Pharos University ME 253 Fluid Mechanics 2 Revision for Mid-Term Exam Dr. A. Shibl
Jan 05, 2016
Pharos UniversityME 253 Fluid Mechanics 2
Revision for Mid-Term ExamDr. A. Shibl
Streamlines• A Streamline is a curve that is
everywhere tangent to the instantaneous local velocity vector.
• Consider an arc length
• must be parallel to the local velocity vector
• Geometric arguments results in the equation for a streamline
dr dxi dyj dzk
dr
V ui vj wk
dr dx dy dz
V u v w
Kinematics of Fluid Flow
1 1 1
2 2 2
w v u w v ui j k
y z z x x y
1xx yy zz
DV u v w
V Dt x y z
1 1 1, ,
2 2 2xy zx yz
u v w u v w
y x x z z y
2
Stream Function forTwo-Dimensional
Incompressible Flow
• Two-Dimensional Flow
Stream Function
Stream Function forTwo-Dimensional
Incompressible Flow
• Cylindrical Coordinates
Stream Function (r,)
Is this a possible flow field
Given the y-component Find the X- Component of the velocity,
Determine the vorticity of flow field described byIs this flow irrotational?
Momentum Equation
• Newtonian Fluid: Navier–Stokes Equations
Example exact solutionPoiseuille 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. Plates are infinite in x and z2. Flow is steady, /t = 03. Parallel flow, V=04. Incompressible, Newtonian, laminar, constant properties5. No pressure gradient6. 2D, W=0, /z = 07. Gravity acts in the -z direction,
– Boundary conditions1. Bottom plate (y=0) : u=0, v=0, w=02. Top plate (y=h) : u=V, v=0, w=0
Fully Developed Couette Flow
• Step 3: Simplify 3 6
Note: these numbers referto the assumptions on the previous slide
This means the flow is “fully developed”or not changing in the direction of flow
Continuity
X-momentum
2 Cont. 3 6 5 7 Cont. 6
Fully Developed Couette Flow
• Step 3: Simplify, cont.Y-momentum
2,3 3 3 3,6 7 3 33
Z-momentum
2,6 6 6 6 7 6 66
Fully Developed Couette Flow
• Step 4: Integrate
Z-momentum
X-momentum
integrate integrate
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. Hydrostatic pressure2. 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 satisfied0 + 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
20
Inviscid Flow for Steady incompressibleInviscid Flow for Steady incompressible
• For steady incompressible flow, the equation reduces to
where = constant.
• Integrate from a reference at along any streamline =C :
gvv p)(0 v
constant22
v 22
gzvp
gzp
21
Two-Dimensional Potential FlowsTwo-Dimensional Potential Flows
• Therefore, there exists a stream function such that in the Cartesian coordinate and
in the cylindrical coordinate
r
,r
v,ur
xy
vu
,,
Potential Flow
23
Two-Dimensional Potential Flows• The potential function and the stream function are conjugate pair
of an analytical function in complex variable analysis.
• The constant potential line and the constant streamline are orthogonal, i.e.,
and
to imply that .
xyyx
and
u,v- v,u
0
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.
yv
xu
xy
vu
,,
Summary• Elementary Potential Flow Solutions
26
Uniform Stream U∞y U∞x
Source/Sink m mln(r)
Vortex -Kln(r) K