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Chapter 9: Problem 44

Water flowing through a garden hose of diameter \(2.74 \mathrm{~cm}\) fills a \(25.0\) - L bucket in \(1.50 \mathrm{~min}\). (a) What is the speed of the water leaving the end of the hose? (b) A nozzle is now attached to the end of the hose. If the nozzle diameter is one-third the diameter of the hose, what is the speed of the water leaving the nozzle?

Short Answer

Expert verified
The speed of the water leaving the hose without the nozzle is approximately 0.503 m/s and after the nozzle is attached is approximately 4.52 m/s.

Step by step solution

01

Determining the Speed of the Water Without Nozzle

First, calculate the volume of the water in cubic meters which is \(25.0 \, L = 25.0 \times 10^{-3} \, m^3 \). Then, calculate the time in seconds which is \(1.50 \, min = 90.0 \, s \). The volume flow rate (Q) is volume/time, so \(Q = \frac{25.0 \times 10^{-3}}{90} \, m^3/s\). The speed of the fluid (v) is given by \(v = \frac{Q}{A} \) where A is the cross-sectional area of the hose. Substituting for A using the formula for the area of a circle (\(A= 蟺 (d/2)^2 \)), where d is the diameter of the hose, yields the speed.
02

Determining the Speed of the Water With Nozzle

Now, calculate the diameter of the nozzle which is \(\frac{d}{3}\). Then calculate the area of the nozzle cross-section using the same formula for the area of a circle as before. Apply the equation of fluid continuity, \(A_1v_1=A_2v_2\), where \(A_1\) is the area of the hose, \(v_1\) is the speed at the hose, \(A_2\) is the area of the nozzle, and \(v_2\) is the speed at the nozzle, to find the speed of the water leaving the nozzle.

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Key Concepts

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

Volume Flow Rate
Volume flow rate is a fundamental concept in fluid dynamics. It describes the quantity of fluid moving through a given area per unit time. Imagine water rushing through a garden hose. Volume flow rate helps us to determine how much water passes a certain point in the hose in a specific time period.

To calculate the volume flow rate (\(Q\)), use the formula:
  • \(Q = \frac{\text{Volume}}{\text{Time}}\)
In the example of water filling a bucket, the volume of water is \(25.0\) liters, which is converted to cubic meters for calculation. And the time taken is \(1.50\) minutes, converted into seconds for uniformity in units.

By understanding volume flow rate, we can determine how quickly fluids are being transported, which is crucial in designing systems that rely on efficient fluid movement.
Continuity Equation
The continuity equation is a principle used in fluid dynamics to understand how fluid flows through varying cross-sections. It's based on the conservation of mass, meaning that the mass of fluid entering a system equals the mass leaving it. This is especially useful when a fluid passes through pipes of different sizes.

The equation itself is expressed as:
  • \(A_1v_1 = A_2v_2\)
Where \(A_1\) and \(v_1\) are the cross-sectional area and fluid velocity at one point, and \(A_2\) and \(v_2\) are those at another point.

In our garden hose example, when a nozzle is attached, the nozzle's reduced diameter changes the cross-sectional area. The continuity equation allows us to calculate the new speed of water exiting the nozzle based on the known speed and area of water before the nozzle was attached.
Cross-Sectional Area
Cross-sectional area is crucial for understanding how fluid flows through a conduit like a hose or a pipe. It refers to the surface area of the opening through which the fluid moves. In a cylindrical object, such as a hose or pipe, this area is shaped like a circle.

To find the cross-sectional area (\(A\)), for objects with circular openings, we use:
  • \(A = \pi \left( \frac{d}{2} \right)^2\)
Where \(d\) is the diameter of the circle. Knowing the cross-sectional area helps find out the speed of the fluid by applying it in the formula for speed: \(v = \frac{Q}{A}\).

Once the nozzle with reduced diameter is attached, the equation must be used again to determine how the reduction affects the speed of the water exiting the hose.

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Most popular questions from this chapter

A wooden block of volume \(5.24 \times 10^{-4} \mathrm{~m}^{3}\) floats in water, and a small steel object of mass \(m\) is placed on top of the block. When \(m=0.310 \mathrm{~kg}\), the system is in equilibrium, and the top of the wooden block is at the level of the water. (a) What is the density of the wood? (b) What happens to the block when the steel object is replaced by a second steel object with a mass less than \(0.310 \mathrm{~kg}\) ? What happens to the block when the steel object is replaced by yet another steel object with a mass greater than \(0.310 \mathrm{~kg}\) ?

To lift a wire ring of radius \(1.75 \mathrm{~cm}\) from the surface of a container of blood plasma, a vertical force of \(1.61 \times 10^{-2} \mathrm{~N}\) greater than the weight of the ring is required. Calculate the surface tension of blood plasma from this information.

A hypodermic needle is \(3.0 \mathrm{~cm}\) in length and \(0.30 \mathrm{~mm}\) in diameter. What pressure difference between the input and output of the needle is required so that the flow rate of water through it will be \(1 \mathrm{~g} / \mathrm{s}\) ? (Use \(1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\) as the viscosity of water.)

Old Faithful geyser in Yellowstone Park erupts at approximately 1-hour intervals, and the height of the fountain reaches \(40.0 \mathrm{~m}\) (Fig. P9.57). (a) Consider the rising stream as a series of separate drops. Analyze the free-fall motion of one of the drops to determine the speed at which the water leaves the ground. (b) Treat the rising stream as an ideal fluid in streamline flow. Use Bernoulli's equation to determine the speed of the water as it leaves ground level. (c) What is the pressure (above atmospheric pressure) in the heated underground chamber \(175 \mathrm{~m}\) below the vent? You may assume the chamber is large compared with the geyser vent.

Suppose a distant world with surface gravity of \(7.44 \mathrm{~m} / \mathrm{s}^{2}\) has an atmospheric pressure of \(8.04 \times 10^{4} \mathrm{~Pa}\) at the surface. (a) What force is exerted by the atmosphere on a disk-shaped region \(2.00 \mathrm{~m}\) in radius at the surface of a methane ocean? (b) What is the weight of a \(10.0-\mathrm{m}\) deep cylindrical column of methane with radius \(2.00 \mathrm{~m}\) ? (c) Calculate the pressure at a depth of \(10.0 \mathrm{~m}\) in the methane ocean. Note: The density of liquid methane is \(415 \mathrm{~kg} / \mathrm{m}^{3}\).
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