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Chapter 14: Problem 51

A garden hose with an internal diameter of \(1.9 \mathrm{~cm}\) is connected to a (stationary) lawn sprinkler that consists merely of a container with 24 holes, each \(0.13 \mathrm{~cm}\) in diameter. If the water in the hose has a speed of \(0.91 \mathrm{~m} / \mathrm{s},\) at what speed does it leave the sprinkler holes?

Short Answer

Expert verified
The water exits the sprinkler holes at a speed of 8.09 m/s.

Step by step solution

01

Identify Known Values

First, let's list all the given data from the problem: - Diameter of the hose: 1.9 cm. - Diameter of each sprinkler hole: 0.13 cm. - Water speed in the hose: 0.91 m/s. - Number of sprinkler holes: 24.
02

Convert Diameter to Radius

To use the area formula, we first need to convert the diameters to radii. - Radius of the hose: \(\frac{1.9}{2} = 0.95\) cm.- Radius of each hole: \(\frac{0.13}{2} = 0.065\) cm.
03

Calculate Cross-Sectional Area

Use the formula for the area of a circle, \( A = \pi r^2 \), to find the cross-sectional areas:- Area of the hose: \( A_h = \pi (0.95)^2 = 2.835\) cm虏.- Area of one hole: \( A_s = \pi (0.065)^2 = 0.0133\) cm虏.
04

Calculate Total Area of All Holes

Multiply the area of one hole by the number of holes:- Total area for all 24 holes: \( A_{total} = 24 \times 0.0133 = 0.3192\) cm虏.
05

Apply the Continuity Equation

Use the principle of conservation of mass, which states that the flow rate must be conserved:\[ A_h \times v_h = A_{total} \times v_s \]Where:\( v_h = 0.91 \text{ m/s}\) is the speed in the hose.\( v_s \) is the speed in the sprinkler holes. Rearrange to solve for \( v_s \):\[ v_s = \frac{A_h \times v_h}{A_{total}} \].
06

Calculate Sprinkler Hole Speed

Now substitute the values into the equation:\[ v_s = \frac{2.835 \times 0.91}{0.3192} = 8.09 \text{ m/s} \].

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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 vital concept in fluid dynamics, embodying the principle of conservation of mass. It states that for an incompressible fluid flowing in a closed system, the mass flow rate must remain constant. This implies that any fluid entering a system must also exit, with no loss or accumulation.
More simply, the product of the cross-sectional area and fluid velocity at any point in the system remains constant:
  • Formula: \( A_1 \times v_1 = A_2 \times v_2 \)
  • \( A_1 \) and \( A_2 \) are the cross-sectional areas at two different points.
  • \( v_1 \) and \( v_2 \) are the fluid velocities at these points.
In this exercise, it helps us find the velocity of water exiting the sprinkler holes, ensuring that the water flow rate remains unchanged from the hose to the sprinkler.
Understanding Cross-Sectional Area
Cross-sectional area refers to the area of a particular slice of a three-dimensional object. In fluid dynamics, it is crucial as it determines how much fluid can pass through a section per unit time. Different shapes require unique calculations for their areas, but circles鈥攃ommon in pipes and hoses鈥攗se the simple formula:
  • Formula: \( A = \pi r^2 \)
  • \( r \) is the radius of the cross section.
The cross-sectional areas of both the hose and sprinkler holes influence the velocity of water flow, based on the continuity equation. Here, knowing the area helps determine changes in flow velocity when the water transitions from the hose to the smaller sprinkler openings.
Breaking Down Flow Rate
Flow rate is the volume of fluid passing through a given cross-sectional area per unit time. It is a fundamental measure in fluid dynamics, usually expressed in cubic meters per second (\( m^3/s \)).
  • Flow rate formula: \( Q = A \times v \)
  • \( Q \) is the flow rate.
  • \( A \) is the cross-sectional area.
  • \( v \) is the velocity of the fluid.
In the problem, since the flow rate stays constant before and after the water sprays from the hose into the air, it dictates the flow speed in the sprinkler holes, based on the areas calculated.
Calculating Speed of Water
The speed of water discharge is a direct result of applying the continuity equation's principles. By keeping the flow rate constant across different sized openings, you need to alter the speed of water. When water flows from a wide hose into small sprinkler holes, the velocity increases to maintain consistent flow rate.
To calculate this,
  • Use \( v_s = \frac{A_h \times v_h}{A_{total}} \).
  • Here, \( A_h \) is the hose's area, and \( v_h \) is the known speed.
  • \( A_{total} \) is the total area of the sprinkler holes.
Thus, these parameters help compute the exit speed of water from the sprinkler holes, as seen in our calculated solution.
Conservation of Mass in Fluid Systems
The principle of conservation of mass asserts that mass cannot be created or destroyed in a closed system. In the context of fluid dynamics, this means the amount of fluid entering a system must equal the amount leaving, assuming no changes in density.
This principle underpins the continuity equation, ensuring that the quantity of water flowing from the hose into the sprinkler remains constant. By maintaining this balance, mass conservation helps us predict how fluids behave when their flow paths or velocities change.
  • It underlines the physical reality ensuring our calculated velocities and flow rates remain consistent and rational.
  • Knowing and applying this law helps us solve fluid problems like finding exit speeds, by using known values others in the system implicitly stay balanced.

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