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Chapter 11: Problem 55

A water line with an internal radius of \(6.5 \times 10^{-3} \mathrm{~m}\) is connected to a shower head that has 12 holes. The speed of the water in the line is \(1.2 \mathrm{~m} / \mathrm{s}\). (a) What is the volume flow rate in the line? (b) At what speed does the water leave one of the holes (effective hole radius \(=4.6 \times 10^{-4} \mathrm{~m}\) ) in the head?

Short Answer

Expert verified
(a) Volume flow rate is approximately 1.592 x 10^{-4} m^3/s. (b) Water leaves one hole at about 20 m/s.

Step by step solution

01

Calculate the Cross-Sectional Area of the Pipe

First, calculate the cross-sectional area of the water line using the internal radius. The area \( A \) is given by the formula for the area of a circle: \[A = \pi r^2\]Where \( r = 6.5 \times 10^{-3} \text{ m} \). So, \[A = \pi (6.5 \times 10^{-3})^2 \approx 1.327 \times 10^{-4} \text{ m}^2\]
02

Compute Volume Flow Rate in the Line

The volume flow rate \( Q \) is the product of the cross-sectional area and the speed of water:\[Q = A \cdot v = 1.327 \times 10^{-4} \text{ m}^2 \times 1.2 \text{ m/s} = 1.592 \times 10^{-4} \text{ m}^3\text{/s}\]
03

Calculate the Cross-Sectional Area of One Hole

Each hole in the shower head is smaller. Calculate the cross-sectional area for one hole using its radius. The area \( A_{hole} \) is given by:\[A_{hole} = \pi (4.6 \times 10^{-4})^2 \approx 6.63 \times 10^{-7} \text{ m}^2\]
04

Determine the Flow Rate through One Hole

Since there are 12 holes, each hole receives an equal share of the total volume flow rate:\[Q_{hole} = \frac{1.592 \times 10^{-4}}{12} \approx 1.327 \times 10^{-5} \text{ m}^3/s\]
05

Calculate Speed of Water Leaving One Hole

To find the speed \( v_{hole} \) of water leaving one hole, use the equation of flow rate:\[Q_{hole} = A_{hole} \cdot v_{hole}\]Rearrange this to solve for \( v_{hole} \):\[v_{hole} = \frac{Q_{hole}}{A_{hole}} = \frac{1.327 \times 10^{-5}}{6.63 \times 10^{-7}} \approx 20 \text{ m/s}\]

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Key Concepts

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

Volume Flow Rate
The volume flow rate is a critical concept in fluid mechanics that tells us how much fluid passes through a specific point in a pipe or channel over a given period of time. Think of it as a measure of how "busy" a water line is, with more volume flow indicating more water moving through. The standard unit for measuring volume flow rate is cubic meters per second (m鲁/s). Understanding volume flow rate helps in ensuring adequate supply and efficient distribution of fluids in systems, such as water through pipes in a household. When you know the volume flow rate, you gain insight into how efficiently the system transports fluids, which is crucial in design and troubleshooting. It takes the cross-sectional area and speed of water into account, showing a whole picture of the flow dynamics.
Cross-Sectional Area
The cross-sectional area is an important factor in determining how much fluid can flow through a pipe at any given moment. Imagine slicing the pipe parallel to its length at any point; the surface area of the circle you see is the cross-sectional area. For circular pipes, this area is calculated using the formula:\[A = \pi r^2\]where \(r\) is the radius of the pipe. The cross-sectional area directly affects the volume flow rate because a larger area means that more fluid can be transported through the pipe, given the same speed of water. Thus, understanding the cross-sectional area is key in designing systems that must handle specific fluid capacities.
Speed of Water
Speed of water refers to how fast the fluid itself is moving through the pipe or channel. It's like looking at how quickly water travels from one end to the other. The speed of water is measured in meters per second (m/s), which tells how many meters the fluid moves forward in one second. This speed is crucial because:
  • It affects how much time water takes to travel through the system, impacting efficiency.
  • It plays a role in the volume flow rate, as higher speeds can increase the total flow through the system, assuming a constant cross-sectional area.
Together with the cross-sectional area, the speed of water determines how effective a flow system is at transporting fluids from one point to another.
Equation of Flow Rate
The equation of flow rate is a straightforward mathematical tool that combines both the cross-sectional area and the speed of water to give you the volume flow rate. It is expressed as:\[Q = A \cdot v\]where \(Q\) is the volume flow rate, \(A\) is the cross-sectional area, and \(v\) is the speed of water. This equation helps engineers and scientists predict how a fluid will behave in a particular system, ensuring that the transport of fluid is efficient and meets required demands.This principle is useful when analyzing situations like determining how fast water will exit a shower head, based on the known speed in the pipe and the cross-sectional area of the outflow holes. Thus, this formula is a cornerstone in fluid mechanics, making it easier to design, evaluate, and improve fluid transport systems.

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Most popular questions from this chapter

Mercury is poured into a tall glass. Ethyl alcohol is then poured on top of the mercury until the height of the ethyl alcohol itself is \(110 \mathrm{~cm}\). The two fluids do not mix, and the air pressure at the top of the ethyl alcohol is one atmosphere. What is the absolute pressure at a point that is \(7.10 \mathrm{~cm}\) below the ethyl alcohol-mercury interface?

The main water line enters a house on the first floor. The line has a gauge pressure of \(1.90 \times 10^{5} \mathrm{~Pa}\). (a) A faucet on the second floor, \(6.50 \mathrm{~m}\) above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it, even if the faucet were open?

Two circular holes, one larger than the other, are cut in the side of a large water tank whose top is open to the atmosphere. The center of one of these holes is located twice as far beneath the surface of the water as the other. The volume flow rate of the water coming out of the holes is the same. (a) Decide which hole is located nearest the surface of the water. (b) Calculate the ratio of the radius of the larger hole to the radius of the smaller hole.

A 1.3-m length of horizontal pipe has a radius of \(6.4 \times 10^{-3} \mathrm{~m}\). Water within the pipe flows with a volume flow rate of \(9.0 \times 10^{-3} \mathrm{~m}^{3} / \mathrm{s}\) out of the right end of the pipe and into the air. What is the pressure in the flowing water at the left end of the pipe if the water behaves as (a) an ideal fluid and (b) a viscous fluid \(\left(\eta=1.00 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\right)\) ?

The Mariana trench is located in the floor of the Pacific Ocean at a depth of about 11000 \(\mathrm{m}\) below the surface of the water. The density of seawater is \(1025 \mathrm{~kg} / \mathrm{m}^{3}\). (a) If an underwater vehicle were to explore such a depth, what force would the water exert on the vehicle's observation window (radius \(=0.10 \mathrm{~m}\) )? (b) For comparison, determine the weight of a jetliner whose mass is \(1.2 \times 10^{5} \mathrm{~kg}\).
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