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Chapter 14: Problem 34

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 \(\mathrm{m} / \mathrm{s}^{3}\) , a How fast will it shoot out of a hole 4.50 \(\mathrm{cm}\) in diameter? (b) At what speed will it shoot out if the diameter of the hole is three times as large?

Short Answer

Expert verified
a) 471.65 m/s; b) 52.38 m/s.

Step by step solution

01

Understand the Problem

We're given a volume flow rate of water, \( Q = 0.750 \ \text{m}^3/\text{s} \), and need to find the velocity of water shooting out of a hole. We'll use the formula relating flow rate to velocity, \( Q = A \cdot v \), where \( A \) is the cross-sectional area of the hole and \( v \) is the velocity.
02

Convert the Diameter into Radius

The diameter of the hole is given as 4.50 cm, which is equivalent to 0.045 m. The radius \( r \) is half of the diameter, so: \( r = \frac{0.045}{2} = 0.0225 \ \text{m} \).
03

Calculate the Area of the Hole

The area \( A \) of the circular hole is calculated using the formula for the area of a circle, \( A = \pi r^2 \). Substituting the radius: \( A = \pi (0.0225)^2 = 0.00159043 \ \text{m}^2 \).
04

Determine the Velocity for Part (a)

Using the relation \( Q = A \cdot v \), we can solve for velocity \( v \): \( v = \frac{Q}{A} = \frac{0.750}{0.00159043} \approx 471.65 \ \text{m/s} \).
05

Calculate New Diameter for Part (b)

If the diameter of the hole is three times as large, the new diameter is \( 3 \times 4.50 = 13.50 \ \text{cm} = 0.135 \ \text{m} \), and so the new radius is \( 0.135/2 = 0.0675 \ \text{m} \).
06

Calculate New Area for Part (b)

Using the new radius, calculate the new area: \( A = \pi (0.0675)^2 = 0.014314 \ \text{m}^2 \).
07

Determine the Velocity for Part (b)

Again, using \( v = \frac{Q}{A} \), substitute the new area: \( v = \frac{0.750}{0.014314} \approx 52.38 \ \text{m/s} \).

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Key Concepts

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

Volume Flow Rate
Volume flow rate is a key concept in fluid dynamics that describes how much volume of a fluid passes through a given point over a specified period of time. It is often symbolized by the letter \( Q \). In the context of our problem, the given volume flow rate \( Q \) is 0.750 \( \, \text{m}^3 / \text{s} \), meaning that 0.750 cubic meters of water flow into the fountain every second.

Understanding volume flow rate is crucial for applications involving fluid transportation, such as in pipes, fountains, and irrigation systems. The formula that expresses this relationship is:
  • \( Q = A \cdot v \)
Here, \( A \) represents the cross-sectional area through which the fluid moves, and \( v \) is the fluid velocity. This formula highlights how the flow rate is affected directly by both the area of the passage and the velocity at which the fluid travels.
Cross-Sectional Area Calculation
Cross-sectional area is important for determining how a fluid will flow through openings like pipes and holes. For any circular opening, such as in this exercise, the cross-sectional area \( A \) can be calculated using the formula for the area of a circle, which is:
  • \( A = \pi r^2 \)
In our problem, we start by converting the given diameter of 4.50 cm into meters by dividing by 100 to get 0.045 meters. The radius \( r \) is half of the diameter, so \( r = 0.0225 \) m.

Plugging this radius into the area formula, we achieve:
  • \( A = \pi (0.0225)^2 \approx 0.00159043 \, \text{m}^2 \)
This calculated cross-sectional area is then utilized in determining the velocity of the fluid.
Velocity Calculation
Velocity calculation requires combining our knowledge of volume flow rate and cross-sectional area. According to the formula \( Q = A \cdot v \), where \( Q \) is the volume flow rate, \( A \) is the cross-sectional area, and \( v \) is velocity, we can solve for velocity:
  • \( v = \frac{Q}{A} \)

In the initial setup (part a of the problem), using the calculated cross-sectional area \( A = 0.00159043 \, \text{m}^2 \) and the given flow rate \( Q = 0.750 \, \text{m}^3/\text{s} \), the velocity becomes:
  • \( v = \frac{0.750}{0.00159043} \approx 471.65 \, \text{m/s} \)
For part b, when the hole's diameter is tripled, we recalculate the radius and area, concluding with a new velocity. Regardless of changes in size, the fundamental relationship between flow rate, area, and velocity remains consistent in fluid dynamics.

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Most popular questions from this chapter

Black Smokers. Black smokers are hot volcanic vents that emit smoke deep in the ocean floor. Many of them teem with exotic creatures, and some biologists think that life on earth may have begun around such vents. The vents range in depth from about 1500 \(\mathrm{m}\) to 3200 \(\mathrm{m}\) below the surface. What is the gauge pressure at a \(3200-\mathrm{m}\) deep vent, assuming the density of water does not vary? Express your answer in pascals and atmospheres.

Hydraulic Lift II. The piston of a hydraulic automobile lift is 0.30 \(\mathrm{m}\) in diameter. What gauge pressure, in pascals, is required to lift a car with a mass of 1200 \(\mathrm{kg}\) ? Also express this pressure in atmospheres.

In intravenous feeding, a needle is inserted in a vein in the patient's arm and a tube leads from the needle to a reservoir of fluid (density 1050 \(\mathrm{kg} / \mathrm{m}^{3} )\) located at height \(h\) above the arm. The top of the reservoir is open to the air. If the gauge pressure inside the vein is 5980 \(\mathrm{Pa}\) , what is the minimum value of \(h\) that allows fluid to enter the vein? Assume the needle diameter is large enough that you can ignore the viscosity (see Section 14.6 ) of the fluid.

The Great Molasses Flood. On the afternoon of January \(15,1919,\) an unusually warm day in Boston, a \(27.4-\mathrm{m}\) -high 27.4 - diameter cylindrical metal tank used for storing molasses ruptured. Molasses flooded into the streets in a 9 -m-deep stream, killing pedestrians and horses, and knocking down buildings. The molasses had a density of \(1600 \mathrm{kg} / \mathrm{m}^{3} .\) If the tank was full before the accident, what was the total outward force the molasses exerted on its sides? (Hint: Consider the outward force on a circular ring of the tank wall of width \(d y\) and at a depth \(y\) below the surface. Integrate to find the total outward force. Assume that before the tank ruptured, the pressure at the surface of the molasses was equal to the air pressure outside the tank.)

A tall cylinder with a cross-sectional area 12.0 \(\mathrm{cm}^{2}\) is partially filled with mercury; the surface of the mercury is 5.00 \(\mathrm{cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?
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