CCS355 Neural Network & Deep Learning UNIT III notes and Question bank .pdf
Β
009a (PPT) Viscous Flow-1 New.pdf .
1. 1
Fluid Mechanics
(MME-2252)
VISCOUS FLOW
IV Semester, Mechanical Engineering
Department of Mechanical and Manufacturing Engineering
Manipal Institute of Technology, MAHE, Manipal
Dr. Vijay G. S.
Professor, Dept. of Mech. & Ind. Engg., MIT, Manipal
email: vijay.gs@manipal.edu
Mob.: 9980032104
2. Topics
2
ο Reynoldβs experiment, laminar and turbulent flows
ο Viscous flow through a circular pipe (Hagen Poiseuille flow)
ο Viscous flow between two parallel plates (Plain Couette flow)
1) The shear stress distribution
2) The velocity distribution
3) The ratio of maximum velocity to average velocity
4) The drop of pressure for a given length
3. 3
Viscous Flow
Difference in Ideal and Real fluid Flow in Pipe
V
Figure (3): Velocity Profile in Real Fluid Flow
V
D V
Figure (2): Velocity Profile in Ideal Fluid Flow
β’ Viscosity offers to viscous shear resistance (Ο = Β΅ du/dy) and opposes the
motion between the layers
β’ This gives rise to change in velocity between layers i.e. velocity gradient
(du/dy) will appear normal to the flow direction
β’ Due to the resistance to motion, power is required to maintain flow of real fluids
β’ Hence, In fluid flow problems, effect of viscosity cannot be neglected
Why this change is appearing?
4. Laminar and Turbulent Flows
Depending upon the relative magnitudes of the viscous forces and inertia forces, flow can exist in
two types- Laminar Flow and Turbulent Flow
Laminar Flow: All particles move in well defined paths and
individual particles do not cross the paths taken by other particle.
Flow takes place in number of sheets or laminae. This type of flow
is also known as streamline or viscous flow.
Turbulent Flow: Particles move in zig-zag way in a flow field i.e.
the flow is disturbed and inter-mixing of particles takes place.
4
In this chapter, Our Focus is in laminar flow i.e. viscous effects are dominant and flow is well defined
5. 5
1) Reynoldsβs Experiment:
O. Reynold, a British professor demonstrated that a laminar flow changes to turbulent flow
Figure: Reynolds Dye Experimental setup and Observed dye streak lines [1]
Q V
6. Observations from the Reynoldβs Dye Experiment
6
For general case, transition of flow is function of
β’ Velocity of fluid (V)
β’ Diameter of the pipe (V)
β’ Viscosity of the fluid (π)
β’ Density of Fluid (Ο)
Figure: Time dependence of fluid velocity at a point [1]
ππ΄ = π’ ΖΈ
π
ππ΄ = π’ ΖΈ
π + π£ ΖΈ
π + π€ΰ·
π
7. Ο = Density of Fluid,
Vavg= Average Velocity,
D= Diameter of Pipe,
Β΅= Dynamic Viscosity
π π =
ππ
ππ£ππ·
π
For a flow in circular pipe,
If Re < 2000, the flow is said to be laminar
If Re > 4000, the flow is said to be turbulent
If 2000 < Re < 4000, in transition from laminar to turbulent
Identifying the type of flow from Reynolds Number
Critical Reynolds Number: It is magnitude of Reynolds Number below
which the flow is definitely laminar.
7
Figure: Laminar and Turbulent
flow regimes in candle smoke [3]
8. 2. Flow of Viscous Fluid Through a Circular Pipe
Explanation:
β’ Consider a horizontal circular cross section pipe of radius βRβ, through which a viscous fluid is flowing
from left to right
β’ Consider a cylindrical fluid element of radius r, thickness dr and length οx sliding inside a cylindrical
fluid element of radius (r + dr).
β’ Forces acting on element: A pressure forces are acting normal to the faces of element, and shear forces
are acting on the surface
Direction
of Flow
r
dr
A
B C
D
r
x
β³x
R
ππ ππ
π +
π π
π π
βπ π ππ
π Γ ππ π Γ βπ
8
9. 2.1 Shear stress distribution:
As there is no acceleration, the algebraic summation of
all the forces acting on the fluid element must be zero.
π Γ ππ2 β π +
ππ
ππ₯
βπ₯ ππ2 β π Γ 2ππ Γ βπ₯ = 0
β
ππ
ππ₯
βπ₯ ππ2
β π Γ 2ππ Γ βπ₯ = 0
π = β
ππ
ππ₯
π
2
(1)
9
ΰ·
β+
π π₯ = 0
10. β’ Shear stress distribution is function of radius of pipe π β π
β’ Shear stress varies linearly from zero at center of pipe to maximum at the boundary (r = R)
β’ The negative sign for pressure gradient indicates a decrease in the pressure in the direction of fluid flow.
β’ As potential and kinetic energies remain constant, the pressure force is the only means to compensate for
the resistance to flow.
Note: Pressure gradient remains constant
ππππ₯ = β
ππ
ππ₯
π
2
π = β
ππ
ππ₯
π
2
10
Meaning of Mathematical Expression ππππ₯
R
Figure: Shear stress distribution profile
A
A
r
x
11. 2.2 Velocity Distribution
To get the velocity distribution, substitute for shear stress in equation (1)
From Newtonβs law of viscosity,
In the above mathematical expression, y is measured from the pipe wall,
Substituting in the expression of Newtonβs law of viscosity,
βΈ« the expression for y can be written as
π = π
ππ’
ππ¦
At pipe wall, π¦ = 0,
At pipe center, π¦ = π
π¦ = π β π
π = βπ
ππ’
ππ
11
dy = -dr
π = π
π = 0
&
(2)
12. Integrating the equation we get,
Where, C is the integration constant and its value can be determined from the boundary conditions.
u = 0 at r = R (velocity is zero at pipe wall). Substituting the boundary condition in equation (3)
βπ
ππ’
ππ
= β
ππ
ππ₯
π
2
1
2
ππ’
ππ
=
1
2π
ππ
ππ₯
Γ π
ΰΆ± ππ’ = ΰΆ±
1
2π
ππ
ππ₯
π. ππ
π’ =
1
2π
Γ
ππ
ππ₯
Γ
π2
2
+ πΆ
β΄ π’ =
1
4π
Γ
ππ
ππ₯
Γ π2 + πΆ
Equating expressions (1) and (2) 20
(3)
13. β’ The velocity distribution of a fully developed laminar flow in a circular pipe must be parabolic.
β’ At the center of the pipe, the velocity is maximum. i.e. at r = 0, u = Umax
Note: In equation (4), Β΅,
ππ
ππ₯
and R are constants.
13
0 =
1
4π
Γ
ππ
ππ₯
Γ π 2 + πΆ
πΆ = β
1
4π
Γ
ππ
ππ₯
Γ π 2
Back substitution of C in (3) gives,
π’ =
1
4π
Γ
ππ
ππ₯
Γ π2 β
1
4π
Γ
ππ
ππ₯
Γ π 2
π’ = β
1
4π
Γ
ππ
ππ₯
Γ π 2 β π2
R r
(4)
14. Substituting r = 0 in equation (4),
Equation (4) can be written as
Substituting (5) in (6),
14
ππππ₯ = β
1
4π
Γ
ππ
ππ₯
Γ π 2
π’ = β
1
4π
Γ
ππ
ππ₯
Γ π 2
1 β
π2
π 2
π’ = ππππ₯ 1 β
π2
π 2
ππππ₯
R
Figure: Shear stress distribution and
Velocity profile
r ππππ₯
(5)
(6)
(7)
15. 15
Figure: Changes in velocity profile and pressure changes along the flow in pipe [2]
16. 16
we need to derive mathematical expressions for the following
1. The shear stress distribution
2. The velocity distribution
3. The ratio of maximum velocity to average velocity
4. The drop of pressure for a given length
Remember the objectives,
17. 3.3 Ratio of Maximum velocity to Average Velocity
17
The average velocity of the fluid flow can be determined by dividing the discharge Q by the
area of the pipe (ΟR2).
dQ = (Area of the ring element) ο΄ (Velocity at radius r) = π’ Γ 2ππ Γ ππ
Substitute for βuβ from equation (4),
ππ = β
1
4π
Γ
ππ
ππ₯
Γ π 2
β π2
Γ 2ππ Γ ππ
βΈ« The total discharge through the pipe is
π = ΰΆ±
0
π
ππ = ΰΆ±
0
π
β
1
4π
Γ
ππ
ππ₯
Γ π 2
β π2
Γ 2ππ Γ ππ
23
r
dr
The discharge βdQβ through the circular ring element of radius βrβ and thickness βdrβ is,
19. 3.4 Drop of Pressure Over a Given Length of Pipe:
19
As the fluid flows through the circular pipe, its pressure reduces along the direction of flow. This
is due to the resistance at the pipe wall that has to be compensated by the pressure energy.
r
x p1 p2
x1
x2
L
1 2
Figure: Pressure drop in pipe flow
Direction
of Flow
We have expression for average velocity,
ΰ΄₯
π =
π
π΄πππ
=
1
8π
β
ππ
ππ₯
π 2
β
ππ
ππ₯
=
8π ΰ΄₯
π
π 2
25
20. Integrating over the length of pipe between points 1 and 2, shown in figure.
β ΰΆ±
π1
π2
ππ = ΰΆ±
1
2
8π ΰ΄₯
π
π 2
ππ₯
β π2 β π1 =
8π ΰ΄₯
π
π 2
π₯2 β π₯1
π1 β π2 =
32π ΰ΄₯
ππΏ
π·2 Where, πΏ = π₯2 β π₯1, π =
π·
2
The βpressure head lost/dropβ, which is also known as the βfrictional head lossβ can be expressed in the
following form
(p1 β p2) = loss or drop in pressure
βπ =
π1 β π2
ππ
=
32π ΰ΄₯
ππΏ
πππ·2
By Substituting for ΰ΄₯
π = Ξ€
π ππ·2/4 βπ =
128πππΏ
ππππ·4
βHagen-Poiseuille Formulaβ
26
21. 21
1) The Power required to maintain the laminar flow in a pipe can be determined by the equation,
πππ€ππ = πΉ Γ ΰ΄€
π’ Where,
F= Pressure Force across the
length of the pipe
Q= ππ 2
Γ ΰ΄€
π’
πππ€ππ = π1 β π2 Γ ππ 2 Γ ΰ΄€
π’
β΄ πππ€ππ = π1 β π2 Γ π
2) Evaluate the total drag force on pipe of length L due to the laminar flow through it.
The drag force is appearing because of wall shear stress acting over the peripheral area of pipe.
π·πππ πΉππππ = ππππ₯ Γ 2ππ πΏ
ππππ₯
R
Figure: Shear stress distribution
profile
22. 22
Numerical -1
Calculate the loss of head in a pipe having a diameter of 15cm and length of 2 km. It carries oil
of specific gravity 0.85 and kinematic viscosity 6 stokes at the rate of 30.48 lps (Assume
Laminar Flow).
Given Information
D= 15 cm = 0.15 m
L= 2 km = 2000 m
S = 0.85
π = 6 stokes = 6 x 10-4 m2/s
Q = 30.48 lps = 30.48 x 103 cm3/s
= 30.48 x 10-3 m3/s
Ο = 850 kg/m3
Β΅= π x Ο = 0.51 N-s/m2
ΰ΄₯
π =
π
π΄
= 1.7246 π/π
Other required Information
βπ =
π1 β π2
ππ
=
32π ΰ΄₯
ππΏ
πππ·2
βπ =
32 Γ 0.51 Γ 1.7246 Γ 2000
850 Γ 9.81 Γ 0.152
βπ =300.03 m
Objective
Evaluate hf
Solution:
23. 23
Calculate the power required to maintain a laminar flow of an oil of viscosity 10 poise through a
pipe of 100 mm diameter at the rate of 10 lps, if the length of pipe is 1 km (Assume Laminar Flow).
Numerical -2
Given Information
D= 100 mm = 0.1 m
L= 1 km = 1000 m
π = 10 poise = 10 x 0.1= 1 N-s/m2
Q = 10 lps = 10 x 103 cm3/s
= 10 x 10-3 m3/s
Objective
Evaluate Power Required
πππ€ππ = π1 β π2 Γ π
βπ =
π1 β π2
ππ
=
32π ΰ΄₯
ππΏ
πππ·2
π1 β π2 =
32π ΰ΄₯
ππΏ
π·2
Required information
A= 7.855x 10-3 m2
ΰ΄₯
π =
π
π΄
= 1.2731 π/π
π1 β π2 = 407.38 πππ
Solution:
β΄ πππ€ππ = 407.38 Γ 10 x 10β3
πππ€ππ = 40.74 ππ
24. 24
Oil of viscosity 8 poise and specific gravity 1.2 flows through a horizontal pipe 80 mm in dimeter. If the
pressure drop in 100 m length of the pipe is 1500 kN/m2, determine rate of flow of oil in lpm, maximum
velocity, total frictional drag over 100 m length of pipe, power required to maintain flow, the velocity
gradient at the pipe wall, the velocity and shear stress at 10 mm from the wall.
Numerical -3
Given Information
π = 8 poise = 8 x 0.1= 0.8 N-s/m2
S= 1.2
D= 80 mm = 0.08 m
β³P(100m)= 1500 kN/m2
Objectives
Q (lpm), Umax, FD ( for 100 m),
Power, du/dr, u and Ο (at 10 mm)
βπ =
π1 β π2
ππ
=
32π ΰ΄₯
ππΏ
πππ·2
Solution: Flow Rate
π1 β π2 =
32π ΰ΄₯
ππΏ
π·2
ΰ΄₯
π = 1500 Γ 1000 Γ
0.082
32 Γ 0.8 Γ 100
= 3.75 π/π
π = π΄ΰ΄₯
π = 5.0272 Γ 10β3
Γ 3.75 = 0.018852 π3/π
π = 18.852 πππ
26. 26
Velocity at 10 mm from the wall
π’ = β
1
4π
Γ
ππ
ππ₯
Γ π 2
β π2
π’ =
1
4 Γ 0.8
Γ
π1 β π2
πΏ
π 2 β π2
y=10 mm,
y = R-r
r= 30 mm = 0.03 m
π’ =
1
4 Γ 0.8
Γ
1500 Γ 1000
100
0.042
β 0.032
u= 3.28125 m/s
Shear Stress at 10 mm from the wall
π = β
ππ
ππ₯
π
2
π =
π1 β π2
πΏ
π
2
π =
1500 Γ 1000
100
0.03
2
π = 225 π/π2
27. 27
Numerical -4
Oil is transported from a tanker to the shore at the rate of 0.6 m3/s using a pipe of 32 cm
diameter for a distance of 20 km. If the oil has viscosity of 0.1 N-s/m2 and density of 900
kg/m3, calculate the power necessary to maintain flow.
Given Information
Q = 0.6 m3/s
D = 32 cm= 0.32 m
L = 20 km
π = 0.1 N-s/m2
Ο = 900 kg/m3
Objectives
Power required
Solution: Power Required
πππ€ππ = π1 β π2 Γ π
βπ =
π1 β π2
ππ
=
32π ΰ΄₯
ππΏ
πππ·2
π1 β π2 =
32π ΰ΄₯
ππΏ
π·2
ΰ΄₯
π =
π
π΄
= 7.46 π/π
π1 β π2 =
32π ΰ΄₯
ππΏ
π·2
=
32 Γ 0.1 Γ 7.46 Γ 20000
0.322
π1 β π2 = 4.6625 πππ
β΄ πππ€ππ = π1 β π2 Γ π
πππ€ππ = 4.6625 Γ 106 Γ 0.6
πππ€ππ = 2.7975 ππ€
28. 28
End of Session-1
Figure: Shear stress distribution and typical velocity profiles for fluid flow in a pipe [1].