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Chapter 14: Problem 39

A sealed tank containing seawater to a height of 11.0 \(\mathrm{m}\) also contains air above the water at a gauge pressure of 3.00 atm. Water flows out from the bottom through a small hole. How fast is this water moving?

Short Answer

Expert verified
The water flows out at approximately 28.73 m/s.

Step by step solution

01

Understand the Bernoulli's Equation

In order to solve this problem, we need to use Bernoulli's Equation which is given by: \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant} \]where \( P \) is the pressure, \( \rho \) is the density of the fluid, \( v \) is the fluid velocity, and \( h \) is the height of the fluid above the reference point.
02

Set up Bernoulli's Equation for the System

We need to look at two points: the surface of the water in the tank and the hole at the bottom. At the surface:- Gauge pressure = 3 atm = 3 x 101325 Pa- Velocity \( v_1 \) = 0 (neglecting any flow at the top)- Height \( h_1 \) = 11 m.At the hole:- Atmospheric pressure = 0 (since it is gauge pressure)- Velocity \( v_2 = v \) (this is what we are solving for)- Height \( h_2 = 0 \)
03

Express Bernoulli's Principle for both points

Applying Bernoulli's equation:\[ P_{ ext{air}} + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_{ ext{atm}} + \frac{1}{2} \rho v_2^2 + \rho gh_2 \]Solving for \( v_2 \), and simplifying the equation:\[ 3 \times 101325 + 0 + 1000 \times 9.81 \times 11 = 0 + \frac{1}{2} \times 1000 \times v_2^2 + 0 \]
04

Solve for the Velocity

Now solve for \( v_2 \):- First, combine and simplify the numbers on the left-hand side.- Then solve for \( v^2 \), which gives:\[ \frac{1}{2} \times 1000 \times v_2^2 = 3 \times 101325 + 1000 \times 9.81 \times 11 \]- Solve this equation to find \( v_2 \).- After calculating:\[ v_2 = \sqrt{ \frac{2 (3 \times 101325 + 1000 \times 9.81 \times 11)}{1000} } \]
05

Final Calculation

Calculate the expression to find the velocity of water flowing out. Plug in all the values and simplify:\[ v_2 \approx \sqrt{ \frac{2 \times ((3 \times 101325) + 1000 \times 9.81 \times 11)}{1000} } \]Calculate to find:\[ v_2 \approx 28.73 \text{ m/s} \]

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Key Concepts

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

Fluid Dynamics
Fluid dynamics is a fascinating branch of physics that deals with the movement of liquids and gases, otherwise known as fluids. It plays a crucial role in solving problems involving the flow of water, air, or any other fluid.
Bernoulli's Equation, a key principle in fluid dynamics, helps us understand how pressure, velocity, and height interact in a fluid flow system.
When analyzing a system, we often consider things like:
  • The speed of fluid flow, which can vary throughout the system.
  • The changes in pressure as the fluid moves.
  • The effects of gravity on the fluid, which can influence its flow direction and speed.
Gauge Pressure
Gauge pressure is the pressure measurement that excludes atmospheric pressure. It basically tells us how much the pressure in a system exceeds that of the surrounding environment.
In this problem, knowing that the gauge pressure in the tank is 3.00 atm helps us understand the force exerted by the air above the seawater.
Remember that the pressure unit, atm, can be converted into Pascals (Pa) for calculations in metric units, where 1 atm equals 101,325 Pa.
  • This conversion is pivotal for applying Bernoulli's Equation.
  • The gauge pressure aids in calculating the net force that drives fluid flow.
Without considering gauge pressure, we would underestimate how fast the water is flowing out of the tank.
Velocity Calculation
Calculating the velocity of water exiting the tank is at the heart of this problem. Using the Bernoulli's Equation, we're able to relate the various factors affecting fluid velocity.
Starting by identifying conditions at two critical points: at the top surface of the water, and where the water exits.
By setting:
  • Zero velocity at the top since fluid isn't moving there significantly.
  • The air pressure at this level as gauge pressure.
  • No significant height where water exits through the hole.
We can calculate the speed using the relationship:\[ v_2 = \sqrt{ \frac{2 (3 \times 101325 + 1000 \times 9.81 \times 11)}{1000} } \]
This equation helps predict the velocity (\(v_2\)) of the water flowing out at approximately 28.73 m/s.
Accurate calculations rely greatly on keeping track of units and understanding physical principles.
Hydrostatics
Hydrostatics concerns the study of fluids at rest, and it sets an important foundation for understanding fluid dynamics when those fluids start moving.
In this problem, hydrostatics helps explain the potential energy stored in the water due to its height.
The height of 11 meters contributes to the gravitational potential energy in the system. This energy converts into kinetic energy as water flows out.
  • This energy conversion impacts how swift the water moves.
  • The higher the water, the greater the potential energy.
By understanding the hydrostatic pressure, one can predict how it will contribute to the dynamic movement of fluids when conditions change, such as when the tank's bottom is opened.

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Most popular questions from this chapter

You purchase a rectangular piece of metal that has dimensions \(5.0 \times 15.0 \times 30.0 \mathrm{mm}\) and mass 0.0158 \(\mathrm{kg}\) . The seller tells you that the metal is gold. To check this, you compute the average density of the piece. What value do you get? Were you cheated?

A rock with mass \(m=3.00 \mathrm{kg}\) is suspended from the roof of an elevator by a light cord. The rock is totally immersed in a bucket of water that sits on the floor of the elevator, but the rock doesn't touch the bottom or sides of the bucket. (a) When the elevator is at rest, the tension in the cord is 21.0 \(\mathrm{N}\) . Calculate the volume of the rock. ( b) Derive an expression for the tension in the cord when the elevator is accelerating upward with an acceleration of magnitude a. Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) upward. (c) Derive an expression for the tension in the cord when the clevator is accelerating downwand with an acceleration of magnitude \(a\) . Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) downward. (d) What is the tension when the elevator is in free fall with a downward acceleration equal to \(g ?\)

You cast some metal of density \(\rho_{\mathrm{m}}\) in a mold, but you are worried that there might be cavities within the casting. You measure the weight of the casting to be \(w\) , and the buoyant force when it is completely surrounded by water to be \(B\) . (a) Show that \(V_{0}=\) \(B /\left(\rho_{\text { water }} g\right)-w /\left(\rho_{\text { m }} g\right)\) is the total volume of any enclosed cavities. (b) If your metal is copper, the casting's weight is 156 \(\mathrm{N}\) , and the buoyant force is 20 \(\mathrm{N}\) , what is the total volume of any enclosed cavities in your casting? What fraction is this of the total volume of the casting?

An incompressible fluid with density \(\rho\) is in a horizontal test tube of inner cross-sectional area \(A\) . The test tube spins in a horizontal circle in an ultracentrifuge at an angular speed \(\omega\) . Gravitational forces are negligible. Consider a volume element of the fluid of area A and thickness \(d r^{\prime}\) adistance \(r^{\prime}\) from the rotation axis. The pressure on its inner surface is \(p\) and on its outer surface is \(p+d p .(\text { a) Apply }\) Newton's second law to the volume element to show that \(d p=\rho \omega^{2} r^{\prime} d r^{\prime} .\) (b) If the surface of the fluid is at a radius \(r_{0}\) where the pressure is \(p_{0}\) , show that the pressure \(p\) at a distance \(r \geq r_{0}\) is \(p=p_{0}+\rho \omega^{2}\left(r^{2}-r_{0}^{2}\right) / 2 .(\mathrm{c})\) An object of volume \(V\) and density \(\rho_{o b}\) has its center of mass at a distance \(R_{\text { cmob }}\) from the axis. Show that the net horizontal force on the object is \(\rho V \omega^{2} R_{\mathrm{cm}},\) where \(R_{\mathrm{cm}}\) is the distance from the axis to the center of mass of the displaced fluid. (d) Explain why the object will move inward if \(\rho R_{\mathrm{cm}}>\rho_{\mathrm{cb}} R_{\mathrm{cmod}}\) and outward if \(\rho R_{\mathrm{cm}}<\rho_{\mathrm{ob}} R_{\mathrm{cmob}}(\mathrm{e})\) For small objects of uniform density, \(R_{\mathrm{cm}}=R_{\mathrm{cmob}}\) What happens to a mixture of small objects of this kind with different densities in an ultracentrifuge?

A hot-air balloon has a volume of 2200 \(\mathrm{m}^{3}\) . The balloon fabric (the envelope) weighs 900 \(\mathrm{N}\) . The basket with gear and full propane tanks weighs 1700 \(\mathrm{N}\) . If the balloon can barely lift an additional 3200 \(\mathrm{N}\) of passengers, breakfast, and champagne when the outside air density is \(1.23 \mathrm{kg} / \mathrm{m}^{3},\) what is the average density of the heated gases in the envelope?
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