The document contains numerical problems related to fluid mechanics concepts like viscosity, surface tension, and capillary action. It provides the formulas and steps to calculate viscosity of fluids given velocity and shear stress between plates, shear stress given viscosity and velocity gradient, viscosity from specific gravity and dynamic viscosity. It also shows calculations for surface tension using pressure difference and circumference of a droplet, and capillary depression in a glass tube based on surface tension, contact angle and dimensions.

1. A S Pawar
Assistant Professor
Module 1
FUNDAMENTAL CONCEPTS

2. 01-01-2024
2
NUMERICAL
A plate 0.0254 mm distant from a fixed plate moves at 61 cm/s and
requires a force of 0.2 kg/m2 to maintain this speed. Determine the
dynamic viscosity of the fluid between the plates.
Given Data:
y = 0.0254 mm = 2.54 X 10-5 m
V = 61 cm/s = 0.61 m/s
 = 0.2 kg/m2
We have,
 = μ
𝑑𝑢
𝑑𝑦
0.2 = μ X
0.61
2.54 𝑋 10−5
μ = 8.328 X 10-6 kg/ms

3. 01-01-2024
3
NUMERICAL
In a stream of glycerine in motion at a certain point, the velocity
gradient is 0.25 s-1. If for fluid, density is 1268.4 kg/m3 and kinematic
viscosity is 6.30 X 10-4 m2/s, calculate the shear stress at that point.
Given Data:
r = 1268.4 kg/m3
𝑑𝑢
𝑑𝑦
= 0.25 s-1
 = 6.30 X 10-4 m2/s
 =
μ
ρ
μ =  x r
= ( 6.30 X 10-4 ) X (1268.4 )
= 0.799 Ns/m2

4. 01-01-2024
4
 = μ
𝑑𝑢
𝑑𝑦
= 0.799 X 0.25
= 0.1997 N/m2

5. 01-01-2024
5
NUMERICAL
Calculate the dynamic viscosity of an oil which is used for lubrication
between a square plate of size 0.8 m x 0.8 m and an inclined plane of
angle of inclination 300 as shown in figure. The weight of square plate
is 300 N and it slides down the inclined plane with a uniform velocity
of 0.3 m/s. The thickness of film is 1.5 mm.
Given Data:

6. 01-01-2024
6
Given Data:
A = 0.8 x 0.8 = 0.64 m2
 = 300
W = 300 N
u = 0.3 m/s
t = dy = 1.5 mm = 1.5 x 10-3 m
Force on bottom of plate
= 300 sin 300 = 150 N
Shear stress,
 =
F
A
=
150
0.64
= 234.375 N/m2

7. 01-01-2024
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We have,
 = μ
𝑑𝑢
𝑑𝑦
234.375 = μ x
(0.3 −0)
1.5 𝑥 10−3
μ = 1.17 Ns/m2
= 1.17 x 10 poise
μ = 11.7 poise

8. 01-01-2024
8
NUMERICAL
A certain liquid has a dynamic viscosity of 0.073 poise and specific
gravity of 0.87. Compute the kinematic viscosity of liquid in units of
stokes and m2/s
Given Data:
μ = 0.073 poise
G = 0.87
We know,
1 kg/ms = 98.1 poise
Therefore, x kg/ms = 0.073 poise
i.e. x x 98.1 = 0.073 x 1
x = 7.44 x 10-4 kg/ms

9. 01-01-2024
9
We have,
G =
rliq
r𝑤
0.87 =
rliq
1000
rliq = 870 kg/m3
Now,
 =
μ
ρ
=
7.44 x 10−4
870
= 8.55 x 10-7 m2/s

10. 01-01-2024
10
 = 8.55 x 10-7 m2/s
= 8.55 x 10-7 x 104 cm2/s
= 0.00855 cm2/s
= 0.00855 stokes

11. 01-01-2024
11
NUMERICAL
If the equation of velocity distribution over a plate is given by :
V = 2y – y2, in which ‘V’ is the velocity in m/s at a distance ‘y’,
measured in ‘m’ above the plate. What is the velocity gradient at the
boundary and at 7.5 cm & 15 cm from it? Also determine the shear
stress at these points if absolute viscosity is 8.60 poise.
Given Data: μ = 8.60 poise
We have,
V = 2y – y2
Derive w.r.t ‘y’
dv
dy
= 2 – 2y …………………Velocity gradient
1) At boundary, y = 0
dv
dy
= 2 – 2(0) = 2 s-1

12. 01-01-2024
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dv
dy
= 2 – 2y …………………Velocity gradient
2) At y = 7.5 cm = 0.075 m
dv
dy
= 2 – 2(0.075)
= 1.85 s-1
3) At y = 15 cm = 0.15 m
dv
dy
= 2 – 2(0.15)
= 1.70 s-1
Now, μ = 8.60 poise
1 Ns/m2 = 10 poise
Therefore, x Ns/m2 = 8.60 poise
i.e x = μ = 0.86 Ns/m2

13. 01-01-2024
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Now,
For
dv
dy
= 2 s-1
 = μ
𝑑𝑢
𝑑𝑦
= 0.86 x 2
= 1.72 Ns/m2
For
dv
dy
= 1.85 s-1
 = μ
𝑑𝑢
𝑑𝑦
= 0.86 x 1.85
= 1.591 Ns/m2

14. 01-01-2024
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Now,
For
dv
dy
= 1.70 s-1
 = μ
𝑑𝑢
𝑑𝑦
= 0.86 x 1.70
= 1.462 N/m2

15. 01-01-2024
15
NUMERICAL
A glass tube 0.25 mm in diameter contains mercury column with air
above the mercury at 200 C. The surface tension of mercury in contact
with air 0.051 N/m. What will be the capillary depression of mercury
if the angle of contact is 1300 and specific gravity of mercury is 13.6
Given Data :
D = 0.25 mm = 0.25 x 10-3 m
σ = 0.051 N/m
θ = 1300
S =
r𝑚𝑒𝑟𝑐𝑢𝑟𝑦
rwater
13.6 =
r𝑚𝑒𝑟𝑐𝑢𝑟𝑦
1000
r𝑚𝑒𝑟𝑐𝑢𝑟𝑦 = 13600 kg/m3

16. 01-01-2024
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Now,
h =
4σ cosθ
r.𝑔.𝐷
=
4 x 0.051 x cos 1300
13600 x 9.81 x 0.25 x 10−3
= 0.0039 m
Hence, capillary depression = 3.9 mm

17. Surface Tension
Surface Tension on liquid Droplet and
Bubble
 Consider a small spherical droplet of a
liquid of radius ‘R'. On the entire
surface of the droplet, the tensile force
due to surface tension will be acting.
 Let σ = surface tension of the liquid
P= Pressure intensity inside the
droplet in excess of the outside
pressure intensity
 Let the droplet is cut into two halves.
The forces acting on one half (say left
half) will be
17

18.  (i) tensile force due to
surface tension acting
around the circumference
of the cut portion as
shown and this is equal to
= σ x Circumference
= σ x 2πR
 (ii) pressure force on the
area
= P x πR2
Surface Tension
18

19.  These two forces will be equal and opposite under
equilibrium conditions, i.e.,
 A hollow bubble like a soap bubble in air, has two
surfaces in contact with air, one inside and other outside.
Thus two surfaces are subjected surface tension.
Surface Tension
19

20. Surface Tension……. Example 1
 Find the surface tension in a soap bubble of 40 mm
diameter when the inside pressure is 2.5 N/m2 above
atmospheric pressure.
20

21. Surface Tension……. Example 2
 The pressure outside the droplet of water of diameter 0.04
mm is 10.32 N/cm2 (atmospheric pressure). Calculate the
pressure within the droplet if surface tension is given as
0.0725 N/m of water.
21

22. 22