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



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



In the Current Lecture…

• In this lecture …

– we will learn the techniques to solve problems

involving fluid flow down an inclined plane

3



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



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



SCENARIO II – Fluid Flow Down anInclined Plane

6



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




•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




• 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

...




• 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




•

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




• 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




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




• Step 9 –

14

Calculate for any parameters required

i. discharge per meter of width of plate, q

ii. maximum velocity, umaxiii. mean velocity, umean




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




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




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




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 :




• By the end of this lecture, you should have …

– learnt the techniques in solving for fluid flow

down an inclined plane

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Flow Over an Inclined Plane