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ME 304 FLUID MECHANICS II Prof. Dr. Haşmet Türkoğlu Çankaya University Faculty of Engineering Mechanical Engineering Department Spring, 2017 du yx dy yx du k dy n du yx k dy ME304 1 n 1 du du dy dy 2 The Fundamental Laws Experience have shown that all fluid motion analysis must be consistent with the following fundamental laws of nature. The law of conservation of mass: Mass can be neither created nor destroyed. It can only be transported or stored. Newton’s three laws of motion: - A mass remains in a state of equilibrium, that is, at rest or moving at constant velocity, unless acted on by an unbalanced force. - The rate of change of linear momentum of mass is equal to the net force acting on the mass. - Any force action has a force reaction equal in magnitude and opposite in direction. The first law of thermodynamics (law of conservation of energy) Energy, like mass, can be neither created nor destroyed. Energy can be transported, changed in form, or stored. The second law of thermodynamics: The entropy of the universe must increase or, in the ideal case, remain constant in all natural processes. The state of postulate (law of property relations): The various properties of a fluid are related. If a certain minimum number (usually two) of fluid’s properties are specified, the remainder of the properties can be determined. Differential versus Integral Formulation We must now consider the level of detail of the resulting flow analysis. We must choose between a detailed point by point description and a global or lumped description. When a point by point (local) description is desired, fundamental laws are applied to an infinitesimal control volume. The result will be a set of differential equations with the fluid velocity and pressure as dependent variables and the location (x, y, z) and time as independent variables. Solution of these differential equations, together with boundary conditions, will be two functions V(x, y, z, t), and P(x, y, z, t) that can tell us the velocity and pressure at every point. When global information such as flow rate, force and temperature change between inlet and outlet is desired, the fundamental laws are applied to a finite control volume. The result will be a set of integral equations. ME304 1 3 FLUID STATICS Basic Equation of fluid statics grad P g 0 p p0 gh dp g dz HYDROSTATIC FORCE ACTING ON A PLANE SUBMERGED SURFACE Integral Method FR pdA A yFR ypdA y A xFR xpdA x A FR FR pdA 1 FR ypdA A 1 xpdA FR A Magnitude of FR A Direction of FR is normal and toward the surface Algebraic method FR ghc Pressure Prism Method FR k dP P k P x' xR x I xyc yc A xc y' yR 1 1 1 xPdA x ghdA FR A P A P I xc yc yc A xd X G P and y 1 1 1 yPdA y ghdA FR A P A P ME304 1 yd YG P 4 BASIC EQUATIONS FOR A FINITE CONTROL VOLUME (Equations In Integral Form) N d V dA Reynolds Transport Equation: t t System C CS d V dA 0 t C CS Equation of Conservation of Mass (Continuity Equation): Linear Momentum Equation: F FS FB V d V V dA t C CS Bernoulli equation subject to restrictions: Vs2 Bernoulli Equation: gz C 1. Steady flow 2 2. No friction p 3. Incompressible flow 4. Flow along a streamline Moment of Momentum Equation: r Fs r gd Tshaft r Vd r VV dA t C C CS Euler Turbine Equation: Energy Equation: Tshaft r2Vt 2 r1Vt1 m Q Ws Wshear Wother ed e pv V dA t C CS 2 2 p1 p2 V V 1 2 Extended Bernouilli Equation: z z 1 1 2 2 hf g 2g 2 g ME304 1 5 HEAD LOSS (PRESSURE DROP) The head loss (pressure loss) in closed conduits can be dived into two part: 1) Major (friction) loss: Losses due to viscous effects on the duct wall. 2) Minor (local) losses): Losses due to the flow through valves, tees, elbows and other non-constant cross-sectional area portions of the system. Major Head Loss (Pressure Drop) L V 2 P f d 2 LV2 hf f d 2g For laminar flow, friction factor, For turbulent flow, friction factor, hf P g 64 f Re f f Re, s / d Friction factors for turbulent flows are given in charts (Mood Diagram) or as correlations. Minor (Local) Head Loss The minor head losses in variable area parts are proportional to the velocity head of the fluid, i.e. V2 hf k 2g Minor head losses for valves, fittings and bends can be calculated using the equivalent length technique, which may be given by the following equation: Le V 2 hf f d 2g Total loss = (Major loss) + (Minor loss) ME304 1 6 INTRODUCTION TO DIFFERENTIAL ANALYSIS OF FLUID MOTION (Chapter 5) In course Fluid Mechanics I, we developed the basic equations in integral form for a finite control volume. The integral equations are particularly useful when we are interested in the gross behavior of a flow and its effect on various devices. However, the integral approach does not enable us to obtain detailed point by point data of the flow field. To obtain this detailed knowledge, we must apply the equations of fluid motion in differential form. In this chapter, we will derive fundamental equations in differential form and apply this equations to simple flow problems. EQUATION OF CONSERVATION OF MASS (CONTINUITY EQUATION) The application of the principle of conservation of mass to fluid flow yields an equation which is referred as the continuity equation. We shall derive the differential equation for conservation of mass in rectangular and in cylindrical coordinates. ME304 1 7 Rectangular Coordinate System The differential form of the continuity equation may be obtained by applying the principle of conservation of mass to an infinitesimal control volume. The sizes of the control volume are dx, dy, and dz. We consider that, at the center, O, of the control volume, the density is and the velocity is V u ı vj wk For the control voule, equation of consevation of mass in integral form is V CS dA t CV dV 0 V dA To evaluate the first term in this equation, we must evaluate the mass CS flow rate over each face of the control volume. To be completed in class The values of the mass fluxes at each of six faces of the control volume may be obtained by using a Taylor series expansion of the density and velocity components about point O. For example, at the right face, x dx 2 2 dx 1 dx 2 x 2 x 2! 2 2 Neglecting higher order terms, we can write and similarly, dx x dx 2 x 2 u dx u x dx u 2 x 2 ME304 1 8 Corresponding terms at the left face are dx dx x dx 2 x 2 x 2 u dx u dx u x dx u u x 2 2 x 2 To be completed in class Table. Mass flux through the control surface of a rectangular differential control volume ME304 1 9 The net rate of mass flux out through control surface is u v w V CS dA x y z dxdydz The rate of change of mass inside the control volume is given by d V dxdydz t CV t Therefore, the continuity equation in rectangular coordinate is u v w 0 x y z t Since the vector operator, , in rectangular coordinates, is given by ı j k x y z V 0 t The continuity equation may be simplified for two special cases. 1. For an incompressible flow, the density is constant, the continuity equation becomes, V 0 2. For a steady flow, the partial derivatives with respect to time are zero, that is _________. Then, ………………………………. ME304 1 10 Example: For a 2-D flow in the xy plane, the velocity component in the y direction is given by v y2 x2 2 y a) b) Determine a possible velocity component in the x direction for steady flow of an incompressible fluid. How many possible x components are there? Is the determined velocity component in the x-direction also valid for unsteady flow of an incompressible fluid? To be completed in class ME304 1 11 Example: A compressible flow field is described by V axi bxyj e kt Determine the rate of change of the density at point x=3 m, y=2 m and z=2 m for t=0. ME304 1 12 Derivation of Continuity Equation Cylindrical Coordinate System In cylindrical coordinates, a suitable differential control volume is shown in the figure. The density at center, O, is and the velocity there is V vr er v e vz ez Figure. Differential control volume in cylindrical coordinates. V dA , we must consider the mass flux through each of To evaluate CS the six faces of the control surface. The properties at each of the six faces of the control surface are obtained from Taylor series expansion about point O. ME304 1 13 Table. Mass flux through the control surface of a cylindrical differential control volume The net rate of mass flux out through the control surface is given by v v v V CS dA vr r r r r z z drddz The rate of change of mass inside the control volume is given by dV rdrddz t CV t In cylindrical coordinates the continuity equation becomes vr v vz vr r r r 0 r z t ME304 1 14 Dividing by r gives vr r vr 1 v vz 0 r r z t or 1 (rvr ) 1 ( v ) ( vz ) 0 r r r z t In cylindrical coordinates the vector operator is given by 1 er e ez r r z Then the continuity equation can be written in vector notation as V 0 t e er Note : r e and er The continuity equation may be simplified for two special cases: 1. For an incompressible flow, the density is constant, i.e., 2. For a steady flow, ME304 1 15 Derivation of Continuity Equation in Cylindrical Coordinate sytem Using Vector Form of the Equation To be completed in class ME304 1 16 Example: Consider one-dimensional radial flow in the r plane, characterized by vr = f(r) and v = 0. Determine the conditions on f(r) required for incompressible flow. ME304 1 17 STREAM FUNCTION FOR TWO-DIMENSIONAL INCOMPRESSIBLE FLOW For a two-dimensional flow in the xy plane of the Cartesian coordinate systems, the continuity equation for an incompressible fluid reduces to u v 0 x y If a continuous function ( x, y, t ) , called stream function, is defined such that u y and v x Then continuity equation is satisfied exactly, since u v 2 2 0 x y xy yx Streamlines are tangent to the direction of flow at every point in the flow field. Thus, if dr is an element of length along a streamline, the equation of streamline is given by V dr 0 (u ı vj ) (dx ı dyj ) (udy vdx)k Substituting for the velocity components of u and v, in terms of the stream function udy vdx 0 (A) At a certain instant of time, t0, the stream function may be expressed as ( x, y, t0 ) . At this instant, the streamfunction dx dy 0 x y (B) Comparing equations (A) and (B), we see that along instantaneous streamline = constant. In the flow field, 2-1, depends only on the end points of integration, since the differential equation of is exact. d dx dy x y d 0 ME304 1 18 Now, consider the two-dimensional flow of an incompressible fluid between two instantaneous streamlines, as shown in the Figure. The volumetric flow rate across areas AB, BC, DE, and DF must be equal, since there can be no flow across a streamline. For a unit depth, the flow rate across AB is Along AB, x = constant and Q y2 y1 d x1 y1 y1 2 dy d 2 1 1 y Along BC, y = constant and d x2 y2 dy y dy . Therefore, y For a unit depth, the flow rate across BC is Q Q y2 Q udy x2 x1 vdx x2 x1 dx x dx . Therefore, x 1 dx d 2 1 2 x Thus, the volumetric flow rate per unit depth between any two streamlines, can be expressed as the difference between constant values of defining the two streamlines. ME304 1 19 In r plane of the cylindrical coordinate system, the incompressible continuity equation reduces to rvr v 0 r The streamfunction (r, ,t) then is defined such that vr 1 r v r Example: Consider the stream function given by = xy. Find the corresponding velocity components and show that they satisfy the differential continuity equation. Then sketch a few streamlines and suggest any practical applications of the resulting flow field. ME304 1 20