/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 A sealed tank containing seawate... [FREE SOLUTION] | 91Ó°ÊÓ

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A sealed tank containing seawater to a height of 11.0 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 is moving at approximately 28.45 m/s.

Step by step solution

01

Identifying Known Values

We are given that the water height in the tank is 11.0 m, and the gauge pressure of the air is 3.00 atm. To solve the problem, it's essential to convert this pressure into pascals: 1 atm = 101325 Pa, so 3 atm = 303975 Pa. We'll also assume the density of seawater is roughly 1025 kg/m³ and the gravitational acceleration is 9.81 m/s².
02

Applying Bernoulli's Equation

Bernoulli's equation states that for an incompressible, non-viscous fluid in a streamline flow: \[ P_1 + \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2 \] where: - \(P_1\) is the pressure at the surface (i.e., 303975 Pa), - \(v_1\) is the velocity at the surface (0 since the surface has negligible motion), - \(g\) is the acceleration due to gravity (9.81 m/s²), - \(h_1\) is the height of water in the tank (11.0 m), - \(P_2\), \(v_2\) are the pressure and velocity at the hole, respectively, and - \(h_2\) is 0 at the hole level.
03

Simplifying Assumptions

Given that the hole is open to the atmosphere, the pressure at the point where water exits can be assumed to be atmospheric, so \(P_2 = 0\). We can also set \(h_2 = 0\) at the nozzle's exit because we're measuring height from this point.
04

Calculating the Exit Velocity

Substitute the known values: \[ 303975 + 0 + 1025 \cdot 9.81 \cdot 11 = 0 + \frac{1}{2} \cdot 1025 \cdot v_2^2 + 0 \]Simplifying gives:\[ 303975 + 110564.25 = \frac{1}{2} \cdot 1025 \cdot v_2^2 \]\[ 414539.25 = \frac{1}{2} \cdot 1025 \cdot v_2^2 \]\[ v_2^2 = \frac{414539.25 \cdot 2}{1025} \]\[ v_2 = \sqrt{\frac{829078.5}{1025}} \approx \sqrt{808.75} \approx 28.45 \text{ m/s} \]
05

Conclusion

The water flows out of the hole with a velocity of approximately 28.45 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 the study of fluids (liquids and gases) in motion. It addresses how forces affect fluid flow and examines how fluids interact with their surroundings. In the world of physics, we often deal with simplifying assumptions to better understand fluid behavior.
Fluid flow can be characterized in various ways:
  • Laminar flow: Smooth and orderly fluid motion, typically in parallel layers without disruption.
  • Turbulent flow: Chaotic and irregular motion, often with swirling vortices and eddies.
  • Streamline flow: An idealization used in Bernoulli’s equation, assumes fluid elements follow a smooth path, representing the average flow direction.
In solving problems involving fluid dynamics, understanding the forces at play, such as pressure differences and gravitational force, is crucial. This knowledge enables us to apply mathematical models, like Bernoulli’s equation, to predict behaviors such as velocity and pressure changes.
Gauge Pressure
Gauge pressure measures the fluid pressure relative to the ambient atmospheric pressure. It differs from absolute pressure, which includes atmospheric pressure in its measurement. In many practical situations, like car tires and experimental setups, gauge pressure is more useful because it provides a measure of pressure above or below the surrounding air pressure.
Crucially, Bernoulli's equation considers pressure as a factor influencing fluid flow. In our original exercise, the tank's contained air exerts a gauge pressure of 3 atm on the water. To use it effectively in fluid equations, it’s converted to pascals ( P_1 = 303975 Pa). Understanding how to work with pressure measurements is vital in experiments and calculations.
Velocity Calculation
Velocity calculation in fluid dynamics often involves Bernoulli's equation, which relates various properties along a streamline. It states that the sum of pressure energy, kinetic energy per unit volume, and potential energy per unit volume of a fluid is constant. In practical problems, this is used to find unknown values such as velocity. The original solution shows how to calculate the velocity of water exiting a tank:
  • Start by identifying changes in height, pressure, and velocities at different points.
  • Simplify using assumptions based on the setup—with open holes, assume atmospheric pressure externally.
  • Apply Bernoulli’s formula and solve for the unknown, substituting appropriate values.
In our case, this calculation revealed an exit velocity of approximately 28.45 m/s for the water. Mastering such calculations is fundamental for predicting fluid movements under various conditions.
Gravitational Acceleration
Gravitational acceleration is a pivotal factor in fluid motion, particularly in vertical setups. It is the acceleration with which objects—and fluids—are pulled towards the center of the Earth. It's usually denoted by the symbol \(g\), with a standardized average value of \(9.81\, \text{m/s}^2\) on the Earth's surface.
In fluid dynamics, gravitational acceleration influences the potential energy term in Bernoulli’s equation, capturing the effect of height differences. For example, in our exercise, the water's height difference adds to the pressure, contributing to fluid flow speed from the tank.
When accounting for gravity, it’s crucial to remember its role in fluid elevation and pressure changes, often transforming potential energy into kinetic energy. This principle not only supports calculations but also enhances understanding of natural and engineered water flow systems.

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Most popular questions from this chapter

A cubical block of wood 0.100 \(\mathrm{m}\) on a side and with a density of 550 \(\mathrm{kg} / \mathrm{m}^{3}\) floats in a jar of water. Oil with a density of 750 \(\mathrm{kg} / \mathrm{m}^{3}\) is poured on the water until the top of the oil layer is 0.035 \(\mathrm{m}\) below the top of the block. (a) How deep is the oil layer? (b) What is the gauge pressure at the block's lower face?

A small circular hole 6.00 mm in diameter is cut in the side of a large water tank, 14.0 m below the water level in the tank. The top of the tank is open to the air. Find (a) the speed of efflux of the water and (b) the volume discharged per second.

Untethered helium balloons, floating in a car that has all the windows rolled up and outside air vents closed, move in the direction of the car's acceleration, but loose balloons filled with air move in the opposite direction. To show why, consider only the horizontal forces acting on the balloons. Let \(a\) be the magnitude of the car's forward acceleration. Consider a horizontal tube of air with a cross-sectional area \(A\) that extends from the windshield, where \(x=0\) and \(p=p_{0},\) back along the \(x\) -axis. Now consider a volume element of thickness \(d x\) in this tube. The pressure on its front surface is \(p\) and the pressure on its rear surface is \(p+d p .\) Assume the air has a constant density \(\rho .\) (a) Apply Newton's second law to the volume element to show that \(d p=\rho a d x .\) (b) Integrate the result of part (a) to find the pressure at the front surface in terms of \(a\) and \(x\) . (c) To show that considering \(\rho\) constant is reasonable, calculate the pressure difference in atm for a distance as long as 2.5 \(\mathrm{m}\) and a large acceleration of 5.0 \(\mathrm{m} / \mathrm{s}^{2}\) . (d) Show that the net horizontal force on a balloon of volume \(V\) is oVa. (e) For negligible friction forces, show that the acceleration of the balloon (average density \(\rho_{\text { bal }}\) is \(\left(\rho / \rho_{\text { bal }}\right) a,\) so that the acceleration relative to the car is \(a_{\mathrm{rcl}}=\left[\left(\rho / \rho_{\mathrm{bal}}\right)-1\right] a\) (f) Use the expression for \(a_{\mathrm{rcl}}\) in part (e) to explain the movement of the balloons.

A slab of ice floats on a freshwater lake. What minimum volume must the slab have for a 45.0 -kg woman to be able to stand on it without getting her feet wet?

An electrical short cuts off all power to a submersible diving vehicle when it is 30 \(\mathrm{m}\) below the surface of the ocean. The crew must push out a hatch of area 0.75 \(\mathrm{m}^{2}\) and weight 300 \(\mathrm{N}\) on the bottom to escape. If the pressure inside is 1.0 atm, what downward force must the crew exert on the hatch to open it?

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