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Chapter 14: Problem 32

Use Bernoulli's Equation to derive Torricelli's Theorem. Assume very large open tank filled with a nonviscous liquid. [Hint: The fluid at the top can be considered to be at rest.]

Short Answer

Expert verified
Torricelli's Theorem states that the efflux speed \( v = \sqrt{2gh} \).

Step by step solution

01

Understand Bernoulli's Equation

Bernoulli's Equation states that for an incompressible, non-viscous fluid flowing in a streamline, the sum of pressure energy, kinetic energy, and gravitational potential energy per unit volume is constant. The equation is given by \( P + \frac{1}{2}\rho v^2 + \rho gh = \, \text{const} \) where \( P \) is pressure, \( \rho \) is the fluid density, \( v \) is the flow velocity, and \( h \) is the height above a reference point.
02

Apply Bernoulli's Equation to the Tank

Consider a large open tank with a small opening at the bottom. Let point 1 be at the surface of the fluid and point 2 be at the opening. At the surface (point 1), the velocity \( v_1 \) is negligible (considered to be at rest), and the pressure is atmospheric. At the opening (point 2), the pressure is also atmospheric as it is exposed to air. Thus, we can cancel the pressure terms when applying Bernoulli's Equation between points 1 and 2.
03

Write the Equation Using Known Values

Using Bernoulli's principle and setting atmospheric pressures equal, the equation at the tank's surface (point 1) and the hole (point 2) is: \( \frac{1}{2}\rho v_1^2 + \rho gh_1 = \frac{1}{2}\rho v_2^2 + \rho gh_2 \). Since \( v_1 \approx 0 \) and \( h_2 = 0 \) (hole at the bottom), the equation simplifies to \( \rho gh_1 = \frac{1}{2}\rho v_2^2 \).
04

Solve for the Flow Velocity

Cancel out \( \rho \) from both sides to get \( gh_1 = \frac{1}{2}v_2^2 \). Solve for \( v_2 \) to find the velocity of the fluid flowing out of the opening. Rearranging gives \( v_2 = \sqrt{2gh_1} \). This is known as Torricelli's Theorem, which states that the speed of efflux of a fluid under gravity through an opening is equivalent to that of a body freely falling from the same height.

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Key Concepts

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

Torricelli's Theorem
Torricelli's Theorem is a vital concept in fluid dynamics, providing insight into how a fluid behaves when it flows out of an opening under the force of gravity. It is closely related to Bernoulli鈥檚 Equation, which describes the conservation of energy in fluid flow. Torricelli's Theorem essentially states that
  • the velocity of fluid efflux through an opening is equivalent to the velocity acquired by an object freely falling from the same vertical height.
To arrive at Torricelli's Theorem, Bernoulli鈥檚 Equation is often applied. When considering a large open tank with a small hole at its bottom, the fluid at the top surface is generally at rest, allowing for simplifications. The theorem is expressed mathematically as: \[ v = \sqrt{2gh} \] Here, \( v \) is the flow velocity, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid column. This theorem assumes that the fluid is incompressible and non-viscous, making it ideal for theoretical calculations.
Fluid Dynamics
Fluid dynamics is the branch of physics concerned with the behavior of fluids in motion. It encompasses the study of how liquids and gases move and the forces acting on them.
  • Key principles such as Bernoulli's Equation play a crucial role in understanding fluid flow.
  • It involves dissecting complex flow patterns into manageable laws and equations.
One essential aspect of fluid dynamics is the prediction of velocity, pressure, and flow patterns under various conditions. Students often consult concepts like Torricelli鈥檚 Theorem within fluid dynamics to predict how fluids exit containers or navigate through complex systems.
Nonviscous Fluid Flow
Nonviscous fluid flow refers to an idealized version of fluid movement where fluid viscosity, or internal friction, is ignored. In reality, all fluids possess some viscosity. However, in many theoretical contexts, assuming the fluid is nonviscous greatly simplifies the analysis and allows the application of specific equations like Bernoulli鈥檚 Equation.
  • In nonviscous fluid flow, there is no energy loss due to internal friction.
  • This assumption makes it easier to apply Bernoulli's Equation effectively.
In such models, fluid flows smoothly and predictably, which helps lay the groundwork for understanding more complex, real-world scenarios. It is crucial for introductory physics and engineering calculations.
Gravitational Potential Energy
Gravitational Potential Energy (GPE) is the energy an object possesses due to its position in a gravitational field. In the context of fluid dynamics and Bernoulli鈥檚 Equation, GPE plays a significant role in determining the energy state of the fluid.
  • Gravitational potential energy is closely tied to the height of the fluid above a reference point.
  • It contributes significantly to the fluid's pressure and velocity as expressed in Bernoulli鈥檚 Equation.
The equation used is typically \( \rho gh \), where \( \rho \) is fluid density, \( g \) is gravitational acceleration, and \( h \) is the height of the fluid. This term is crucial in understanding how potential energy relates to kinetic energy and pressure energy, laying the foundation for critical concepts like Torricelli's Theorem.

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Most popular questions from this chapter

A pump lifts water at the rate of \(9.0\) liters/s from a lake through a 5.0-cm-i.d. pipe and discharges it into the air at a point \(16 \mathrm{~m}\) above the level of the water in the lake. What are the theoretical (a) velocity of the water at the point of discharge and \((b)\) power delivered by the pump.

A wind tunnel is to be used with a 20-cm-high model car to approximately reproduce the situation in which a \(550-\mathrm{cm}\) -high car is moving at \(15 \mathrm{~m} / \mathrm{s}\). What should be the wind speed in the tunnel? Is the flow likely to be turbulent? We want the Reynolds number \(N_{R}\) to be the same in both cases, so that the situations will be similar. That is, $$ N_{R}=\left(\frac{\rho v D}{\eta}\right)_{\text {tunnel }}=\left(\frac{\rho v D}{\eta}\right)_{\text {air }} $$ Both \(\rho\) and \(\eta\) are the same in the two cases, hence, $$ v, D_{i}=v_{a} D_{a} \text { from which } v_{t}=v_{a} \frac{D_{a}}{D_{i}}=(15 \mathrm{~m} / \mathrm{s})(550 / 20)=0.41 \mathrm{~km} / \mathrm{s} $$ To investigate turbulence, evaluate \(N_{R}\) using \(\rho=1.29 \mathrm{~kg} / \mathrm{m}^{3}\) and \(\eta\) \(=1.8 \times 10^{-5} \mathrm{~Pa} \cdot \mathrm{s}\) for air. Consequently \(N_{R}=5.9 \times 10^{6}\), a value far in excess of that required for turbulent flow. The flow will certainly be turbulent.

A venturi meter equipped with a differential mercury manometer is shown in. At the inlet, point-1, the diameter is \(12 \mathrm{~cm}\), while at the throat, point-2, the diameter is \(6.0 \mathrm{~cm}\). What is the flow \(J\) of water through the meter if the mercury manometer reading is 22 \(\mathrm{cm}\) ? The density of mercury is \(13.6 \mathrm{~g} / \mathrm{cm}^{3}\). From the manometer reading (remembering that \(1 \mathrm{~g} / \mathrm{cm}^{3}=1000\) \(\mathrm{kg} / \mathrm{m}^{3}\) ): \(P_{1}-P_{2}=\rho g h=\left(13600 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.22 \mathrm{~m})=2.93 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}\) Since \(J=v_{1} A_{1}=v_{2} A_{2}\), we have \(v_{1}=J / A_{1}\) and \(v_{2}=J / A_{2} .\) Using Bernoulli's Equation with \(h_{1}-h_{2}=0\) gives $$ \begin{array}{c} \left(P_{1}-P_{2}\right)+\frac{1}{2} \rho\left(v_{1}^{2}-v_{2}^{2}\right)=0 \\\ 2.93 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}+\frac{1}{2}\left(1000 \mathrm{~kg} / \mathrm{m}^{3}\right)\left(\frac{1}{A_{1}^{2}}-\frac{1}{A_{2}^{2}}\right) J^{2}=0 \end{array} $$ where \(A_{1}=\pi_{1}^{2}=\pi(0.060)^{2} \mathrm{~m}^{2}=0.01131 \mathrm{~m}^{2} \quad\) and \(\quad A_{2}=\pi / \frac{2}{2}=\pi(0.030)^{2} \mathrm{~m}^{2}=0.0028 \mathrm{~m}^{2}\) Substitution then gives \(J=0.022 \mathrm{~m}^{3} / \mathrm{s}\).

What volume of water will escape per minute from an open-top tank through an opening \(3.0 \mathrm{~cm}\) in diameter that is \(5.0 \mathrm{~m}\) below the water level in the tank?

Find the maximum amount of water that can flow through a 3.0cm-i.d. pipe per minute without turbulence. Take the maximum Reynolds number for nonturbulent flow to be 2000 . For water at \(20^{\circ} \mathrm{C}, \eta=1.0 \times 10^{-3} \mathrm{~Pa} \cdot \mathrm{s}\)
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