009b (PPT) Viscous Flow -2.pdf

1. 1
Fluid Mechanics
(MME-2252)
VISCOUS FLOW
IV Semester, Mechanical Engineering
Department of Mechanical and Manufacturing Engineering
Manipal Institute of Technology, Manipal
Dr. Vijay G. S.
Professor, Dept. of Mech. & Ind. Engg., MIT, Manipal
email: vijay.gs@manipal.edu
Mob.: 9980032104

2. Topics
2
 Reynold’s experiment, laminar and turbulent flows
 Viscous flow through a circular pipe (Hagen Poiseuille flow)
 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. Lecture Overview for Viscous
Flow
3
Session-1: Reynold’s experiment, laminar and turbulent flows
Viscous flow through a circular pipe (Hagen Poiseuille flow)
Session-2: Viscous flow between two parallel plates (Couette flow)
Session-3: Application based problems

4. Examples of Couette Flow
6
Slipping over Oily Surface
Flow in narrow gap of Journal
Bearings [1]
Axial viscous flow through annular space
between two fixed concentric cylinders [3,1]

5. 7
3. Flow of Viscous Fluid Between Two Parallel Plates:
• Consider steady, incompressible, laminar and fully developed flow between two infinite
parallel fixed plates (2D flow), separated by a distance ‘t’ and fluid of viscosity ‘µ’
• Imagine a fluid element of length △x, thickness △y and depth △z
• The element is located at a distance ‘y’ from the bottom fixed plate
A
B C
D
𝒑∆𝒚∆𝒛 𝒑 +
𝝏𝒑
𝝏𝒙
∆𝒙 ∆𝒚∆𝒛
𝝉 +
𝝏𝝉
𝝏𝒚
∆𝒚 △x∆𝒛
𝝉△x∆𝒛
△x
Figure: Viscous flow between two parallel Plates
A D
C
B z
x
y
y

6. 8
Assumptions: The flow is steady and uniform, so there is no acceleration.
෍
→+
𝑓 𝑥 = 0
𝑢 + 𝑑𝑢
𝑢
𝑢 + 2𝑑𝑢
𝑢
𝑢 + 𝑑𝑢
𝑢 + 2𝑑𝑢
So, the resultant force in the direction of flow is zero.
A D
C
B z
x
y

7. 9
−
𝜕𝑝
𝜕𝑥
∆𝑥∆𝑦∆𝑧 +
𝜕𝜏
𝜕𝑦
∆𝑦∆𝑥∆𝑧 = 0
Divide throughout by the volume of the element,
−
𝜕𝑝
𝜕𝑥
+
𝜕𝜏
𝜕𝑦
= 0
∴
𝜕𝑝
𝜕𝑥
=
𝜕𝜏
𝜕𝑦
𝑝∆𝑦∆𝑧 − 𝑝 +
𝜕𝑝
𝜕𝑥
∆𝑥 ∆𝑦∆𝑧 − 𝜏∆𝑥∆𝑧 + 𝜏 +
𝜕𝜏
𝜕𝑦
∆𝑦 ∆𝑥∆𝑧 = 0
(1)
෍
→+
𝑓 𝑥 = 0

8. 10
3.1: Velocity Distribution:
Substitute for τ from Newton’s law of viscosity to get expression for velocity distribution.
𝜕𝑝
𝜕𝑥
=
𝜕𝜏
𝜕𝑦
We have,
𝜕𝑝
𝜕𝑥
=
𝜕
𝜕𝑦
𝜇
𝜕𝑢
𝜕𝑦
𝜕2
𝑢
𝜕𝑦2
=
1
𝜇
𝜕𝑝
𝜕𝑥
Integrating above Equation w.r.t ‘y’,
𝜕𝑢
𝜕𝑦
=
1
𝜇
𝜕𝑝
𝜕𝑥
𝑦 + 𝐶1
Integrating again,
𝑢 =
1
𝜇
𝜕𝑝
𝜕𝑥
𝑦2
2
+ 𝐶1𝑦 + 𝐶2
Note:
𝜕𝑝
𝜕𝑥
is constant at a given section

9. 11
Here, C1 and C2 are integration constants and its values are determined by applying boundary
conditions.
The Boundary conditions are;
(I) y = 0 u = 0
(II) y = t u = 0
Use First Boundary condition,
𝑢 =
1
𝜇
𝜕𝑝
𝜕𝑥
𝑦2
2
+ 𝐶1𝑦 + 𝐶2
𝐶2 = 0
Substitute of Second boundary condition,
0 =
1
𝜇
𝜕𝑝
𝜕𝑥
𝑡2
2
+ 𝐶1𝑡 + 0
𝐶1 = −
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡
Substituting the values of C1 and C2 in equation, we get
𝑢 =
1
𝜇
𝜕𝑝
𝜕𝑥
𝑦2
2
−
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 + 0
𝑢 = −
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 − 𝑦2

10. 12
𝑢 = −
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 − 𝑦2
About mathematical equation,
1) 𝜇,
𝜕𝑝
𝜕𝑥
and t are constants
2) The velocity varies as the square of y, velocity distribution
across the section of parallel plates is parabolic in nature
3.2 Shear Stress Distribution
It is obtained by substituting the value of u into Newton’s law of viscosity
𝜏 = 𝜇
𝜕𝑢
𝜕𝑦
𝜏 = 𝜇
𝜕
𝜕𝑦
−
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 − 𝑦2
𝜏 = −
1
2
𝜕𝑝
𝜕𝑥
𝑡 − 2𝑦
Figure: Velocity profile
𝑈𝑚𝑎𝑥
y
x
ത
𝑢
Parabolic
Curve
(2)

11. 13
𝜏 = −
1
2
𝜕𝑝
𝜕𝑥
𝑡 − 2𝑦
About Mathematical Equation,
• Shear stress varies linearly with y
• Shear stress is maximum at the walls of the plates i.e. y = 0 and y = t
• Shear stress is zero at the center line between the two plates i.e. y = t/2
∴ 𝜏𝑚𝑎𝑥 = −
1
2
𝜕𝑝
𝜕𝑥
𝑡
𝜏𝑚𝑎𝑥
Figure: Shear stress distribution
and Velocity profile
𝑈𝑚𝑎𝑥
y
x
(3)

12. 14
3.3. Ratio of maximum velocity to average velocity
Maximum Velocity,
𝑢 = −
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 − 𝑦2
𝑢𝑚𝑎𝑥 = −
1
8𝜇
𝜕𝑝
𝜕𝑥
𝑡2
At y = t/2,
△x
t
Direction of Flow
y
x
A
B C
D
△y
Average Velocity,
ത
𝑢 =
𝑄
𝐴
Q can be obtained by considering the rate fluid flow through the strip of thickness △y and integrating it.
y

13. 15
dQ = velocity at distance y  Area of the strip
= 𝑢 × ∆𝑦∆𝑧
Substituting the velocity u from equation
𝑑𝑄 = −
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 − 𝑦2 × ∆𝑦∆𝑧
The total discharge Q is
𝑄 = න
0
𝑡
𝑑𝑄 = න
0
𝑡
−
1
2𝜇
𝜕𝑝
𝜕𝑥
𝑡𝑦 − 𝑦2
× ∆𝑦∆𝑧
𝑄 = −
∆𝑧
2𝜇
𝜕𝑝
𝜕𝑥
න
0
𝑡
𝑡𝑦 − 𝑦2
∆𝑦
𝑄 = −
∆𝑧
12𝜇
𝜕𝑝
𝜕𝑥
𝑡3
ത
𝑢 =
𝑄
𝐴𝑟𝑒𝑎
=
−
∆𝑧
12𝜇
𝜕𝑝
𝜕𝑥
𝑡3
𝑡×∆𝑧
ത
𝑢 = −
1
12𝜇
𝜕𝑝
𝜕𝑥
𝑡2
Ratio of maximum velocity to average velocity,
𝑈𝑚𝑎𝑥
ത
𝑢
=
−
1
8𝜇
𝜕𝑝
𝜕𝑥
𝑡2
−
1
12𝜇
𝜕𝑝
𝜕𝑥
𝑡2
𝑈𝑚𝑎𝑥
ത
𝑢
= 3/2
Average velocity,

14. 1
6
3.4) Drop in pressure for a given length of plates
From equation of velocity,
ത
𝑢 = −
1
12𝜇
𝜕𝑝
𝜕𝑥
𝑡2
𝜕𝑝
𝜕𝑥
= −
12𝜇ത
𝑢
𝑡2
Pressure loss between section 1 & 2
y x
x1
x2
L
1 2
Figure: Pressure Drop along the flow
න
1
2
𝑑𝑝 = න
1
2
−
12𝜇ത
𝑢
𝑡2
𝑑𝑥
(𝑝2−𝑝1) = −
12𝜇ത
𝑢
𝑡2
𝑥2 − 𝑥1
The pressure drop, ∆𝑝 =
12𝜇ത
𝑢𝐿
𝑡2
The drop in pressure head (Couette Formula),
ℎ𝑓 =
∆𝑝
𝜌𝑔
=
12𝜇ത
𝑢𝐿
𝜌𝑔𝑡2
(𝑝1−𝑝2) =
12𝜇ത
𝑢
𝑡2
𝐿

15. 17
3.5) The Power required to maintain the laminar flow between two fixed parallel plates
𝑃𝑜𝑤𝑒𝑟 = 𝐹 × ത
𝑢
Where,
F= Pressure Force across
the length of the pipe
Q= 𝜋𝑅2 × ത
𝑢
𝑃𝑜𝑤𝑒𝑟 = 𝑃1 − 𝑃2 × 𝐴𝑟𝑒𝑎 × ത
𝑢
∴ 𝑃𝑜𝑤𝑒𝑟 = 𝑃1 − 𝑃2 × 𝑄
3.6) Evaluate the total drag force due to the laminar flow
The drag force is appearing because of wall shear stress acting over the peripheral area of pipe.
𝐷𝑟𝑎𝑔 𝐹𝑜𝑟𝑐𝑒 = 𝜏𝑚𝑎𝑥 × 2 × 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑃𝑙𝑎𝑡𝑒
𝜏𝑚𝑎𝑥
R
Figure: Shear stress distribution profile

16. 18
The radial clearance between a hydraulic plunger and the cylinder walls is 0.1 mm. The
length of the plunger is 300 mm and diameter 100 mm. Find the velocity of leakage and rate
of leakage past the plunger at an instant when difference of the pressure between the two
ends of the plunger is 9 m of water. Take µ = 0.0127 poise.
Numerical -5
Note: Flow through the clearance area can be considered as flow between two parallel surfaces
Given Information
t = 0.1 mm= 0.0001 m
L= 300 mm= 0.3 m
D= 100 mm= 0.1 m
𝜇 = 0.00127 N-s/m2
𝑝1 − 𝑝2
𝜌𝑔
= 9 𝑚 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟
∆𝑝 = 9 × 1000 × 9.81 = 96236.1 𝑁/𝑚2
Objectives
1) Velocity of Leakage = Mean Velocity
2) Rate of Leakage, Q
Solution: 1) Mean Velocity
ത
𝑢 =
1
12𝜇
∆𝑝
𝐿
𝑡2
ത
𝑢 =
−1
12 × 0.00127
×
96236.1
0.3
× 0.00012
ഥ
𝒖 = 𝟎. 𝟐𝟏
𝒎
𝒔
= 𝟐𝟏 𝒄𝒎/𝒔
ℎ𝑓 =
∆𝑝
𝜌𝑔
=
12𝜇ത
𝑢𝐿
𝜌𝑔𝑡2

17. 19
Solution: 2) Rate of Leakage Q
𝑄 = ത
𝑢 × 𝑎𝑟𝑒𝑎
𝑄 = 0.21 × 𝜋𝐷𝑡
𝑄 = 6.5982 × 10−6 𝑚3/𝑠
𝑄 = 6.5982 × 10−3
𝑙𝑝𝑠

18. 20
Numerical -6
There is horizontal crack 40 mm wide and 2.5 mm deep in a wall of thickness 100 mm. Water
leaks through the crack. Find the rate of leakage through the crack if the difference of pressure
between the two ends of the crack is 0.02943 N/cm2. Take the viscosity of water equal to 0.01
poise.
Given Information
b= 40 mm= 0.04 m
t = 2.5 mm= 0.0025 m
L= 100 mm= 0.1 m
𝑝1 − 𝑝2= 0.02943 N/cm2 =294.3 N/m2
𝜇 = 0.001 N-s/m2
Objective
1) Rate of Leakage, Q
ℎ𝑓 =
∆𝑝
𝜌𝑔
=
12𝜇ത
𝑢𝐿
𝜌𝑔𝑡2
Solution: Using Couette Formula
ത
𝑢 = −
1
12𝜇
𝜕𝑝
𝜕𝑥
𝑡2
or
∆𝑝 =
12𝜇ത
𝑢𝐿
𝑡2
ത
𝑢 =
∆𝑝𝑡2
12𝜇𝐿
=
294.3 × 0.00252
12 × 0.001 × 0.1
= 1.5328 𝑚/𝑠
𝑅𝑎𝑡𝑒 𝑜𝑓 𝐿𝑒𝑎𝑘𝑎𝑔𝑒 𝑄 = ത
𝑢 × 𝑏 × 𝑡
𝑄 = 0.1538 × 10−4
𝑚3
𝑠
= 0.1528 𝑙𝑝𝑠

19. End of Session-2
21
Image Sources [3,2,1]
𝜏𝑚𝑎𝑥
𝑈𝑚𝑎𝑥
y
x