( D^2 \):\[D^2 = \frac{d^2 v_0}{\sqrt{v_0^2 + 2gy}}\]Finally, solve for \( D \):\[D = d \sqrt{\frac{v_0}{\sqrt{v_0^2 + 2gy}}}\]

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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 ensuring that mass is conserved in a fluid flow system. It states that the product of the cross-sectional area of a flow and its velocity is constant. For incompressible and steady-flow fluids, the equation can be represented as \(A_1 v_1 = A_2 v_2\), where \(A \) is the area and \(v \) is the velocity at different points in the flow. This means that if a fluid accelerates, its cross-sectional area must decrease to maintain the flow rate.

In our water stream problem, the water exiting the faucet must maintain the same flow rate as it falls. The significant change is in speed and diameter, linked by the continuity equation \(A_0 v_0 = A v\). Here, the increase in speed as the water falls under gravity means the diameter of the water stream decreases.
Bernoulli's Equation
Bernoulli's equation is a principle of fluid dynamics that describes the conservation of energy in flowing fluids. It delineates how fluid pressure, velocity, and potential energy are related. The equation is typically stated as \[ P + \frac{1}{2} \rho v^2 + \rho gy = \,\text{constant} \]where \(P\) denotes the fluid pressure, \(\rho\) the fluid density, \(v\) the velocity, and \(y\) the height above a reference point.

Applying Bernoulli's equation in the problem helps us understand how the speed of the water stream changes as it falls. Starting with the initial velocity \(v_0\) at the faucet, the water gains kinetic energy as it falls, moving faster. The expression \(v^2 = v_0^2 + 2gy\) is derived, showing that the speed \(v\) at a given height is dependent on both the initial speed and the height it has fallen due to gravity. This change directly affects the stream's diameter as it falls.
Conservation of Mass
The conservation of mass principle is vital in understanding fluid flow problems. It states that for any system closed to all transfers of matter and energy, the mass of the system must remain constant over time. In fluid dynamics, this concept ensures that the mass entering a system is equal to the mass exiting.

For the problem of a water stream, as water flows from a higher to a lower point, it accelerates due to gravity, but the overall quantity of mass in the flow remains unchanged. The continuity equation in this context signifies that the mass flow rate \(\dot{m} = \rho A v\) should remain constant. This leads to changes in stream diameter and velocity to comply with mass conservation principles, which are elegantly summarized in both the continuity and Bernoulli's equation used in analyzing the problem.

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Most popular questions from this chapter

(III) Show that inside a soap bubble, there must be a pressure \(\Delta P\) in excess of that outside equal to \(\Delta P=4 \gamma / r,\) where \(r\) is the radius of the bubble and \(\gamma\) is the surface tension. [Hint: Think of the bubble as two hemispheres in contact with each other; and remember that there are two surfaces to the bubble. Note that this result applies to any kind of membrane, where \(2 \gamma\) is the tension per unit length in that membrane.

Airlines are allowed to maintain a minimum air pressure within the passenger cabin equivalent to that at an altitude of 8000 \(\mathrm{ft}(2400 \mathrm{m})\) to avoid adverse health effects among passengers due to oxygen deprivation. Estimate this minimum pressure (in atm).

(III) A common effect of surface tension is the ability of a liquid to rise up a narrow tube due to what is called capillary action. Show that for a narrow tube of radius \(r\) placed in a liquid of density \(\rho\) and surface tension \(\gamma\), the liquid in the tube will reach a height \(h=2 \gamma / \rho g r\) above the level of the liquid outside the tube, where \(g\) is the gravitational acceleration. Assume that the liquid "wets" the capillary (the liquid surface is vertical at the contact with the inside of the tube).

(II) A 6.0-cm-diameter horizontal pipe gradually narrows to \(4.5 \mathrm{~cm} .\) When water flows through this pipe at a certain rate, the gauge pressure in these two sections is \(32.0 \mathrm{kPa}\) and \(24.0 \mathrm{kPa}\), respectively. What is the volume rate of flow?

A bucket of water is accelerated upward at 1.8 \(\mathrm{g} .\) What is the buoyant force on a 3.0 -kg granite rock \((\mathrm{SG}=2.7)\) submerged in the water? Will the rock float? Why or why not?
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