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Magazine Chapter 12: Problem 41
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 flows out at approximately 28.70 m/s.
Step by step solution
01
Identify Given Values
The problem provides the height of the seawater in the tank as 11.0 meters. Additionally, it gives the gauge pressure of the air above the water as 3.00 atm.
02
Convert Gauge Pressure to Pascals
Since gauge pressure is given in atmospheres but we need it in pascals for calculations, convert 3.00 atm to pascals using the conversion 1 atm = 101325 Pa. Thus, the gauge pressure in pascals is \(3.00 \times 101325 = 303975\, \text{Pa}.\)
03
Apply Bernoulli's Equation
Bernoulli's equation relates the pressure, kinetic energy per unit volume, and potential energy per unit volume for an incompressible, non-viscous fluid in steady flow. The equation is given by: \[ P_1 + \frac{1}{2}\rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho gh_2 \] where \(P\) is pressure, \(\rho\) is the density of water (approximately 1000 kg/m鲁), \(v\) is velocity, \(g\) is acceleration due to gravity (9.81 m/s虏), and \(h\) is height.
04
Assume Conditions at Top and Bottom of Tank
At the top of the tank (where the air compresses water), the velocity \(v_1\) is approximately zero because the water is not moving there. At the bottom, just outside the hole, we want to find the velocity \(v_2\). The pressure at the top is the gauge pressure \(P_1 = 303975\, \text{Pa}\), and at the bottom outside the hole the pressure is atmospheric pressure \(P_2 = 0\) (since it's gauge pressure).
05
Simplify Bernoulli's Equation for the tank
Set \(P_1 = 303975\, \text{Pa}\), \(v_1 = 0\), \(h_1 = 11.0\, \text{m}\), \(P_2 = 0\), and \(h_2 = 0\) (the bottom of the tank, just outside the hole). Substitute these into Bernoulli's equation: \[ 303975 + \rho g \times 11 = \frac{1}{2}\rho v_2^2 \] where \(\rho = 1000\, \text{kg/m}^3\).
06
Solve for Water Velocity
Rearrange the simplified Bernoulli's equation to solve for \(v_2\): \[ \frac{1}{2} \cdot 1000 \cdot v_2^2 = 303975 + 1000 \times 9.81 \times 11 \] Calculate \(1000 \times 9.81 \times 11\) to get \(107910\, \text{Pa}\). So, \[ \frac{1}{2} \cdot 1000 \cdot v_2^2 = 303975 + 107910 = 411885 \] Divide both sides by 500 to solve for \(v_2^2\): \[ v_2^2 = \frac{411885}{500} = 823.77 \] Finally, take the square root to find \(v_2\): \(v_2 = \sqrt{823.77} \approx 28.70\, \text{m/s}\).
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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 essential in understanding how fluids behave, especially when in motion. A fluid can be a liquid or a gas, and its movement involves complex properties like flow rate and velocity. In this context, we encounter key principles that govern how the fluid moves, such as pressure differences and energy conservation.
In fluid dynamics, fluids are often assumed to be incompressible and non-viscous, meaning they have constant density and ignore internal friction. This simplifies calculations significantly and allows us to use Bernoulli鈥檚 equation to solve problems.
In scenarios like a tank with water exiting through a hole, properties such as the fluid's height contribute to its potential energy, affecting the speed of flow. Understanding these concepts aids in predicting the behavior of many real-world fluid systems, such as pipelines, atmosphere, and even blood flow in the body.
In fluid dynamics, fluids are often assumed to be incompressible and non-viscous, meaning they have constant density and ignore internal friction. This simplifies calculations significantly and allows us to use Bernoulli鈥檚 equation to solve problems.
In scenarios like a tank with water exiting through a hole, properties such as the fluid's height contribute to its potential energy, affecting the speed of flow. Understanding these concepts aids in predicting the behavior of many real-world fluid systems, such as pipelines, atmosphere, and even blood flow in the body.
Gauge Pressure
Gauge pressure is the pressure measured relative to ambient atmospheric pressure. In scientific terms, it represents the pressure in a system above the atmospheric pressure. For example, when the problem states a gauge pressure of 3.00 atm, it implies the air pressure in the tank is 3.00 atm above the atmospheric pressure.
The conversion from atmosphere to pascals is crucial for accurate calculations, because pascals are the SI unit for pressure. In the given problem, the conversion from 3.00 atm to pascals is necessary since Bernoulli鈥檚 equation uses pascal units for consistent calculations.
This pressure affects how fluids behave, particularly in closed systems where pressure can significantly influence the fluid's speed as it exits the system. Correctly understanding and converting gauge pressure is a foundational skill in fluid dynamics that allows for more precise predictions in engineering and scientific calculations.
The conversion from atmosphere to pascals is crucial for accurate calculations, because pascals are the SI unit for pressure. In the given problem, the conversion from 3.00 atm to pascals is necessary since Bernoulli鈥檚 equation uses pascal units for consistent calculations.
This pressure affects how fluids behave, particularly in closed systems where pressure can significantly influence the fluid's speed as it exits the system. Correctly understanding and converting gauge pressure is a foundational skill in fluid dynamics that allows for more precise predictions in engineering and scientific calculations.
Kinetic Energy in Fluids
Kinetic energy in fluids refers to the energy possessed by a fluid due to its motion. This is one of the forms of energy included in Bernoulli's equation, which describes the principle of energy conservation in fluid flow. According to this principle, as potential energy due to gravity decreases when a fluid flows downwards, kinetic energy must increase if no external work is done.
In detail, the kinetic energy per unit volume for a fluid can be expressed as \( \frac{1}{2} \rho v^2 \). Here, \( \rho \) represents fluid density, and \( v \) is fluid velocity. As seen in the exercise solution, the fluid's velocity at the exit point of a tank can be determined by applying Bernoulli's equation, illustrating the interplay between energy forms.
Understanding kinetic energy in the context of fluids allows us to predict speeds at different points in a fluid system. This plays a critical role in designing systems like water networks or aerodynamics, where predicting fluid behavior is essential.
In detail, the kinetic energy per unit volume for a fluid can be expressed as \( \frac{1}{2} \rho v^2 \). Here, \( \rho \) represents fluid density, and \( v \) is fluid velocity. As seen in the exercise solution, the fluid's velocity at the exit point of a tank can be determined by applying Bernoulli's equation, illustrating the interplay between energy forms.
Understanding kinetic energy in the context of fluids allows us to predict speeds at different points in a fluid system. This plays a critical role in designing systems like water networks or aerodynamics, where predicting fluid behavior is essential.
