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Pharos University ME 253 Fluid Mechanics 2 Revision for Mid-Term Exam Dr. A. Shibl Streamlines • A Streamline is a curve that is everywhere tangent to the instantaneous local velocity vector. • Consider an arc length dr dxi dyj dzk • dr must be parallel to the local velocity vector V ui vj wk • Geometric arguments results in the equation for a streamline dr dx dy dz V u v w Kinematics of Fluid Flow 1 w v 1 u w 1 v u i j k 2 y z 2 z x 2 x y 2 1 DV u v w xx yy zz V Dt x y z 1 u v 1 w u 1 v w xy , zx , yz 2 y x 2 x z 2 z y Stream Function for Two-Dimensional Incompressible Flow • Two-Dimensional Flow Stream Function y Stream Function for Two-Dimensional Incompressible Flow • Cylindrical Coordinates Stream Function y(r,q) Is this a possible flow field Given the y-component Find the X- Component of the velocity, Determine the vorticity of flow field described by Is this flow irrotational? Momentum Equation • Newtonian Fluid: Navier–Stokes Equations Example exact solution Poiseuille Flow Example exact solution Fully Developed Couette Flow • For the given geometry and BC’s, calculate the velocity and pressure fields, and estimate the shear force per unit area acting on the bottom plate • Step 1: Geometry, dimensions, and properties Fully Developed Couette Flow • Step 2: Assumptions and BC’s – Assumptions 1. 2. 3. 4. 5. 6. 7. Plates are infinite in x and z Flow is steady, /t = 0 Parallel flow, V=0 Incompressible, Newtonian, laminar, constant properties No pressure gradient 2D, W=0, /z = 0 Gravity acts in the -z direction, – Boundary conditions 1. Bottom plate (y=0) : u=0, v=0, w=0 2. Top plate (y=h) : u=V, v=0, w=0 Fully Developed Couette Flow • Step 3: Simplify 3 Note: these numbers refer to the assumptions on the previous slide 6 Continuity This means the flow is “fully developed” or not changing in the direction of flow X-momentum 2 Cont. 3 6 5 7 Cont. 6 Fully Developed Couette Flow • Step 3: Simplify, cont. Y-momentum 2,3 3 3,6 3 7 3 3 3 Z-momentum 2,6 6 6 6 7 6 6 6 Fully Developed Couette Flow • Step 4: Integrate X-momentum integrate integrate Z-momentum integrate Fully Developed Couette Flow • Step 5: Apply BC’s – y=0, u=0=C1(0) + C2 C2 = 0 – y=h, u=V=C1h C1 = V/h – This gives – For pressure, no explicit BC, therefore C3 can remain an arbitrary constant (recall only P appears in NSE). • Let p = p0 at z = 0 (C3 renamed p0) 1. 2. Hydrostatic pressure Pressure acts independently of flow Fully Developed Couette Flow • Step 6: Verify solution by back-substituting into differential equations – Given the solution (u,v,w)=(Vy/h, 0, 0) – Continuity is satisfied 0+0+0=0 – X-momentum is satisfied Fully Developed Couette Flow • Finally, calculate shear force on bottom plate Shear force per unit area acting on the wall Note that w is equal and opposite to the shear stress acting on the fluid yx (Newton’s third law). Momentum Equation • Special Case: Euler’s Equation Inviscid Flow for Steady incompressible • For steady incompressible flow, the equation reduces to v 0 ( v ) v p g where = constant. • Integrate from a reference at along any streamline y=C : p 2 2 v p v gz gz constant 2 2 20 Two-Dimensional Potential Flows • Therefore, there exists a stream function such that y in the Cartesian coordinate and y y u, v , x y in the cylindrical coordinate y y ur ,vq , r rq 21 Potential Flow Two-Dimensional Potential Flows • The potential function and the stream functiony are conjugate pair of an analytical function in complex variable analysis. y x y and y y x • The constant potential line and the constant streamline are orthogonal, i.e., u ,v to imply that . and y - v,u y 0 23 Stream and Potential Functions • If a stream function exists for the velocity field u = a(x2 -- y2) & v = - 2axy & w = 0 Find it, plot it, and interpret it. • If a velocity potential exists for this velocity field. Find it, and plot it. y y u, v , x y u x v y Summary • Elementary Potential Flow Solutions y Uniform Stream U∞y U∞x Source/Sink mq mln(r) -Kln(r) Kq Vortex 26