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A shower head has 20 circular openings, each with radius 1.0 \(\mathrm{mm}\) . The shower head is connected to a pipe with radius 0.80 \(\mathrm{cm} .\) If the speed of water in the pipe is \(3.0 \mathrm{m} / \mathrm{s},\) what is its speed as it exits the shower-head openings?

Short Answer

Expert verified
The speed of water as it exits the shower openings is 9.6 m/s.

Step by step solution

01

Convert Units

The opening radius of each shower head hole is given as 1.0 mm. First, convert this to centimeters: \[ 1.0 \text{ mm} = 0.1 \text{ cm} \] The pipe radius is already in centimeters, so no conversion is needed for that.
02

Calculate Area of Pipe

The cross-sectional area of the pipe can be calculated using the formula for the area of a circle, \[ A = \pi r^2 \] where \( r = 0.80 \text{ cm} \). Thus, \[ A_{pipe} = \pi (0.80)^2 = \pi \times 0.64 \]},
03

Calculate Total Area of Shower Openings

The area of one shower opening is calculated with \[ A_{opening} = \pi r^2 \] with \( r = 0.1 \text{ cm} \). Thus, \[ A_{opening} = \pi (0.1)^2 = \pi \times 0.01 \] Since there are 20 openings, the total area is:\[ A_{total} = 20 \times \pi \times 0.01 = 0.2\pi \]
04

Apply Continuity Equation

According to the continuity equation for incompressible fluid flow, \[ A_{pipe} \cdot v_{pipe} = A_{total} \cdot v_{openings} \] We know:- \( v_{pipe} = 3.0 \text{ m/s} \)- \( A_{pipe} = \pi \times 0.64 \)- \( A_{total} = 0.2\pi \)We'll solve for \( v_{openings} \).
05

Solve for Speed at Openings

Substitute the known values into the continuity equation to solve for \( v_{openings} \):\[ \pi \times 0.64 \times 3.0 = 0.2\pi \times v_{openings} \]\[ 1.92 = 0.2 \times v_{openings} \]\[ v_{openings} = \frac{1.92}{0.2} = 9.6 \text{ m/s} \]

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Key Concepts

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

Continuity Equation
The continuity equation is a fundamental principle in fluid dynamics. It describes how fluid flows are conserved in systems, ensuring that the mass flow rate remains constant from one cross-section of a pipe or channel to another. The continuity equation is expressed mathematically as:\[ A_1 \cdot v_1 = A_2 \cdot v_2 \]where:
  • \(A_1\) and \(A_2\) are the cross-sectional areas at two different points in the flow.
  • \(v_1\) and \(v_2\) are the fluid velocities at these points.
This equation holds true for incompressible fluids, such as water, assuming the fluid density does not change. In our example, the flow in the pipe is steady, with water moving from a larger cross-sectional area (pipe) to multiple smaller ones (shower openings). By knowing the pipe's area and velocity, we can find the velocity of water exiting the shower by applying the continuity equation. This ensures that the same volume of water flows through both sections without any loss.
Incompressible Fluid Flow
Incompressible fluid flow refers to the flow of a fluid whose density remains constant throughout. Most liquids, like water, are considered incompressible for practical purposes because their density does not significantly change with pressure variations. This is an essential assumption in fluid dynamics problems where the continuity equation is applied.For the exercise in question, we assume that the water flowing through the pipe and then through the shower openings is incompressible. This allows us to apply the continuity equation straightforwardly because the same volume per unit time enters and exits.When dealing with incompressible flow:
  • Density (\(\rho\)) is constant across the flow.
  • The mass flow rate is conserved across any cross-section.
This concept simplifies many calculations in fluid mechanics by allowing us to focus on the volume of the fluid rather than its density changes.
Cross-sectional Area
The cross-sectional area of a pipe or channel is crucial in determining the fluid flow characteristics. It is the area of the internal surface through which the fluid can flow and is especially relevant when applying the continuity equation. The cross-sectional area for a circular pipe is calculated using the formula:\[ A = \pi r^2 \]where \(r\) is the radius of the pipe.In this problem, the pipe has a larger cross-sectional area compared to each of the showerhead's openings. Calculating each area allows us to understand how much the fluid's velocity will change as it flows from the pipe through to the smaller openings.
  • Larger area in the pipe means slower velocity initially.
  • Smaller total area in the openings means higher velocity upon exit.
Understanding how the cross-sectional area affects fluid speed helps in designing systems where consistent and predictable fluid flow is needed.
Unit Conversion
Unit conversion is a vital skill in fluid dynamics and engineering calculations. In this exercise, unit conversion is needed to ensure consistency between different measurements.For instance, the radius of the shower openings is given in millimeters, while the radius of the pipe is in centimeters. Before performing any calculations, it's important to convert all units so they match.Here are the conversion steps for this exercise:
  • Convert the radius from millimeters to centimeters: \(1.0\, \text{mm} = 0.1\, \text{cm}\).
  • Ensure all areas and velocities use compatible units (e.g., meters, centimeters).
Consistent units across calculations prevent errors and ensure the accurate application of formulas, such as the continuity equation. Keeping units consistent is one of the simplest ways to avoid mistakes in solving physics problems, especially in topics like fluid dynamics.

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Most popular questions from this chapter

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