Momentum Analysis of Flow Systems

5 Key excerpts on "Momentum Analysis of Flow Systems"

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  eBook - ePub

  Basic Mechanics with Engineering Applications

  • J. Jones, J. Burdess, J.N. Fawcett, J Jones, J Burdess, J Fawcett(Authors)
  • 2012(Publication Date)
  • Routledge

    (Publisher)

  7

  Momentum and mass transfer

  Basic Theory

  In Chapter 6 , Section 6.1 , we developed an expression for the linear momentum of a closed system of particles and showed that the rate of change of linear momentum of the system was equal to the sum of the external forces applied to the system, i.e. F = dp/dt. Since this system was assumed to be closed no mass could, by definition, enter or leave the system across its boundaries. As the system boundary moved the mass moved with it. Although this formulation is very useful for systems built from an assembly of rigid bodies it is not a convenient way of dealing with devices, such as pumps, compressors, turbines and jet engines, which require a flow of fluid into and out of the system.

  7.1   Variable mass and fluid flow systems

  We shall now show how the results of Section 6.1 can be extended to take into account the transfer of mass across the system boundary. Figure 7.1(a) shows a system of particles such as P

  i

  contained within a boundary S at time t. The particle P

  i

  is shown situated in the centre region of the system whilst the particle at the point Q on the boundary is just inside the boundary and represents a small mass which is just about to leave the system.

  FIG . 7.1

  Let us assume that this particle has mass δmQ and is moving with velocity which has components . It is assumed that the velocity of this particle is different from the velocity of the coincident point Q on the system boundary. Q on the boundary therefore has coordinates which at time t are the same as the coordinates xQ , yQ of the coincident particle. The asterisk will be used to denote the displacements and velocities of points on the boundary.

  Outside the boundary of the system there are other particles. Let us consider the particle at the point Q'. This particle has mass δmQ and is just outside the system boundary. It therefore must be excluded from the system at time t. It is assumed that this mass moves with velocity which is different from the velocity

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  eBook - ePub

  Fluid Mechanics, Hydraulics, Hydrology and Water Resources for Civil Engineers

  • Amithirigala Widhanelage Jayawardena(Author)
  • 2021(Publication Date)
  • CRC Press

    (Publisher)

  Chapter 13 Applications of basic fluid flow equations

  13.1 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 is

  V 2

  2 g

  per unit weight, but is different from

  ∫

  cross   section

  v 2

  2 g

  . Therefore, a correction factor, α , is used for

  V 2

  / 2 g

  so that

  α

  V 2

  2 g

  is the average kinetic energy for unit weight, passing the section.

  Referring to Figure 13.1 , the kinetic energy passing

  δ A

  per unit time is

  Figure 13.1 Velocity profile in pipe flow.

  ∫

  v 2

  2

  (

  ρ v d A

  )

  where ρvdA is the mass.

  Therefore,

  α

  V 2

  2

  ρ V A =

  ∫ A

  v 2

  2

  ( ρ v

  d A )

  which leads to

  α =

  1 A

  ∫ A

  (

  v V

  )

  3

  d A

  (13.1)

  The Bernoulli equation then becomes

  z +

  p

  ρ g

  + α

  V 2

  2 g

  = C o n s t a n t

  For laminar flow in pipes

  α = 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.

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  Computational Methods for Process Simulation

  • W. Fred Ramirez(Author)
  • 1997(Publication Date)
  • Butterworth-Heinemann

    (Publisher)

  The first class of problems, calculating the viscous frictional losses, involves the use of the mechanical energy equation since these terms appear explicitly in that equation. The second class of problems, calculating a force, involves the use of the momentum balance since forces enter the momentum balance explicitly. Class three problems, calculating the pressure, can be approached by using either the momentum or mechanical energy equations since pressure terms appear in both equations. The choice of which equation to use depends upon what type of information is specified. If viscous frictional losses are given, then the mechanical energy balance is used. If force information is specified, then the momentum balance is used.

  EXAMPLE 1.7 Horizontal Converging Nozzle:: Water is flowing at 9.45 × 10−3 m3 /s through a horizontal converging nozzle. The upstream ID is 7.62 cm (3 in.), and the downstream ID is 1 in. Calculate the resultant force on the nozzle when it discharges to the atmosphere. Consider that the nozzle is attached at its upstream end and that frictional forces are negligible.

  The control volume for this system as shown in Figure 1.13 includes the walls of the pipe since we are interested in computing the structural or resultant force.

  Figure 1.13 Horizontal Converging Nozzle.

  The momentum balance equation (1.7.7) for this system is

  (1.7.8)

  The pressure force is given as difference between the pressure p 1 acting on area A 1 and the sum of the atmospheric pressure,

  pa

  , acting on the area A 2 and the difference of the areas (A 1 − A 2 ). Essentially, the atmospheric pressure acts on the entire area A 1 when the control volume includes the walls of the pipe.

  (1.7.9)

  or

  (1.7.10)

  where

  pa

  is the atmospheric pressure. We will use equation (1.7.8) to compute the resultant force

  rx

  .

  To compute the pressure drop, we use the mechanical energy balance equation (1.6.18) , which becomes at steady state and with negligible viscous frictional losses

  (1.7.11)

  The information flow diagram for this system is shown in Figure 1.14 . The numerical answer for the problem is

  rx

  = −625 N

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  eBook - ePub

  Introduction to Fluid Mechanics, Sixth Edition

  • William S. Janna(Author)
  • 2020(Publication Date)
  • CRC Press

    (Publisher)

  1   Fundamental Concepts

  Fluid mechanics is the branch of engineering that deals with the study of fluids—both liquids and gases. Such a study is important because of the prevalence of fluids and our dependence on them. The air we breathe, the liquids we drink, the water transported through pipes, and the blood in our veins are examples of common fluids. Further, fluids in motion are potential sources of energy that can be converted into useful work—for example, by a waterwheel or a windmill. Clearly, fluids are important, and a study of them is essential to the engineer.

  After completing this chapter, you should be able to:

  • Describe commonly used unit systems;
  • Define a fluid;
  • Discuss common properties of fluids;
  • Establish features that distinguish liquids from gases; and
  • Present the concept of a continuum.

  1.1 DIMENSIONS AND UNITS

  Before we begin the exciting study of fluid mechanics, it is prudent to discuss dimensions and units. In this text, we use two unit systems: the British gravitational system and the international system (SI). Whatever the unit system, dimensions can be considered as either fundamental or derived. In the British system, the fundamental dimensions are length, time, and force. The units for each dimension are given in the following table:

  British Gravitational System
  Dimension	Abbreviation	Unit
  Length	L	foot (ft)
  Time	T	second (s)
  Force	F	pound-force (lbf)

  Mass is a derived dimension with units of slug and defined in terms of the primary dimensions as

  1   slug = 1 1 bf ⋅ s 2 ft

  (1.1)

  Converting from the unit of mass to the unit of force is readily accomplished because the slug is defined in terms of the lbf (pound-force).

  Example 1.1 An individual weighs 150 lbf.

  1. a. What is the person’s mass at a location where the acceleration due to gravity is 32.2 ft/s2 ?
  2. b. On the moon, the acceleration due to gravity is one-sixth of that on earth. What is the weight of this person on the moon?

  Solution

  1. a. Applying Newton’s law, we write

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  eBook - ePub

  Aerodynamics of Wind Turbines

  • Martin Hansen(Author)
  • 2015(Publication Date)
  • Taylor & Francis

    (Publisher)

  equation (A.17) becomes:

  where M is an unknown torque acting on the fluid in the control volume and r is the radius from the cylindrical axis. If the flow is uniform at the inlet and exit of the control volume and the only non-zero component of M is in the flow direction z, Euler’s turbine equation (A.18) can be derived from Equation A.17 :

  P is power removed from the flow on a mechanical shaft, ω is the rotational speed of the shaft, Vϑ is the tangential velocity component, ṁ is the mass flow through the control volumes and subscripts 1 and 2 denote the inlet and exit of the control volume, respectively.

  Another important equation is the integral conservation of energy or the first law of thermodynamics for a control volume, which for steady flow is Equation A.19 :

  where P and Q are the mechanical power and the rate of heat transfer added to the control volume and ui

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