Physics
Application of Bernoulli's Equation
The application of Bernoulli's equation involves using the principle of conservation of energy to analyze the behavior of fluids in various situations. It is commonly used to study the flow of fluids through pipes, in airfoil design for aircraft, and in understanding the dynamics of blood flow in the human body. By applying Bernoulli's equation, engineers and scientists can make predictions about fluid behavior and design efficient systems.
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- eBook - ePub
- B.H Brown, R.H Smallwood, D.C. Barber, P.V Lawford, D.R Hose(Authors)
- 2017(Publication Date)
- CRC Press(Publisher)
Euler’s equation. In two and three dimensions Euler’s equation can be used to investigate the pressure drag associated with external flow over objects. Analytical solutions are available for some simple geometries, and we can write relatively straightforward computational algorithms to study more general shapes. We might use such a program to look at problems such as the design of a cycle helmet.Viscous forces within a fluid act to dissipate energy, and thus in a viscous fluid-dynamic system mechanical energy is not conserved. For inviscid flows the balance between potential and kinetic energy yields Bernoulli’s equation. We can usefully apply Bernoulli’s equation, in conjunction with the continuity equation, to get first estimates of the pressure changes associated with area changes in vessels. We can also use it to investigate problems such as the speed of emptying of pressurized vessels. Bernoulli’s equation is in fact a form of Euler’s equation, integrated along a streamline to make it appear one dimensional. In biofluid mechanics we are concerned mostly with flow in tubes and we might use the one-dimensional Bernoulli equation to obtain initial estimates of the pressure gradients in branching tube systems. We sometimes modify the Bernoulli equation to include a factor to account for viscous losses.Laminar versus turbulent
In the Navier–Stokes equation the acceleration of the fluid is related to the variations of the pressure and to the viscosity of the fluid. There are thus two contributions to the pressure gradients: one from the accelerations (inertial terms) and one from the viscous stresses (viscous terms). In steady flow the velocity at a point in space is fixed, and the only acceleration terms are associated with convection. The relative magnitudes of the inertial and viscous terms can be related to a dimensionless parameter called the Reynolds number, - Amithirigala Widhanelage Jayawardena(Author)
- 2021(Publication Date)
- CRC Press(Publisher)
Chapter 13 Applications of basic fluid flow equations13.1 Introduction
There are many problems in nature that involve principles of fluid mechanics. Solutions to such problems are obtained by applying the governing equations of fluid flow or their approximations. The three basic equations involved are the continuity equation, the momentum equation and the energy equation. Assumptions and/or simplifications necessary to apply the basic equations to certain types of practical problems include incompressibility, steady state condition, ideal fluid, and sometimes reducing the dimensionality of the problem. In this chapter, how the governing equations of fluid flow are applied to some typical practical problems in nature are highlighted.13.2 Kinetic energy correction factor
When flows in open channels or pipes are considered, it is generally assumed to be one-dimensional with an average velocity at each section. The kinetic energy isper unit weight, but is different fromV 22 g. Therefore, a correction factor, α , is used for∫cross sectionv 22 gso thatV 2/ 2 gis the average kinetic energy for unit weight, passing the section.αV 22 gReferring to Figure 13.1 , the kinetic energy passingδ Aper unit time isFigure 13.1 Velocity profile in pipe flow.∫v 22(ρ v d A)where ρvdA is the mass.Therefore,αρ V A =V 22d A )∫ A( ρ vv 22α =(13.1)1 Ad A∫ A(3v V)z +p+ αρ g= C o n s t a n tV 22 gFor laminar flow in pipesα = 2; for turbulent flows in pipes, α varies from 1.01 to 1.10 and is usually ignored.All terms in the Bernoulli equation are available energy, and, for real fluids flowing through a system, the available energy decreases in the downstream direction. It is the energy that it is available for doing work, as in hydropower generation.- eBook - ePub
- Colin Salter(Author)
- 2021(Publication Date)
- Pavilion(Publisher)
Euler and Bernoulli worked together on the relationship between blood pressure and the speed of blood flow. Daniel’s particular interest was the theory of conservation of energy – he noted the principle of a change from kinetic to potential energy in a moving body which accompanied a gain in height, and applied it to fluids, in which kinetic energy was exchanged for pressure.Daniel Bernoulli published his magnum opus, Hydrodynamica , in 1738. The name was a word of his own invention and was adopted for the new field of engineering, hydrodynamics, which the book opened up. In it he took the conservation of energy as his starting point and considered the efficiency of hydraulic machines. It contains the first exposition of the kinetic theory of gases.Daniel Bernoulli was actually born in the Netherlands, then under Spanish rule. The talented family of mathematicians moved to Basel in Switzerland to escape Spanish persecution of Protestants.Above all it includes Bernoulli’s Principle, the rule that an increase in the speed of a fluid goes hand in hand with a decrease in pressure or a decrease in the fluid’s potential energy. Leonhard Euler devised the equation that goes with it. Besides its application to hydrodynamic engineering, Bernoulli’s Principle is central to aerodynamics. It is the reason that aeroplane wings have their distinctive cross-section, which encourages lift and flight.Daniel’s father was so jealous of Hydrodynamica that he plagiarized it in his own 1739 book, Hydraulica , which he backdated to 1732 so that it appeared that he, Johann Bernoulli, had thought of Bernoulli’s Principle first. Johann harboured this resentment of his son’s success until the day he died.A pattern of air flow demonstrating the Bernoulli Principle, which states that the internal pressure of a gas is lowered the faster it travels. This principle gives an aeroplane wing lift.Bernoulli’s great work was shamelessly plagiarized by his father. - eBook - ePub
Introduction to Engineering Mechanics
A Continuum Approach, Second Edition
- Jenn Stroud Rossmann, Clive L. Dym, Lori Bassman(Authors)
- 2015(Publication Date)
- CRC Press(Publisher)
a along a streamline, may be integrated to yield the following equation:p ρ+ g z +1 2V 2= constant along a streamline ,(18.34) where we have assumed that gravity acts in the negative z-direction, and where V is the velocity in the s direction, simply the magnitude of the velocity vector since V is in the s direction. This equation is known as the Bernoulli equation, and it is true for steady flow of an incompressible fluid under inviscid conditions. For convenience, we write the equation together with its restrictions:p ρ+ g z +1 2V 2= constant• On a streamline• For steady flow• For incompressible fluid• If viscous effects neglectedMany problems can be solved using the Bernoulli equation, allowing us to dodge having to solve the full Euler or Navier–Stokes equations. It should not escape our notice that the Bernoulli equation, derived from ∑F = ma, looks like an energy conservation equation. This is even easier to see if we multiply through by the (assumed constant) density: Equation 18.34 becomesp + ρ g z +1 2ρV 2= constant ,(18.35) and we can think of pressure p as a measure of flow work, ρgz as a gravitational potential energy, andas a kinetic energy, all per unit volume of fluid. Daniel Bernoulli actually first arrived at Equation 18.34 by performing an energy balance, even though the concept of energy was still a bit fuzzy in 1738.1 2ρV 2One of the most useful applications of the Bernoulli equation is a device known as a Pitot* tube, and its cousin the Pitot-static tube, used to measure flow velocities. The tube (Figure 18.9a ) contains a column of air. When an oncoming fluid flow impinges on the nose of the Pitot tube, it displaces this air. As we know from hydrostatics, the displacement will be proportional to the pressure at the stagnation point on the Pitot tube nose. The stagnation point is at the divide between the flow that goes up-and-over and that which goes down-and-under (imagining a flow in the plane of the page for simplicity) and there the velocity must be zero. The difference between this “stagnation pressure” (where the fluid has speed V = 0) and the “static pressure” elsewhere in the flow (where the fluid has average speed V∞ - eBook - ePub
- William S. Janna(Author)
- 2020(Publication Date)
- CRC Press(Publisher)
We conclude that for both equations to be identical, any change in internal energy of the fluid must equal the amount of heat transferred. It can be seen that for an incompressible flow with no work, no heat transfer, and no changes in internal energy, the energy equation and the Bernoulli equation derived from the momentum equation become identical. Thus, under certain flow conditions, the energy and momentum equations reduce to the same expression. Hence, the Bernoulli equation is referred to as the mechanical energy equation. For many flow problems, only the continuity and Bernoulli equations are required for a description of the flow. Example 3.13 A water jet issues from a sink and falls vertically downward. The water flow rate is such that it will fill a 250 ml cup in 8 seconds. The faucet is 30 cm above the sink, and at the point of impact, the jet diameter is 0.3 cm. What is the jet diameter at the faucet exit? Solution: A jet exiting a faucet and impacting a flat surface; we locate section 1 at the faucet exit and section 2 at the sink. Continuity Q = A 1 V 1 = A 2 V 2 assuming one-dimensional, steady flow. The volume flow rate is Q = 0.251 Liters/s 8 s = 0.031 25 × 10 − 3 m 3 /s Substituting for flow rate and area gives the velocity at each section. as V 1 = Q A 1 = 4 Q π D 1 2 = 4 0.031 25 × 10 − 3 π D 1 2 or V 1 = 3.98 × 10 − 5 D 1 2 or V 2 = 4 0.031 25 × 10 − 3 π 0.003 2 = 4.42 m/s The Bernoulli equation applied to these two sections. is p 1 ρ + V 1 2 2 g c + g z 1 g c = p 2 ρ + V 2 2 2 g c + g z 2 g c Evaluating properties: p 1 = p 2 = p atm z 1 = 0.3 m z 2 = 0 The Bernoulli equation becomes, after. simplification, V 1 2 2 g + z 1 = V 2 2 2 g Substituting, 3.98 × 10 − 5 D 1 2 2 1 2 9.81 + 0.3 = 4.42 2 2 9.81 Solving, D 1 4 = 1.16 × 10 − 10 and D 1 = 3.28 × 10 − 3 m = 0.328 cm ¯ Example 3.14 Consider the flow of water through a venturi meter, as shown in Figure 3.24 - eBook - ePub
- Rose G Davies(Author)
- 2020(Publication Date)
- CRC Press(Publisher)
Bernoulli’s equation . The conditions required to obtain this equation can be summarized as follows: incompressible fluid particle flows along a streamline; the flow is in steady state; and consider no friction.Each term of the equations represents a type of energy per unit mass (2.10a), or per unit volume (2.10b) of the fluid along the streamline. The statement of the Bernoulli’s equation is that for a flow of an incompressible ideal fluid along its streamline the sum of the potential energy (ghρ ), kinetic energy (ρ () and the capability (p ) to produce work due to pressure at any point on the streamline remains constant – Bernoulli’s Theorem .v 2/ 2 )Take the fluid flow shown in Figure 2.1 as an example, there are two labeled sections: upstream section 1 and downstream section 2. p 1 , and p 2 are the pressure at these two sections, v 1 , and v 2 are the speed of the fluid, and h 1 , and h 2 are the height at these two sections respectively. According to Bernoulli’s Theorem, the sum of the energies mentioned above remains constant; therefore, we can obtain:gh 1ρ +p 1+ρ 2v 1 2= gh 2ρ +p 2+ρ 2v 2 2(2.11) From Equation (2.11) we learn that the speed of the fluid flow will decrease if the fluid flows up if the fluid pressure is the same at both sections; or when flow speed increases, the pressure will decrease if these two sections are at the same level.Each term of Bernoulli’s Equation (2.11) has the unit of pressure, [Pa]. We can use P as the constant in Equation (2.10):g h ρ + p +ρ 2v 2= P(2.12) ghρ – “Potential” pressure ; p – Static pressure ;– Dynamic pressure ;P – Total pressure. (Sometimes pρ 2v 2tis used)Divide each term of Equation (2.12) by gρ , assuming the fluid is incompressible, and it becomes:h +p+g ρ= Hv 22 g(2.13) h – (“Natural ” height) Head ;p– Static headg ρ - eBook - ePub
- Roger Legg(Author)
- 2017(Publication Date)
- Butterworth-Heinemann(Publisher)
Chapter 13Fluid Flow: General Principles
Abstract
In this chapter, the general principles of airflow in ducts are explained, the fluid being treated as incompressible. The relevant equations are given and the general characteristics of flow described. The methods of calculating pressure losses in straight ducts and in fittings are illustrated, together with the pressure distributions. The concept of resistance is explained, and the chapter concludes by giving some relevant methods for measuring flow rates. An understanding of these topics is important before going on in subsequent chapters to consider the design and sizing of ductwork systems, fan selection, and on-site balancing procedures.Keywords
Bernoulli equation; Conservation of energy; Measurement of flow rate; Moody chart; Orifice plate; Pitot-static tube; Pressure losses in fittings; Pressure losses in straight ducts; Resistance; Reynolds numberSymbolsA duct or pipe cross sectionA oorifice plate areab duct breadthC calibration coefficient for Pitot-static tubeD diameterD eequivalent diameter of rectangular ductd diameter of orifice or conical inletf friction factorK pressure loss coefficientK bbend pressure loss coefficientK eexpansion pressure loss coefficientK fstraight duct pressure loss coefficientk roughness coefficient of duct wallL duct lengthmass flow ratem .p atatmospheric pressurep sstatic pressure