CP502 Advanced Fluid Mechanics Flow of Viscous Fluids Set 03.

CP502 Advanced Fluid Mechanics

Flow of Viscous Fluids

Set 03

R. Shanthini 05 April 2012

Continuity and Navier-Stokes equations for incompressible flow of Newtonian fluid

ρυ


x

yz

θ

direction of flow

Exercise 1:

Show that, for steady, fully developed laminar flow down the slope (shown in the figure), the Navier-Stokes equations reduces to

sin2

2 g

dy

ud

where u is the velocity in the x-direction, ρ is the density, μ is the dynamic viscosity, g is acceleration due to gravity, and θ is the angle of the plane to the horizontal. Solve the above equation to obtain the velocity profile u and obtain the expression for the volumetric flow rate for a flowing film of thickness h. Exercise 2:

If there is another solid boundary instead of the free-surface at y = h and the flow occurs with no pressure gradient, what will be the volumetric flow rate?

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid down an inclined plane under gravity


Navier-Stokes equation is already chosen since the system considered is incompressible flow of a Newtonian fluid.

Step 2: Choose the coordinate system

Cartesian coordinate system is already chosen.

Step 1: Choose the equation to describe the flow

x

yz

θ

direction of flowStep 3: Decide upon the functional dependence of the velocity components

Steady, fully developed flow and therefore no change in time and in the flow direction. Channel is not bounded in the z-direction and therefore nothing happens in the z-direction.

}(1)),,,(:direction zyxtfunctionux ),,,(:direction zyxtfunctionvy ),,,(:direction zyxtfunctionwz

}(2))(:direction yfunctionux )(:direction yfunctionvy

0:direction wz


Step 4: Use the continuity equation in Cartesian coordinates

0

z

w

y

v

x

u0

y

v

0orconstant vv

vFlow geometry shows that v can not be a constant, and therefore we choose

v

0v

x

yz

θ

direction of flow


}y direction: v = 0

z direction: w = 0

x direction: u = function of (y) (3)

The functional dependence of the velocity components therefore reduces to

Step 5: Using the N-S equation, we get

x - component:

y - component:

z - component:


N-S equation therefore reduces to

02

2

xgy

u

x

p

0

ygy

p

0

zgz

p

No applied pressure gradient to drive the flow. Flow is driven by gravity alone. Therefore, we get

0

z

p p is not a function of z

(4)

x - component:

y - component:

z - component:

x - component:

y - component:

z - component:

x

yz

θ

direction of flow

xgy

u

2

2

sin2

2 g

dy

ud

cosggy

py

What was asked to

be derived in

Exercise 1


x

yz

θ

direction of flow

Exercise 1:

Show that, for steady, fully developed laminar flow down the slope (shown in the figure), the Navier-Stokes equation reduces to

sin2

2 g

dy

ud




√done


(4)

x

yz

θ

sin2

2 g

dy

ud

Equation (4) is a second order equation in u with respect to y. Therefore, we require two boundary conditions (BC) of u with respect to y.

h

BC 1: At y = 0, u = 0 (no-slip boundary condition)BC 2: At y = h, (free-surface boundary condition)

0dy

du

Integrating equation (4), we get Ayg

dy

du

sin

Applying BC 2, we get hg

A

sin

(5)

(6)

Combining equations (5) and (6), we get yhg

dy

du

sin (7)


x

yz

θ

h

Integrating equation (7), we get

Applying BC 1, we get B = 0

(8)

(9)

Combining equations (8) and (9), we get (10)

By

hyg

u

2sin

2

2sin

2yhy

gu


Volumetric flow rate through one unit width fluid film along the z-direction is given by

(10)

2sin

2yhy

gu

dyuQh

0

dyy

hyg

Qh

0

2

2

sin

sin362

sin62

sin333

0

32 ghhhgyyh

gQ

h

(11)

x

yz

θ

h


x

yz

θ

direction of flow

Exercise 1:

Show that, for steady, fully developed laminar flow down the slope (shown in the figure), the Navier-Stokes equation reduces to

sin2

2 g

dy

ud




√done

√done


(4)

x

yz

θ

sin2

2 g

dy

ud

hBC 1: At y = 0, u = 0 (no-slip boundary condition)

BC 2: At y = h, (free-surface boundary condition)

0dy

du

Integrating equation (4), we get Ayg

dy

du

sin (12)

Equation does not change.BCs change.

u = 0 (no-slip boundary condition)

BAyyg

u

2sin

2

Integrating equation (12), we get (13)

Applying the BCs in (13), we get B = 0 and 2

sinhg

A


x

yz

θ

h

Therefore, equation (13) becomes

(14)

22sin

2yhygu


dyuQh

0

(15)

dyyhyg

Qh

0

2

22

sin

sin1264

sin64

sin333

0

32 ghhhgyhygQ

h


x

yz

θ

h

(14)

22sin

2yhygu

(10)

2sin

2yhy

gu

(15)

sin

12

3ghQ

sin3

3ghQ (11)

Free surface gravity flow Gravity flow through two planes

Summary of Exercises 1 and 2

Why the volumetric flow rate of the free surface gravity flow is 4 times larger than the gravity flow through two

planes?

x

yz

θ

h


Any clarification?


Exercise 3:

A viscous film of liquid draining down the side of a wide vertical wall is shown in the figure. At some distance down the wall, the film approaches steady conditions with fully developed flow. The thickness of the film is h. Assuming that the atmosphere offers no shear resistance to the motion of the film, obtain an expression for the velocity distribution across the film and show that

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid down a vertical plane under gravity

x

yz

h

where ν is the kinematic viscosity of the liquid, Q is the volumetric flow rate per unit width of the plate and g is acceleration due to gravity.

)3/1(3

g

Qh


Workout Exercise 3 in 5 minutes!


Oil Skimmer Example

An oil skimmer uses a 5 m wide x 6 m long moving belt above a fixed platform (= 60º) to skim oil off of rivers (T = 10ºC). The belt travels at 3 m/s. The distance between the belt and the fixed platform is 2 mm. The belt discharges into an open container on the ship. The fluid is actually a mixture of oil and water. To simplify the analysis, assume crude oil dominates. Find the discharge and the power required to move the belt.

hl

= 1x10-2 Ns/m2

= 860 kg/m3

30ºg

xy

U


N-S equation reduces to

02

2

xgy

u

x

p

0

ygy

p

0

zgz

p

No applied pressure gradient to drive the flow. Flow is driven by gravity alone. Therefore, we get

0

z

p p is not a function of z

(16)

x - component:

y - component:

z - component:

x - component:

y - component:

z - component:

xgy

u

2

2

cosggy

py

sin2

2 g

dy

ud

Oil Skimmer Discharge = ?


(16)

BC 1: At y = 0, u = 0 (no-slip boundary condition)

BC 2: At y = h, (free-surface boundary condition)

0dy

du


Sign changes in the equation

u = U (no-slip boundary condition)


Applying the BCs in (18), we get B = 0 and h

UhgA

2sin

sin2

2 g

dy

ud

Ayg

dy

du

sin

BAyyg

u

2sin

2


Therefore, equation (18) becomes

(19)yh

Uyhygu

22sin

2


(20)

dyh

Udy

yhygdyuQ

hhh

00

2

0

y 22

sin

2sin

12

3 UhghQ

2

m) 002.0(m/s) 3()30sin(

)Ns/m 10)(12(

m)002.0()m/s 806.9()kg/m 860( o22

323

Q

per unit width of the belt/sm 0027.0/sm 003.0/sm 000281.0 222 Q

/sm 0135.0m) 5/s)(m 0027.0( 32 Q


Oil Skimmer Power Requirements = ?

How do we get the power requirement?

What is the force acting on the belt?

Equation for shear?

Power = Force x Velocity [N·m/s]

Shear force (·L · W)

=(du/dy)


Evaluate=(du/dy) at the moving belt

(19)yh

Uyhygu

22sin

2

h

Uy

hg

dy

du

2sin

At the moving belt

h

Uhg

dy

du

hy

2sin

at belt at the

m) 002.0(

m/s) 3()Ns/m 10(

2

m002.0)(0.5)m/s 806.9()kg/m 860(

2223

belt at the

222belt at the kg/m.s 21.19kg/m.s 15kg/m.s 21.4 = 19.21 N/m2


Power = shear force at the belt * L * W * U

= (19.21 N/m2) (6 m) (5 m) (3 m/s)

= 1.73 kW

To reduce the power requirement, decrease the shear force


Exercise 4:

An incompressible, viscous fluid (of kinematic viscosity ν) flows between two straight walls at a distance h apart. One wall is moving at a constant velocity U in x-direction while the other is at rest as shown in the figure. The flow is caused by the movement of the wall. The walls are porous and a steady uniform flow is imposed across the walls to create a constant velocity V through the walls. Assuming fully developed flow, show that the velocity profile is given by

Also, show that

(i) u approaches Uy/h for small V, and

(ii) u approaches for very

large Vh/ν.

Steady, fully developed, laminar, incompressible flow of a Newtonian fluid over a porous plate sucking the fluid

dy

duV

dy

ud

2

2

/exp yhVU

x

y

z

U

VU

uv

h

UVh

Vyu

)/exp(1

)/exp(1


Step 2: Choose the coordinate system

Step 1: Choose the equation to describe the flow

Step 3: Decide upon the functional dependence of the velocity components

Steady, fully developed flow and therefore no change in time and in the flow direction. Channel is not bounded in the z-direction and therefore nothing happens in the z-direction.

}(1))(:direction yfunctionux )(:direction yfunctionvy

0:direction wz

done

done

Step 4: Use the continuity equation in Cartesian coordinates

0y

v

0or constant vv

x

y

z

U

VU

uv

h

Vv

0

z

w

y

v

x

u


}y direction: v = V

z direction: w = 0

x direction: u = function of (y) (2)

The functional dependence of the velocity components therefore reduces to

Step 5: Using the N-S equation, we get

x - component:

y - component:

z - component:


N-S equation therefore reduces to

No applied pressure gradient to drive the flow. Flow is caused by the movement of the wall. Therefore, we get

(3)

x - component:

y - component:

z - component:

x - component:

2

2

y

u

x

p

y

uV

gy

p

0

z

p

dy

duV

dy

ud

2

2

x

y

z

U

VU

uv

h


(3)

Equation (3) is a second order equation in u with respect to y. Therefore, we require two boundary conditions (BC) of u with respect to y. BC 1: At y = 0, u = 0 (no-slip boundary condition)BC 2: At y = h, u = U (no-slip boundary condition)



Applying the BCs in equation (5), we get

V

dy

du

dy

duV

dy

ud where

2

2

)exp( Aydy

du

BAyu )exp(1

BA )exp(1

0

BAhU )exp(1

(6)

(7)


From equations (6) and (7), we get

(8)

1)exp()exp(

h

UA

1)exp()exp(

1

h

UAB

Substituting the above in equation (5), we get

Uh

y

h

Uy

h

Uu

)exp(1

)exp(1

1)exp()exp(

1)exp(

UVh

Vyu

)/exp(1

)/exp(1

x

y

z

U

VU

uv

h


(8)UVh

Vyu

)/exp(1

)/exp(1

x

y

z

U

VU

uv

h

(i) For small V, expand exp(Vy/ν) and exp(Vh/ν) using Taylor series as follows:

For small V, we can ignore the terms with power. We then get

UVhVh

Vh

VyVyVy

u

!3

)/(

!2

)/()/(11

!3

)/(

!2

)/()/(11

22

22

Uh

yU

Vh

Vyu

/

/

Could you recognize the above profile?


(8)UVh

Vyu

)/exp(1

)/exp(1

x

y

z

U

VU

uv

h

For very large Vh/ν, exp(Vh/ν) goes to infinity. Therefore. Divide equation (8) by exp(Vh/ν). We then get

1)/exp(

)/exp()/exp()/exp(

Vh

VhVyVhu

For very large Vh/ν, exp(-Vh/ν) goes to zero. Therefore, we get

/exp yhVUu

UVhVy

u)1(

)/exp()/exp(

CP502 Advanced Fluid Mechanics Flow of Viscous Fluids Set 03.

Documents

Transcript of CP502 Advanced Fluid Mechanics Flow of Viscous Fluids Set 03.