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Mm302e Fluid Mechanics Ii ME 304 FLUID MECHANICS II Prof. Dr. Haşmet Türkoğlu Çankaya University Faculty of Engineering Mechanical Engineering Department Spring, 2018 du yx dy n n1 du du du du yx k yx k dy dy dy dy ME304 1 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 dp p p gh grad P g 0 g 0 dz HYDROSTATIC FORCE ACTING ON A PLANE SUBMERGED SURFACE Integral Method F pdA R A 1 yF ypdA y ypdA R A FR A 1 xF xpdA x xpdA R A FR A F F pdA Magnitudeof F R R R A Direction of FR is normal and towardthesurface Algebraic method I xyc I xc x' xR xc y' yR yc FR ghc yc A yc A Pressure Prism Method 1 1 1 x xPdA xghdA xd X G FR A P A P F k d k P R P P and P 1 1 1 y yPdA yghdA yd Y F G R A P A P P ME304 1 4 BASIC EQUATIONS FOR A FINITE CONTROL VOLUME (Equations In Integral Form) N Reynolds Transport Equation: d V dA t System t C CS Equation of Conservation of Mass (Continuity Equation): d V dA 0 t C CS Linear Momentum Equation: F FS FB Vd VV dA t C CS p V 2 Bernoulli equation subject to restrictions: Bernoulli Equation: s gz C 1. Steady flow 2 2. No friction 3. Incompressible flow 4. Flow along a streamline Moment of Momentum Equation: r Fs r gd Tshaft r Vd r VV dA C t C CS Euler Turbine Equation: T r V rV m shaft 2 t2 1 t1 Q W W W ed e pv V dA Energy Equation: s shear other t C CS p V 2 p V 2 Extended Bernouilli Equation: 1 1 z 2 2 z h 1 1 2 2 f g 2g g 2 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. Total loss = (Major loss) + (Minor loss) Major Head Loss (Pressure Drop) 2 L V 2 L V P P f hf f hf d 2 d 2g g 64 For laminar flow, friction factor, f Re For turbulent flow, friction factor, 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. V 2 h k f 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: L V 2 h f e f d 2g 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 uı vj wk For the control voule, equation of consevation of mass in integral form is V dA dV 0 CS t CV 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, 2 2 dx 1 dx dx x 2 2 x 2 x 2! 2 dx dx Neglecting higher order terms, we can write x 2 x 2 u dx dx and similarly, ux u 2 x 2 ME304 1 8 Corresponding terms at the left face are dx dx u dx u dx dx dx x ux u u 2 x 2 x 2 2 x 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 dA dxdydz x y z CS The rate of change of mass inside the control volume is given by dV dxdydz t CV t Therefore, the continuity equation in rectangular coordinates 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 2 2 v y x 2y a) Determine a possible velocity component in the x direction for steady flow of an incompressible fluid. How many possible x components are there? b) 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 kt V axi bxyje Determine the rate of change of the density at point x=3 m, y=2 m and z=2 m for t=0. To be completed in class 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 vrer v e vzez Figure. Differential control volume in cylindrical coordinates. V dA CS To evaluate , we must consider the mass flux through each of 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 vr v vz V dA vr r r drddz r z CS The rate of change of mass inside the control volume is given by dV rdrddz t t CV In cylindrical coordinates the continuity equation becomes v v v v r r r z r 0 r r z t ME304 1 14 Dividing by r gives v v 1 v v r r z 0 r r r z t or 1 (rv ) 1 (v ) (v ) r z 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 er e V 0 Note: e and er t r The continuity equation may be simplified for two special cases: 1.
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