COURSE NUMBER: ME 321 Fluid Mechanics I 3 credit hour Basic Equations in fluid Dynamics Course teacher Dr. M. Mahbubur Razzaque Professor Department of Mechanical Engineering BUET 1
Description of Fluid Motion In the description of a flow field, it is convenient to think of individual particles each of which is considered to be a small mass of fluid, consisting of a large number of molecules, that occupies a small volume that moves with the flow. If the fluid is incompressible, the volume does not change in magnitude but may deform. If the fluid is compressible, as the volume deforms, it also changes its magnitude. In both cases the particles are considered to move through a flow field as an entity. There are two approaches of mathematical description of the physical quantities of a fluid motion as a function of space and time coordinates : Lagrangian Descriptions of Motion and Eulerian Descriptions of Motion 2
Lagrangian Descriptions of Motion In the study of particle mechanics, where attention is focused on individual particles, motion is observed as a function of time. The position, velocity, and acceleration of each particle are listed as s(x0, y0, z0, t), V(x0, y0, z0, t), and a(x0, y0, z0, t), and quantities of interest can be calculated. The point (x0, y0, z0) locates the starting point the name of each particle. This is the Lagrangian description, named after Joseph L. Lagrange (1736 1813), of motion that is used in a course on dynamics. In the Lagrangian description many particles can be followed and their influence on one another noted. This becomes, however, a difficult task as the number of particles becomes extremely large in even the simplest fluid flow. 3
Eulerian Descriptions of Motion The region of flow that is being considered is called a flow field. In Eulerian description, points in a flow field is identified in space and then the velocity of fluid particles passing each point is observed; we can observe the rate of change of velocity as the particles pass each point, that is, and we can observe if the velocity is changing with time at each particular point, that is,. In this Eulerian description, named after Leonhard Euler (17071783), of motion, the flow properties, such as velocity, are functions of both space and time. In Cartesian coordinates the velocity is expressed as V = V(x, y, z, t). 4
Lagrangian and Eulerian Descriptions An example may clarify these two ways of describing motion. An engineering firm is hired to make recommendations that would improve the traffic flow in a large city. The engineering firm has two alternatives: Hire students to travel in automobiles throughout the city recording the appropriate observations (the Lagrangian approach), or hire students to stand at the intersections and record the required information (the Eulerian approach). A correct interpretation of each set of data would lead to the same set of recommendations, that is, the same solution. In an introductory course in fluids, however, the Eulerian description is used exclusively since the physical laws using the Eulerian description are easier to apply to actual situations. 5
Acceleration 6
Acceleration 7
Acceleration This derivative is called the substantial derivative, or material derivative. It is given a special name and special symbol (D/Dt instead of d/dt) because we followed a particular fluid particle, that is, we followed the substance (or material). It represents the relationship between a Lagrangian derivative in which a quantity depends on time t and an Eulerian derivative in which a quantity depends on position (x, y, z) and time t. 8
Acceleration The time-derivative term on the right side of Eqs. 3.2.8 for the acceleration is called the local acceleration and the remaining terms on the right side in each equation form the convective acceleration. Hence the acceleration of a fluid particle is the sum of the local acceleration and convective acceleration. In a pipe, local acceleration results if, for example, a valve is being opened or closed; and convective acceleration occurs in the vicinity of a change in the pipe geometry, such as a pipe contraction or an elbow. In both cases fluid particles change speed, but for very different reasons. 9
Basic Laws in Fluid Dynamics The basic laws in fluid dynamics are expressed either in terms of a system or a control volume. A system is a fixed collection of material particles. For example, if we consider flow through a pipe, we could identify a fixed quantity of fluid at time t as the system (Fig. 4.1); this system would then move due to velocity to a downstream location at time t + Dt. Any of the basic laws could be applied to this system. 10
Our interest is most often focused on a device, or a region of space, into which fluid enters and/or from which fluid leaves; we identify this region as a control volume. An example of a fixed control volume is shown in Fig. 4.2a. A control volume need not be fixed; it could deform. However, only fixed control volumes will considered from now on. The difference between a control volume and a system is illustrated in Fig. 4.2b. 11
System to control volume transformation System to control-volume transformation, or equivalently, the Reynolds transport theorem is written as, Here h represents the intensive property (the property of the system per unit mass) associated with Nsys. The relation between h and Nsys is expressed as The first integral of Eq. (4.2.9) represents the rate of change of the extensive property in the control volume. The second integral represents the flux of the extensive property across the control surface; it may be nonzero only where fluid crosses the control surface. 12
CONSERVATION OF MASS A system is a given collection of fluid particles; hence its mass remains fixed: Using Reynolds transport theorem and recognizing that Nsys = msys i.e. the mass of the system (an extensive property for which h =1), we can write This is called the continuity equation. If the flow is steady, there results 13
CONSERVATION OF MASS For a uniform flow with one entrance and one exit, Eq. (4.3.4) takes the form Where we have used If the density is constant in the control volume, the continuity equation reduces to This form of the continuity equation is used quite often, particularly with liquids and low-speed gas flows. The above derivation assumed uniform velocities at the inlet and exit sections. What happens when the velocities at the inlet and the exit sections are non uniform? 14
Continuity equation Suppose that the velocity profiles at the entrance and the exit are not uniform, such as sketched in Fig. 4.7. Furthermore, suppose that the density is uniform over each area. Then the continuity equation takes the form Where, are the average velocities at sections I and 2, respectively. In examples and problems the overbar is often omitted. It should be kept in mind, however, that actual velocity profiles are usually not uniform. 15
Continuity equation Two fluxes are useful in specifying the quantity of flow. The mass flux or the mass rate of flow is and has units of kg/s; Vn is the normal component of velocity. The flow rate Q, the volume rate of flow, is and has units of m3/s. The mass flux is usually used in specifying the quantity of flow for a compressible flow and the flow rate for an incompressible flow. In terms of average velocity, we have Where uniform density and normal velocity is assumed. 16
Example 4.1 Water flows at a uniform velocity of 3 m/s into a nozzle that reduces the diameter from 10 cm to 2 cm. Calculate the water s velocity leaving the nozzle and the flow rate. Solution The control volume is selected to be the inside of the nozzle as shown. Flow enters the control volume at section 1 and leaves at section 2. The simplified continuity equation (4.4.6) is used since the density of water is assumed constant and the velocity profiles are uniform: The flow rate or discharge is found to be: 17
or 18
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Continuity equation in differential form Consider the mass flux through each face of the infinitesimal control volume shown in Fig. 5.1. We set the net flux of mass entering the element equal to the rate of change of the mass of the element; that is, 20
This takes the form 21
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Then the continuity equation becomes, Or, in vector form, The continuity equation in cylindrical coordinate is 23
Navier-Stokes equations In an incompressible flow the density remains constant or the change in density is very negligible. In an incompressible flow of a Newtonian Fluid with constant viscosity, the momentum equations (derived from Newton s second law of motion) are: These are the called Navier-Stokes equations, named after C. I. M. H. Navier (l785-1836) and Sir George G. Stokes (1819-1903), who are credited with their derivation. These are second-order nonlinear partial differential equations. 24
Navier-Stokes equations. In vector notations, the Navier-Stokes equations may be written as: Which may be interpreted as: Inertia force per unit vol. = gravity force per unit vol. + pressure force per unit vol. + viscous force per unit volume There are four unknowns: p, u, v, and w. So, for solution, it should be combined with the incompressible continuity relation to form four equations in these four unknowns. The continuity equation for incompressible flow in steady state is: Or, 25
Incompressible Inviscid flow. Computation Fluid Dynamics (CFD) is the branch of fluid mechanics that uses numerical methods to solve the Navier-Stokes equations and continuity equation with appropriate boundary conditions and without any analytical simplification. However, analytical solution for many viscous flow problems are also possible through some simplifying assumptions. If viscosity is assumed to be zero (inviscid flow approximation), the second order terms in the Navier-Stokes equations are lost and we get the Euler s Equation, as: Integration of this equation along a streamline results in Bernoulli s equation. 26
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Energy Equation Many problems involving fluid motion demand that the energy equation, be used to relate quantities of interest. If the heat transferred to a device (a boiler or compressor), or the work done by a device (a pump or turbine), is desired, the energy equation is needed. It is also used to relate pressures and velocities when Bernoulli s equation is not applicable; this is the case whenever viscous effects cannot be neglected, such as flow through a piping system. Let us express the energy equation in control volume form. For a system it is 30
where the specific energy e is, In terms of a control volume, Eq. 4.5.1 becomes This can be put in simplified forms for certain restricted flows. The rate-of-heat transfer term Q represents the rate-of-energy transfer across the control surface due to a temperature difference. (Do not confuse this term with the flow rate Q.) The work-rate term results from work being done by the system. Work of interest in fluid mechanics is due to a force moving through a distance while it acts on the control volume. The The work-rate term is discussed in detail in the following section. 31
Considering different situations in fluid problems, the work-rate term becomes: We should note that third and fourth components in the work-rate term are seldom encountered in problems in an introductory course 32
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In many fluid flows, useful forms of energy (kinetic energy and potential energy) and flow work are converted into unusable energy forms (internal energy or heat transfer). However, in an introductory fluid mechanics course the sum of these effects is lumped together. We define losses as the sum of all the terms representing unusable forms of energy: For a steady-flow situation in which there is one entrance and one exit across which uniform profiles can be assumed, the energy equation becomes. 34
Here, Ws is shaft work and hl is head loss. If the losses are negligible, if there is no shaft work or heat flow and if the flow is incompressible, we can write the energy equation in its most useful form as, Note that this form of energy equation is identical with Bernoulli's equation. However, we must understand that the Bernoulli's equation is a momentum equation applicable along a stream line and the equation above is an energy equation applied between two sections of a flow. When the energy equation is applied to any steady uniform flow, the control volume is usually selected such that the entrance and exit sections have a uniform total head. 35
For example, if the energy equation is applied to a flow passing a gate, the appropriate control volume may be chosen as shown below The total head at the entrance and exit can be evaluated at an point at the entrance and exit, respectively. However, a convenient choice would be the points at the water surface. Thus the energy equation becomes 36
where we have introduced the head loss hl, often written in terms of a loss coefficient K as A final note to this section regards nomenclature for pumps and turbines in a flow system. It is often conventional to call the energy term associated with a pump the pump head Hp, and the term associated with a turbine the turbine head HT. Then the energy equation takes the form In this form we have equated the energy at the inlet plus added energy to the energy at the exit plus extracted energy (energy per unit weight, of course). If any of the quantities is zero (e.g., there is no pump), the appropriate term is simply omitted. 37
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The power generated by a turbine with an efficiency of ht is The power required by a pump with an efficiency of hp would be Usually power in expressed in kilowatts or horsepower. One horsepower is equivalent to 0.746 kw. 40
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Momentum Equation Newton's second law, often called the momentum equation, states that the resultant force acting on a system equals the rate of change of momentum of the system. If a device has entrances and exits across which the velocity is assumed to be uniform and if the flow is steady, the momentum equation can then be written as Where, If we consider the nozzle shown and determine the xcomponent of the force of the joint on the nozzle, the momentum equation xdirection becomes 43
An example of a free-surface flow in a rectangular channel is shown in Fig. 4.12. If we want to determine the force of the gate on the flow, the following expression can be derived from the momentum equation: where F1 and F2 are pressure forces. Momentum Equation Applied to Deflectors The application of the momentum equation to deflectors forms an integral part of the analysis of many turbomachines, such as turbines and pumps. 44
Momentum Equation Applied to Deflectors The analysis is separated into two parts: jets deflected by stationary deflectors and jets deflected by moving deflectors. For both problems we will assume the following: -The pressure external to the jets is everywhere constant so that the pressure in the fluid entering the deflector is the same as that in the fluid exiting the deflector. -The frictional resistance due to the fluid-deflector interaction is negligible so that the relative speed between the deflector surface and the jet stream remains unchanged. - Lateral spreading of a plane jet is neglected. -Body forces are small and will be neglected. 45
Stationary Deflector Let us first consider the stationary deflector, illustrated in Fig. 4.13. Bernoulli's equation allows us to conclude that V2 = V1 since the pressure is assumed to be constant external to the fluid jet and elevation changes are negligible. Assuming steady, uniform flow the x- and y- components of the momentum equation takes the form of Eq. 4.5.15. 46
For given jet conditions the reactive force components can be calculated. Moving Deflectors The situation involving a moving deflector depends on whether a single deflector is moving (a water scoop used to slow a high-speed train) or whether a series of deflectors is moving (the vanes on a turbine). 47
Moving Deflectors Let us first consider that the single deflector shown in Fig. 4.14 is moving in the positive x-direction with the speed VB. In a reference frame attached to the stationary nozzle, from which the fluid jet issues, the flow is unsteady; that is, at a particular point in space, the flow situation varies with time. A steady flow is observed, however, from a reference frame attached the deflector. 48
From this reference frame, moving with the constant velocity VB, we observe the relative speed Vr entering the control volume to be V1 VB, as shown in Fig. 4.14. It is this speed that remains constant as the fluid flows relative to the deflector. Hence the momentum equation takes the forms Where represents only that part of the mass flux exiting the fixed jet that has its momentum changed. 49
Since the deflector moves away from the jet some of the fluid that exits the fixed jet never experiences a momentum change; this fluid is represented by the distance VB Dt, shown in Fig. 4.14. Hence Where the relative speed (V1 - VB) is used in the calculation; the mass flux ravb is subtracted from the exiting mass flux pav1 to provide the mass flux. 50
Series of vanes For a series of vanes (a cascade) the jets may be oriented to the side, as shown in Fig. 4.15. The actual force on a particular vane would be zero until the jet strikes the vane; then the force would increase to a maximum and decrease to zero as the vane leaves the jet. 51
Assume that, on the average, the jet is deflected by the vanes as shown as viewed from a stationary reference frame; the fluid jet enters the vanes with an angle b1 and exits with an angle b2. What is desired, however, is that the relative velocity enter the vanes tangent to the leading edge of the vanes, that is, Vr1 in Fig. 4.16b is at the angle a1. The relative speed then remains constant as it travels over the vane with the exiting relative velocity Vr2 leaving with the vane angle a2.the relative and absolute velocities are related with the velocity polygons of Fig. 4.16 b and c. 52
Assuming that all of the mass exiting the fixed jet has its momentum changed, we can write the momentum equation as Interest is usually focused on the x-component of force since it is this component that is related to the power output (or requirement). The power would be found by multiplying the x-component force by the blade speed for each jet; this takes the form where N represents the number of jets. 53
EXAMPLE 4.11 Water flows through a horizontal pipe bend and exits into the atmosphere. The flow rate is 0.01 m3/s. Calculate the force in each of the rods holding the pipe bend in position. Neglect body forces and viscous effects. Solution We have selected a control volume that surrounds the bend as shown. Since the rods have been cut, the forces that the rods exert on the control volume are included. The pressure forces at the entrance and exit of the control volume are also shown. 54
The flexible section is capable of resisting the interior pressure but it transmits no axial force or moment. The body force (weight of the control volume does not act in the x- or y- direction but normal to it. Therefore, no other forces are shown. The average velocities are found to be Before we can calculate the forces Rx and Ry we need to find the pressures p1 and p2. The pressure p2 is zero because the flow exits into the atmosphere. The pressure at section 1 can be determined using the energy equation or the Bernoulli equation. 55
Neglecting losses between sections 1 and 2, the energy equation gives Now we can apply the momentum equation in the x-direction to find Rx and in the y- direction to find Ry: Note that we have assumed uniform profiles and steady flow. These are the usual assumptions if information is not given otherwise. 56
EXAMPLE 4.12 A deflector turns a sheet of water through an angle of 30o as shown. What force per unit width is necessary to hold the deflector in place if the mass flux is 32kg/s? Solution The control volume we have selected includes the deflector and the water adjacent to it. The only force that is acting on the control volume is due to the support. This force has been decomposed into Rx and Ry. 57
The velocity V1 is found to be Bernoulli s equation shows that if the pressure does not change, then the magnitude of the velocity does not change, provided that there is no significant change in elevation and that viscous effects are negligible; thus we can conclude that V2 = V1 since p2 = p1. 58
Next, the momentum equation is applied in the x-direction to find Rx and then in the y-direction for Ry: 59
EXAMPLE 4.13 The deflector shown moves to the right at 3 m/s while the nozzle remains stationary. Determine (a) the force components needed to move the deflector, (b) V2 as observed from a fixed observer, and (c) the power generated by the vane. The jet velocity is 8 m/s. Solution (a) To solve the problem of a moving deflector, we will observe the flow from a reference frame attached to the deflector. In this moving reference frame the flow is steady and Bernoulli's equation can then be used to show that Vr1 = Vr2 = 5 m/s, the velocity of the sheet of water as observed from the deflector. Note that we cannot apply Bernoulli's equation in a fixed reference frame since the flow would not be steady. 60
Applying the momentum equation to the moving control volume, which is indicated again by the dashed line, we obtain the following: When calculating the mass flux, we must use only that water which has its momentum changed; hence the velocity used is 5 m/s. 61
(b) Observed from a fixed observer the velocity V2 of the fluid after the deflection is V2 = Vr2 + VB, where Vr2 is directed tangential to the deflector at the exit and has a magnitude equal to Vr1 (see the velocity diagram above). Thus Finally, (c) The power generated by the moving vane is equal to the velocity of the vane times the force the vane exerts in the direction of the motion. Therefore, = 3 x 26.8 = 80.4 W 62
EXAMPLE 4.17 Find an expression for the head loss in a sudden expansion in a pipe in terms of V1 and the area ratio. Assume uniform velocity profiles and assume that the pressure at the sudden enlargement is p1. Solution A sketch is shown of a sudden expansion with the diameter changing from d1 to d2. The pressure at the sudden enlargement is p1, so that the force acting on the left end shown is p1a2. Newton's second law applied to the control volume yields, assuming uniform profiles, 63
The energy equation provides To express this in terms of only V1, we can use continuity and relate Then the expression above for the head loss becomes 64