Flow Over an Inclined Plane
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Transcript of Flow Over an Inclined Plane
7/23/2019 Flow Over an Inclined Plane
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Fluid Mechanics II
MBB 2063 & MCB 2053
Introduction to Navier-Stokes Equations
7/23/2019 Flow Over an Inclined Plane
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Review of previous lecture…
• In the previous lecture, we have seen
– how the continuity equation and Navier-Stokes equations
are applied for 2D fluid flow problems – how to solve problems with one fixed boundary and one
moving plate
2
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In the Current Lecture…
• In this lecture …
– we will learn the techniques to solve problems
involving fluid flow down an inclined plane
3
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Intro to Navier-Stokes EquationsSummary of steps involved in solving the 3 specific scenarios:
1. Sketch diagram, label axes on diagram, label boundary conditions ondiagram
2. List down all assumptions involved, e.g. for steady flow , for
incompressible fluid etc
3. Apply continuity equation
4. Apply y momentum equation
5. Apply x momentum equation
6. Solve the resulting differential equations by double integration
7. Apply boundary conditions to get the 2 constants resulted from the
double integration
8. Substitute the 2 constants into the velocity equation (u equation) to
get the equation for velocity distribution
9. Calculate for any parameters required
4
0
t t cons tan
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Intro to Navier-Stokes Equations
The 3 specific scenarios that will be considered:
1. Fluid flow produced by a moving plate (seen in last lecture)
2. Fluid flow down an inclined plane
3. Flow between two stationary parallel plates
5
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SCENARIO II – Fluid Flow Down anInclined Plane
6
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Intro to Navier-Stokes Equations• Scenario II – Fluid flow down an inclined plane
Figure 2.2.1 Uniform Flow of a Crude Oil Layer on an Inclined Surface
Description of the problem:
A fluid is flowing down an inclined long plate. The fluid is crude oil with
viscosity, μ, and specific gravity, SG. The depth of the fluid layer is d . The
layer of fluid has a free surface on top. The shear stress at the free surface
between the air and fluid is small and negligible. The pressure does not
change in the streamline direction.
Determine the velocity profile. 7
BC2: y=d, du/dy=0
BC1: y=0, u=0
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Intro to Navier-Stokes Equations
•Step 1 –
• Step 2 –
• Step 3 –
8
done, as in Figure 2.2.1assumptions made:
Steady flow,
Incompressible fluid, ρ is constant
Viscous fluid,
Long plate,
Pressure does not change in the streamline direction,
0
t
0
0v0
x
p
Apply continuity equation
0
y
v
x
u
t
...
0
x
u
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Intro to Navier-Stokes Equations
• Step 4 –
• Step 5 –
9
Apply y momentum equation
cos2
2
2
2
g y
v
x
v
y
p
y
vv
x
vu
t
v
Apply x momentum equation
sin
2
2
2
2
g y
u
x
u
x
p
y
u
v x
u
ut
u
cos g y
p
sin2
2
g y
u
sin1
2
2
g
dy
ud
...
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Intro to Navier-Stokes Equations
• Step 6 –
10
Solve differential equations by doubleintegration
dy g dydy
ud
sin
12
2
sin1
2
2
g dy
ud
dyC y g dydy
du1sin
1
1sin
1C y g
dy
du
21
2
2sin
1C yC
y g u
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Intro to Navier-Stokes Equations
•
Step 7 –
11
Apply boundary conditions (BC) to getC1 and C2
BC1: y=0, u=0
0
2sin
1
2
21
2
C
C yC y
g u
BC2: y=d, 0
dy
du
d g C
C y g dy
du
sin1
sin1
1
1
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Intro to Navier-Stokes Equations
• Step 8 –
12
Substitute the C1 and C2 found into u
equation to get velocity profile
21
2
2sin
1C yC
y g u
Quadratic velocity profile is obtained
0
sin1
;
2
1
C
d g C with
yd y g u
yd g y
g u
2sin1
sin1
2sin
1
2
2
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Intro to Navier-Stokes Equations
The crude oil involved has kinematic viscosity, υ of 9.3 x 10-5 m2/s
and specific gravity, SG of 0.92. The depth of the fluid layer, d , is
6 mm. If the inclination is very small, with value of the slope, S o of
0.02, find:
i. discharge per meter of width of plate, q
ii. maximum velocity, umax of the flow
iii. mean velocity , umean of the flow
13
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Intro to Navier-Stokes Equations
• Step 9 –
14
Calculate for any parameters required
i. discharge per meter of width of plate, q
ii. maximum velocity, umaxiii. mean velocity, umean
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Intro to Navier-Stokes Equations
i. Discharge per meter of width of plate, q
15
d
dyuq0
dy y y g
q
006.0
2
sin 2
006.0
0
smq /1052.1 24
006.0
0
23
2
006.0
23
sin
y
y
y y g q
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Intro to Navier-Stokes Equations
ii. maximum velocity of the flow, umax
16
yd
y g u
2sin
1 2
;0dydu
d y g dy
du
d g y g dy
du
C y g dydu
sin1
sin1
sin1
sin1 1
d y
d y
dy
du
0
0
We have obtained a quadratic velocity profile earlier on :
to find the maximum velocity, let
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Intro to Navier-Stokes Equations
umax occurs when y=d =0.006 m
17
yd
y g u
2sin
1 2
smu
S Slope
ninclinatio small for
o
/038.0006.0
2
006.0
103.9
02.081.9
02.0tan
tansin
22
5max
006.0006.02
006.0sin
006.02
sin1
2
max
2
g u
y y
g u
Substitute y=0.006 m to the velocity equation and knowing depth of fluid
layer, d , is 0.006
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Intro to Navier-Stokes Equations
iii. mean velocity of the flow , umean
18
d
qumean
sm
m
smumean
/0253.0
006.0
/1052.1 24
the mean velocity can be calculated by division of q with cross-sectional
area, d :
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Intro to Navier-Stokes Equations
• By the end of this lecture, you should have …
– learnt the techniques in solving for fluid flow
down an inclined plane
19