Fluid Mechanics

Fluid Mechanics

Fluid as a continuum or continuum based approach

Fluid is made of molecules. However, for most of the engineering applications, when we speak of fluid's properties such as density, or conditions such as pressure and temperature, we do not imply such properties or conditions of individual molecules, but those of “fluid” as a whole.  In other words, we refer to the average or macroscopic aggregate effects of the fluid- molecules, reflected in pressure, temperature, density, etc.  Such an approach to treating a fluid is called continuum based approach. In other words, fluid is treated as continuum.  However, there is a restriction. The continuum approach can be applied only when the mean free path of the fluid (largely, gas) is smaller (actually much smaller!!) than the physical characteristic length of the system under consideration, say, the diameter of the tube in which the gas flows, or size of a container in which gas is stored.

(Fig. 3a)

The continuum approach is usually not valid when the gas pressure is very small (few milli-torr like in a vacuum), or the aperture size is small (like in an orifice)  Mathematically, for the continuum approach based model to hold good, where  is the mean free path of the gas molecule and  is the characteristic length of the system. Alternatively, Knudsen defined as . Fluid Statics  Concerned with forces in a stationary fluid  Recall. In a stationary fluid, shear stress . However, fluid can sustain the normal stress.  Fluid pressure: It is a normal force per unit area of the fluid element, acting inward onto the element. Consider a fluid element, ABC:  (Fig. 3b) Pressure acts inward on all three faces of the element.  Consider another fluid element:  (Fig. 3c) Pressure also acts inward on the differential element dA Pascal's Theorem-Pressure at any point within a static fluid is the same in all directions. Let us prove it. Consider a triangular fluid element ABC in a 2-D(x-z) scenario. Pressure acts on all 3-faces of the element inward and normal to the surface.  (Fig. 3d) (Fig. 3e) Force balance:  Taking limit  The basic equation of fluid statics  First, let us calculate force on a fluid element because of pressure variation. Consider an elemental fluid volume bounded by (ABCD)-(A'B'C'D') surfaces.  (Fig. 3f) Pressure at   is  Volume of the fluid under consideration is  Force acting on the face AA'C'C:   (+ ve direction) Force acting on the face   (-ve direction) (Neglecting the higher order terms or assuming linear variation over ) Differential force,  (acting along –ve x-direction) Similarly,  Therefore, we have expression for differential pressure–force or,   If there is no pressure variation or pressure is uniform around an element, the net force is zero, or  The basic equation of fluid statics (continued)  2-Now, let us calculate pressure–gradient in a static fluid.  Identify all forces: pressure forces, and external body force which is gravity in the present case.Consider a differential fluid elemental volume, .  Force Balance (in vector form ) (pressure force + gravitational force) = 0  or,  or, (form previous lecture) or,  This is the basic (vector) equation for fluid statics.  The equation in the scalar form: : x direction : y direction : z direction  Where,  If gravity acts in the negative z-direction   or  where  Notes:  Pressure gradient in a static fluid is zero if there is no gravity. Alternatively, the net force acting on a fluid volume is zero because pressure–force balances the force due to gravity. Alternatively, pressure-variation occurs in a static fluid because of gravity only.  If the fluid is water, static pressure is often referred as ‘hydrostatic' pressure.   If a body is submerged in a fluid, which is in contact with (or open to) atmosphere, the atmospheric pressure acts uniformly on the body. For most of the engineering applications one is interested in calculating pressure due to the fluid only, ignoring the atmospheric pressure. In such case, the pressure is specified as a gauge pressure (above the atmosphere pressure): . Therefore, the gauge pressure may be negative if the pressure in the fluid is sub or below atmospheric. Absolute pressure is always positive Pressure in the atmosphere may vary, of course, over a relatively longer altitude, because the density of air is small (1 kg/m3 in comparison to 1000 kg/m3 for water). Consider an atmosphere consisting of ideal gases. Static pressure gradient  And,  Substituting,  (Fig. 4a) or,  If  or an isothermal atmosphere  And,  or,  The pressure varies (decreases) exponentially with altitude, if temperature variation is considered negligible. Engineering calculations based on the fluid static equation  Pressure variation in an incompressible fluid Integrating,  (Fig. 4b) Therefore, pressure decreases lineally with altitude, or increases with depth in the fluid  (Fig. 4c)  or,  (Fig. 4d)  (, where  is the depth of the fluid measured from the top )  (2) Pressure- variation in two-immiscible liquids (one is heavier than the other)  (Fig. 5a) Consider two liquids of densities . Depth (measured form the top) of the top-fluid is   and that of the bottom-fluid is  Put the co-ordinate axis at the bottom (left). The governing equation for the static fluid is   or    On integration,    Similarly,    or  One can get similar expression by putting the co-ordinate axis at the top of the surface, with z-axis pointing vertically downward (right):   or,    On integration  Similarly,   or,   Note:  Both expressions are same! (3) Barometer (Instrument to measure atmospheric pressure)  (Fig. 5b) Neglect surface tension or capillary effects  Height of the liquid in the vertical tube is h measured from the free surface.     On integration,   But, at the same level  and  Therefore, .  is nothing but the vapor pressure of Hg in the space over the liquid Hg in the tube. Generally,   (very small at Room temperature). Hence,. One can experimentally measure to show that There are two problems in using water-barometer  Tube is long! Vapor pressure of water is significantly larger than that of Hg, i.e.,  and should be considered in the calculation!! 4) Monometer (To measure pressure of the fluid in a device or container) (Fig. 5c) On integration between  On integration between C and D  But,  Therefore,  Therefore,  However,  ( NOTE    at sea - level) In this lecture we discuss steady state/unsteady state condition for velocity fields.  Steady state  The steady state condition for a flow-field implies that the velocity field and any property associated with the flow field remain unchanged with time.In other words, local derivative of the velocity is zero.  Mathematically, 0. The concentration and temperature fields, if associated will also be under steady–state:  As explained below,  may or may not be zero.  Consider the flow of a fluid through a convergent nozzle. If the velocity field is steady, or the flow–field is under steady-state conditions, . However, an observer moving with the flow field will experience ‘acceleration' as he moves from the larger diameter-section to the smaller diameter-section of the nozzle. This is because the velocity increases as the diameter of the nozzle decreases. Therefore, velocity remains unchanged with time anywhere in the flow- field. However, it has a spatial variation, or  (Fig. 10a) , , (between 1-2)  The last expression is tantamount to saying that the velocity of a labeled or marked material-particle is not constant, as it moves from the larger section to the smaller section of the nozzle. Example 1: At a certain time, the concentration of  measured at a location near a power-plant (A) is 200 ppm (parts per million). The  concentration at ‘B', 1 km downstream, is measured to be 10 ppm. At this instance, the average wind velocity is 1 km/hr. The wind is blowing in the direction of the higher (A) to lower concentration (B). If a balloon-sensor floating with the wind records a decrease of 50 ppm per hour, as it flows past ‘B', determine the rate of increase of the concentration measured by a stationary observer at ‘B'.  (Fig. 10b) (Assume that the concentration gradient between two locations is linear) Example 2: A boat has a thermo-server attached to it. While floating with the fluid, it measures the temperature of the fluid. Both flow and temperature are under unsteady-state condition: Determine the rate of change of the temperature recorded by the sensor at t = 1, when the boat flows past the location, whose spatial co-ordinates are .  ;;  Here,  Substituting;  (Note:   Mathematical analysis of fluid motion and associated properties  There are two ways to carry out the analysis: Control mass: In such an analysis, a fixed mass of fluid element in the flow-field is identified and conservation equations for properties such as momentum, energy or concentration are written. The identified mass moves around in the flow-field. Its property may change from one location to another; however, the property must correspond to the same contents of the identified fluid element. In general, such approach is mathematically or experimentally difficult to apply. Control volume: This approach is popular and widely applied in the analysis. An arbitrary fixed volume located at a certain place in the flow-field is identified and the conservation equations are written. The property under consideration or analysis may change with time. If  is a general property associated with fluid, then  and   may be considered to describe change in the property,  of the fluid, following the control mass (CM) and control volume (CV)approach, respectively.  The surface which bounds CV is called control surface (CS). In the CV approach, the co-ordinate axis is first fixed. A ‘CV’ is then marked in the flow-field. Choice of location and shape of CV are important for mathematical formulation. Reynolds Transport Theorem  We exclude the derivation of the theorem from the present course; however discuss the significance and the application of the same. The theorem relates the rate of change of a property associated with a CV to the material or particle rate of change of that property. Consider a CV bound by the CS through which a fluid flows, described by the flow field,   relative to co-ordinates . As per the definition, CV is fixed in the flow field: (Fig. 11a) N is a property associated with flow field in CV. N could be either mass or momentum or energy or concentration of a species dispersed in the fluid. The theorem states:                                     where  is the specific property, N/M, where M = mass of fluid in volume . or,       Note that if the property is mass, . If the property is momentum, . If the property is kinetic energy, . Let us give physical interpretation to the above three forms of the theorem. The first term is the rate of change in the property contained in the CV, which is fixed in the flow field. This term also describes if the property is under steady state or not, or if there is any accumulation of the property, N in the CV.  The 2nd term represents the material rate of change of the property and may be seen as the rate of generation of the property. Note that if because mass of the fluid particles initially marked in CV cannot change, nor it can be generated or destroyed, as the particles move in the flow field. The 3rd term represents the net rate of change in the property because of the flow of the fluid in and out of the CV through CS.  is the rate of volumetric flux in and out of dA, a differential area on CS.  is the rate of mass flux.  is the rate of the property-flux in and out of the differential area   is the rate of the total property over the entire CV. Take a simple CV in the flow field, shown by the dotted line. (Fig. 11b) It is easier to show that the last term  In such case, the fluid enters and leaves the CS only through  and . The remaining surface of CS is impervious to the mass.   Conservation of mass N = property =M; Therefore,  Apply Reynold’s Transport Theorem: (Fig. 12a) (Particle-mass cannot be created or destroyed)  The 1st term of the right-hand-side is zero:  particle-mass cannot be created or destroyed. Therefore,  : conservation of mass equation Note that  of a fluid in a CV may change with time, for example, in a tank filled with air and water. If water is drained out,  of the mixture (air + water) will vary with time. Message: Be careful while evaluating the integral, whether  under evaluation refers to the density of a pure fluid or fluids-mixture. The RHS of the above equation is the integral (over the entire cross-section) of the differential volumetric flow rate:  (Fig. 12d)     Or   Ans. Apply the conservation law again over larger     to show that,                 Ans. Further,  (for the 2nd section)  (for the 1st section)     or,   Re-visit,   or,   (volumetric flow rate) or, ,   average velocity in the section Compare to  to show that . Differential form of mass-conservation In the previous lecture, we considered the conservation of total mass over a CV, by applying the general Reynolds Transport theorem. Now, we obtain the differential form of the equation, also known as the continuity equation. Consider CV of            (Fig. 13a)     CV is bound by six CSs. fluid- density  Consider CS marked 1: Rate of mass-in:    Differential form of mass-conservation In the previous lecture, we considered the conservation of total mass over a CV, by applying the general Reynolds Transport theorem. Now, we obtain the differential form of the equation, also known as the continuity equation. Consider CV of            (Fig. 13a)     CV is bound by six CSs. fluid- density  Consider CS marked 1: Rate of mass-in:    (It is equivalent to to  of the Reynold’s transport theorem) CS marked 2: Rate of mass-out:  Net rate of change in mass in x-direction  Similar expressions can be obtained for the other directions, for examples,  Net rate of accumulation of mass in CV  Therefore,  The significance of this expression is the same as that for the total mass-conservation rule applied to a CV in the earlier lecture Re-arrange:  This is the vector-form of continuity-equation and readily applied for all co-ordinate systems. In this course we will focus mostly on Cartesian and cylindrical 1D or 2D geometry. To this end, the above conservation equation may also be written as:                  Re-call the difference between  if the flow is steady  if the fluid is incompressible,  In other words, local density my change with time ,as the fluid enters and leaves a ‘CV’ at different rates resulting in the accumulation or depletion of mass of the matter per unit volume of the CV. This is referred as the unsteady-state. However, if the fluid is incompressible, its density will be constant and the material rate of change of density will be zero.   (incompressible fluid) can be written for 2D geometries as :    : Cartesian  : Cylindrical  Pay special attention to the last term of the above-equation. For 1D radial flow, .  (Fig. 13b)   const under steady-state. To sum-up, total mass flow rate   is constant at all locations in the r-direction. However, mass flux  changes because the differential area changes in the r-direction B. Momentum - conservation equation (Fig. 13c) Reynolds Transport Theorem: Property,    1st Term: local rate of change of momentum in CV 2nd Term: Net rate of momentum in and out of CV through CS. 3rd Term: material or partial –rate of change of momentum, and it may be seen as the external body-force or the source term for the generation of momentum by the external forces. Mathematically, Re-write Momentum-Theorem (Example)  Consider the steady-state flow of water though a  reducing bend. The water-velocity is uniform at the inlet and outlet cross-section . If the water enters the bend at  and is discharged at   to the atmosphere, determine the force required to hold the bend in place or the force acting on the bend. Neglect the weight of the bend and water in the bend.  (Fig. 14a) Choose CV so that CS is perpendicular to the velocity field at the inlet and the outlet. Apply momentum theorem:  Under steady-state, 1st term of the RHS is zero. Note that , which follows from continuity. Apply continuity under SS:  Therefore,          Let us make free-body- diagram on CV (Fig. 14b) There are pressure forces acting on  by the water; reaction forces on CV, Rx and Ry (reaction forces from the support to hold bend in place). Atmospheric pressure acts uniformly over the entire CV, except on  where the pressure is (absolute). Therefore, add and substract  to make the contribution of atmosphere zero on CV. In other words, .   .   to hold the bend in place    Consider the flow of an incompressible fluid under steady-state in an expander: (Fig. 14c) Fluid-velocity is uniform at the entrance  and at the outlet . Pressures are considered also uniform at sections 1 and 2 of the nozzle. Atmospheric pressure acts uniformly on the nozzle. Calculate the pressure drop .    Choose CV as shown below: (Fig. 14d)   (velocity everywhere in the smaller cross-section   Apply continuity over the same CV: Or,    Substitute,               neglecting viscous forces (friction on wall), neglecting viscous forces.        Equating, Ans. Choice of CV is important. If CV is chosen so that the CS cuts the nozzle, then there will be reaction forces, which will be unknown: (Fig. 14e) ; However there is no viscous force as in the previous case!  Weight of the nozzle and the water inside. The latter–CV may be chosen when the question posed is: determine the force to hold the nozzle in place. Equation of motion (Differential form of momentum conservation rate). Before we apply Reynolds Transport theorem, it is important to understand the forces acting on a fluid-element. We have earlier noted that the only force acting on a static fluid is the normal force or the pressure acting normal to the surface of the fluid element. Static fluid:   (Fig. 15a) In flowing fluid, there is a tangential force or shear force acting parallel to the surface of the fluid-element. Thus, there are two types of stress developed in a flowing fluid: one is the normal stress,  and the other is the shear stress,  :  (Fig. 15b) It is shown that,  where is the strain rate. In the case of a static fluid,  at any location in the fluid. In a moving fluid, however, :  (Fig. 15c) If fluid is inviscid,  And, , as in the case of a static fluid  For a 3D flow, there are two  and one  on a plane, as shown below  (Fig. 15d) There are sign conventions:  x: direction of normal to the plane or the direction of the area y: direction of shear stress    Therefore, for a 3D fluid-element, there are six shear stresses and 3 normal stresses  (Fig. 15e) The stresses are represented as It can be shown that shear stresses are symmetric: etc, at any location in the flow-field The Newton’s law of viscosity relates shear-stress to velocity gradient via the viscosity of a fluid. For 1D flow of a Newtonian fluid,   is the velocity gradient and can also be represented as the strain rate. In general, when shear stress is applied on a fluid-volume, there are translation, deformation, dilation, rotation and distortion, mathematically expressed by strain rates. Analogous to the representation of shear stresses, strain rates are represented as the combination of the following velocity gradients:  The Newton’s extended law of viscosity relates . For example,  Again, it is obvious that if a fluid is stationary or inviscid    Equation of motion (conservation of momentum in differential form)  Consider  flow of an incompressible fluid   (Fig. 15f) X-momentum balance:  (Fig. 15g) See the figure above.   (due to flow) = Differential force acting on  = shear-stress  normal stress      F body force =   (due to gravity)  Newton’s 2nd law of motion can be applied on CV  or,    or,   Insert the expressions for    and simplify for an incompressible fluid  Similarly, one can write (or develop) y-momentum balance equation       In vector form  1st term: Accumulation or transient or unsteady state 2nd term: inertial (convective momentum flux) 3rd term: pressure force 4th term: viscous forces 5th term: External body-force We call this equation Navier-Stokes (NS) equation Assumptions:  (1)  (incompressible fluid)                          (2) Newtonian fluid                          (3) Laminar flow NS equation and examples Before we apply the NS equation, let us make a note that the NS equation is a 2nd order PDE. Therefore, 2 BCs are required for  along each direction to solve for the velocity-field. We should note the common boundary conditions: No slip condition at the solid surface: it is an experimental observation that the relative velocity at the surface is zero. So,    at a stationary solid surface At the interface of liquid-gas or liquid–liquid (immiscible fluids), there is no jump in the velocity, shear stress and pressure  (Fig. 16a) (gradients maybe different)    Symmetric BC:    At the center- line of tube or rectangular channel, velocity- gradient is zero. (Fig. 16b) Ex.1 Couette–flow Consider the steady-state 2D-flow of an incompressible Newtonian fluid in a long horizontal rectangular channel. The bottom surface is stationary, whereas the top surface is moved horizontally at the constant velocity, . Determine the velocity field in the channel. Assume fully developed flow. (Fig. 16c) Assumption:   is applicable      Fully developed flow:   (Note: this statement is equivalent to saying, neglect the end effects) Apply continuity:    For fully developed flow, first term is zero.  is constant along y direction. But,  at y = ±  (no- slip condition) So,  everywhere.  Apply NS in x-direction:    1st term is zero for SS for fully developed flow; . Horizontal  Therefore, the equation is simplified as   No external pressure was applied and in addition, it is a long plate     Integrate twice,   Apply BC (1)   (no slip condition)  BC (2)  Solve for  to obtain  This is the expression for velocity profiles in a Couette flow  (Note: at  ).  (Fig. 16d) Example 2: Suppose that an external pressure-drop along x-direction is applied across the channel, in addition to the conditions mentioned in example 1. Now, calculate the velocity profiles in the channel.  X-momentum balance is now modified to    Y–momentum balance: But is a function of y only. This implies that  (constant) Integrating twice,  Apply BCs to obtain  You should check the two BCs. Special case     (Same as in example 1)    ,    where,    This is called Poiseuille-flow (velocity profile is parabolic) (Fig. 16e) Determine the average velocity in the channel    Simplify to show that    or,   Next, plot   for Poiseuille and Couette flow :  Poiseuille flow (Fig. 16f) (Fig. 16g)   Steady-state, laminar flow through a horizontal circular pipe (Fig. 17a) Assumptions:   (no swirling, circulation) We are interested in solving    Continuity:     Fully developed flow:    Therefore, r   But   (No-slip condition)   everywhere. So, the problem is reduced to solving . NS (X-direction):   From the formulation of the problem, 1st, 2nd, and 3rd terms on the LHS are zero.      r-direction:  O-direction:  Note:  Re-look at last two equations.  and  depend on gravity. There is no viscous effects   depends on viscous forces only. There are no gravitational effects. Therefore, define non-gravitational pressure, so that or    Check: this definition of   satisfies the above momentum-balance equation. Therefore,   Substitute,    Say,  (const ) integrate twice to obtain Apply BC1:  (symmetric BC)   (No slip BC) where, Note that velocity profile is parabolic in the tube. Calculate, volumetric flow rate  You should be able to show that Or,  Re–arrange, This equation to calculate pressure-drop in a horizontal pipe of Length L and inside-diameter D, for a viscous incompressible fluid flowing under steady-state fully developed laminar condition, is known as Hagen-Poiseuille equation Newtonian  and non-Newtonian fluid - macroscopic pressure-drop balance In the previous lecture, we applied the NS equation to obtain an expression for the pressure–drop in a tube for the condition of the steady-state laminar flow of a Newtonian fluid. We can also derive the same expression, the Hagen- poiseuille equation, by making an integral (macroscopic) force balance. (Fig. 18a)  If the flow is laminar (Re<2100) the velocity profile is parabolic: For the Newtonian fluid,   and,  Therefore,  : shear-stress varies linearly in r-direction of the tube. Now, make a force balance over  which is the same as that obtained by applying differential momentum balance (NS equation) and then integrating with BCs. From the above expression, one can also determine viscosity  of a fluid by measuring pressure- drop across a tube for different flowrates: (Fig. 18b)  The above-plot is known as pseudo-shear diagram, which is generally used to determine the viscosity of a slurry-mixture. A similar approach can be followed to determine the viscosity or effective viscosity of Non-Newtonian fluids. Bingham plastic fluid  Such a fluid has the following shear-stress  vs strain rate  characteristics: -----------(1) where  is known as the yield stress. Fluid is supposed to be ‘frozen’ if  or,      If the flow is laminar, the velocity profile is parabolic near the walls and is uniform in the central core of the tube. Shear-stress in such fluid also varies linearly with r: (Fig. 18c)       Equation (1) can be integrated with   to obtain:     Therefore,  The uniform velocity within the core of the tube   Again,    Integrate as an exercise to obtain  (Fig. 18d)  Energy Conservation Equation  Similar to the derivation of continuity and momentum conservation equations, one can also apply Reynolds Transport equation to obtain the energy-conservation equation. In this course, we will, however, not derive the expression and restrict our discussion to the salient features of the equation and its application to the fluid-flow. For a flow-system,  (specific property) for energy consists of three terms :  where, N = total property   Re-call,   for N= mass                                    for N = momentum   for N= energy (J) KE      PE   IE  The three terms may be recognized as kinetic energy, potential energy, and internal energy, respectively.  Reynolds transport equation:   has two components; (1) surface-work because of flow  and (2) shaft-work.  , surface-work also consists of two terms, or is contributed by two forces (shear and normal )  (Fig. 19a)  If  (stagnant fluid) there is no surface work,     is the shaft- work, for example, due to a turbine or a pump or blower, which delivers energy to the fluid or vice-versa.  if the machine is turbine   if the machine is pump or blower   can also be written as     Note that  is zero if   (inviscid fluid)  (tangential velocity) = 0    (choice of CS)  The general energy-balance equation can be re-arranged to obtain the following expression for a CV: The terms of the equation are accumulation, conduction, shaft-work, viscous work, and energy efflux across CS,  including pressure-energy                                                           . For most engineering applications of steady-state, two–port system (one inlet and one outlet), uniform flow over the ports (1-D flow), and  being perpendicular to CS (by a suitable choice), one can obtain the following expression for energy- conservation: Moreover, if the flow is under isothermal condition without exchange of heat with surrounding (insulated system)  and   Note: unit of each term is or  If  (no pump or turbine)      over two ports system, for example, in a tube or pipe or a tank:  (Fig. 19b)  The above conservation equation is also known as mechanical energy balance, consisting of KE, PE, and pressure-energy. Each term is also known as ‘head’.  We re-note the assumptions made in obtaining the above ME balance equation. (1) SS (2)  isothermal (3) No shaft work  (4) negligible viscous work, .  It is important to note that viscous work can be neglected if the fluid is inviscid or  a suitable choice of CS so that  is perpendicular to CS. Bernoulli equation and applications In the previous lecture, we obtained an expression for the conservation of mechanical energy in a flowing system. We will like to apply the same expression between two points or location in the flow- field. Before that, let us understand ‘streamlines’ and “stream tubes”. Streamline – it represents a line drawn in the flow field such that tangent drawn at every point of it is in the direction of the local velocity vector,    (Fig. 20a) Note that streamline will change in the unsteady-state flow field. Also, note that laminar flow is represented/ characterized by streamlines (turbulent flow is characterized by eddy, irregular, unstable flow patterns). If we inject dyes or color at a certain location in the laminar flow, we can track the path of dyes and visualize streamlines. Similarly, if we inject or sprinkle several tiny (mass-less) needles in the fluid under steady- state laminar flow conditions, the needles will align themselves along the fluid- flow path. A hypothetical line connecting the head of the needles may be considered to be ‘streamline. Two streamlines cannot cross each other, because there will be two velocities at the point of intersection, which is not possible. On similar note, mass cannot cross a streamline. Based on the understanding of streamlines, one can visualize a stream tube as a hypothetical 3D tube encompassing streamlines inside, whose surface also consists of streamlines.  Note that at the entry and exit of the stream tube (marked in bold lines), velocity is perpendicular to the CS. (Fig. 20b) It can be said that fluid cannot cross the CS of the stream tube, and therefore, the outer boundary or CS of the tube may be replaced with a solid wall! We can now apply the mechanical energy balance equation to the CV of a stream tube between ports 1 and 2:   constant, with the assumptions  (a)  (2)  (3) laminar flow (4) inviscid fluid or no viscous loss, and (5)isothermal.  Under the extreme case of zero width or diameter of 3D stream tube, the equation may be applied between points 1 and 2 along a streamline   (Fig. 20c) The same equation gets a name, the famous “Bernoulli equation” with the additional constraint (6) that this equation must be applied along a streamline. See the figure below. The BE cannot be applied between ‘1’ and ’3’, because they do not lie on a streamline, but can be applied between ‘1’ and ‘2’. (Fig. 20d) It is interesting to mention that if there is a pump which delivers energy to the fluid and if there are viscous losses because of friction with the walls or expansion or contraction, the BE is modified as an engineering approximation :         Example:  An inverse U-tube is used to drain water from a tank or reservoir. The bend of the inverted tube is 1m above the free surface of the water in the tank and 10 m above the ground where the water is discharged. Water exits the tube at atmospheric pressure. Determine the pressure in the bend. (Fig. 20e) To apply BE, assume , and neglect viscous losses (losses in the bend or due to wall-friction)  Apply BE between (1) and (B). (It is obvious that we have chosen a streamline along (1) and (B), shown as a dotted line)   , where    and  are the elevations from a reference line.  (considering that the tank area is significantly larger than the tube’s size, which is always true). Mathematically, one can apply continuity in such case: . Therefore,    Obtain . How? Apply continuity between ‘B' and ‘2’   Obtain . How? Apply BE between ‘1’and ‘2’ Substitute to obtain   Substitute  to obtain   x  Energy balance equation: Minor losses The energy balance equation contains a term, ‘’ which represents irreversible conversion of viscous work to thermal energy and is always ‘+ ve’:  (neglecting any shaft work)(Note: it has been shown that   Minor losses refer to the losses in pipe bends, fittings, valves, and nozzles, etc, whereas major losses refer to the same in a long pipe or tube. We will first minor losses. Example1 Consider the following system of pipe flow, consisting of several fittings such as pipe bends, and valves:  (Fig. 21a) As the fluid flows through pipe and associated fittings, including valves, there is invariably loss of mechanical energy, which may be mathematically quantified as: , where,  is often calculated at the inlet to the fitting or valves and K is known as loss coefficient and depends on Reynolds number and type of fittings. The manufacturers of valves and fittings provide ‘K’ value. For example: Fitting            K Globe - valve (fully-open) 10     half open     20  Gate- value (fully- open) 0.3     half open    5 In general, loss in the globe valves is more than that in gate valves, because of the fact that fluid changes its direction sharply in the former. The globe valves are preferred over gate valves because of better flow-control. Losses also occur in flange or weld joints connecting two pipes or fittings. Losses are also accounted for in diverging or converging nozzles or expanders, or where there are geometrical changes: (Fig. 21b) Apply mechanical energy balance between 1 and 2:  (assuming velocity profiles are uniform over the cross- sections at 1 and 2)   One can also apply continuity and momentum-balance equations between ‘1’ and ‘2’:  (1)  (2)  (CV is marked with dotted line) (Fig. 21c) It is an experimental observation that . Therefore,  Or,  (It is left as an exercise to the students to obtain the above–derivation from combined mass, moments, and energy conservations) Thus, we have shown that, For a converging channel, viscous loss,  is also estimated as Note that  is the velocity in the converging section and  is the cross-sectional area at the section where “Vena-Contracta" (or, the minimum flow area) occurs: (Fig. 21d) We revert to “Vena–Contracta” later in the course. Before we take–up the topic on flow measuring devices, which are related to the application of mechanical energy balance, we address kinetic energy correction factor: Consider the flow of viscous fluid in a tube. The flow is under laminar conditions and the velocity profile is parabolic: (Fig. 21e) We are interested in evaluating total KE across the section: One can also evaluate KE based on the average velocity,  as   where   Kinetic energy correction factor is defined as    or  It is left as an exercise to show that   If the flow is turbulent,  is approximated between  The advantage of defining  is now clear. For a tubular laminar flow, one can write the energy conservator equation in terms of average velocity, , without evaluating the complex integral:  = constant KE correction factor  The energy balance equation–flow measuring devices In this lecture, we discuss a few common devices, which are mostly based on the principle of general mechanical energy balance, to measure the volumetric flow rate of a fluid: Venturimeter  Orificemeter  Pitot tube  Rotameter*  The first three devices are based on the mechanical energy balance. *The last instrument is based on a simple force balance. (a) Venturimeter: it is a simple instrument having a converging–diverging section installed in a pipe. Pressure-drop is measured across two sections and is correlated to the flow rate, accounting for the viscous (minor) loss. (Fig. 22a)  (Neglecting minor-loss) (Minor–loss can be minimized by slowly or gradually expanding the diverging section. That is why, the venturimeters are long, bulky, and expensive)   (without loss)  For Venturimeters,   or actual flowrate is calculated as,    where  is called the discharge coefficient and accounts for minor–loss or viscous–loss, because of non- uniform flow– patterns in the converging–diverging sections and at throat. The supplier of the instrument provides a plot to calculate :  (Fig. 22b) (b) Orifice meter: The instrument has an orifice–plate (circular plate with a small hole)  inserted across the section of the pipe or tube:  (Fig. 22c) In such a device, the flow from the upstream–section accelerates as the flow-area decreases from the section 1 to 2 at the orifice. The flow continues to accelerate, or the main flow area continues to decrease till the section 2 and further downstream of the orifice at C, before it starts increasing downstream. The section at C contains the minimum flow area, known as vena–contracta. Similar to the analysis carried out for venturimeter, one can calculate the ideal or  theoretical flow rate: To account for viscous-losses, a parameter   is defined, so that     and  C is known as the discharge coefficient and   is the velocity-of-approach factor. A single factor, flow coefficient   is used to express K is reported in a graph supplied by the manufactures. (Fig. 22d) (C) Pitot tube: In such a device, part of the flow is made stagnant so that the entire KE of the fluid is converted into pressure–energy. A small opening is made in the device, parallel to the flow, where the pressure is the same as the fluid pressure.  (Fig. 22e) Apply Bernoulli’s equation between 1 and 1’       (along the same–streamline; flow is parallel to opening )  Static pressure Apply Bernoulli’s equation between 1and 2 But,  Therefore,  By measuring the difference between stagnation and static pressures, one can calculate the velocity and the volumetric flow rate. In earlier lecture, we obtained an expression for pressure-drop in a pipe or tube for the flow of a fluid under laminar conditions: The expression was obtained analytically by applying the NS equation and integrating the same with appropriate boundary conditions. The expression (also known as the Hagen–Poiseuille equation) was also derived by making the force-balance over a CV in the tube. It is important to note that such mathematical treatment can be carried out, only if the flow is laminar. Laminar flow refers to the flow which can be characterized by streamlines. The flow is controlled by viscous effects and the fluid velocity is relatively smaller. A parabolic-velocity profile obtained in a tube for a fluid–flow at small velocity is a good example of laminar-flow conditions.  (Fig. 23a) Turbulent flow: Such flow occurs at relatively larger velocities and is characterized by chaotic behavior, irregular motion, large mixing, and eddies. For such flow, inertial effects are more pronounced than viscous effects. Mathematically, velocity field is represented as  , or the velocity fluctuates at small time scales around a large time-averaged velocity. Similarly,    etc In the latter lectures, we will see that a parameter called Reynolds number   is used to characterize laminar flow vis a vis turbulent flow. If  for a tubular flow, the flow- characteristic is observed to be laminar and , the flow is turbulent. Re- visiting the Hagen–Poiseuille equation for pressure- drop: or, It follows that, if we apply the mechanical energy balance equation between two sections of a horizontal pipe through which there is a laminar flow under steady-state conditions, we can show that  where,   Thus, we have obtained an expression to calculate , major loss in the pipe because of viscous- effect.    Considering that, for turbulent flow the velocity or pressure–fields may not be exactly (analytically) represented, one resorts to dimensional analysis. (Fig. 23b) The experimental observations suggest that pressure drop in a pipe–flow under turbulent conditions depends on Reynolds number and surface roughness. Taking an analogy from the major–loss for a laminar–flow wl =   At this point, an engineering parameter called friction–factor, f is defined as   so that   for a laminar–pipe flow, and   in general. This equation is known as Fanning equation. Here, f is experimentally shown to be dependent on the surface roughness   and Reynolds number, and is obtained from a plot called Moody’s plot: (Fig. 23c) The bottom most curve shown for   represents the friction factor for nearly hydraulically smooth tube or pipe, for which the friction factor has reached the smallest value with increasing smoothness. Now, we have an expression to calculate energy loss    for a pipe–flow under both laminar and turbulent conditions: where,       A common empirical formula to calculate friction factor is proposed in literature:   Centrifugal pump The fluid turbo machinery essentially consists of an impeller rotating in a casing. Fluid enters the eye of the impeller (the center of the impellers) and exits though the space between the impeller blades to the space between the impeller and casing walls. The velocity of fluid elements is in both tangential and radial directions, as the impeller rotates. The velocity as well as the pressure, both increase, as the fluid flows through the impeller    (Fig. 25a) Efficiency of pump Difference between heads at location (A) and (B) may be calculated as  where the terms in each parenthesis consist of pressure, kinetic energy and potential energy. The increase or difference between the kinetic and potential heads is usually negligible. Therefore,  The efficiency of the pump  is defined as  where the term in the numerator represents power delivered by the pump because of the pressure-developed. BHP is the brake–horse power; required to drive the pump. BHP depends upon the speed of the pump, vane angle (design of the impeller) and the flow rate of the fluid. Net positive – suction head (NPSH) Defined as the net head developed at the suction port of the pump, in excess of the head due to the vapor pressure of the liquid at the temperature in the pump. NPSH must be positive for preventing the liquid from boiling. Boiling or cavitations may damage the pump. where,  is the vapor–pressure of the liquid. If the pump is placed at a height  above the free surface of a liquid where the atmosphere pressure is  the NPSH may be evaluated by writing the Bernoulli’s equation between the free surface and the suction port of the pump as  where  frictional loss in the suction pipe between the liquid–surface and the pump. Therefore, it is obvious that for NPSH to be positive or maximum, Z and  should be minimum.  Most ideally, for maximum NPSH,  and  Performance characteristics of a pump There are there performance characteristics of pump: Head developed by the pump (H)  Brake horse power (BHP) Efficiency of the pump  -all plotted against the flow rate. (Fig. 25b) As seen from the graph, head decreases with increasing flow rate till the pump cannot deliver the fluid at . Horsepower increases with flow rate. Efficiency of the pump initially increases with increasing flow rate, reaches a maximum, and then decreases to zero at . Pump is operated at . Creeping, potential and boundary–layer flows In the previous lectures, we took-up couple of examples on the application of the NS-equation, in particular, the flow in a circular tube. You must have realized that the NS–equation is non–linear on velocity, and the velocity and pressure–fields are coupled. In general, it is difficult to solve the NS equation, and more often numerical techniques are used to solve the equation. In this context. approximations are made to simplify the equation for solution. In this lecture, we will make use of the dimensional analysis to non–dimensionalize the NS equation, and then, explore the possibility of approximation.   x-momentum     Non–dimensionalize the variables as follows:  , The characteristic variables are generally chosen to make the dimensionless quantities vary between 0-1. Therefore, L may be the length of a tube or channel.   is the average velocity of the fluid, and  and  are the up-and downstream  pressures. On substitution, the following equation is obtained: Let us define,          The non–dimensionalized equation is, therefore, written as      From the above-equation, it is clear that several approximations can be made to drop–out one or more terms from the equation, depending upon the range of , the three dimensionless groups obtained for the NS-equation. We will continue our discussion in this lecture to only two approximations of common engineering application:  Low Re High Re  Low Reynolds-number Flow: Under this condition, the viscous effects dominate the inertial effects.  Mathematically  and   Such flow is called as creeping-flow One of the most common examples of creeping flow is the flow past a spherical object at low Reynolds number.  We will not derive the expression for velocity fields in this lecture. Readers can refer to an advanced book on fluid mechanics. However, it may be mentioned that the total force comprising of normal and shear can be combined to obtain the well–known Stoke’s law for the drag (force acting along the flow–direction) on a sphere:    Where,                           Fluid Viscosity  = Diameter of the particle  =Relative velocity of the particle with respect to the fluid–velocity  This equation is valid for  One of the applications of Stoke's law is to calculate terminal velocity of a sphere in a quiescent fluid. Let us consider a sphere falling under gravity in stagnant fluid :  (Fig. 26a) There are 3 forcess acting on the sphere Gravity Buoyancy drag (viscous) forces  Therefore,   When the particle acquires a steady-velocity or it reaches a constant (terminal) velocity       Therefore,    Flow at high Reynolds number (Boundary layer theory) Re-visit the non–dimensionalized NS equation: At high Reynolds number, viscous effects are considered negligible in the flow away from a solid surface. In other words, fluid is supposed to be inviscid or .  In such case the equation is simplified to   , or in the dimensional form  Neglecting the gravitational effect, one can state that the inertial force is balanced by pressure force. Such flow is termed as potential flow.  Assuming 1-D flow, the equation can be integrated to obtain  Re-call. We have previously obtained the above equation as Bernoulli’s equation. It is important to note that the potential theory or inviscid flow–condition predicts ‘zero–drag’ on a solid surface. The common examples of potential flow are source/sink, free vortex, doublet, and Rankine’s half–body. We skip the analysis of such flows. To overcome the paradox of zero drag on a solid surface, Prandtl (1903) came–up with the boundary–layer theory. As per the main postulate of the theory, the viscous term cannot be dropped from the NS equation for the high–Reynolds number flow. Physically, viscous effects are important at or near the solid surface, even at high Reynolds number.   By choosing the appropriate characteristic variables, it can be shown that one of the viscous terms, namely   must be retained. This is possible if we assume that the characteristic length in y–direction is much smaller than that in the x–direction.  Re–defining variables as   where,  one obtains the following dimensionless form of the NS equation: At the limit of  , the 1st viscous term is neglected. However, the 2nd term is retained because  The dimensional form of the NS equation (for x-direction) may now be written for boundary–layer flow (at high Reynolds number) as:  momentum  Blausius has proposed an approximate solution to the above equation for the flow near the solid surface. From the solution, ,  the gradient at or near the surface may be calculated  to obtain drag  (Fig. 27a) For the flow over a flat plate, the local shear stress has been calculated as  where,   Defining drag coefficient,   , where  is the drag on the plate (or any solid surface) and ‘A’ is the plate-area,  is calculated as    -----(A)  At higher Reynolds number, turbulence is observed. For a wide-range of Reynolds number, plots are available to calculate  for the flow over a flat plate or past a sphere.  (Fig. 27b) (Fig. 27c) In general,   , where , the drag–coefficient can be read from the plots. Empirical equations are also available. There are several features of fluid flow at high Reynolds number, such as boundary layer separation, eddies, and adverse pressure gradients. These topics and the other related topics such as bluff body, streamline body, Von–Karman Street, etc, are excluded from discussion in this course.