Unit I discusses fluid properties and flow characteristics. It covers fluid density, viscosity, surface tension, compressibility, and vapor pressure. It also discusses the concepts of control volume and the application of continuity, energy, and momentum equations to fluid flow. Specific topics covered include density, specific gravity, viscosity, surface tension, capillarity, compressibility, and vapor pressure. Several sample problems are worked through to demonstrate calculations for properties like density, specific weight, viscosity, surface tension, and capillary rise.

1. UNIT I - FLUID PROPERTIES &
FLOW CHARACTERISTICS

2. UNIT I FLUID PROPERTIES AND FLOW CHARACTERISTICS
Units and dimensions - Properties of fluids- mass density, specific weight,
specific volume, specific gravity, viscosity, compressibility, vapor pressure,
surface tension and capillarity. Flow characteristics – concept of control
volume - application of continuity equation, energy equation and
momentum equation.

3. Fluid Properties:
1. Density
2. Specific Gravity
3. Viscosity
4. Surface Tension
5. Capillarity
6. Compressibility
7. Bulk modulus
8. Vapour pressure

4. Density:
Mass density or specific mass: 
The density of a fluid is defined as its mass per unit volume at
standard temperature and pressure.
𝜌 =
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
Where,
 =
𝑚
𝑣
unit - kg/m3
For water,  = 1000 kg/m3

5. Specific Weight or weight density: w
It is defined as Weight per unit volume at standard temperature and pressure
Where,
w = g ; Unit - N/m3 ;
wwater = 9.81 kN/m3
𝑤 =
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
=
𝑀𝑎𝑠𝑠 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
× 𝑔
Specific Volume: v
It is defined as Volume per unit mass of the fluid.
v = V/m = 1/
Unit - m3/kg; 1 m3 = 1000 litres

6. Specific Gravity : S
It is the ratio of the specific weight of the liquid to the
specific weight of the standard fluid.
𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 =
specific weight of the liquid
Specific weight of the pure water
Where,
Weight Density of Liquid = S X Weight density of water
= S X 1000 X 9.81 N/m3
Density of Liquid = S X density of water
= S X 1000 kg/m3

7. Problem 1: Calculate the density, specific weight and weight of one litre of
petrol of specific gravity = 0.7
Given:
𝑉 = 1 𝑙𝑖𝑡𝑟𝑒 = 1 𝑥 1000 𝑐𝑚3 =
1000
106 m3 = 0.001 m3
S = 0.7
Solution:
(i) Density, ρ = S x 1000 kg/m3 = 0.7 x 1000 = 700 kg/m3
(ii) Specifc weight, w = ρ x g = 700 x 9.81 N/m3 = 6867 N/m3
(iii) Weight, W =
𝑊𝑒𝑖𝑔ℎ𝑡
𝑉𝑜𝑙𝑢𝑚𝑒
w =
𝑊
0.001
or 6867 =
𝑊
0.001
W = 6867 x 0.001
W = 6.867 N

8. Kinematic viscosity:
It is defined as the ratio between the dynamic viscosity and mass
density of fluid.
Units: m2/s ; 1 Stoke = 10-4 m2/s
Viscosity:
Dynamic viscosity:
Viscosity is defined as the property of fluid which offers resistance to
flow.
Unit - Ns/m2 ; 1 Poise = (1/10) Ns/m2 ; 1 Centi poise= (1/100) Poise
𝜏 𝛼
𝑑𝑢
𝑑𝑦
, 𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
,
𝜈 =
𝜇
𝜌

9. Dynamic Viscosity
Force
Area
Shear
stress
(
)
Shear rate (du/dy)
Slope of line
 =
𝜇 =
𝜏
ൗ
𝑑𝑢
𝑑𝑦

10. Problem 2: A Plate 0.025 mm distant from a fixed plate, moves at 60 cm/s and
requires a force of 2N per unit area i.e., 2 N/m2 to maintain this speed.
Determine the fluid viscosity between the plates.
Given:
dy = 0.025 mm = 0.000025 m
u = 60 cm/s = 0.6 m/s
du = u – 0 = 0.6 m/s
𝜏 = 2 N/m2
Solution:
µ =
𝜏
𝑑𝑢
𝑑𝑦
=
2
0.6
0.000025
µ = 8.33 x 10-5 Ns/m2 (or) 8.33 x 10-4 poise

11. Problem 3: Determine the intensity of shear of an oil having viscosity = 1 poise. The oil
is used for lubricating the clearance between a shaft of diameter 10 cm and its journal
bearing. The clearance is 1.5 mm and the shaft rotates at 150 r.p.m.
Given:
µ = 1 poise =
1
10
𝑁 𝑠
𝑚2
D = 10 cm = 0.1 m
Dy = 1.5 mm = 0.0015 m
N = 150 r.p.m.
Solution:
u =
𝜋𝐷 𝑁
60
=
𝜋 ∗0.1 ∗150
60
= 0.785 m/s
du = u – 0 => u = 0.785 m/s
𝜏 = µ
𝑑𝑢
𝑑𝑦
=
1
10
x
0.785
1.5 𝑥 10−3
𝝉 = 52.33 N/m2

12. Problem 4: If the velocity distribution over a plate is given by u= 3/4 y – y2 in which u is
the velocity in mertre per sec at a distance y metre above the plate. Determine the shear
stress at y = 0 and y = 0.15 m. Take dynamic viscosity of fluid as 8.5 poise.
Solution:
u =
3
4
y – y 2
𝑑𝑢
𝑑𝑦
=
3
4
− 2 y
At y = 0.15,
𝑑𝑢
𝑑𝑦
=
3
4
− 2 ∗ 0.15 = 0.75 − 0.30 = 0.45
Viscosity, µ = 8.5 poise = 0.85 Ns/m2
We know,
𝜏 = µ
𝑑𝑢
𝑑𝑦
= 0.85 ∗ 0.45
𝝉 = 0.3825 N/ m2

13. Problem 5: A 400 mm diameter shaft is rotating at 200 r.p.m. in a bearing of length 120 mm. If
the thickness of oil film is 1.5 mm and the dynamic viscosity of the oil is 0.7 N.s/m2 determine:
(i) Torque required to overcome friction in bearing (ii) Power utilised in overcoming viscous
resistance.
Given: D = 400 mm = 0.4 m
N = 200 rpm
L = 120 mm = 0.12 m
dy = 1.5 mm
µ = 0.7 Ns/m2
To find:
(i) Torque
(ii) Power
Solution:
u1 = 0; u2 =
𝜋𝐷 𝑁
60
=
𝜋 ∗ 0.4 ∗ 120
60
= 2.5 m/s
du = u2 – u1 = 2.5 m/s

14. µ =
𝜏
𝑑𝑢
𝑑𝑦
=> 0.7 =
𝜏
2.5
0.0015
=> 𝜏 = 1172.26 N/m2
Fshear = 𝜏 * A = 𝜏 * 𝜋 * D * L
= 1172.26 * 𝜋 * 0.4 * 0.12
Fshear = 176.77 N
T = F Shear * r
= 176.77 * 0.2
T = 35.2 N m
P =
2𝜋 𝑁𝑇
60
=
2𝜋 120∗35.2
60
P = 737.2 W

15. Problem 6: Calculate the dynamic viscosity of an oil, which is used for lubrication between a
square plate of slice 0.8 m x 0.8 m and an inclined plane with angle of inclination 30o as shown
in fig. The weight of the square plate is 300 N and it slides down the inclined plane with a
uniform velocity of 0.3 m/s. The thickness of oil film is 1.5 mm.
Given:
A = 0.8 x 0.8 = 0.64 m2
θ = 30o
W = 300 N
u = 0.3 m/s
du = u – 0 = 0.3 m/s
t = dy = 1.5 mm = 1.5 x 10-3 m
To find:
µ

16. Solution:
Component of weight W, along the plane = W cos 60o
= 300 cos 60o = 150 N
Shear force F, on the bottom surface of the plate = 150 N
𝜏 =
𝐹
𝐴𝑟𝑒𝑎
=
150
0.64
= 234.375
Using Newton’s law,
µ =
𝜏
𝑑𝑢
𝑑𝑦
µ =
234.375
0.3
0.0015
= 1.17 Ns/m2
µ = 11.7 poise.
30o
60o
W

17. Home work: Two large plane surfaces are 2.4 cm apart. The space between the surfaces is filled
with glycerine. What force is required to drag a very thin plate of surface area 0.5 square metre
between the two large plane surfaces at a speed of 0.6 m/s, if (i) the thin plate is in the middle of
the two plane surfaces, and (ii) the thin plate is at a distance of 0.8 cm from one of the plane
surfaces? Take the dynamic viscosity of glycerine = 8.10 x 10-1 Ns/m2

18. Types of Fluids:
Shear
stress
(𝝉)
Shear rate (du/dy)
Ideal
Solid
Ideal Fluid

19. Ideal Fluid:
A fluid, which is incompressible and is having no viscosity, is known as Ideal
fluid. It is only imaginary Fluid
Real Fluid:
A fluid, which is having viscosity, is known as Real fluid. All fluids are Real
fluids.
Newtonian fluids:
These fluids follow Newton’s viscosity equation. For such fluids  does not
change with rate of deformation. Exp. Water, Kerosene etc.,
Non Newtonian fluids:
There fluids which do not follow the linear relationship between the shear
stress and the rate of deformation. Exp. Polymer solutions, Blood etc.,

20. Surface Tension of Liquids
The phenomenon of surface tension arises due to the two kinds
of intermolecular forces. Surface Tension is caused by the force of
cohesion at the free surface.
(i) Cohesion : The force of attraction between the molecules of the
same liquid. Surface tension is due to cohesion between particles at
the free surface.
(ii) Adhesion : The force of attraction between unlike molecules of a
solid boundary surface in contact with the liquid

21. Case I - Water droplet:
Let, P = Pressure inside the droplet above outside pressure
D = Diameter of the droplet
 = Surface tension of the liquid
i. Pressure force = 𝑃 ×
𝜋
4
𝑑2 and
ii. Surface tension force acting around the circumference =   d
Under equilibrium conditions these two forces will be equal and opposite,
𝑃 ×
𝜋
4
𝑑2 = 𝜎 × 𝜋𝑑
𝑝 =
4𝜎
𝑑

22. Case II - Soap bubble:
i. Pressure force = 𝑃 ×
𝜋
4
𝑑2
, and
ii. Surface tension force acting around the circumference =2(  d)
𝑃 ×
𝜋
4
𝑑2
= 2 𝜎 × 𝜋𝑑
𝑝 =
8𝜎
𝑑
Case III- A liquid Jet:
i. Pressure force = p l d , and
ii. Surface tension force acting around the circumference = 2 l
𝑃 × 𝑙 × 𝑑 = 𝜎 × 2𝑙
𝑝 =
2𝜎
𝑑

23. Surface Tension Formulae:
S. No. Pressure force Surface Tension Pressure
Water Jet p l d  2 l 𝑝 =
2𝜎
𝑑
Water Droplet 𝑃 ×
𝜋
4
𝑑2
  d 𝑝 =
4𝜎
𝑑
Soap Bubble 𝑃 ×
𝜋
4
𝑑2
2 (  d) 𝑝 =
8𝜎
𝑑

24. Problem 7: Find the surface tension in a soap bubble of 40 mm diameter when
the inside pressure is 2.5 N/m2 above atmospheric pressure.
Given:
d = 40 mm = 0.04m
p = 2.5 N/m2
To find :
Surface Tension (σ)
Solution:
p =
8𝜎
𝑑
σ =
p ∗ d
8
=>
2.5 ∗ 0.04
8
σ = 0.0125 N/m

25. Capillarity:
➢Capillarity is a phenomenon by which a liquid rises into a thin glass tube
above or below its general level. surface relative to the adjacent level of
the fluid is called capillarity.
➢This phenomenon is due to the combined effect of cohesion and adhesion
of liquid particles.
For water  = 0
For mercury  = 1400

27. Expression for Capillary Rise:
 = Surface tension of the liquid
 = Angle of contact between the liquid and glass tube
 = Density of the liquid
The weight of the liquid of height h in the tube = (Area of the tube x h) x  x g
=
𝜋
4
𝑑2
× ℎ × 𝜌 × 𝑔
Vertical component of the surface tensile force= (𝜎 × 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒) × cos 𝜃
= 𝜎 × 𝜋𝑑 × cos 𝜃
Equating these two eqn.,
𝜋
4
𝑑2
× ℎ × 𝜌 × 𝑔 = 𝜎 × 𝜋𝑑 × cos 𝜃
ℎ =
𝜎 × 𝜋𝑑 × cos 𝜃
𝜋
4
𝑑2 × 𝜌 × 𝑔
=
4𝜎 cos 𝜃
𝜌 × 𝑔 × 𝑑
ℎ =
4𝜎
𝜌 × 𝑔 × 𝑑
If  is equal to zero

28. Expression for Capillary Fall:
 = Surface tension of the liquid
 = Angle of contact between the liquid and glass tube
 = Density of the liquid
Intensity of the pressure at the depth = (Area of the tube x h) x  x g
=
𝜋
4
𝑑2 × ℎ × 𝜌 × 𝑔
Vertical component of the surface tensile force = (𝜎 × 𝐶𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒) × cos 𝜃
= 𝜎 × 𝜋𝑑 × cos 𝜃
Equating these two eqn.,
𝜋
4
𝑑2
× ℎ × 𝜌 × 𝑔 = 𝜎 × 𝜋𝑑 × cos 𝜃
ℎ =
𝜎 × 𝜋𝑑 × cos 𝜃
𝜋
4
𝑑2 × 𝜌 × 𝑔
ℎ =
4𝜎 cos 𝜃
𝜌 × 𝑔 × 𝑑
Note:  For mercury
and glass tube is 138o

29. Problem 8: Calculate the capillary rise in a glass tube of 4 mm diameter, when
immersed in (i)water, and (ii)mercury.the temperature of the liquid is 20oC and the
values of the surface tension of water and mercury at 20oC in contact with air are
0.073575 N/m & 0.51 N/m respectively. The angle of contact for water is zero that for
mercury 130o. Take density of water at 20oC as equal to 998 kg/m3.
Given:
d = 4 mm = 0.004 m
Solution:
(i) Capillary effect for water
σ = 0.073575 N/m, θ = 0o
ρ = 998 kg/m3 @ 20 oC
h =
4 σ cos θ
ρ 𝑔 𝑑
=
4 ∗ 0.073575 ∗ cos 0
998 ∗ 9.81 ∗ 0.004
h = 0.00751 m or 7.51 mm

30. Contd…
(i) Capillary effect for mercury
σ = 0.51 N/m, θ = 130o
ρ = 13600 kg/m3 (ρ = S * 1000 => 13.6 * 1000)
h =
4 σ cos θ
ρ 𝑔 𝑑
=
4 ∗ 0.51 ∗ cos 130
13600 ∗ 9.81 ∗ 0.004
h = - 0.00246 m or – 2.46 mm

31. Bulk modulus:
It is defined as the ratio of compressive stress to volumetric
strain.
Unit - Τ
𝑁
𝑚2
The negative sign indicates the volume decreases as pressure
increases.
Compressibility:
When a fluid is subjected to a pressure increase the
volume of the fluid decreases. It is known as compressibility. It
is reciprocal of bulk modulus.
Compressibility = 1/k
𝑘 = −
𝑑𝑝
Τ
𝑑𝑣 𝑣

32. Vapour pressure:
All liquids have a tendency to evaporate when exposed
to a gaseous atmosphere. The vapour molecules exert a partial
pressure in the space above the liquid, known as vapour
pressure.

33. Gas Laws:
Boyle’s Law
At a given temperature for a given quantity of gas, the product of
the pressure and the volume is a constant.
P1V1= P2V2
Ideal Gas Law:
Ideal Gas Law relates pressure to Temp for a gas
P = RT
Where, T = Absolute temperature in oK units
R= Gas constant = 287 Joule / Kg-oK

34. Universal Gas constant:
PV = mRT
Where,
m = Mass of gas in kg;
v = Specific Volume.
p = Absolute pressure ;
T = Absolute temperature

35. Pressure in a Fluid
𝑝 = 𝜌𝑔ℎ

36. Pressure:
Force per unit area: 1 N/m2 = Pascal = Pa
Standard Atmosphere P = 101.33 kPa = 101.3 kN/m2 =10.3 m of mg

37. Thank you