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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 openings is approximately 9.6 m/s.

Step by step solution

01

Convert Units

First, we need to ensure that all measurements are in the same unit system. The radius of the openings is given in mm, so we convert it to meters: \(1.0\, \text{mm} = 0.001\, \text{m}\). Similarly, the pipe radius is given in cm, which we convert to meters: \(0.80\, \text{cm} = 0.008\, \text{m}\).
02

Calculate Areas

Calculate the cross-sectional areas:1. For the pipe: \(A_p = \pi \times (0.008\, \text{m})^2 = \pi \times 6.4 \times 10^{-5} \) square meters.2. For one opening: \(A_o = \pi \times (0.001\, \text{m})^2 = \pi \times 10^{-6} \) square meters.3. Total area of all openings: \(A_{total} = 20 \times A_o = 20 \pi \times 10^{-6} \) square meters.
03

Apply Continuity Equation

According to the principle of conservation of mass (continuity equation), the flow rate must remain constant. This can be expressed as: \[ A_p \times v_p = A_{total} \times v_o \]Where \(v_p = 3.0\, \text{m/s}\) is the speed of water in the pipe and \(v_o\) is the exit speed of water through the openings.
04

Solve for Exit Speed

Rearrange the continuity equation to solve for the speed of water as it exits:\[ v_o = \frac{A_p \times v_p}{A_{total}} \]Substitute the known values:\[ v_o = \frac{\pi \times 6.4 \times 10^{-5} \text{ m}^2 \times 3.0 \text{ m/s}}{20 \pi \times 10^{-6} \text{ m}^2} \approx \frac{19.2 \times 10^{-5} \text{ m}^3 / \text{s}}{20 \times 10^{-6}} = 9.6 \text{ m/s} \]
05

Verification

Check if the units are consistent and the calculations are correct. All inputs are in meters and speeds are in meters per second, confirming the methodology and final answer.

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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 pivotal. It states that in any steady-state process, the mass flow rate at different points in a system remains constant. Imagine the pipe and the shower head in a balance. The water entering the pipe equals the water exiting the shower head, ensuring no water magically appears or disappears. The equation can be written as:
  • \( A_1 \times v_1 = A_2 \times v_2 \)
Here, \( A \) represents the cross-sectional area, and \( v \) represents the fluid speed. In our shower example, \( A_1 \) and \( v_1 \) are the area of the pipe and speed inside it. Meanwhile, \( A_2 \) and \( v_2 \) pertain to the total area of the openings and the speed of the water as it exits. This principle is crucial in ensuring that the calculations follow the fundamental laws of physics.
Unit Conversion
Unit conversion often seems trivial but is essential for solving many engineering problems accurately. When working with quantities like metric units of length, a mismatch can lead to errors. Ensure everything is expressed in the same unit before computations. In this exercise, radii are provided in different units, millimeters, and centimeters. Correct conversion guarantees that all further calculations, like the areas, are accurately performed. Practice converting:
  • \( 1.0 \text{ mm} = 0.001 \text{ m} \)
  • \( 0.80 \text{ cm} = 0.008 \text{ m} \)
Always remember: consistent units equal reliable results. By adopting a consistent metric system unit, you establish a strong foundation for accurate problem-solving.
Cross-Sectional Area
Calculating cross-sectional areas is a fundamental step in applying the continuity equation. It involves determining the size of any slice perpendicular to the flow. For a circular opening, this is easy; simply use the formula for the area of a circle:
  • \( A = \pi r^2 \)
Concretely:
  • The pipe's area: \( A_p = \pi \times (0.008)^2 = \pi \times 6.4 \times 10^{-5} \text{ m}^2 \)
  • Each shower head opening: \( A_o = \pi \times (0.001)^2 = \pi \times 10^{-6} \text{ m}^2 \)
  • Total for 20 openings: \( A_{total} = 20 \times A_o \)
This step sets the stage for using the continuity equation, as it directly ties into how flow transitions from the pipe into the dispersing openings.
Speed Calculation
With unit conversion and area determinations settled, calculating the speed of the fluid through any part of the system can proceed. Examining the shower head case, the continuity equation helps find how fast water exits the openings once you know inputs from the pipe:
  • \( v_o = \frac{A_p \times v_p}{A_{total}} \)
Substitute known factors:
  • \( v_p = 3.0 \text{ m/s} \)
  • \( A_p = \pi \times 6.4 \times 10^{-5} \text{ m}^2 \)
  • \( A_{total} = 20 \pi \times 10^{-6} \text{ m}^2 \)
  • Resulting in \( v_o = 9.6 \text{ m/s} \)
Such calculations ensure a clear understanding of how fluid velocities adjust based on the restrictions and expansions in the flow path.

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