Numerical on fundamental concepts from hydraulics 1

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
7
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
12
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
13
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
14
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
16
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