CP502 Advanced Fluid Mechanics Flow of Viscous Fluids Set 03.
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Transcript of 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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
}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:
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
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
√done
R. Shanthini 05 April 2012
(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)
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
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
√done
√done
R. Shanthini 05 April 2012
(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
R. Shanthini 05 April 2012
x
yz
θ
h
Therefore, equation (13) becomes
(14)
22sin
2yhygu
Volumetric flow rate through one unit width fluid film along the z-direction is given by
dyuQh
0
(15)
dyyhyg
Qh
0
2
22
sin
sin1264
sin64
sin333
0
32 ghhhgyhygQ
h
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
Any clarification?
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
Workout Exercise 3 in 5 minutes!
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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 = ?
R. Shanthini 05 April 2012
(16)
BC 1: At y = 0, u = 0 (no-slip boundary condition)
BC 2: At y = h, (free-surface boundary condition)
0dy
du
Integrating equation (16), we get (17)
Sign changes in the equation
u = U (no-slip boundary condition)
Integrating equation (17), we get (18)
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
R. Shanthini 05 April 2012
Therefore, equation (18) becomes
(19)yh
Uyhygu
22sin
2
Volumetric flow rate through one unit width fluid film along the z-direction is given by
(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
R. Shanthini 05 April 2012
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)
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
}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:
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
(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)
Integrating equation (3), we get (4)
Integrating equation (4), we get (5)
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)
R. Shanthini 05 April 2012
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
R. Shanthini 05 April 2012
(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?
R. Shanthini 05 April 2012
(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(