Fluid mechanics notes for gate

voddepalli soumith
FLUID MECHANICS

FLUID KINEMATICS...............................................................................................................................................................................................2
BUOYANCY & FLOATATION ...............................................................................................................................................................................9
PRESSURE MEASUREMENT............................................................................................................................................................................12
FLUID MECHANICS..............................................................................................................................................................................................15
FLUID DYNAMICS ................................................................................................................................................................................................22
LAMINAR FLOW ...................................................................................................................................................................................................31
BOUNDARY LAYER THEORY...........................................................................................................................................................................35
HYDROSTATIC FORCES.....................................................................................................................................................................................46
DIMENSIONAL ANALYSIS.................................................................................................................................................................................50

FLUID KINEMATICS
Kinematics deal with motion of fluid without any reference to cause of motion i.e., force.
The fluid flow is analysed by using two techniques.
1. Langrangian technique
2. Eulerian technique
 In Langrangian technique, single fluid particle is taken and the behaviour of this particle is analysed at different
instances of time.
 In Eulerian technique, certain section is taken and fluid flow is analysed at that section.
Different types of fluid flow
Steady & Unsteady flow
A flow is said to be steady flow when fluid properties do not change at any cross section at any given time,
otherwise flow is unsteady.
𝐹𝑜𝑟 𝑆𝑡𝑒𝑎𝑑𝑦 𝑓𝑙𝑜𝑤 → [
𝑑𝑣
𝑑𝑡
]
𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
= 0 & [
𝑑𝜌
𝑑𝑡
]
𝑔𝑖𝑣𝑒𝑛 𝑠𝑒𝑐𝑡𝑖𝑜𝑛
= 0
Uniform & non-uniform flow
A flow is said to be uniform when fluid properties especially velocity don’t change with space at any given instant
of time, otherwise the flow is non-uniform.
𝐹𝑜𝑟 𝑈𝑛𝑖𝑓𝑜𝑟𝑚 𝑓𝑙𝑜𝑤 → [
𝑑𝑣
𝑑𝑠(𝑥, 𝑦, 𝑧)
]
𝑔𝑖𝑣𝑒𝑛 𝑡𝑖𝑚𝑒
= 0 {𝑠 = 𝑠𝑝𝑎𝑐𝑒(𝑥, 𝑦, 𝑧)}
Laminar & Turbulent flow
In laminar flow fluid particles move in the form of layers, with one layer sliding over the other layer. Laminar flow
generally occurs at low velocities.
In turbulent flow, fluid particles move in highly disorganized manner, leading to rapid mixing of particles.
Turbulent flow generally occurs at high velocities.
Rotational & Irrotational flow
A flow is said to be rotational flow when fluid particles rotate about their own mass centres, otherwise the flow is
irrotational.
Rotation is possible when there is a tangential force, these tangential forces are associated with viscous fluids.
Therefore, real fluids are generally rotational fluids and ideal fluids are irrotational fluids.
Internal & External flows
When the fluid flows through confined passage (Ex- flow of fluid through pipes, ducts) then it is internal flow.
When the fluid flow through unconfined passage (Ex- Flow of fluid (air) over aircraft wing) then flow is external
flow.
Categorization of flow
1. One-dimensional flow
2. Two-dimensional flow
3. Three-dimensional flow
Flow can never be 1-D, because of viscosity.
Stream line
It is an imaginary line or curve drawn in space such that a tangent drawn to it at any point gives velocity vector.
Stream line gives direction of flow as there is no component of velocity in perpendicular direction there is no flow
across the stream line, there is flow only along the stream line. Stream line gives instantaneous snapshot of a flow
pattern. It has no time history. No two stream lines can intersect because velocity is unique at any given instant of
time at a particular time.

Equation of Stream line
In 3-D
𝑣⃗ = 𝑢𝑖̂ + 𝑣𝑗̂ + 𝑤𝑘̂
𝑣 = √ 𝑢2 + 𝑣2 + 𝑤2
In 2-D
𝑣⃗ = 𝑢𝑖̂ + 𝑣𝑗̂
𝒖 =
𝑑𝑥
𝑑𝑡
⇒ 𝑑𝑡 =
𝑑𝑥
𝑢
& 𝒗 =
𝑑𝑦
𝑑𝑡
⇒ 𝑑𝑡 =
𝑑𝑦
𝑣
𝒅𝒙
𝒖
=
𝒅𝒚
𝒗
→ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑟𝑎𝑚 𝑙𝑖𝑛𝑒 𝑖𝑛 2 − 𝐷
𝒅𝒙
𝒖
=
𝒅𝒚
𝒗
=
𝒅𝒛
𝒘
→ 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑟𝑎𝑚 𝑙𝑖𝑛𝑒 𝑖𝑛 3 − 𝐷
Path line
It is the locus of single fluid particle at different instances of time. It follows
Langrangian approach. A path line can intersect with itself.
Streak line
It is the locus of different fluid particles through a fixed point.

Unsteady Flow
11:00―11:30  North to South
11:30―12:00 East to West
Steady Flow
11:00 ―12:00North to South
In a steady flow stream lines, streak lines & path lines are identical, whereas in unsteady flow they are different.
Stream lines intersect at stagnation point.
Conservation of mass (Continuity equation)
Generalized Continuity equation
𝑚 = 𝜌 ⋅ 𝑉 → 𝑙𝑛 𝑚 = 𝑙𝑛 𝜌 + 𝑙𝑛 𝑉
Differentiating above equation and simplifying gives
Every fluid flow must satisfy mass conservation or continuity equation. If the fluid flow doesn’t
satisfy continuity equation, then that flow is not possible.
This equation is applicable for any type of fluid flow.
Case-A (Steady flow)
𝐹𝑜𝑟 𝑠𝑡𝑒𝑎𝑑𝑦 𝑓𝑙𝑜𝑤 →
𝜕𝜌
𝜕𝑡
= 0
𝑆𝑜,
𝝏
𝝏𝒙
(𝝆𝒖) +
𝝏
𝝏𝒚
(𝝆𝒗) +
𝝏
𝝏𝒛
(𝝆𝒘) = 𝟎
Case-B (Incompressible flow)
𝐹𝑜𝑟 𝑖𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑙𝑜𝑤 → 𝜌 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡,
𝜕𝜌
𝜕𝑡
= 0
0 +
𝜕
𝜕𝑥
(𝜌𝑢) +
𝜕
𝜕𝑦
(𝜌𝑣) +
𝜕
𝜕𝑧
(𝜌𝑤) = 0 → 𝜌 (
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
) = 0
𝝏𝒖
𝝏𝒙
+
𝝏𝒗
𝝏𝒚
+
𝝏𝒘
𝝏𝒛
= 𝟎
This equation is applicable for any type of incompressible flow. (Steady or unsteady)
𝑇ℎ𝑒 𝑐𝑜𝑛𝑡𝑖𝑛𝑢𝑖𝑡𝑦 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑓𝑜𝑟 2 − 𝐷 𝑖𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠
𝝏𝒖
𝝏𝒙
+
𝝏𝒗
𝝏𝒚
= 𝟎
Continuity equation for steady 1-Dimensional flow
Flow through pipes, nozzles & diffusers etc…
𝜌 =
𝑚𝑎𝑠𝑠
𝑣𝑜𝑙𝑢𝑚𝑒
𝑚𝑎𝑠𝑠(𝒎) = 𝜌 × 𝑣𝑜𝑙𝑢𝑚𝑒(𝒗)
𝑚̇ =
𝑚
𝑡
=
𝜌 × 𝑣
𝑡
=
𝜌 × (𝐴 × 𝑙)
𝑡
𝒎̇ = 𝝆 × 𝑨 × 𝒗
𝐹𝑜𝑟 𝑠𝑡𝑒𝑎𝑑𝑦 𝑓𝑙𝑜𝑤 → 𝑚1 = 𝑚2
𝝆 𝟏 𝑨 𝟏 𝒗 𝟏 = 𝝆 𝟐 𝑨 𝟐 𝒗 𝟐
𝐼𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑖𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒 → 𝜌1 = 𝜌2
𝐼𝑓 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑆𝑡𝑒𝑎𝑑𝑦 & 𝐼𝑛𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑖𝑏𝑙𝑒,
𝑨 𝟏 𝒗 𝟏 = 𝑨 𝟐 𝒗 𝟐 (𝜌1 = 𝜌2)
𝜕𝜌
𝜕𝑡
+
𝜕
𝜕𝑥
(𝜌𝑢) +
𝜕
𝜕𝑦
(𝜌𝑣) +
𝜕
𝜕𝑧
(𝜌𝑤) = 0

Discharge (Q)
Volume flow rate is known as discharge.
𝑄 =
𝑉𝑜𝑙𝑢𝑚𝑒 𝑓𝑙𝑜𝑤𝑖𝑛𝑔 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑎 𝑐𝑟𝑜𝑠𝑠𝑒𝑐𝑡𝑖𝑜𝑛
𝑇𝑖𝑚𝑒
=
𝐴 × 𝑙
𝑡
→ 𝑄 = 𝐴 × 𝑣⃗
In a steady 1-D incompressible flow, discharge remains constant.
Acceleration of a fluid particle
𝑢 = 𝑓(𝑥, 𝑦, 𝑧, 𝑡), 𝑣 = 𝑓(𝑥, 𝑦, 𝑧, 𝑡), 𝑤 = 𝑓(𝑥, 𝑦, 𝑧, 𝑡)
𝑎 =
𝑑𝑣⃗
𝑑𝑡
=
𝑑𝑢
𝑑𝑡
𝑖̂ +
𝑑𝑣
𝑑𝑡
𝑗̂ +
𝑑𝑤
𝑑𝑡
𝑘̂
𝑎 𝑥 =
𝑑𝑢
𝑑𝑡
, 𝑎 𝑦 =
𝑑𝑣
𝑑𝑡
, 𝑎 𝑧 =
𝑑𝑤
𝑑𝑡
𝑎 𝑥 =
𝑑𝑢
𝑑𝑡
=
𝜕𝑢
𝜕𝑥
×
𝜕𝑥
𝜕𝑡
+
𝜕𝑢
𝜕𝑦
×
𝜕𝑦
𝜕𝑡
+
𝜕𝑢
𝜕𝑧
×
𝜕𝑧
𝜕𝑡
+
𝜕𝑢
𝜕𝑡
𝑎 𝑦 =
𝑑𝑣
𝑑𝑡
=
𝜕𝑣
𝜕𝑥
×
𝜕𝑥
𝜕𝑡
+
𝜕𝑣
𝜕𝑦
×
𝜕𝑦
𝜕𝑡
+
𝜕𝑣
𝜕𝑧
×
𝜕𝑧
𝜕𝑡
+
𝜕𝑣
𝜕𝑡
𝑎 𝑧 =
𝑑𝑤
𝑑𝑡
=
𝜕𝑤
𝜕𝑥
×
𝜕𝑥
𝜕𝑡
+
𝜕𝑤
𝜕𝑦
×
𝜕𝑦
𝜕𝑡
+
𝜕𝑤
𝜕𝑧
×
𝜕𝑧
𝜕𝑡
+
𝜕𝑤
𝜕𝑡
Convective Acceleration
The acceleration due to change of velocity with space is known as convective acceleration. For uniform flow
convective acceleration is zero.
Temporal or Local Acceleration
The acceleration due to change of velocity with respective to time is known as temporal acceleration. For steady
flow temporal acceleration is zero.
Type of flow Convective Acceleration Temporal Acceleration
Steady & uniform 0 0
Steady & Non-uniform exists 0
Unsteady & uniform 0 exists
Unsteady & Non-uniform exists exists
Steady flow, 1-Dimensional & incompressible
𝑇𝑒𝑚𝑝𝑜𝑟𝑎𝑙 = 0
𝐴1 𝑣1 = 𝐴2 𝑣2 ⇒ 𝒗 𝟏 = 𝒗 𝟐
Velocity is not changing w.r.t time.
𝐴1 𝑣1 = 𝐴2 𝑣2
𝐴2 < 𝐴1
𝑣2 > 𝑣1
`Stream lines are converging Convective acceleration
𝐴1 𝑣1 = 𝐴2 𝑣2
𝐴2 > 𝐴1
𝑣2 < 𝑣1
Stream lines are diverging deceleration

Rotational Components
𝑡𝑎𝑛 𝑑𝛼 =
𝜕𝑣
𝜕𝑥
𝑑𝑥 ⋅ 𝑑𝑡
𝑑𝑥
𝑑𝛼 =
𝜕𝑣
𝜕𝑥
⋅ 𝑑𝑡 (𝑑𝛼 ≈ 0 → 𝑡𝑎𝑛 𝑑𝛼 ≅ 𝑑𝛼)
𝑑𝛼
𝑑𝑡
=
𝜕𝑣
𝜕𝑥
⟲→anti-clockwise (+ve rotation) ⟳→clockwise (-ve rotation)
𝑡𝑎𝑛 𝑑𝛽 =
𝜕𝑢
𝜕𝑦
𝑑𝑥 ⋅ 𝑑𝑡
𝑑𝑦
⇒ 𝑑𝛽 =
𝜕𝑢
𝜕𝑦
⋅ 𝑑𝑡
𝑑𝛽
𝑑𝑡
= −
𝜕𝑢
𝜕𝑦
(⟳→ 𝑐𝑙𝑜𝑐𝑘𝑤𝑖𝑠𝑒 (−𝑣𝑒 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛))
In fluid mechanics, angular velocity is defined as average angular
velocity of initially 2 perpendicular line segments.
𝜔 𝑧 =
1
2
(
𝑑𝛼
𝑑𝑡
+
𝑑𝛽
𝑑𝑡
) =
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
𝜔 = 𝜔 𝑥 𝑖̂ + 𝜔 𝑦 𝑗̂ + 𝜔 𝑧 𝑘̂
𝜔 = |
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑢 𝑣 𝑤
|
𝜔 𝑥 =
1
2
(
𝜕𝑤
𝜕𝑦
−
𝜕𝑣
𝜕𝑧
) 𝜔 𝑦 =
1
2
(
𝜕𝑢
𝜕𝑧
−
𝜕𝑤
𝜕𝑥
) 𝜔 𝑧 =
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
Condition for irrotational flow
𝜔 = 𝜔 𝑥 𝑖̂ + 𝜔 𝑦 𝑗̂ + 𝜔 𝑧 𝑘̂
For irrotational flow,
𝜔 = 0 → 𝜔 𝑥 = 𝜔 𝑦 = 𝜔 𝑧 = 0
𝜔 𝑧 = 0 →
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
= 0 ⇒
𝝏𝒗
𝝏𝒙
=
𝝏𝒖
𝝏𝒚
Vorticity
Twice the rotation (2ω) is known as vorticity.
𝑽𝒐𝒓𝒕𝒊𝒄𝒊𝒕𝒚 = 𝟐𝝎 = |
𝑖̂ 𝑗̂ 𝑘̂
𝜕
𝜕𝑥
𝜕
𝜕𝑦
𝜕
𝜕𝑧
𝑢 𝑣 𝑤
|
For irrotational flow, vorticity is zero.

Circulation (Γ)
It is the line integral of tangential component of velocity taken
around a closed curve.
𝛤 = 𝑢 ⋅ 𝑑𝑥 + (𝑣 +
𝜕𝑣
𝜕𝑥
𝑑𝑥) ⋅ 𝑑𝑦 − (𝑢 +
𝜕𝑢
𝜕𝑦
𝑑𝑦) ⋅ 𝑑𝑥 − 𝑣 ⋅ 𝑑𝑦
𝛤 = (
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
) 𝑑𝑥 ⋅ 𝑑𝑦
[𝐴𝑟𝑒𝑎 = 𝑑𝑥 ⋅ 𝑑𝑦]
𝐶𝑖𝑟𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 (𝛤) = 𝑉𝑜𝑟𝑡𝑖𝑐𝑖𝑡𝑦 (2𝜔 𝑧) × 𝐴𝑟𝑒𝑎
In case of irrotational flow, vorticity is zero & circulation is zero.
Velocity Potential function (ϕ)
It is a function of space & time defined in such a manner, that its negative derivative w.r.t space gives velocity in
that direction. The negative sign is taken as the flow is in the direction of decreasing potential.
−
𝜕𝜙
𝜕𝑥
= 𝑢 −
𝜕𝜙
𝜕𝑦
= 𝑣 −
𝜕𝜙
𝜕𝑧
= 𝑤
Velocity potential function can be defined for 2-Dimensional flow
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑥
=
𝜕
𝜕𝑥
(−
𝜕𝜙
𝜕𝑥
) +
𝜕
𝜕𝑦
(−
𝜕𝜙
𝜕𝑦
) = − (
𝜕2
𝜙
𝜕𝑥2
+
𝜕2
𝜙
𝜕𝑦2
)
Case 1
𝐼𝑓
𝜕2
𝜙
𝜕𝑥2
+
𝜕2
𝜙
𝜕𝑦2
= 0, 𝝓 𝒔𝒂𝒕𝒊𝒔𝒇𝒊𝒆𝒔 𝑳𝒂𝒑𝒍𝒂𝒄𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑎𝑠 (
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑥
= 0)
→ Continuity equation is satisfied and flow is possible.
Case 2
𝐼𝑓
𝜕2
𝜙
𝜕𝑥2
+
𝜕2
𝜙
𝜕𝑦2
≠ 0, 𝝓 𝒅𝒐𝒆𝒔𝒏′
𝒕 𝒔𝒂𝒕𝒊𝒔𝒇𝒊𝒆𝒔 𝑳𝒂𝒑𝒍𝒂𝒄𝒆 𝒆𝒒𝒖𝒂𝒕𝒊𝒐𝒏 𝑎𝑠 (
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑥
≠ 0)
→ Continuity equation is not satisfied and flow is not possible.
Case 3
𝜔 𝑧 =
1
2
(
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
) =
1
2
(−
𝜕2
𝜙
𝜕𝑥 ⋅ 𝜕𝑦
+
𝜕2
𝜙
𝜕𝑦 ⋅ 𝜕𝑥
)
𝜔 𝑧 = 0 → 𝐼𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑙𝑜𝑤
Velocity Potential function exits only for Irrotational flow i.e., the existence of velocity potential function
implies the flow is irrotational. Sometimes irrotational flow are also known as Potential flow.

Stream function (Ψ)
It is a function of space & time defined in such a manner that it satisfies continuity equation.
𝑢 = −
𝜕𝜓
𝜕𝑦
𝑣 =
𝜕𝜓
𝜕𝑥
Note Though velocity potential function can be defined for 3-Dimensional flows, it is difficult to define stream
function in 3-Dimensional flows. Therefore, stream functions are generally defined for 2-D flows.
𝜔 𝑧 =
1
2
(
𝜕𝑣
𝜕𝑥
−
𝜕𝑢
𝜕𝑦
) =
1
2
(
𝜕2
𝜓
𝜕𝑥2
+
𝜕2
𝜓
𝜕𝑦2
)
Case 1
𝐼𝑓 𝜓 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
= 0) &
𝜕2
𝜓
𝜕𝑥2
+
𝜕2
𝜓
𝜕𝑦2
= 0 ⇒ 𝜔𝑧 = 0 → 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑖𝑟𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙
Case 2
𝐼𝑓 𝜓 𝑑𝑜𝑒𝑠𝑛′
𝑡 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝐿𝑎𝑝𝑙𝑎𝑐𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 (
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
≠ 0) &
𝜕2
𝜓
𝜕𝑥2
+
𝜕2
𝜓
𝜕𝑦2
≠ 0 ⇒ 𝜔𝑧 ≠ 0 → 𝑡ℎ𝑒 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙
Velocity potential function exists only for rotational flow whereas stream function exists for both rotational &
irrotational flow.
If stream function satisfies Laplace equation, then flow is irrotational.
Significance of Stream Function
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑆𝑡𝑟𝑒𝑎𝑚 𝑙𝑖𝑛𝑒 →
𝑑𝑥
𝑢
=
𝑑𝑦
𝑣
→ 𝑣 ⋅ 𝑑𝑥 = 𝑢 ⋅ 𝑑𝑦 ⇒ 𝑣 ⋅ 𝑑𝑥 − 𝑢 ⋅ 𝑑𝑦 = 0
𝑢 = −
𝜕𝜓
𝜕𝑦
𝑣 =
𝜕𝜓
𝜕𝑥
Substituting we get,
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 − (−
𝜕𝜓
𝜕𝑦
) ⋅ 𝑑𝑦 = 0 ⇒
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 +
𝜕𝜓
𝜕𝑦
⋅ 𝑑𝑦 = 0 → ① → 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑎 𝑝𝑎𝑟𝑡𝑖𝑐𝑢𝑙𝑎𝑟 𝑠𝑡𝑟𝑒𝑎𝑚 𝑙𝑖𝑛𝑒
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 +
𝜕𝜓
𝜕𝑦
⋅ 𝑑𝑦 = 𝜵𝝍 = 𝑑𝜓 = 0
𝐴𝑠 𝒅𝝍 = 0 → 𝝍 = 𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕
For a particular stream line, Stream function remains constant.
𝑄 = 𝐴 ⋅ 𝑣 = (𝑑𝑥 ⋅ 1) ⋅ 𝑣 = 𝑣 ⋅ 𝑑𝑥 =
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥
𝑄 =
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 ①
𝑑𝜓 =
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 +
𝜕𝜓
𝜕𝑦
⋅ 𝑑𝑦 (𝑑𝑦 = 0)
𝑑𝜓 =
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 ②
𝐴𝑠 ① = ②,
𝑸 = 𝒅𝝍 (𝐷𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑖𝑛 𝑆𝑡𝑟𝑒𝑎𝑚 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛)
The difference in stream function gives discharge per unit width.
Relationship between Equipotential lines & Constant Stream Function lines
𝜙(𝑥, 𝑦) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝐸𝑞𝑢𝑖𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑙𝑖𝑛𝑒𝑠 (𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑙𝑖𝑛𝑒𝑠 ℎ𝑎𝑣𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙)
𝑑𝜙 =
𝜕𝜙
𝜕𝑥
⋅ 𝑑𝑥 +
𝜕𝜙
𝜕𝑦
⋅ 𝑑𝑦 = 0 ⇒
𝜕𝜙
𝜕𝑥
⋅ 𝑑𝑥 = −
𝜕𝜙
𝜕𝑦
⋅ 𝑑𝑦 ⇒ −𝑢 ⋅ 𝑑𝑥 = 𝑣 ⋅ 𝑑𝑦 (𝑢 = −
𝜕𝜙
𝜕𝑥
𝑣 = −
𝜕𝜙
𝜕𝑦
)
(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐸𝑞𝑢𝑖𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑙𝑖𝑛𝑒)
𝑑𝑦
𝑑𝑥
= −
𝑢
𝑣
=
𝜕𝜙
𝜕𝑥
−
𝜕𝜙
𝜕𝑦
⁄

𝑑𝜓 =
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 +
𝜕𝜓
𝜕𝑦
⋅ 𝑑𝑦 = 0 ⇒
𝜕𝜓
𝜕𝑥
⋅ 𝑑𝑥 = −
𝜕𝜓
𝜕𝑦
⋅ 𝑑𝑦 ⇒ 𝑣 ⋅ 𝑑𝑥 = 𝑢 ⋅ 𝑑𝑦
(𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑆𝑡𝑟𝑒𝑎𝑚 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑙𝑖𝑛𝑒𝑠)
𝑑𝑦
𝑑𝑥
=
𝑣
𝑢
𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝐸𝑞𝑢𝑖𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝑙𝑖𝑛𝑒 × 𝑆𝑙𝑜𝑝𝑒 𝑜𝑓 𝑆𝑡𝑟𝑒𝑎𝑚 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑙𝑖𝑛𝑒 = −
𝑢
𝑣
×
𝑣
𝑢
= −1
Equipotential lines & Constant Stream function lines are perpendicular to
each other.
Cauchy―Reimann Equations
𝑢 = −
𝜕𝜙
𝜕𝑥
= −
𝜕𝜓
𝜕𝑦
& 𝑣 = −
𝜕𝜙
𝜕𝑦
=
𝜕𝜓
𝜕𝑥
𝝏𝝓
𝝏𝒙
=
𝝏𝝍
𝝏𝒚
& −
𝝏𝝓
𝝏𝒚
=
𝝏𝝍
𝝏𝒙
BUOYANCY & FLOATATION
Archimedes principle
When a body is submerged either partially or completely, the net vertical upward force exerted by the fluid on the
body is known as buoyancy force (Fb), this buoyancy force is equal to weight of the fluid displaced and this is
known as Archimedes principle.
𝑉𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 𝑉𝑓𝑑 = 𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 = 𝐴(𝑥2 − 𝑥1)
(𝑁𝑒𝑡 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑢𝑝𝑤𝑎𝑟𝑑 𝑓𝑜𝑟𝑐𝑒 𝑒𝑥𝑒𝑟𝑡𝑒𝑑 𝑏𝑦 𝑓𝑙𝑢𝑖𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦) 𝑭 𝒗⋅𝑵𝒆𝒕
= 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 𝜌 𝑓 ⋅ 𝑔 ⋅ (𝑥2 − 𝑥1) ⋅ 𝐴 = 𝜌 𝑓 ⋅ 𝑔 ⋅ 𝑉𝑓𝑑
𝑭 𝒗⋅𝑵𝒆𝒕 = 𝑭 𝒃𝒖𝒐𝒖𝒂𝒏𝒄𝒚 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 𝜌 𝑓 ⋅ 𝑔 ⋅ 𝑉𝑓𝑑
Centre of Buoyancy (B)
It is the point which the Buoyancy force is supposed to be acting, and this buoyancy
force will act at the centroid of the displacement volume. Therefore, centre of buoyancy will lie at the centroid of
displaced volume.
Note
 When a homogenous body is completely submerged, then the centre of gravity of body & centre of
buoyancy coincide.
 For a floating homogenous body, centre of buoyancy is below the centre of gravity.
 For a non-homogenous body (heterogenous), centre of buoyancy and centre of gravity may not coincide
even if it’s completely submerged.
Principle of Flotation
For a floating body to be in equilibrium, Weight of the body must be EQUAL to Weight of fluid displaced and the
line of action of these 2 forces must be same.
𝑊𝑏𝑜𝑑𝑦 = 𝐹𝑏 → 𝐹𝑏 = 𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑏𝑜𝑑𝑦
𝑊𝑏𝑜𝑑𝑦 = 𝑊𝑓𝑙𝑢𝑖𝑑 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑 → 𝑏𝑜𝑑𝑦 𝑖𝑠 𝑖𝑛 𝑒𝑞𝑢𝑙𝑖𝑏𝑟𝑖𝑢𝑚
Types of Equilibrium

Stability conditions for completely submerged bodies
A completely submerged body will be in stable
equilibrium, when the centre of buoyancy is above centre
of gravity.
A completely submerged body will be in unstable
equilibrium when the centre of buoyancy is below centre
of gravity.
A completely submerged body will be in neutral
equilibrium, when centre of buoyancy coincides with
centre of gravity.
Metacentre (M)
It’s the point of intersection normal axis of the body to the new line of action of buoyancy force when the body is
tilted.
Metacentric height
The distance between centre of gravity and Metacentre (M) measured along the normal axis is called as
Metacentric height.
For stable equilibrium Metacentric height is positive, unstable negative.
Stability conditions for partially submerged/floating bodies
A floating body will be in stable
equilibrium, when metacentre is above
centre of gravity.
A floating body is said to be in unstable
equilibrium when the meta centre is
below centre of gravity.
A floating body is said to be in neutral
equilibrium when the meta centre
coincides with centre of gravity.
Mathematical condition for Stable equilibrium
For more stable equilibrium conditions, BM or GM must be as large as possible.

𝐼𝑙𝑙 =
𝑙𝑏3
12
𝐼𝑡𝑡 =
𝑏𝑙3
12
𝑙 > 𝑏 → 𝐼𝑡𝑡 > 𝐼𝑙𝑙
𝐵𝑀 =
𝐼
𝑉𝑓⋅𝑑
𝐵𝑀𝑙−𝑙 =
𝐼𝑙𝑙
𝑉𝑓⋅𝑑
𝐵𝑀𝑡−𝑡 =
𝐼𝑡𝑡
𝑉𝑓⋅𝑑
𝑩𝑴𝒍−𝒍 < 𝑩𝑴𝒕−𝒕 (𝐼𝑡𝑡 > 𝐼𝑙𝑙)
From design point of view the least BM is calculated, i.e., BM about longitudinal axis is calculated. As BMt-t>BMl-l
the body will be more stable when ot oscillates about transverse axis (t-t) than longitudinal axis (l-l).
Oscillation about longitudinal axis are known as Rolling and
transverse axis is known as Pitching.
𝐵𝑀𝑟𝑜𝑙𝑙𝑖𝑛𝑔 < 𝐵𝑀 𝑝𝑖𝑡𝑐ℎ𝑖𝑛𝑔
If rolling is taken care of, then pitching is already taken care of.
Time period of Oscillation
𝑇 = 2𝜋√
𝑘 𝑔
2
𝑔(𝐺𝑀)
(𝑘 𝑔 = 𝑅𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑔𝑦𝑟𝑎𝑡𝑖𝑜𝑛 = √
𝐼
𝐴
)
For more stable equilibrium conditions, metacentric height must be larger, but larger GM results in smaller time
period of oscillation i.e., more number of oscillations in a given time. Therfore passengers are not comfortable
under such conditions. Therefore, for passenger ships, metacentric height is not very high. In case of war ships
stability of ship is more important than comfort, so metacentric height is larger than passenger ships.
Weight lost due to Buoyancy
𝑊𝑒𝑖𝑔ℎ𝑡 𝑙𝑜𝑠𝑠 = 𝑇 − 𝑇1 = 𝑊 − (𝑊 − 𝐹𝑏) = 𝐹𝑏
𝑊𝑒𝑖𝑔ℎ𝑡 𝑙𝑜𝑠𝑠 = 𝐵𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑓𝑜𝑟𝑐𝑒
As density of air is very small, the buoyancy effects are negligible in air.
Therefore, correct weight of body is obtained when it is submerged in air.

PRESSURE MEASUREMENT
Pressure
It is defined as external normal force exerted in unit area. The area can be real or imaginary.
The unit of pressure is Newton (N)/mm2.
Pressure is a representative of no. of collisions per second.
Mohr’s circle for a Static fluid
For a static fluid there is no shear stress and there are only normal forces
(pressure). Therefore, Mohr’s circle is a point as shown in figure.
Pascals Law
According to Pascal’s Law, pressure at any point in a static fluid is equal in all directions. Conversely if pressure is
applied in a static fluid it is transmitted equally in all directions.
Applications― Hydraulic Lift, Hydraulic brakes etc…
𝐹
𝑎
=
𝑊
𝐴
⇒
𝐴
𝑎
=
𝑊
𝐹
> 1
𝑾 > 𝑭 𝑎𝑠 𝐴 > 𝑎
As W>F, by applying small force large weights can be raised. This doesn’t
mean energy conservation is violated because smaller force moves
through larger distance and larger force moves through smaller distance.
Atmospheric Pressure
Pressure exerted by environmental mass is known as atmospheric pressure. It is around 1.013 bar.
Gauge Atmospheric pressure (Pguage)
The pressure measured w.r.t atmospheric pressure is known as Gauge Pressure.
Absolute Pressure
The pressure measured w.r.t zero pressure is known as absolute pressure.
Vacuum Pressure
The pressure less than atmospheric pressure is known as vacuum pressure. There can be positive gauge or
negative gauge pressure, but there can’t be negative absolute pressure.
𝑉𝑎𝑐𝑢𝑢𝑚 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑃𝑎𝑡𝑚 − 𝑃𝑎𝑏𝑠𝑜𝑙𝑢𝑡𝑒

Hydrostatic Law
𝑃 ⋅ 𝑑𝐴 + 𝜌𝑔 ⋅ 𝑑𝐴 ⋅ 𝑑ℎ = (𝑃 + 𝑑𝑃) ⋅ 𝑑𝐴
𝑃 + 𝜌𝑔 ⋅ 𝑑ℎ = 𝑃 + 𝑑𝑃
𝑑𝑃
𝑑ℎ
= 𝜌𝑔 → 𝐻𝑦𝑑𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝐿𝑎𝑤
Hydrostatic Law gives variation of pressure in the vertical direction. For a static fluid, the forces acting on liquid
element are pressure & gravity forces.
𝐼𝑓 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑖𝑠 𝑡𝑎𝑘𝑒𝑛 𝑖𝑛 𝑢𝑝𝑤𝑎𝑟𝑑 𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛, 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑤𝑖𝑡ℎ ℎ𝑒𝑖𝑔ℎ𝑡,
𝑑𝑃
𝑑ℎ
= −𝑤 = −𝜌𝑔
Pressure at any depth h
𝐴𝑡 𝑓𝑟𝑒𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 (ℎ = 0)𝑃 = 𝑃𝑎𝑡𝑚
𝑑𝑃
𝑑ℎ
= 𝑤 → 𝑑𝑃 = 𝑤 ⋅ 𝑑ℎ → 𝑃 = 𝑤 ⋅ ℎ + 𝑐
𝑃 = 𝑤 ⋅ ℎ + 𝑃𝑎𝑡𝑚 (ℎ = 0 → 𝑃 = 𝑃𝑎𝑡𝑚)
𝑷 𝒈𝒂𝒖𝒈𝒆 = 𝒘 ⋅ 𝒉 = 𝝆𝒈𝒉 (𝑃𝑎𝑡𝑚 = 0 𝑓𝑜𝑟 𝑔𝑎𝑢𝑔𝑒 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒)
𝑃 = 𝜌𝑔ℎ 𝑖𝑠 𝑏𝑎𝑠𝑒𝑑 𝑜𝑛 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑑𝑒𝑛𝑠𝑖𝑡𝑦(𝜌)𝑖𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
Sometimes the pressure is expressed in height column (h) because ρ & g are almost constants and pressure vary
directly with height column.
Barometer
Barometer is used for measuring Atmospheric pressure.
𝑃𝑎𝑡𝑚 = 0 + 𝜌𝑔ℎ ⇒ 𝑃𝑎𝑡𝑚 = 𝜌𝑔ℎ
h calculated is found to be 0.76m.
𝑃𝑎𝑡𝑚 = 𝜌𝑔ℎ = 13.6 × 9.81 × 0.76 = 1.01325 × 105
𝑁
𝑚𝑚2
= 1.01325 𝑏𝑎𝑟
If water is used instead of mercury, the corresponding height will be 10.3 metres. Mercury is used because of its
high density.
Conversion of one fluid column into other fluid column
𝑃1 = 𝑃2 ⇒ 𝜌1 ⋅ 𝑔 ⋅ ℎ1 = 𝜌2 ⋅ 𝑔 ⋅ ℎ2 ⇒ 𝝆 𝟏 ⋅ 𝒉 𝟏 = 𝝆 𝟐 ⋅ 𝒉 𝟐
Assume both are liquids
𝜌1ℎ1
𝜌 𝐻2 𝑂
=
𝜌2ℎ2
𝜌 𝐻2 𝑂
⇒ 𝒔 𝟏 ⋅ 𝒉 𝟏 = 𝒔 𝟐 ⋅ 𝒉 𝟐
If both are gases
𝜌1ℎ1
𝜌 𝑎𝑖𝑟
=
𝜌2ℎ2
𝜌 𝑎𝑖𝑟
⇒ 𝒔 𝟏 ⋅ 𝒉 𝟏 = 𝒔 𝟐 ⋅ 𝒉 𝟐
ℎ2 =
𝑠1
𝑠2
⋅ ℎ1

Piezometers
It is a device which is open at both the ends with one end connected at a point where pressure
is to be calculated and another end is open to atmosphere.
𝑃𝑔 𝑎𝑢𝑔𝑒 = 𝜌𝑔ℎ
Piezometers are not suitable for measuring high pressures like gas at high pressures. They are
suitable for moderate liquid pressures
Manometer
They are used for measuring pressure, they are based on balancing of liquid column.
They are divided into 2 types
1. Simple U-Tube (Pressure at a point)
2. Differential (Measure Pressure differences)
Simple U-tube Manometer
Jumping of fluid technique
𝑃 + 𝜌𝑔𝑦 − 𝜌 𝐻𝑔 𝑔𝑥 − 𝑃𝑎𝑡𝑚 = 0
𝑃𝑔 𝑎𝑢𝑔𝑒 = 𝜌 𝐻𝑔 𝑔𝑥 − 𝜌𝑔𝑦
Datum line technique
𝑃𝐴 = 𝑃𝐵
𝑃𝐴 = 𝑃 + 𝜌𝑔𝑦 𝑃𝐵 = 𝜌 𝐻𝑔 𝑔𝑥
𝑃 + 𝜌𝑔𝑦 = 𝜌 𝐻𝑔 𝑔𝑥
𝑃 = 𝜌 𝐻𝑔 𝑔𝑥 − 𝜌𝑔𝑦
Multi U-tube manometers are used for measuring High Pressures

FLUID MECHANICS
Fluid
Fluid is a substance which is capable of moving or deforming under the action of shear force.
As long as there is shear force, the fluid flows or deforms continuously.
Examples- Liquids, Gases etc…
Difference between Solids & Fluids
In case of solids under the action of shear force, there is deformation and this deformation doesn’t change with
time. Therefore, deformation dθ is important when this shear force is removed, solids will try to come back to its
original position.
In case of fluids, the deformation is continuous as long as there is shear force, this deformation changes with time.
In fluids the rate if deformation (dθ/dt) is important than dθ. After the removal of shear force the fluid will never
try to come back to its original position. For a static fluid shear force is zero.
Fluid Properties
Density
It is defined as ratio of mass of fluid to its volume. It actually represents the quantity of matter in a given volume.
Its unit is Kg/m3.
Density of water for all calculation purposes is taken as 1000 Kg/m3.
Density depends on temperature and Pressure. As temperature increases density decreases and as pressure
increases density increases.
Specific Weight/ Weight density (w)
It is defined as the ratio of weight of the fluid to its volume. Its unit is N/mm3.
𝑤 =
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑙𝑢𝑖𝑑
𝑣𝑜𝑙𝑢𝑚𝑒
=
𝑚𝑔
𝑉
= 𝜌𝑔
𝑤 = 𝜌𝑔
Density is an absolute quantity, whereas specific weight is not an absolute quantity, because it varies from
location to location.
Specific gravity
It is defined as the ratio of density of fluid to density of standard fluid.
In case of liquids the standard fluid is water.
In case if gases the standard fluid is either Hydrogen or air at a given temperature and pressure.
Specific gravity of water is one. If the specific gravity of a liquid is less than one it is lighter than water, if greater
than one liquid is heavier than water.
Relative density
It is the ratio of density of one fluid to other fluid.
All Specific gravities are relative densities, but not all relative densities are not specific gravities.
Compressibility (β)
It is the measure of the change of volume (or) change of density w.r.t pressure on a given mass of fluid.
Mathematically it is defined as reciprocal of Bulk Modulus (K)
𝛽 =
1
𝐾

𝐾 =
𝑑𝑃
−𝑑𝑉
𝑉
= −𝑉 ⋅
𝑑𝑃
𝑑𝑉
= 𝜌 ⋅
𝑑𝑃
𝑑𝜌
(
−𝑑𝑉
𝑉
=
𝜌
𝑑𝜌
) → 𝐾 =
𝜌 ⋅ 𝑑𝑃
𝑑𝜌
𝛽 =
𝑑𝜌
𝜌 ⋅ 𝑑𝑃
Liquids are generally treated as incompressible and gases as compressible.
Isothermal compressibility of Ideal gas
𝑃 = 𝜌 ⋅ 𝑅 ⋅ 𝑇 (𝑇 = 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡)
𝑑𝑃
𝑑𝜌
= 𝑅𝑇
𝐾 𝑇(𝑇 − 𝑐𝑜𝑛𝑠𝑡. ) =
𝜌 ⋅ 𝑑𝑃
𝑑𝜌
= 𝜌 ⋅
𝑑𝑃
𝑑𝜌
= 𝜌 ⋅ 𝑅 ⋅ 𝑇 = 𝑃
𝑲 𝑻 = 𝑷
Adiabatic Bulk Modulus/ Isentropic Bulk Modulus of an Ideal gas
𝑃𝑉 𝛾
= 𝐶1
𝑃𝑉 𝛾
= 𝑃 (
𝑚
𝜌
)
𝛾
= 𝐶1 ⇒
𝑃
𝜌 𝛾
=
𝐶1
𝑚 𝛾
= 𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 𝐶 ⇒ 𝑃 = 𝐶𝜌 𝛾
𝑑𝑃
𝑑𝜌
= 𝐶 ⋅ 𝛾 ⋅ 𝜌 𝛾−1
𝐾𝑎(𝐴𝑑𝑖𝑎𝑏𝑎𝑡𝑖𝑐) = 𝜌 ⋅
𝑑𝑃
𝑑𝜌
= 𝜌 ⋅ (𝐶 ⋅ 𝛾 ⋅ 𝜌 𝛾−1) = 𝛾 ⋅ 𝐶 ⋅ 𝜌 𝛾
= 𝛾𝑃
𝑲 𝒂 = 𝜸 ⋅ 𝑷
As γ>1, Adiabatic bulk modulus is greater than isothermal bulk modulus.
Bulk Modulus is not constant and it increases with increase in Pressure,
because at higher pressure the fluid offers more resistance to further
compression.
Viscosity
The Internal resistance offered by one layer of fluid to the adjacent layer is called Viscosity.
Need to define viscosity
Though the density of oil and water are almost same, their flow behaviour is not same and hence a property is
required to define flow behaviour. This property for defining flow behaviour is Viscosity.
𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡 =
𝑑𝑢
𝑑𝑦
𝑑𝜃
𝑑𝑡
=
𝑑𝑢
𝑑𝑦
𝜏 = 𝜇
𝑑𝜃
𝑑𝑡
= 𝜇
𝑑𝑢
𝑑𝑦

→ 𝜇 =
𝜏
𝑑𝜃
𝑑𝑡
=
𝜏
𝑑𝑢
𝑑𝑦
𝐼𝑓
𝑑𝜃
𝑑𝑡
𝑖𝑠 𝑙𝑎𝑟𝑔𝑒, 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑒𝑎𝑠𝑦 𝑎𝑠 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑖𝑠 𝑙𝑒𝑠𝑠 𝑖. 𝑒. , 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑙𝑒𝑠𝑠
𝐼𝑓
𝑑𝜃
𝑑𝑡
𝑖𝑠 𝑠𝑚𝑎𝑙𝑙, 𝑓𝑙𝑜𝑤 𝑖𝑠 𝑑𝑖𝑓𝑓𝑖𝑐𝑢𝑙𝑡 𝑎𝑠 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦 𝑖𝑠 ℎ𝑖𝑔ℎ 𝑖. 𝑒. , 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑜 𝑓𝑙𝑜𝑤 𝑖𝑠 ℎ𝑖𝑔ℎ
⇒
𝑑𝜃
𝑑𝑡
= 𝑅𝑎𝑡𝑒 𝑜𝑓 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑑𝑒𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 (𝑜𝑟)𝑅𝑎𝑡𝑒 𝑜𝑓 𝑆ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑎𝑖𝑛
Unit of Viscosity is Newton Second/ square metre (N⋅ s/m2)
Unit of Viscosity in CGS system is poise.
1 poise = 0.1 N⋅ s/m2
Variation of Viscosity with temperature
In case of liquids, the intermolecular distance is small, hence cohesive forces are large. In case of gases,
intermolecular distance is small and cohesive forces are negligible.
With increase in temperature, cohesive forces decrease and resistance to flow also decreases. Therefore, viscosity
of liquid decreases with increase in temperature.
With increase in temperature molecular disturbance increases and resistance to flow increases. Viscosity of gases
increase with temperature.
Newtonian Fluid
Fluids which obey Newton’s law of viscosity are known as Newtonian fluids. According to Newton’s Law of
viscosity, shear stress is directly proportional to rate of shear strain.
𝜏 ∝
𝑑𝜃
𝑑𝑡
⇒ 𝜏 ∝
𝑑𝑢
𝑑𝑦
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
→ 𝑁𝑒𝑤𝑡𝑜𝑛𝑖𝑎𝑛 𝐹𝑙𝑢𝑖𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
μ is the slope in the graph.
Examples- Air, Water, Diesel, Kerosene, Oils, Mercury etc...
Note
For a Newtonian fluid, Viscosity doesn’t change with rate of deformation.

Non-Newtonian Fluids
Fluids which don’t obey Newton’s law of viscosity are known as Non-Newtonian fluids.
𝑇ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑙 𝑟𝑒𝑙𝑎𝑡𝑖𝑜𝑛 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 (𝜏) & 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑔𝑟𝑎𝑑𝑖𝑒𝑛𝑡 (
𝑑𝑢
𝑑𝑦
) 𝑖𝑠 𝜏 = 𝐴 ⋅ (
𝑑𝑢
𝑑𝑦
)
𝑛
+ 𝐵
Case 1
𝐵 = 0; 𝑛 > 1 𝐷𝑖𝑙𝑎𝑡𝑎𝑛𝑡 𝑓𝑙𝑢𝑖𝑑
A fluid is said to be dilatant fluid for which the apparent viscosity increases with rate of deformation.
Examples- Rice Starch, Sugar in water
As the apparent viscosity is increasing with deformation, these fluids are known
as Shear Thickening Fluids.
𝜏 = 𝐴 ⋅ (
𝑑𝑢
𝑑𝑦
)
𝑛
+ 0 = 𝐴 ⋅ (
𝑑𝑢
𝑑𝑦
)
𝑛
𝜏 = 𝐴 ⋅ (
𝑑𝑢
𝑑𝑦
)
𝑛−1
⋅
𝑑𝑢
𝑑𝑦
= 𝜇 𝑎𝑝𝑝𝑎𝑟𝑒𝑛𝑡 ⋅
𝑑𝑢
𝑑𝑦
Case 2
𝐵 = 0; 𝑛 < 1 𝑃𝑠𝑒𝑢𝑑𝑜 𝑝𝑙𝑎𝑡𝑖𝑐 𝑓𝑙𝑢𝑖𝑑𝑠
For a pseudo plastic fluid, apparent viscosity decreases with rate of deformation.
Examples- Blood, milk, Colloidal Solutions etc…
As the apparent viscosity is decreasing with deformation, these fluids are known
as Shear Thinning Fluids.
Case 3
𝐵 ≠ 0; 𝑛 = 1 𝐵𝑖𝑛𝑔ℎ𝑎𝑚 𝑃𝑙𝑎𝑠𝑡𝑖𝑐
Example- Toothpaste
In case of Bingham plastic fluids, certain minimum shear stress is required for
causing the flow of fluid. Below this shear stress there is no flow and therefore it
acts like a solid. After that it behaves like a fluid. Such substances that behaves
as both solids and fluids are known as Rheological substances and the study of
these substances is known as Rheology.
Ideal Fluid
A fluid which is non-viscous and incompressible is known as ideal fluid though
there’s no ideal fluid, it’s introduced to bring simplicity to analysis.
𝜇 𝐻2 𝑂, 20℃ = 1 𝐶𝑒𝑛𝑡𝑖 𝑃𝑜𝑖𝑠𝑒 = 10−3
𝐾𝑔 𝑚 − 𝑠⁄ 𝜇 𝐻𝑔 = 1.55 𝐶𝑒𝑛𝑡𝑖𝑝𝑜𝑖𝑠𝑒
𝜇 𝐻2 𝑂 = (50 − 55)𝜇 𝑎𝑖𝑟

Equation for Linear Velocity Profile
The velocity profile can be approximated as a linear velocity profile, if the gap between plates is very small
(narrow passages).
𝑡𝑎𝑛 𝜃 =
𝑑𝑢 ⋅ 𝑑𝑡
𝑑𝑦
=
𝑣 ⋅ 𝑑𝑡
𝑦
𝑑𝑢
𝑑𝑦
=
𝑣
𝑦
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
= 𝜇
𝑣
𝑦
𝐹 = 𝜏 ⋅ 𝐴 = 𝜇 ⋅ 𝐴 ⋅
𝑣
𝑦
Kinematic Viscosity (υ)
In fluid mechanics the term (μ/ρ) appears frequently and for convenience this term is taken as Kinematic
viscosity.
𝜐 =
𝜇
𝜌
𝐼𝑡𝑠 𝑢𝑛𝑖𝑡𝑠 𝑎𝑟𝑒
𝑚2
𝑠
𝑖𝑛 𝑆. 𝐼 &
𝑐𝑚2
𝑠
𝑜𝑟 𝑆𝑡𝑜𝑘𝑒.
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
=
𝜇
𝜌
𝑑(𝜌𝑢)
𝑑𝑦
= 𝜐
𝑑(𝜌𝑢)
𝑑𝑦
Significance of Kinematic Viscosity
Kinematic viscosity represents the ability of fluid to resist momentum. Therefore, it’s a measure of momentum
diffusivity.
Surface Tension
Consider the molecule A which is below the surface of liquid, this molecule is surrounded by various corresponding
molecules and hence under the influence of various cohesive forces, it will be in equilibrium.
Now consider molecule B which is on the surface of liquid, this molecule is under the influence of net downward
force, because of this there seems to be a layer formed which can resist small tensile loads.
This phenomenon is known as Surface Tension. It’s a line force i.e., it acts normal to the line drawn on the surface
and it lies in the plane of surface.
It is denoted by σ.
As Surface tension is basically due to unbalanced cohesive forces and with increase in temperature, cohesive
forces decrease decreasing Surface tension. At critical point surface tension is zero. Surface tension is very small,
so it is neglected in further fluid mechanics analysis.
Surface tension for water air interface at 20ᵒC is 0.0736 N/m.
Liquid droplets assume Spherical shape due to surface tension.∆P

Pressure in liquid drop in excess of Atmospheric pressure
𝐹𝑝 = 𝛥𝑃 ⋅ 𝐴 = 𝛥𝑃 ⋅
𝜋
4
𝑑2
𝜎 =
𝐹𝑠
𝐿
⇒ 𝐹𝑠 = 𝜎 ⋅ 𝐿 = 𝜎 ⋅ 𝜋𝑑
𝐹𝑜𝑟 𝐸𝑞𝑢𝑙𝑖𝑏𝑟𝑖𝑢𝑚 ⇒ 𝐹𝑝 = 𝐹𝑠 ⇒ 𝛥𝑃 ⋅
𝜋
4
𝑑2
= 𝜎 ⋅ 𝜋𝑑
Pressure forces tries to separate the droplet whereas surface tension tries to contract the droplet i.e., surface
tension force tries to minimize surface area.
Droplets take spherical shape because sphere has minimum surface area for a given volume.
Capillarity
Capillarity is the effect of surface tension. It’s not a property.
The rise or fall of a liquid when a small diameter tube is introduced in it is known as capillarity.
The capillary rise is due to adhesion and capillary rise is due to cohesion. Water is an example for adhesion and
mercury for cohesion.
Expression for capillary rise/fall in a glass tube
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑙𝑖𝑞𝑢𝑖𝑑 𝑏𝑟𝑜𝑢𝑔ℎ𝑡 𝑢𝑝 = 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑠
𝑊𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑙𝑖𝑞𝑢𝑖𝑑 = 𝑆𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑤𝑒𝑖𝑔ℎ𝑡 × 𝑉𝑜𝑙𝑢𝑚𝑒
𝑊𝑒𝑖𝑔ℎ𝑡 = 𝓌 × 𝑉 = 𝓌 ×
𝜋𝑑2
ℎ
4
𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡 𝑜𝑓 𝐹𝑠 = 𝐹𝑠 ⋅ 𝑐𝑜𝑠 𝜃 = 𝜎𝜋𝑑 ⋅ 𝑐𝑜𝑠 𝜃
𝓌 ×
𝜋𝑑2
ℎ
4
= 𝐹𝑠 ⋅ 𝑐𝑜𝑠 𝜃
ℎ =
4𝜎 𝑐𝑜𝑠 𝜃
𝓌𝑑
𝜟𝑷 =
𝟒𝝈
𝒅
→ 𝑳𝒊𝒒𝒖𝒊𝒅 𝒅𝒓𝒐𝒑
𝜟𝑷 =
𝟖𝝈
𝒅
→ 𝑺𝒐𝒂𝒑 𝑩𝒖𝒃𝒃𝒍𝒆
𝜟𝑷 =
𝟐𝝈
𝒅
→ 𝑳𝒊𝒒𝒖𝒊𝒅 𝒋𝒆𝒕

Expression for capillary rise in the annulus of 2 concentric tubes
𝓌 ×
𝜋
4
(𝑑 𝑜
2
− 𝑑𝑖
2
)ℎ = 𝜎𝜋(𝑑 𝑜 + 𝑑𝑖) ⋅ 𝑐𝑜𝑠 𝜃
ℎ =
4𝜎 𝑐𝑜𝑠 𝜃
𝓌(𝑑 𝑜 − 𝑑𝑖)
Expression for capillary rise between two parallel plates
𝑤𝑒𝑖𝑔ℎ𝑡 = 𝓌 × 𝑏ℎ𝑡
𝐹𝑠 = 𝜎(𝑏 + 𝑏) = 2𝜎𝑏
𝓌 × 𝑏ℎ𝑡 = 2𝜎𝑏 ⋅ 𝑐𝑜𝑠 𝜃
ℎ =
2𝜎 ⋅ 𝑐𝑜𝑠 𝜃
𝓌𝑡
Work done in stretching a surface
𝑊𝑜𝑟𝑘 = 𝐹𝑜𝑟𝑐𝑒 × 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 = (𝜎 ⋅ 𝐿) × 𝑥 = 𝜎 × 𝐼𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝐴𝑟𝑒𝑎 (𝐿 ⋅ 𝑥)
𝑊𝑜𝑟𝑘 = 𝜎 × 𝛥𝐴
Note
 The angle of contact between water & glass is 22ᵒ.
 The angle of contact between pure water & clean glass tube is 0ᵒ.
 The angle of contact between mercury & glass is 130ᵒ.
If the height of capillary tube is not sufficient for possible rise, the liquid will rise up to
top and stops because for further rise there is no glass molecules so, it stops.
If the top of the capillary tube is close, then capillary rise will decrease because the air
trapped at top exerts pressure in the downward direction.
Vapour Pressure
Let us consider a closed container with liquid partially filled in it. The surface molecules due to additional energy
overcomes cohesive forces of liquid below surface. This process occurs until the space above the liquid is
saturated. Under equilibrium the no. of molecules leaving the surface is equal to no. of molecules joining surface.
Under these conditions the pressure exerted by vapour on surface of
liquid is called Vapour pressure.
Vapour pressure increases with increases in temperature because at
higher temperature molecular activity is high.
High Volatile liquids (petrol) have high vapour pressure. Mercury has
least vapour pressure and because of this property it is used in
Manometers.

FLUID DYNAMICS
Generally, the forces acting on fluid element are pressure force Fp, gravity force Fg & viscous force Fv.
In Navier stokes equation all these forces are taken into consideration. In Euler’s analysis viscous forces are
neglected, only pressure & gravity forces are taken into consideration.
Navier stokes equation is momentum conservation equation.
Euler’s Equation
Assumption- Flow is Non-viscous
𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑 = 𝓌 × 𝑉 = 𝜌𝑔 × 𝑑𝐴 ⋅ 𝑑𝑆
𝑤 = 𝑚𝑔 → 𝑚 = 𝜌𝑉 = 𝜌 ∙ 𝑑𝐴 ⋅ 𝑑𝑆
→ 𝑤 = 𝜌𝑔 × 𝑑𝐴 ⋅ 𝑑𝑆
𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑡𝑟𝑒𝑎𝑚 = 𝑎 𝑠 = 𝑣
𝜕𝑣
𝜕𝑠
+
𝜕𝑣
𝜕𝑡
𝑆𝑡𝑟𝑒𝑎𝑚 𝐹𝑜𝑟𝑐𝑒 (𝐹𝑠) = 𝑚 ∙ 𝑎 𝑠 = 𝜌 ∙ 𝑑𝐴 ⋅ 𝑑𝑆 × 𝑎 𝑠
𝐹𝑠 = 𝜌 ∙ 𝑑𝐴 ⋅ 𝑑𝑆 × (𝑣
𝜕𝑣
𝜕𝑠
+
𝜕𝑣
𝜕𝑡
)
𝑐𝑜𝑠 𝜃 =
𝑑𝑧
𝑑𝑆
⇒ 𝑑𝑧 = 𝑑𝑆 ⋅ 𝑐𝑜𝑠 𝜃
𝑤 ⋅ 𝑐𝑜𝑠 𝜃 = 𝜌𝑔 ∙ 𝑑𝐴 ⋅ 𝑑𝑆 ⋅ 𝑐𝑜𝑠 𝜃 = 𝜌𝑔 ∙ 𝑑𝐴 ⋅ 𝑑𝑧
𝑆𝑡𝑟𝑒𝑎𝑚 𝐹𝑜𝑟𝑐𝑒 (𝐹𝑠) = 𝑃 ⋅ 𝑑𝐴 − (𝑃 + 𝑑𝑃) ⋅ 𝑑𝐴 − 𝑤 ⋅ 𝑐𝑜𝑠 𝜃
𝜌 ∙ 𝑑𝐴 ⋅ 𝑑𝑆 × (𝑣
𝜕𝑣
𝜕𝑠
+
𝜕𝑣
𝜕𝑡
) = 𝑃 ⋅ 𝑑𝐴 − (𝑃 + 𝑑𝑃) ⋅ 𝑑𝐴 − 𝜌𝑔 ∙ 𝑑𝐴 ⋅ 𝑑𝑧
The above equation is Euler’s Equation.
Bernoulli’s Equation (Conservation of Energy equation)
Assumptions
1. Flow is non-viscous
2. Flow is along a stream line
3. No energy is supplied and no energy is taken out from the fluid during the flow
4. Steady flow & Incompressible
𝑑𝑃 + 𝜌𝑔 ∙ 𝑑𝑧 + 𝜌 ⋅ 𝑑𝑆 × (𝑣
𝜕𝑣
𝜕𝑠
+
𝜕𝑣
𝜕𝑡
) = 0 → 𝑑𝑃 + 𝜌𝑔 ∙ 𝑑𝑧 + 𝜌 ⋅ 𝑑𝑆 × 𝑣
𝑑𝑣
𝑑𝑠
= 0 → 𝑑𝑃 + 𝜌𝑔 ∙ 𝑑𝑧 + 𝜌𝑣 ∙ 𝑑𝑣 = 0
→
𝑑𝑃
𝜌
+ 𝑔 ⋅ 𝑑𝑧 + 𝑣 ⋅ 𝑑𝑣 = 0
After Integration,
→
1
𝜌
∫ 𝑑𝑃 + 𝑔 ∫ 𝑑𝑧 + 𝑣 ∫ 𝑑𝑣 = ∫ 0
𝑃
𝜌
+ 𝑔𝑧 +
𝑣2
2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝐶𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖’𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
In above equation, each term represents energy of the fluid per unit mass.
𝑑𝑃 + 𝜌𝑔 ∙ 𝑑𝑧 + 𝜌 ⋅ 𝑑𝑆 × (𝑣
𝜕𝑣
𝜕𝑠
+
𝜕𝑣
𝜕𝑡
) = 0
Steady flow

Bernoulli’s Theorem
In a steady incompressible non-viscous flow along a stream line, the sum of pressure, kinetic & potential energy is
constant.
𝐸𝑛𝑒𝑟𝑔𝑦
𝑚𝑎𝑠𝑠
→
𝑃
𝜌
+
𝑣2
2
+ 𝑔𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑚𝑎𝑠𝑠 × 𝑔
→
𝑤𝑒𝑖𝑔ℎ𝑡
→
𝑃
𝜌𝑔
+
𝑣2
2𝑔
+ 𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑃
𝓌
+
𝑣2
2𝑔
+ 𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
In this equation, each term represents energy per unit weight.
Various heads in Fluid mechanics
Pressure Head
The height by which fluid rises due to pressure when a piezometer is
connected is known as pressure head.
𝑃 = 0 + 𝜌𝑔ℎ = 𝓌ℎ
ℎ =
𝑃
𝓌
= 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 ℎ𝑒𝑎𝑑
Velocity Head/ Kinematic energy head
It’s the height by which fluid falls in a frictionless environment to reach to a particular height.
𝑣 = √2𝑔ℎ
ℎ =
𝑣2
2𝑔
= 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 ℎ𝑒𝑎𝑑
Potential Energy head (z)
It’s the vertical distance with respect to a reference line.
Piezometric Head
The sum of pressure and potential energy is known as piezometric head.
𝑃𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑎𝑑 =
𝑃
𝓌
+ 𝑧
Relationship between first law of thermodynamics & Bernoulli’s equation
ℎ1 +
𝑣1
2
2
+ 𝑧1 𝑔 + 𝑞 = ℎ2 +
𝑣2
2
2
+ 𝑧2 𝑔 + 𝑤
ℎ = 𝑠𝑝𝑒𝑐𝑖𝑓𝑖𝑐 𝑒𝑛𝑡ℎ𝑎𝑙ℎ𝑦, 𝑞 =
ℎ𝑒𝑎𝑡 𝑡𝑟𝑎𝑛𝑠𝑓𝑒𝑟
𝑚𝑎𝑠𝑠
, 𝑤 =
𝑊𝑜𝑟𝑘
𝑚𝑎𝑠𝑠
Assumptions
1. Steady flow & incompressible
2. No heat transfer & work transfer
3. No change in Internal energy
ℎ = 𝑢 + 𝑃𝓋 = 𝑢 +
𝑃
𝜌
𝑢1 +
𝑃1
𝜌
+
𝑣1
2
2
+ 𝑧1 𝑔 + 𝑞 = 𝑢2 +
𝑃2
𝜌
+
𝑣2
2
2
+ 𝑧2 𝑔 + 𝑤
(𝑢1 = 𝑢2, 𝑞 = 𝑤 = 0)
𝑃
𝓌
+
𝑣2
2𝑔
+ 𝑧 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖′𝑠𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛

Bernoulli’s equation for a Horizontal Stream line
𝑧1 = 𝑧2
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
=
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
Bernoulli’s equation for a real fluid flow problem
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + ℎ 𝐿
(ℎ 𝐿 = 𝐻𝑒𝑎𝑑 𝐿𝑜𝑠𝑠)
In case of irrotational flow, Bernoulli’s equation can be applied between any 2 points (throughout the flow field),
because the stream line constants are same for different streamlines in irrotational flow.
In case of rotational flow, Bernoulli’s must be applied only for a particular stream line, because the stream line
constants are different for different stream lines.
Bernoulli’s equation is not the total energy conservation equation because heat transfer and work transfer are
not taken into consideration. Therefore, Bernoulli’s equation is known as Mechanical Energy Conservation.
Applications of Bernoulli’s Equation
Venturimeter
It’s used for calculating Discharge.
𝑃1
𝓌
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝓌
+
𝑣2
2
2𝑔
+ 𝑧2
(
𝑃1
𝓌
−
𝑃2
𝓌
) + (𝑧1 − 𝑧2) =
𝑣2
2
2𝑔
−
𝑣1
2
2𝑔
= ℎ (𝑃𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑖𝑔ℎ𝑡)
(
𝑃1
𝓌
+ 𝑧1) − (
𝑃2
𝓌
+ 𝑧2) =
𝑣2
2
2𝑔
−
𝑣1
2
2𝑔
= ℎ
𝐹𝑜𝑟 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝐶𝑎𝑠𝑒(𝑧1 = 𝑧2),
(
𝑃1
𝓌
−
𝑃2
𝓌
) =
𝑣2
2
2𝑔
−
𝑣1
2
2𝑔
= ℎ → 𝑣1
2
− 𝑣2
2
= 2𝑔ℎ
𝑄 = 𝐴1 ⋅ 𝑣1 = 𝐴2 ⋅ 𝑣2 → 𝑣1 =
𝑄
𝐴1
→ 𝑣2 =
𝑄
𝐴2

𝑣1
2
− 𝑣2
2
= (
𝑄
𝐴1
)
2
− (
𝑄
𝐴2
)
2
= 2𝑔ℎ → 𝑄2
(
1
𝐴1
2 −
1
𝐴2
2) = 2𝑔ℎ
𝑄 =
𝐴1 ⋅ 𝐴2 √2𝑔ℎ
√𝐴1
2
− 𝐴2
2
As no losses were assumed while deriving this equation, this discharge is known as ideal
discharge or theoretical discharge.
𝑄𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 =
𝐴1 ⋅ 𝐴2 √2𝑔ℎ
√𝐴1
2
− 𝐴2
2
→
𝑃1
𝓌
+ 𝐻 + 𝑥 −
𝑥 ⋅ 𝑠 𝐻𝑔
𝑠
− 𝐻 =
𝑃2
𝓌
→
𝑃1
𝓌
−
𝑃2
𝓌
= 𝑥 ⋅ (
𝑠 𝐻𝑔
𝑠
− 1) = ℎ (𝑃𝑖𝑒𝑧𝑜 ℎ𝑡. )
→ 𝐴1 𝑣1 = 𝐴2 𝑣2 → 𝐴2 < 𝐴1 ⇒ 𝑣2 > 𝑣1
→
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
+
𝑣2
2
2𝑔
→ 𝑣2 > 𝑣1 ⇒ 𝑃2 < 𝑃1
Principle of Venturimeter
By reducing the area in a steady incompressible flow, velocity increases. This results in decrease of pressure. Due
to this pressure difference, there will be manometric deflection when differential manometer is connected. By
measuring the 𝑥, the discharge will be calculated.
Coefficient of discharge (Cd)
It is defined as the ratio of actual discharge to theoretical discharge.
Cd depends on type of flow (Reynolds no.) and area ratio.
As Venturimeter is gradually converging and diverging device, losses are less and hence Cd is 0.94―0.98.
𝑄𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 =
𝐴1 ⋅ 𝐴2 √2𝑔ℎ
√𝐴1
2
− 𝐴2
2
𝑄 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐶 𝑑 ×
𝐴1 ⋅ 𝐴2 √2𝑔ℎ
√𝐴1
2
− 𝐴2
2
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
+
𝑣2
2
2𝑔
+ ℎ𝑙𝑜𝑠𝑠𝑒𝑠 → (
𝑃1
𝓌
−
𝑃2
𝓌
) − ℎ𝑙 =
𝑣2
2
2𝑔
−
𝑣1
2
2𝑔
ℎ − ℎ𝑙 =
𝑣2
2
− 𝑣1
2
2𝑔
⇒ 𝑣2
2
− 𝑣1
2
= 2𝑔(ℎ − ℎ𝑙)
(
𝑄
𝐴1
)
2
− (
𝑄
𝐴2
)
2
= 2𝑔(ℎ − ℎ𝑙) (𝑣1 =
𝑄
𝑎1
, 𝑣2 =
𝑄
𝑎2
)
𝑄 𝑎𝑐𝑡𝑢𝑎𝑙 =
𝐴1 ⋅ 𝐴2 √2𝑔(ℎ − ℎ𝑙)
√𝐴1
2
− 𝐴2
2

𝐶 𝑑 ×
𝐴1 ⋅ 𝐴2 √2𝑔ℎ
√𝐴1
2
− 𝐴2
2
=
𝐴1 ⋅ 𝐴2 √2𝑔(ℎ − ℎ𝑙)
√𝐴1
2
− 𝐴2
2
𝐶 𝑑 = √
ℎ − ℎ𝑙
ℎ
General Properties of a Venturimeter
𝑑2 = (
1
3
𝑡𝑜
1
2
) · 𝑑1
𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 = 20ᵒ − 22ᵒ
𝐴𝑛𝑔𝑙𝑒 𝑜𝑓 𝑑𝑖𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 = 7ᵒ
The angle of divergence is generally kept less than 7ᵒ, in order to avoid flow separation.
Orifice meter
The device is used for finding out discharge and it is the cheapest measurement for calculating discharge.
It is based on same principle as Venturimeter.
It’s a circular disc with a circular hole.
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐶𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 ( 𝐶𝑐) =
𝐴2
𝐴 𝑜
=
𝑉𝑒𝑛𝑎 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑎 𝐴𝑟𝑒𝑎
𝑂𝑟𝑖𝑓𝑖𝑐𝑒 𝐴𝑟𝑒𝑎
𝐶𝑐 =
𝐴2
𝐴 𝑜
→ 𝐴2 = 𝐶𝑐 ⋅ 𝐴 𝑜
𝑄 = 𝐴1 ⋅ 𝑣1 = 𝐴2 ⋅ 𝑣2 → 𝑣1 =
𝐴2 ⋅ 𝑣2
𝐴1
𝑣1 =
𝐶 𝑐 ⋅ 𝐴 𝑜 ⋅ 𝑣2
𝐴1
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
+
𝑣2
2
2𝑔
→ (
𝑃1
𝓌
−
𝑃2
𝓌
) =
𝑣2
2
2𝑔
−
𝑣1
2
2𝑔
= ℎ → 𝑣2
2
− 𝑣1
2
= 2𝑔ℎ
𝑣2
2
− (
𝐶 𝑐 ⋅ 𝐴 𝑜 ⋅ 𝑣2
𝐴1
)
2
= 2𝑔ℎ ⇒ 𝑣2
2
(1 − (
𝐶 𝑐 ⋅ 𝐴 𝑜
𝐴1
)
2
) = 2𝑔ℎ
𝑣2 =
√2𝑔ℎ
√1 −
𝐶𝑐
2 ⋅ 𝐴 𝑜
2
𝐴1
2
𝑄 = 𝐴2 ⋅ 𝑣2 = 𝐶𝑐 ⋅ 𝐴 𝑜 ⋅
√2𝑔ℎ
√1 −
𝐶𝑐
2 ⋅ 𝐴 𝑜
2
𝐴1
2
= 𝐶𝑐 ⋅ 𝐴 𝑜 ⋅
√2𝑔ℎ
√1 −
𝐶𝑐
2 ⋅ 𝐴 𝑜
2
𝐴1
2
×
√1 −
𝐴 𝑜
2
𝐴1
2
√1 −
𝐴 𝑜
2
𝐴1
2
(𝑀𝑢𝑙𝑡𝑝𝑙𝑖𝑛𝑔 𝑎𝑛𝑑 𝑑𝑖𝑣𝑖𝑑𝑖𝑛𝑔 𝑏𝑦 √1 −
𝐴 𝑜
2
𝐴1
2 𝑎𝑛𝑑 𝑟𝑒𝑎𝑟𝑟𝑎𝑛𝑔𝑖𝑛𝑔)

→ 𝑄 =
𝐶𝑐 ⋅ √1 −
𝐴 𝑜
2
𝐴1
2
√1 −
𝐶𝑐
2 ⋅ 𝐴 𝑜
2
𝐴1
2
×
𝐴 𝑜 ⋅ √2𝑔ℎ
√1 −
𝐴 𝑜
2
𝐴1
2
→ 𝑄 = 𝐶 𝑑 ×
𝐴 𝑜 ⋅ √2𝑔ℎ
√1 −
𝐴 𝑜
2
𝐴1
2
(
𝐶 𝑑 =
𝐶𝑐 ⋅ √1 −
𝐴 𝑜
2
𝐴1
2
√1 −
𝐶𝑐
2 ⋅ 𝐴 𝑜
2
𝐴1
2
)
→ 𝑄 =
𝐶 𝑑 ⋅ 𝐴 𝑜 ⋅ 𝐴1 ⋅ √2𝑔ℎ
√𝐴1
2
− 𝐴 𝑜
2
As the area reduction is sudden in orifice meter, losses are more and hence Cd of orifice meter is
less. (Cd⇒ 0.68―0.76)
Pilot tube
It’s used for finding velocity of the flow.
Case 1: - Velocity in Open Chamber
𝑣2 = 0
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
+
𝑣2
2
2𝑔
⇒
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
𝑃1
𝓌
= 𝑆𝑡𝑎𝑡𝑖𝑐 𝐻𝑒𝑎𝑑,
𝑣1
2
2𝑔
= 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝐻𝑒𝑎𝑑,
𝑃2
𝓌
= 𝑆𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑
𝑆𝑡𝑎𝑡𝑖𝑐 𝐻𝑒𝑎𝑑 + 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝐻𝑒𝑎𝑑 = 𝑆𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 𝐻𝑒𝑎𝑑
𝑃1 = 0 + 𝜌𝑔ℎ 𝑜 →
𝑃1
𝜌𝑔
=
𝑃1
𝓌
= ℎ 𝑜
𝑃2 = 0 + 𝜌𝑔(ℎ + ℎ 𝑜) →
𝑃2
𝜌𝑔
=
𝑃2
𝓌
= ℎ + ℎ 𝑜
Substituting we get,
ℎ 𝑜 +
𝑣1
2
2𝑔
= (ℎ + ℎ 𝑜) → ℎ =
𝑣1
2
2𝑔
= 𝐷𝑦𝑛𝑎𝑚𝑖𝑐 𝐻𝑒𝑎𝑑
𝑣1 = √2𝑔ℎ = √2𝑔 ⋅ (𝐷𝑦𝑛𝑎𝑚𝑖𝑐 ℎ𝑒𝑎𝑑)
𝑣1 = √2𝑔ℎ = √2𝑔 ⋅ ( 𝑆𝑡𝑎𝑔𝑛𝑎𝑡𝑖𝑜𝑛 ℎ𝑒𝑎𝑑 − 𝑆𝑡𝑎𝑡𝑖𝑐 ℎ𝑒𝑎𝑑)
Case 2: - Velocity in Pipes
𝑣2 = 0
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
+
𝑣2
2
2𝑔
⇒
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
→
𝑣1
2
2𝑔
=
𝑃2 − 𝑃1
𝓌
𝑃1
𝓌
+ 𝐻 +
𝑥 ⋅ 𝑆 𝐻𝑔
𝑆
− 𝑥 − 𝐻 =
𝑃2
𝓌
⟹
𝑃2 − 𝑃1
𝓌
= 𝑥 (
𝑆 𝐻𝑔
𝑆
) − 𝑥
Equating above equations, we get
𝑣1
2
2𝑔
=
𝑥 ⋅ 𝑆 𝐻𝑔
𝑆
− 𝑥
𝑣1 = √2𝑔𝑥 ⋅ (
𝑆 𝐻𝑔
𝑆
− 1)

If the specific gravity of manometric fluid is less than specific gravity of following fluid, inverted differential
U―tube manometer is used.
𝑣 𝑎𝑐𝑡𝑢𝑎𝑙 = 𝐶𝑣 ⋅ √2𝑔𝑥 ⋅ (
𝑆 𝐻𝑔
𝑆
− 1) = 𝐶𝑣 ⋅ 𝑣𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙
Device Shape Losses Cd Cost
Venturimeter Low High High
Flow Nozzle Medium Medium Medium
Orifice meter High Low Cheap
Force on Pipe Bends
Momentum Equation
𝛴𝐹 = 𝑚 ⋅ 𝑎 = 𝑚 (
𝑣 − 𝑢
𝑡
) = 𝑚̇ ⋅ (𝑣 − 𝑢)
(𝑚̇ = 𝜌 ⋅ 𝐴 ⋅ 𝑣⃗)
𝛴𝐹 = 𝜌𝑄(𝑣 − 𝑢)  Momentum Equation
Applying Momentum equation in x-direction,
𝑃1 ⋅ 𝐴1 + 𝐹𝑥 − 𝑃2 ⋅ 𝐴2 ⋅ 𝑐𝑜𝑠 𝜃 = 𝜌𝑄(𝑣2 ⋅ 𝑐𝑜𝑠 𝜃 − 𝑣1)
Momentum equation in y-direction,
𝐹𝑦 − 𝑃2 ⋅ 𝐴2 ⋅ 𝑠𝑖𝑛 𝜃 = 𝜌𝑄(𝑣2 ⋅ 𝑠𝑖𝑛 𝜃 − 0)

VORTEX MOTION
The motion of the fluid along the curved path is known as vortex motion.
1. Vortex motion is of 2 types, Forced Motion & Free Vortex
Forced Vortex Motion
The motion if a fluid in a curved path under the influence of external agency is known as forced vortex motion. As
there is a continuous expenditure of energy in forced vortex motion, Bernoulli’s equation is not applicable. The
equation 𝑣=r⋅ω, is applicable for forced vortex motion.
Example- Liquid in a container when rotated, motion of fluid in impeller of a centrifugal pump.
Forced vortex motion is Rotational Flow.
Free Vortex motion
In free vortex motion, the fluid moves in curved path due to internal fluid action. But not due to external torque.
As there is no expenditure of energy. Therefore, Bernoulli’s equation is applicable for free vortex motion.
𝑑
𝑑𝑡
(𝑚𝑣𝑟) = 𝑇 = 0 → 𝑚𝑣𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 → 𝑣𝑟 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑣 ⋅ 𝑟 = 𝐾 → 𝑓𝑟𝑒𝑒 𝑣𝑜𝑟𝑡𝑒𝑥 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
Example – Motion of fluid in diffuser of centrifugal pump. Flow of fluid in pipe bends, whirl pool, flow of liquid in wash
basin.
Free Vortex is an Irrotational flow
Generalized equation for Vortex Motion
𝑉𝑜𝑙𝑢𝑚𝑒 = 𝑑𝐴 ⋅ 𝑑𝑟 𝑀𝑎𝑠𝑠 = 𝑣𝑜𝑙𝑢𝑚𝑒 ⋅ 𝜌
𝑚 = 𝜌 ⋅ 𝑑𝐴 ⋅ 𝑑𝑟
𝑃 ⋅ 𝑑𝐴 +
𝜌 ⋅ 𝑑𝐴 ⋅ 𝑑𝑟 ⋅ 𝑣2
𝑟
= (𝑃 +
𝜕𝑃
𝜕𝑟
𝑑𝑟) 𝑑𝐴
𝑃 +
𝜌 ⋅ 𝑑𝑟 ⋅ 𝑣2
𝑟
= 𝑃 +
𝜕𝑃
𝜕𝑟
𝑑𝑟
𝜌 ⋅ 𝑣2
𝑟
𝑑𝑟 =
𝜕𝑃
𝜕𝑟
𝑑𝑟
𝜌 ⋅ 𝑣2
𝑟
=
𝜕𝑃
𝜕𝑟
This equation gives the variation of pressure in radial direction.
𝜕𝑃
𝜕𝑧
= −𝓌 = −𝜌𝑔
𝑑𝑃 =
𝜕𝑃
𝜕𝑟
⋅ 𝑑𝑟 +
𝜕𝑃
𝜕𝑧
⋅ 𝑑𝑧
𝑑𝑃 =
𝜌 ⋅ 𝑣2
𝑟
⋅ 𝑑𝑟 − 𝜌𝑔 ⋅ 𝑑𝑧

Free Vortex Motion Equation
𝑑𝑃 =
𝜌 ⋅ 𝑣2
𝑟
𝐹𝑜𝑟 𝑓𝑟𝑒𝑒 𝑣𝑜𝑟𝑡𝑒𝑥 𝑚𝑜𝑡𝑖𝑜𝑛, 𝑣𝑟 = 𝑐 → 𝑣 =
𝑐
𝑟
→ ∫ 𝑑𝑃
𝑃2
𝑃1
= ∫
𝜌
𝑟
⋅
𝑐2
𝑟2
⋅ 𝑑𝑟
𝑟2
𝑟1
− ∫ 𝜌𝑔 ⋅ 𝑑𝑧
𝑧2
𝑧1
𝑃2 − 𝑃1 = 𝜌 ⋅ 𝑐2
⋅ (
1
𝑟1
2 −
1
𝑟2
2) − 𝜌𝑔(𝑧2 − 𝑧1) → 𝑃2 − 𝑃1 = 𝜌 ⋅
𝑐2
𝑟1
2 − 𝜌 ⋅
𝑐2
𝑟2
2 − 𝜌𝑔𝑧2 − 𝜌𝑔𝑧1
𝑃2 − 𝑃1 = 𝜌 ⋅
𝑣1
2
2
− 𝜌 ⋅
𝑣2
2
2
− 𝜌𝑔𝑧2 − 𝜌𝑔𝑧1
𝑃2 +
𝜌𝑣1
2
2
+ 𝜌𝑔𝑧1 = 𝑃1 +
𝜌𝑣2
2
2
+ 𝜌𝑔𝑧2 → 𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖′
𝑠 𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛
Bernoulli’s equation is applicable for free vortex of motion.
Forced Vortex Motion Equation
𝑑𝑃 =
𝜌 ⋅ 𝑣2
𝑟
𝐹𝑜𝑟 𝑓𝑜𝑟𝑐𝑒𝑑 𝑣𝑜𝑟𝑡𝑒𝑥 𝑚𝑜𝑡𝑖𝑜𝑛, 𝑣 = 𝑟𝜔
𝑑𝑃 =
𝜌 ⋅ (𝑟𝜔)2
𝑟
⋅ 𝑑𝑟 − 𝜌𝑔 ⋅ 𝑑𝑧 → 𝑑𝑃 = 𝜌 ⋅ 𝑟 ⋅ 𝜔2
∫ 𝑑𝑃
𝑃2
𝑃1
= ∫ 𝜌 ⋅ 𝑟 ⋅ 𝜔2
⋅ 𝑑𝑟
𝑟2
𝑟1
− ∫ 𝜌𝑔 ⋅ 𝑑𝑧
𝑧2
𝑧1
𝑃2 − 𝑃1 =
𝜌 ⋅ 𝜔2
2
⋅ ( 𝑟2
2
− 𝑟1
2) − 𝜌𝑔( 𝑧2 − 𝑧1)
𝐿𝑒𝑡 𝑢𝑠 𝑛𝑜𝑤 𝑠𝑒𝑙𝑒𝑐𝑡 𝑡𝑤𝑜 𝑝𝑜𝑖𝑛𝑡𝑠 ① & ② 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒. → (𝑷 𝟏 = 𝑷 𝟐)
Substituting in above eqn (Bernoulli’s)
0 =
𝜌 ⋅ 𝜔2
2
⋅ (𝑟2
2
− 𝑟1
2) − 𝜌𝑔(𝑧2 − 𝑧1)
𝑧2 − 𝑧1 =
𝜔2
2𝑔
⋅ (𝑟2
2
− 𝑟1
2)
𝐼𝑓 𝑝𝑜𝑖𝑛𝑡 ① 𝑖𝑠 𝑡𝑎𝑘𝑒𝑛 𝑜𝑛 𝑎𝑥𝑖𝑠, 𝑟1 = 0
𝑧2 − 𝑧1 = 𝑧 =
𝜔2
2𝑔
⋅ 𝑟2
2
𝐼𝑓 𝑟2 = 𝑅, → 𝑧 = 𝐻𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑝𝑎𝑟𝑎𝑏𝑜𝑙𝑙𝑖𝑜𝑑 (𝐻) =
𝜔2
⋅ 𝑅2
2𝑔
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑃𝑎𝑟𝑎𝑏𝑜𝑙𝑙𝑖𝑜𝑑 =
𝜋𝑅2
𝐻
2

LAMINAR FLOW
(Viscous flow of incompressible fluids)
Reynolds Number
It is the ratio of inertia force to viscous force.
𝑅 𝑒 =
𝜌𝑣𝐿
𝜇
𝐿 → 𝑐ℎ𝑎𝑟𝑒𝑐𝑡𝑒𝑟𝑖𝑠𝑡𝑖𝑐 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛
Significance of L
It is such a dimension over which significant changes in properties occur.
For flow through pipes, characteristic dimension is pipe diameter. For flow over a flat plate, characteristic
dimension is distance from leading edge (𝑥).
Reynold found from his experiment for flow through pipes,
Re < 2000  Laminar
2000 < Re <4000 Transition
Re > 4000  Turbulent
𝑃1
𝜌𝑔
+
𝑣1
2
2𝑔
+ 𝑧1 =
𝑃2
𝜌𝑔
+
𝑣2
2
2𝑔
+ 𝑧2 + ℎ 𝐿
𝑃1
𝜌𝑔
=
𝑃2
𝜌𝑔
+ ℎ 𝐿 (𝑣1 = 𝑣2, 𝑧1 = 𝑧2)
𝑃1 − 𝑃2
𝜌𝑔
= ℎ 𝐿
The pressure decreases in the direction of flow in order to overcome loses i.e.,
pressure gradient is negative in the direction of flow.
Darcy-Weisbach equation
This equation is used for calculating head loss due to friction.
→ ℎ 𝐿 =
𝑓𝐿𝑣2
2𝑔𝐷
𝑓 → 𝐷𝑎𝑟𝑐𝑦 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 (𝑜𝑟)𝑀𝑜𝑜𝑑𝑦 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
→ ℎ 𝐿 =
4𝑓′𝐿𝑣2
2𝑔𝐷
𝑓′
→ 𝐹𝑎𝑛𝑛𝑖𝑛𝑔𝑠 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟
𝑓 = 4𝑓′
This equation is applicable for Laminar or turbulent flow, horizontal, inclined or vertical pipes, but the flow must
be steady.
Fully developed flow
A flow is said to be a fully developed flow if the velocity profile doesn’t change in longitudinal direction and
pressure gradient (dP/dx) remains constant.

Laminar flow through Circular pipes (Hagen- Poiseuille flow)
Assumptions-
1. Steady flow
2. Fully developed flow
𝛴𝐹 = 𝑚𝑎 = 0
(𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑎𝑠𝑠𝑢𝑚𝑝𝑡𝑖𝑜𝑛 𝑆𝑝𝑎𝑡𝑖𝑎𝑙 𝑎𝑛𝑑 𝑡𝑒𝑚𝑝𝑜𝑟𝑎𝑙 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑎𝑟𝑒 𝑧𝑒𝑟𝑜)
𝑃 ⋅ 𝜋𝑟2
− (𝑃 +
𝜕𝑃
𝜕𝑥
𝑑𝑥) 𝜋𝑟2
− 𝜏 ⋅ 2𝜋𝑟 ⋅ 𝑑𝑥 = 0
𝑃 ⋅ 𝑟 − (𝑃 +
𝜕𝑃
𝜕𝑥
𝑑𝑥) 𝑟 − 2𝜏 ⋅ 𝑑𝑥 = 0 ⟶ −
𝜕𝑃
𝜕𝑥
𝑑𝑥 = 2𝜏 ⋅ 𝑑𝑥
𝜏 = −
𝜕𝑃
𝜕𝑥
⋅
𝑟
2
𝐹𝑜𝑟 𝑓𝑢𝑙𝑙𝑦 𝑑𝑒𝑣𝑒𝑙𝑜𝑝𝑒𝑑 𝑓𝑙𝑜𝑤,
𝜕𝑃
𝜕𝑥
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑆𝑜, 𝜏 ∝ 𝑟
As the shear stress is zero at the centre of pipe, therefore, viscous forces are zero at the
centre and hence Bernoulli’s equation can be applied along the axis of the pipe.
In a Laminar flow through pipes, shear stress varies linearly from zero at the centre to
the maximum at the pipe wall.
Velocity Distribution
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
→ 𝜏 = 𝜇
𝑑𝑢
−𝑑𝑟
= −𝜇
𝑑𝑢
𝑑𝑟
(𝑟 + 𝑦 = 𝑅 → 𝑑𝑦 = −𝑑𝑟)
𝜏 = −
𝜕𝑃
𝜕𝑥
⋅
𝑟
2
Equating we get,
−
𝜕𝑃
𝜕𝑥
⋅
𝑟
2
= −𝜇
𝑑𝑢
𝑑𝑟
𝑑𝑢
𝑑𝑟
=
𝑟
2𝜇
(
𝜕𝑃
𝜕𝑥
) ⟶ 𝑑𝑢 =
1
2𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑟 ⋅ 𝑑𝑟
Integrating we get,
𝑢 =
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑟2
+ 𝐶
𝐴𝑡 𝑡ℎ𝑒 𝑃𝑖𝑝𝑒 𝑤𝑎𝑙𝑙, 𝑟 = 𝑅 & 𝑢 = 0
0 =
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2
+ 𝐶 ⟶ 𝐶 = −
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2

𝑢 =
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑟2
−
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2
𝑢 = −
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) [𝑅2
− 𝑟2]
(𝑙𝑜𝑐𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)𝑢 = −
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2
⋅ [1 −
𝑟2
𝑅2
]
𝑊𝑒 𝑔𝑒𝑡 𝑢 𝑚𝑎𝑥 𝑤ℎ𝑒𝑛 𝑟 = 0,
𝑢 𝑚𝑎𝑥 = −
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2
𝑢 = 𝑢 𝑚𝑎𝑥 ⋅ [1 −
𝑟2
𝑅2
]
The velocity distribution is parabolic in Laminar flow through pipes.
Discharge
Let us calculate discharge through elemental ring,
𝑑𝑄 = 𝑢 ⋅ 2𝜋𝑟 ⋅ 𝑑𝑟
𝑄 = ∫ 𝑢 ⋅ 2𝜋𝑟 ⋅ 𝑑𝑟
𝑅
0
= ∫ 𝑢 𝑚𝑎𝑥 ⋅ [1 −
𝑟2
𝑅2
] ⋅ 2𝜋𝑟 ⋅ 𝑑𝑟
𝑅
0
= 2𝜋 ⋅ 𝑢 𝑚𝑎𝑥 [
𝑟2
2
−
𝑟4
4𝑅2
]
0
𝑅
𝑄 = 2𝜋 ⋅ 𝑢 𝑚𝑎𝑥 [
𝑅2
2
−
𝑅2
4
] = 2𝜋 ⋅ 𝑢 𝑚𝑎𝑥 ⋅
𝑅2
4
𝑄 =
𝑢 𝑚𝑎𝑥 ⋅ 𝜋𝑅2
2
𝑄 =
𝜋 ⋅ 𝑢 𝑚𝑎𝑥 ⋅ 𝑅2
2
=
𝜋 ⋅ 𝑅2
2
(−
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2
)
𝑄 = −
𝜋
8𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅4

Average Velocity (𝑣)
𝑄 =
2
𝑄 = 𝐴 ⋅ 𝑣 = 𝜋𝑅2
⋅ 𝑣
Equating we get,
𝜋𝑅2
⋅ 𝑣 =
2
𝑣 =
𝑢 𝑚𝑎𝑥
2
Pressure drop in a given length L
𝑣 =
𝑢 𝑚𝑎𝑥
2
=
1
2
(−
1
4𝜇
(
𝜕𝑃
𝜕𝑥
) ⋅ 𝑅2
)
8𝜇𝑣
𝑅2
𝜕𝑥 = −𝜕𝑃
Integrating we get,
⟶ ∫
8𝜇𝑣
𝑅2
𝜕𝑥
𝑥1
𝑥2
= ∫ −𝜕𝑃
𝑃2
𝑃1
⟶
8𝜇𝑣
𝑅2
(𝑥2 − 𝑥1) = −(𝑃2 − 𝑃1) ⟶
8𝜇𝑣
𝑅2
⋅ 𝐿 = 𝑃1 − 𝑃2
(𝑥2 − 𝑥1 = 𝐿)
𝑃1 − 𝑃2 =
8𝜇𝑣𝐿
𝑅2
=
8𝜇𝑣𝐿
(𝐷 2⁄ )2
𝑃1 − 𝑃2 =
32𝜇𝑣𝐿
𝐷2
(𝜇 = 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦, 𝑣 = 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦, 𝐿 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏 𝑤⁄ 𝑝𝑜𝑖𝑛𝑡𝑠, 𝑅 = 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑝𝑖𝑝𝑒)
𝑃1 − 𝑃2
𝓌
= ℎ 𝐿 =
𝑓 ⋅ 𝐿 ⋅ 𝑣2
2𝑔𝐷
→ 𝑃1 − 𝑃2 =
𝜌 ⋅ 𝑓 ⋅ 𝐿 ⋅ 𝑣2
2𝐷
𝑃1 − 𝑃2 = 𝓌 ⋅ ℎ 𝐿 =
32𝜇𝑣𝐿
𝐷2
(𝑓𝑟𝑜𝑚 𝑝𝑟𝑒𝑣𝑖𝑜𝑢𝑠 𝑑𝑒𝑟𝑖𝑣𝑒𝑑 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛)
Equating we get,
32𝜇𝑣𝐿
𝐷2
=
𝜌 ⋅ 𝑓 ⋅ 𝐿 ⋅ 𝑣2
2𝐷
𝑓 =
64
𝜌𝑣𝐷
𝜇
=
64
𝑅𝑒
In Laminar flow through pipes, friction factor (f) will depend on Reynold’s Number.

Shear Factor (V*)
𝜏 = −
𝜕𝑃
𝜕𝑥
⋅
𝑟
2
𝜏 𝑜 = −
𝜕𝑃
𝜕𝑥
⋅
𝑅
2
= −
(𝑃2 − 𝑃1)
(𝑥2 − 𝑥1)
⋅
𝐷
4
=
𝑃2 − 𝑃1
𝐿
⋅
𝐷
4
𝑃1 − 𝑃2 = 𝓌 ⋅ ℎ 𝐿 =
𝜌 ⋅ 𝑓 ⋅ 𝐿 ⋅ 𝑣2
2𝐷
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 (𝑃2 − 𝑃1) 𝑖𝑛 𝜏 𝑜,
𝜏 𝑜 =
𝜌 ⋅ 𝑓 ⋅ 𝐿 ⋅ 𝑣2
2𝐷
⋅
𝐷
4𝐿
𝜏 𝑜 =
𝜌 ⋅ 𝑓 ⋅ 𝑣2
8
𝜏 𝑜
𝜌
=
𝑓 ⋅ 𝑣2
8
√
𝜏 𝑜
𝜌
= √
𝑓
8
⋅ 𝑣
𝑆ℎ𝑒𝑎𝑟 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (√
𝜏 𝑜
𝜌
) = 𝑉∗
𝑉∗
= √
𝑓
8
⋅ 𝑣
BOUNDARY LAYER THEORY
Boundary layer theory was proposed by Prandtl in 1904.
When a real fluid flows past a solid object, the velocity of the fluid will be same that of object when it comes in
contact with the object. If the object is stationary, the fluid will also have zero velocity. Away from the object the
fluid velocity increases and at some distance from the object, the fluid velocity will be free stream velocity. This
distance from the object where there are velocity gradients is known as Boundary layer thickness and this region
is known as boundary layer region.
In the boundary layer region, the flow is viscous & rotational, as the flow is viscous in boundary layer region. As
the flow is non-viscous outside the boundary layer region, the Bernoulli’s equation can be applied.

Development of Boundary Layer over a flat Plate
When a real fluid flows past a flat plate, the velocity of fluid on plate will be same as that of the plate velocity. If
the plate is at rest, the fluid will also have zero velocity. The boundary layer thickness grows as the distance from
the leading-edge increases. Up to a certain distance from the leading edge the flow in Boundary layer is laminar.
As the laminar boundary layer grows instability occurs and the flow changes from laminar to turbulent through
transition. It’s found that even in turbulent boundary layer region close to the plate, the flow is laminar, this
region is known as laminar sub-layer region. Laminar sublayer region exists in turbulent boundary region.
Boundary Conditions
𝐴𝑡 𝑥 = 0 → 𝛿 = 0
𝐴𝑡 𝑦 = 0 → 𝑢 = 0
𝐴𝑡 𝑦 = 𝛿 → 𝑢 = 𝑢∞
𝐴𝑡 𝑦 = 𝛿 →
𝑑𝑢
𝑑𝑦
= 0
Boundary Layer thickness (δ)
It is the distance from the boundary to the point in y-direction, where the velocity is 99% of free stream velocity.
For all calculations, 𝐴𝑡 𝑦 = 𝛿 → 𝑢 = 𝑢∞
Displacement thickness (δ*)
𝑚̇ 𝑖𝑑𝑒𝑎𝑙 = 𝜌𝐴𝑣⃗ = 𝜌 ⋅ (𝑑𝑦 ⋅ 1) ⋅ 𝑢∞
𝑚̇ 𝑟𝑒𝑎𝑙 = 𝜌 ⋅ (𝑑𝑦 ⋅ 1) ⋅ 𝑢
𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 𝑖𝑛 𝑚𝑎𝑠𝑠 𝑓𝑙𝑜𝑤 𝑟𝑎𝑡𝑒 𝑑𝑢𝑒 𝑡𝑜 𝐵𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑙𝑎𝑦𝑒𝑟 𝑔𝑟𝑜𝑤𝑡ℎ = 𝑚̇ 𝑖𝑑𝑒𝑎𝑙 − 𝑚̇ 𝑟𝑒𝑎𝑙
𝑚̇ 𝑖𝑑𝑒𝑎𝑙 − 𝑚̇ 𝑟𝑒𝑎𝑙 = 𝜌 ⋅ 𝑑𝑦 ⋅ 𝑢∞ − 𝜌 ⋅ 𝑑𝑦 ⋅ 𝑢 = 𝜌 ⋅ (𝑢∞ − 𝑢) ⋅ 𝑑𝑦
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = ∫ 𝜌 ⋅ (𝑢∞ − 𝑢) ⋅ 𝑑𝑦
𝛿
0
Displacement thickness is the thickness by which boundary should be displaced in order to compensate for mass
flow rate due to boundary layer growth.
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑟𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛 = ∫ 𝜌 ⋅ (𝑢∞ − 𝑢) ⋅ 𝑑𝑦
𝛿
0
= 20

𝜌 ⋅ (𝛿∗
⋅ 1) ⋅ 𝑢∞ = 20
𝜌 ⋅ (𝛿∗
⋅ 1) ∙ 𝑢∞ = ∫ 𝜌 ⋅ (𝑢∞ − 𝑢) ⋅ 𝑑𝑦
𝛿
0
𝛿∗
= ∫ (1 −
𝑢
𝑢∞
) ⋅ 𝑑𝑦
𝛿
0
Momentum thickness (θ)
It is the distance by which boundary should be displaced in order to compensate the momentum due to boundary
layer growth.
𝜃 = ∫
𝑢
𝑢∞
⋅ (1 −
𝑢
𝑢∞
) ⋅ 𝑑𝑦
𝛿
0
Energy thickness (δE)
It is the distance by which boundary should be displaced in order to compensate for the reduction in Kinetic
energy due to boundary layer growth.
𝛿 𝐸 = ∫
𝑢
𝑢∞
⋅ (1 −
𝑢2
𝑢∞
2
) ⋅ 𝑑𝑦
𝛿
0
𝑄) 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛 𝑖𝑛 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑙𝑎𝑦𝑒𝑟 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦
𝑢
𝑢∞
=
𝑦
𝛿
.
Displacement thickness(δ*)
𝛿∗
= ∫ (1 −
𝑢
𝑢∞
) ⋅ 𝑑𝑦
𝛿
0
= ∫ (1 −
𝑦
𝛿
) ⋅ 𝑑𝑦
𝛿
0
= ∫ 1 ⋅ 𝑑𝑦
𝛿
0
− ∫
𝑦
𝛿
⋅ 𝑑𝑦
𝛿
0
= (𝛿 − 0) −
1
𝛿
⋅
(𝛿2
− 0)
2
= 𝛿 −
𝛿
2
=
𝛿
2
𝛿∗
=
𝛿
2
Momentum thickness
𝜃 = ∫
𝑢
𝑢∞
(1 −
𝑢
𝑢∞
) 𝑑𝑦
𝛿
0
= ∫
𝑦
𝛿
(1 −
𝑦
𝛿
) 𝑑𝑦
𝛿
0
= ∫
𝑦
𝛿
𝑑𝑦
𝛿
0
− ∫
𝑦2
𝛿2
𝑑𝑦
𝛿
0
=
1
𝛿
⋅
(𝛿2
− 0)
2
−
1
𝛿2
⋅
(𝛿3
− 0)
3
=
𝛿
2
−
𝛿
3
=
𝛿
6
𝜃 =
𝛿
6
𝛿 > 𝛿∗
> 𝜃
Note
𝑇ℎ𝑒 𝑠ℎ𝑎𝑝𝑒 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑎 𝑏𝑜𝑢𝑛𝑑𝑎𝑟𝑦 𝑙𝑎𝑦𝑒𝑟 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜 𝐻 =
𝛿⋆
𝜃
, 𝑡ℎ𝑖𝑠 𝑡𝑒𝑟𝑚 𝑖𝑠 𝑢𝑠𝑒𝑑 𝑖𝑛 𝑎𝑛𝑎𝑙𝑦𝑠𝑖𝑠 𝑜𝑓 𝑓𝑙𝑜𝑤 𝑠𝑒𝑝𝑒𝑟𝑎𝑡𝑖𝑜𝑛.
For linear velocity profiles, the shape factor is 3.
Drag force (FD)
It is the force exerted by fluid on plate in direction parallel to relative motion. When angle of incidence of plate is
zero, then drag is due to shear only.

Von-Karman Integral equation
Assumptions
1. Steady flow
2. Incompressible flow
3. 2―D flow
4.
𝑑𝑃
𝑑𝑥
= 0 (𝑇ℎ𝑖𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛 𝑖𝑠 𝑣𝑎𝑙𝑖𝑑 𝑜𝑛𝑙𝑦 𝑓𝑜𝑟 𝑒𝑥𝑡𝑒𝑟𝑛𝑎𝑙 𝑓𝑙𝑜𝑤𝑠)
From Newton second law of motion, Von-Karman equation can be derived.
𝜏 𝑜
𝜌𝑢∞
2
=
𝑑𝜃
𝑑𝑥
→ 𝑉𝑜𝑛 − 𝐾𝑎𝑟𝑚𝑎𝑛 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
( 𝜏 𝑜 → 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒, 𝜃 → 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠, 𝑥 → 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑒𝑑𝑔𝑒)
Significance of Van-Karman equation
1. With the help of Van-Karman equation, Boundary layer thickness δ can be calculated.
2. The shear stress on the surface of plate can be calculated.
3. The drag force on the plate can be calculated.
𝑅𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟 𝑖𝑠
𝜌𝑢∞ 𝑥
𝜇
(𝑥 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑙𝑒𝑎𝑑𝑖𝑛𝑔 𝑒𝑑𝑔𝑒)
For flow over the flat plate, if the Reynolds number is less than
5⨯105, then the flow is taken as Laminar.
If the flow is greater than 5⨯105, then the flow is taken as
turbulent.
Average Drag force coefficient (Cd)
𝐶 𝑑 =
𝐹𝐷
1
2
𝜌 ⋅ 𝐴 ⋅ 𝑢∞
2
With the help of average drag force coefficient, drag force can be calculated.
Local drag coefficient (or) Skin friction coefficient (Cfx)
𝐶𝑓𝑥 =
𝜏 𝑜
1
2
𝜌 ⋅ 𝑢∞
2
𝑄) 𝑭𝒐𝒓 𝒂 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚 𝒑𝒓𝒐𝒇𝒊𝒍𝒆 𝒇𝒐𝒓 𝒂 𝒍𝒂𝒎𝒊𝒏𝒂𝒓 𝒃𝒐𝒖𝒏𝒅𝒂𝒓𝒚 𝒍𝒂𝒚𝒆𝒓
𝒖
𝒖∞
=
𝟑𝒚
𝟐𝜹
−
𝒚 𝟑
𝟐𝜹 𝟑
Find boundary Layer thickness (δ), shear stress on surface of plate, Drag force, Average drag coefficient
in terms of Reynolds number?
→ 𝜃 = ∫
𝑢
𝑢∞
⋅ (1 −
𝑢
𝑢∞
) ⋅ 𝑑𝑦
𝛿
0
→ 𝜃 = ∫ (
3𝑦
2𝛿
−
𝑦3
2𝛿3
) ⋅ (1 − (
3𝑦
2𝛿
−
𝑦3
2𝛿3
)) ⋅ 𝑑𝑦
𝛿
0
⇒ 𝜃 =
39𝛿
280
→
𝜏 𝑜
𝜌𝑢∞
2
=
𝑑𝜃
𝑑𝑥
=
𝑑
𝑑𝑥
(
39𝛿
280
) =
39
280
𝑑𝛿
𝑑𝑥
→ 𝜏 𝑜 = 𝜌𝑢∞
2
⋅
39
280
𝑑𝛿
𝑑𝑥
①
𝜏 𝑜 → 𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒 (𝑎𝑡 𝑦 = 0)
𝜏(𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠) = 𝜇
𝑑𝑢
𝑑𝑦
→ 𝜏 𝑜(𝑠ℎ𝑒𝑎𝑟 𝑠𝑡𝑟𝑒𝑠𝑠 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑜𝑓 𝑝𝑙𝑎𝑡𝑒) = 𝜇
𝑑𝑢
𝑑𝑦
|
𝑦=0

𝑢 = 𝑢∞ ⋅ (
3𝑦
2𝛿
−
𝑦3
2𝛿3
)
𝑑𝑢
𝑑𝑦
= 𝑢∞ ⋅
𝑑
𝑑𝑦
(
3𝑦
2𝛿
−
𝑦3
2𝛿3
) = 𝑢∞ ⋅ (
3
2𝛿
−
3𝑦2
2𝛿3
)
𝜏 𝑜 = 𝜇 ⋅
𝑑𝑢
𝑑𝑦
|
𝑦=0
= 𝜇 ⋅ 𝑢∞ ⋅ (
3
2𝛿
−
3𝑦2
2𝛿3
)|
𝑦=0
= 𝜇 ⋅ 𝑢∞ ⋅
3
2𝛿
𝜏 𝑜 = 𝜇 ⋅ 𝑢∞ ⋅
3
2𝛿
②
𝐹𝑟𝑜𝑚 ① & ②, (𝜏 𝑜 = 𝜌𝑢∞
2
⋅
39
280
𝑑𝛿
𝑑𝑥
& 𝜏 𝑜 = 𝜇 ⋅ 𝑢∞ ⋅
3
2𝛿
)
→ 𝜇 ⋅ 𝑢∞ ⋅
3
2𝛿
= 𝜌𝑢∞
2
⋅
39
280
𝑑𝛿
𝑑𝑥
→ 𝛿 ⋅ 𝑑𝛿 =
140 ⋅ 𝜇
13 ⋅ 𝜌 ⋅ 𝑢∞
⋅ 𝑑𝑥
Integrating we get,
∫ 𝛿 ⋅ 𝑑𝛿 = ∫
140 ⋅ 𝜇
13 ⋅ 𝜌 ⋅ 𝑢∞
⋅ 𝑑𝑥 →
𝛿2
2
=
140 ⋅ 𝜇 ⋅ 𝑥
13 ⋅ 𝜌 ⋅ 𝑢∞
+ 𝐶
𝐴𝑡 𝑥 = 0 → 𝛿 = 0, 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔 𝑤𝑒 𝑔𝑒𝑡 𝐶 = 0
→
𝛿2
2
=
140 ⋅ 𝜇 ⋅ 𝑥
13 ⋅ 𝜌 ⋅ 𝑢∞
⇒ 𝛿 = √
280
13
⋅
𝜇 ⋅ 𝑥
𝜌 ⋅ 𝑢∞
= √
280
13
⋅
𝑥2
𝜌 ⋅ 𝑢∞. 𝑥
𝜇
= √
280
13
⋅
𝑥
√𝑅𝑒
=
4.64 ⋅ 𝑥
√𝑅𝑒
𝛿 =
4.64 ⋅ 𝑥
√𝑅𝑒
𝐹𝑟𝑜𝑚 𝛿 = √
280
13
⋅
𝜇 ⋅ 𝑥
𝜌 ⋅ 𝑢∞
→ 𝛿 ∝ √ 𝑥 →
𝛿1
𝛿2
=
√ 𝑥1
√ 𝑥2
As x increases → δ increases (As the distance from leading edge is increasing, Boundary layer thickness is also
increasing).
𝜏 𝑜 = 𝜇 ⋅ 𝑢∞ ⋅
3
𝟐𝜹
→ 𝜏 𝑜 = 𝜇 ⋅ 𝑢∞ ⋅
3
2 × (
4.64 ⋅ 𝑥
√𝑅𝑒
)
=
0.323 ⋅ 𝜇 ⋅ 𝑢∞
𝑥
⋅ √𝑅𝑒 =
0.323 ⋅ 𝜇 ⋅ 𝑢∞
𝑥
⋅ √
𝜌 ⋅ 𝑢∞. 𝑥
𝜇
→ 𝜏 𝑜 = 0.323 ⋅ √
𝜇 ⋅ 𝜌 ⋅ 𝑢∞
𝑥
𝜏 𝑜 =
0.323 ⋅ 𝜇 ⋅ 𝑢∞
𝑥
⋅ √𝑅𝑒 → 𝐼𝑛 𝑡𝑒𝑟𝑚𝑠 𝑜𝑓 𝑟𝑒𝑦𝑛𝑜𝑙𝑑𝑠 𝑛𝑢𝑚𝑏𝑒𝑟
𝜏 𝑜 ∝
1
√ 𝑥
→
𝜏1
𝜏2
=
√ 𝑥2
√ 𝑥1
As the distance from the leading edge increases the shear stress decreases.
𝐹𝐷 → 𝐷𝑟𝑎𝑔 𝑓𝑜𝑟𝑐𝑒 = 𝑆ℎ𝑒𝑎𝑟 𝑓𝑜𝑟𝑐𝑒 × 𝐴𝑟𝑒𝑎
𝑑𝐹𝐷 = 𝜏 𝑜 × 𝐵 ⋅ 𝑑𝑥

→ 𝐹𝐷 = ∫ 𝜏 𝑜 × 𝐵 ⋅ 𝑑𝑥
𝐿
0
= ∫
0.323 ⋅ 𝜇 ⋅ 𝑢∞
𝑥
√𝑅𝑒 𝑥 × 𝐵 ⋅ 𝑑𝑥
𝐿
0
= ∫
0.323 ⋅ 𝜇 ⋅ 𝑢∞
𝑥
√
𝜌 ⋅ 𝑢∞. 𝑥
𝜇
× 𝐵 ⋅ 𝑑𝑥
𝐿
0
⇒ ∫
0.323𝜇𝑢∞ 𝐵
√ 𝑥
√
𝜌𝑢∞
𝜇
𝑑𝑥
𝐿
0
= 0.323 ⋅ 𝜇 ⋅ 𝑢∞√
𝜌𝑢∞
𝜇
(√𝐿 − 0) = 0.646 ⋅ √
𝜌 ⋅ 𝑢∞. 𝐿
𝜇
⋅ 𝐵 ⋅ 𝜇 ⋅ 𝑢∞
𝐹𝐷 = 0.646 ⋅ √𝑅𝑒 𝐿 ⋅ 𝐵 ⋅ 𝜇 ⋅ 𝑢∞ (𝑅𝑒 𝑥 =
𝜌 ⋅ 𝑢∞. 𝑥
𝜇
, 𝑅𝑒 𝐿 =
𝜌 ⋅ 𝑢∞. 𝐿
𝜇
)
𝐹𝐷 = 0.646 ⋅ √𝑅𝑒 𝐿 ⋅ 𝐵 ⋅ 𝜇 ⋅ 𝑢∞
𝐶 𝑑 =
𝐹𝐷
1
2
𝜌 ⋅ 𝐴 ⋅ 𝑢∞
2
=
0.646 ⋅ √𝑅𝑒 𝐿 ⋅ 𝐵 ⋅ 𝜇 ⋅ 𝑢∞
1
2
𝜌 ⋅ (𝐵 ⋅ 𝐿) ⋅ 𝑢∞
2
=
0.646 ⋅ √
𝜌 ⋅ 𝑢∞. 𝐿
𝜇
⋅ 𝐵 ⋅ 𝜇 ⋅ 𝑢∞
1
2
𝜌 ⋅ (𝐵 ⋅ 𝐿) ⋅ 𝑢∞
2
=
1.292
√
𝜌 ⋅ 𝑢∞. 𝐿
𝜇
𝐶 𝑑 =
1.292
√ 𝑅𝑒 𝐿
Boundary Layer Separation
When fluid flows through converging passage, velocity increase and pressure decrease i.e., fluid flows under
negative pressure gradient (favourable pressure gradient), this flow is also known as accelerating flow. The
boundary layer thickness decreases in this region due to increase in the velocity.
When fluid flows in diverging passage, velocity decreases and pressure increase i.e., fluid flows under positive
pressure gradient. If the angle of divergence is large, the retardation of fluid particles will be more and at some
point, the fluid particles may not support the flow and fluid may separate from its boundary and may reverse the
flow, this is known as Boundary Layer Separation.
As the velocity gradient is zero at separation point, the shear stress is zero at separation point.

Blasius Equation
Blasius developed non-linear third order ordinary differential equations for obtaining boundary layer solutions.
Laminar Turbulent
𝛿 =
5𝑥
√𝑅𝑒 𝑥
𝛿 =
5𝑥
(𝑅𝑒 𝑥)1 5⁄
𝐶𝑓𝑥 =
0.664
√𝑅𝑒 𝑥
𝐶𝑓𝑥 =
0.558
(𝑅𝑒 𝑥)1 5⁄
𝐶 𝐷 =
1.328
√𝑅𝑒 𝐿
𝐶 𝐷 =
0.074
(𝑅𝑒 𝑥)1 5⁄
Flow through pipes
When fluid flows through pipes it encounters various losses, these losses are classified into Major loss & minor loss.
Major loss
The head loss due to friction is known as major loss. It is given by Darcy- Weisbach equation.
ℎ 𝐿 =
𝑓𝐿𝑣2
2𝑔𝐷
(𝑄 = 𝐴 ⋅ 𝑣 =
𝜋𝐷2
4
⋅ 𝑣 → 𝑣 =
4 ⋅ 𝑄
𝜋 ⋅ 𝐷2
)
ℎ 𝐿 =
𝑓 ⋅ 𝐿
2𝑔𝐷
⋅ (
4𝑄
𝜋𝐷2
)
2
=
𝑓 ⋅ 𝐿
2𝑔
⋅
16 ⋅ 𝑄2
𝜋2 ⋅ 𝐷5
ℎ 𝐿 =
𝑓 ⋅ 𝐿 ⋅ 𝑄2
12 ⋅ 𝐷5
𝑀𝑎𝑗𝑜𝑟 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑟𝑒 𝑎𝑙𝑠𝑜 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑒𝑑 𝑏𝑦 𝐶ℎ𝑒𝑧𝑦’𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎 → 𝑣 = 𝑐√𝓂 ⋅ 𝑖
(𝓂 → 𝐻𝑦𝑑𝑎𝑢𝑙𝑖𝑐 𝑚𝑒𝑎𝑛 𝑑𝑒𝑝𝑡ℎ, 𝑖 → 𝐻𝑦𝑑𝑟𝑎𝑢𝑙𝑖𝑐 𝑠𝑙𝑜𝑝𝑒)
𝓂(𝐻𝑦𝑑𝑎𝑢𝑙𝑖𝑐 𝑚𝑒𝑎𝑛 𝑑𝑒𝑝𝑡ℎ) =
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑓𝑙𝑜𝑤
𝑊𝑒𝑡𝑡𝑒𝑑 𝑃𝑒𝑟𝑖𝑚𝑒𝑡𝑒𝑟
=
𝐴
𝑝
=
𝜋𝐷2
4
𝜋 ⋅ 𝐷
=
𝐷
4
→ 𝓂 =
𝐷
4
𝑖(𝐻𝑦𝑑𝑎𝑢𝑙𝑖𝑐 𝑠𝑙𝑜𝑝𝑒) = 𝑡𝑎𝑛 𝜃 =
ℎ 𝐿
𝐿
→ 𝑖 =
ℎ 𝐿
𝐿
𝑣 = 𝑐√𝓂 ⋅ 𝑖 = 𝑐 ⋅ √
𝐷
4
⋅
ℎ 𝐿
𝐿
→ 𝑣2
= 𝑐2
⋅
𝐷 × ℎ 𝐿
4 ⋅ 𝐿
ℎ 𝐿 =
4𝐿𝑣2
𝑐2 𝐷

ℎ 𝐿 =
4𝐿𝑣2
𝑐2 𝐷
→ 𝐶ℎ𝑒𝑧𝑦′
𝑠𝑓𝑜𝑟𝑚𝑢𝑙𝑎 , ℎ 𝐿 =
𝑓𝐿𝑣2
2𝑔𝐷
→ 𝐷𝑎𝑟𝑐𝑦 − 𝑊𝑒𝑖𝑛𝑏𝑎𝑐ℎ 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
Equating Darcy & Chezy’s equations,
𝑓𝐿𝑣2
2𝑔𝐷
=
4𝐿𝑣2
𝑐2 𝐷
𝑐 = √
8𝑔
𝑓
Minor loss
Loss due to sudden expansion or sudden contraction, bend loss, entrance loss,
exit loss are known as Minor losses.
→
𝑃1
𝓌
+
𝑣1
2
2𝑔
=
𝑃2
𝓌
+
𝑣2
2
2𝑔
+ ℎ 𝐿
→
𝑃1 − 𝑃2
𝓌
+
𝑣1
2
− 𝑣2
2
2𝑔
= ℎ 𝐿𝑜𝑠𝑠 𝑑𝑢𝑒 𝑡𝑜 𝑒𝑥𝑝𝑎𝑛𝑠𝑖𝑜𝑛 ①
Assumptions
The pressure in the eddy region is assumed to equal to upstream pressure.
→ 𝛴𝐹 = 𝑚 (
𝑣−𝑢
𝑡
) = 𝑚̇ ⋅ (𝑣 − 𝑢) = 𝜌 ⋅ 𝑄 ⋅ (𝑣 − 𝑢)
𝑃1 ⋅ 𝐴1 + 𝑃1 ⋅ (𝐴2 − 𝐴1) − 𝑃2 ⋅ 𝐴2 = 𝜌 ⋅ 𝑄 ⋅ (𝑣2 − 𝑣1)
𝐴2(𝑃1 − 𝑃2) = 𝜌 ⋅ 𝑄 ⋅ (𝑣2 − 𝑣1) →
𝑃1 − 𝑃2
𝜌
=
𝑄 ⋅ (𝑣2 − 𝑣1)
𝐴2
𝑃1 − 𝑃2
𝜌
=
𝑄
𝐴2
(𝑣2 − 𝑣1) = 𝑣2(𝑣2 − 𝑣1)
Dividing both sides by g,
𝑃1 − 𝑃2
𝜌𝑔
=
𝑣2 ⋅ (𝑣2 − 𝑣1)
𝑔
⇒
𝑃1 − 𝑃2
𝓌
=
𝑣2 ⋅ (𝑣2 − 𝑣1)
𝑔
… … ②
𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑖𝑛𝑔
𝑃1 − 𝑃2
𝓌
𝑖𝑛 ①,
𝑣2 ⋅ (𝑣2 − 𝑣1)
𝑔
+
𝑣1
2
− 𝑣2
2
2𝑔
= ℎ 𝐿
→ ℎ 𝐿 =
(𝑣1 − 𝑣2)2
2𝑔
In deriving this equation Bernoulli’s equation, Momentum equation, Continuity equation is used.
ℎ 𝐿⋅𝑒𝑥𝑝𝑛 =
(𝑣1 − 𝑣2)2
2𝑔
=
𝑣1
2
2𝑔
(1 −
𝑣2
𝑣1
)
2
=
𝑣1
2
2𝑔
(1 −
𝐴1
𝐴2
)
2
(𝐴1 ⋅ 𝑣1 = 𝐴2 ⋅ 𝑣2 →
𝑣2
𝑣1
=
𝐴1
𝐴2
)
→ ℎ 𝐿⋅𝑒𝑥𝑝𝑛 =
𝑣1
2
2𝑔
(1 −
𝐴1
𝐴2
)
2

Exit Loss
It’s similar to sudden expansion, with A2 → ∞.
ℎ 𝐿 =
𝑣1
2
2𝑔
(1 −
𝐴1
𝐴2
)
2
ℎ 𝐿⋅𝑒𝑥𝑖𝑡 =
𝑣1
2
2𝑔
(1 −
𝐴1
∞
)
2
=
𝑣1
2
2𝑔
→ ℎ 𝐿⋅𝑒𝑥𝑖𝑡 =
𝑢2
2𝑔
Sudden Contraction
𝐶𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝐶𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝐶𝑐) =
𝐴 𝑐
𝐴2
ℎ 𝐿 =
(𝑣1 − 𝑣2)2
2𝑔
ℎ 𝐿⋅𝐶𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =
(𝑣𝑐 − 𝑣2)2
2𝑔
=
𝑣2
2
2𝑔
(
𝑣 𝑐
𝑣2
− 1)
2
=
𝑣2
2
2𝑔
(
𝐴2
𝐴 𝑐
− 1)
2
(𝐴 𝑐 ⋅ 𝑣𝑐 = 𝐴2 ⋅ 𝑣2)
(𝐴2 = 𝐶𝑐 × 𝐴 𝑐 →
𝐴2
𝐴 𝑐
=
1
𝐶𝑐
)
ℎ 𝐿⋅𝐶𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 =
𝑣2
2
2𝑔
(
1
𝐶𝑐
− 1)
2
𝐼𝑓 𝑡ℎ𝑒 𝑐𝑜𝑒𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑡 𝑜𝑓 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 (𝐶𝑐)𝑖𝑠 𝑛𝑜𝑡 𝑔𝑖𝑣𝑒𝑛 𝑡ℎ𝑒𝑛 𝑠𝑢𝑑𝑑𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑙𝑜𝑠𝑠𝑒𝑠 𝑎𝑟𝑒 𝑡𝑎𝑘𝑒𝑛 𝑎𝑠
0.5 ⋅ 𝑣2
2
2𝑔
,
𝑤ℎ𝑒𝑟𝑒 𝑣2 𝑖𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑠𝑚𝑎𝑙𝑙𝑒𝑟 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟 𝑝𝑖𝑝𝑒.
Entrance Loss
𝐼𝑡 𝑖𝑠 𝑠𝑖𝑚𝑖𝑙𝑎𝑟 𝑡𝑜 𝑠𝑢𝑑𝑑𝑒𝑛 𝑐𝑜𝑛𝑡𝑟𝑎𝑐𝑡𝑖𝑜𝑛, 𝑡ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑒𝑛𝑡𝑟𝑎𝑛𝑐𝑒 𝑙𝑜𝑠𝑠 𝑖𝑠 𝑒𝑞𝑢𝑎𝑙 𝑡𝑜
0.5 ⋅ 𝑣2
2𝑔
, 𝑣 → 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑖𝑛 𝑝𝑖𝑝𝑒.
Bend Loss
𝐵𝑒𝑛𝑑 𝐿𝑜𝑠𝑠𝑒𝑠 𝑎𝑟𝑒 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦 (ℎ 𝐿⋅𝐵𝑒𝑛𝑑 =
𝐾𝑣2
2𝑔
),
𝑊ℎ𝑒𝑟𝑒 𝑣 → 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑜𝑓 𝑝𝑖𝑝𝑒 𝑎𝑛𝑑 𝐾 → 𝑑𝑒𝑝𝑒𝑛𝑑𝑠 𝑜𝑛 𝑎𝑛𝑔𝑙𝑒 𝑜𝑓 𝑏𝑒𝑛𝑑 𝑎𝑛𝑑 𝑟𝑎𝑑𝑖𝑢𝑠 𝑜𝑓 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒 𝑜𝑓 𝑏𝑒𝑛𝑑.
Hydraulic Gradient Line & Total Gradient Line (HGL & UGL)
Hydraulic Gradient Line
The line joining piezometric heads at various points in a flow is known as hydraulic gradient line. If the pipe is
horizontal and of uniform diameter. HGL represents pressure variation.
𝑃𝑖𝑒𝑧𝑜𝑚𝑒𝑡𝑟𝑖𝑐 ℎ𝑒𝑎𝑑 =
𝑃
𝓌
+ 𝑧

Total Gradient Line
The line joining total heads at various points in a flow is known as total energy line.
𝑇𝑜𝑡𝑎𝑙 ℎ𝑒𝑎𝑑 =
𝑃
𝓌
+ 𝑧 +
𝑣2
2𝑔
The distance between TEL & HGL gives velocity head.
In a flow hydraulic gradient line can rise or fall, but total energy line can never rise as long as there I no external
energy input i.e., total energy line will rise in case of pumps & compressors.
Pipes in Series
𝑄1 = 𝑄2 = 𝑄3 = 𝑄4 = 𝑄
ℎ 𝐿 =
𝑓 ⋅ 𝑙1 ⋅ 𝑄2
12 ⋅ 𝑑1
5 +
𝑓 ⋅ 𝑙2 ⋅ 𝑄2
12 ⋅ 𝑑2
5 +
𝑓 ⋅ 𝑙3 ⋅ 𝑄2
12 ⋅ 𝑑3
5 +
𝑓 ⋅ 𝑙4 ⋅ 𝑄2
12 ⋅ 𝑑4
5 + ⋯
ℎ 𝐿 = ℎ𝑙1
+ ℎ𝑙2
+ ℎ𝑙3
+ ℎ𝑙4
+ ⋯

Equivalent Pipe
A pipe of uniform diameter is set to be equivalent to compound pipe, when the discharge and the head losses are
same in both pipes.
𝑓 ⋅ 𝑙 𝑒 ⋅ 𝑄2
12 ⋅ 𝑑 𝑒
5 =
𝑓 ⋅ 𝑙1 ⋅ 𝑄2
12 ⋅ 𝑑1
5 +
𝑓 ⋅ 𝑙2 ⋅ 𝑄2
12 ⋅ 𝑑2
5 +
𝑓 ⋅ 𝑙3 ⋅ 𝑄2
12 ⋅ 𝑑3
5 + ⋯
(𝑙 𝑒 → 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑙𝑒𝑛𝑔𝑡ℎ, 𝑑 𝑒 → 𝐸𝑞𝑢𝑖𝑣𝑎𝑙𝑒𝑛𝑡 𝑑𝑖𝑎𝑚𝑒𝑡𝑒𝑟)
𝑙 𝑒
𝑑 𝑒
5 =
𝑙1
𝑑1
5 +
𝑙2
𝑑2
5 +
𝑙3
𝑑3
5 + ⋯ → 𝐷𝑢𝑝𝑖𝑡𝑠 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛
In Dupits equation minor losses are neglected.
Pipes in Parallel
Parallel connection is used for increasing discharge
𝑃𝑎
𝓌
+
𝑣 𝑎
2
2𝑔
+ 𝑍 𝑎 =
𝑃𝑏
𝓌
+
𝑣 𝑏
2
2𝑔
+ 𝑍 𝑏 + ℎ 𝐿1
𝑃𝑎
𝓌
+
𝑣 𝑎
2
2𝑔
+ 𝑍 𝑎 =
𝑃𝑏
𝓌
+
𝑣 𝑏
2
2𝑔
+ 𝑍 𝑏 + ℎ 𝐿2
ℎ 𝐿1 = ℎ 𝐿2
Equivalent Pipe (Parallel)
ℎ 𝐿⋅𝑒 =
𝑓 ⋅ 𝑙 𝑒 ⋅ 𝑄2
12 ⋅ 𝑑 𝑒
5
‘n’ Similar pipes are connected in parallel.
ℎ 𝐿 = 𝑓 ⋅ 𝐿 ⋅ (
𝑄
𝑛
)
2
⋅
1
12 ⋅ 𝑑5
=
𝑓 ⋅ 𝑙 ⋅ 𝑄2
12 ⋅ 𝑛2 ⋅ 𝑑5
ℎ 𝐿⋅𝑒 = ℎ 𝐿
𝑓 ⋅ 𝑙 𝑒 ⋅ 𝑄2
12 ⋅ 𝑑 𝑒
5 =
𝑓 ⋅ 𝑙 ⋅ 𝑄2
12 ⋅ 𝑛2 ⋅ 𝑑5
→
𝑙 𝑒
𝑑 𝑒
5 =
𝑙
𝑛2 ⋅ 𝑑5
(𝑙 𝑒 = 𝑙)
𝑑 𝑒
5
= 𝑛2
⋅ 𝑑5
→ 𝑑 𝑒 = 𝑛2/5
⋅ 𝑑

Power transmission through Pipes
𝑃𝑡ℎ𝑒𝑜𝑟𝑒𝑡𝑖𝑐𝑎𝑙 = 𝓌 ⋅ 𝑄 ⋅ 𝐻
𝑃𝑎𝑐𝑡𝑢𝑎𝑙 = 𝓌 ⋅ 𝑄 ⋅ (𝐻 − ℎ 𝐿)
𝜂 =
𝑃𝑎𝑐𝑡
𝑃𝑡ℎ
𝜂 =
𝓌. 𝑄 ⋅ (𝐻 − ℎ 𝐿)
𝓌 ⋅ 𝑄 ⋅ 𝐻
→ 𝜂 =
𝐻 − ℎ 𝐿
𝐻
Conditions for Maximum Power transmission
→ 𝑃𝑎𝑐𝑡𝑢𝑎𝑙 = 𝓌 ⋅ 𝑄 ⋅ (𝐻 − ℎ 𝐿)
→ 𝑃𝑎𝑐𝑡𝑢𝑎𝑙 = 𝓌 ⋅ 𝑄 ⋅ (𝐻 − ℎ 𝐿) = 𝓌 ⋅ 𝑄 ⋅ (𝐻 −
𝑓 ⋅ 𝑙 ⋅ 𝑄2
12 ⋅ 𝑑5
) = 𝓌 ⋅ (𝑄 ⋅ 𝐻 −
𝑓 ⋅ 𝑙 ⋅ 𝑄3
12 ⋅ 𝑑5
)
𝐹𝑜𝑟 𝑀𝑎𝑥𝑖𝑚𝑢𝑚 𝑃𝑜𝑤𝑒𝑟,
𝑑𝑃𝑎𝑐𝑡
𝑑𝑄
= 0
𝑑𝑃𝑎𝑐𝑡
𝑑𝑄
= 𝓌 ⋅ (1 ⋅ 𝐻 −
𝑓 ⋅ 𝑙 ⋅ 3 ⋅ 𝑄2
12 ⋅ 𝑑5
) = 𝓌 ⋅ (𝐻 −
𝑓 ⋅ 𝑙 ⋅ 𝑄2
4 ⋅ 𝑑5
) = 0
𝐻 − 3. (
𝑓 ⋅ 𝑙 ⋅ 𝑄2
12 ⋅ 𝑑5
) = 0 → 𝐻 − 3 ⋅ ℎ 𝐿 = 0
𝐻 = 3 ⋅ ℎ 𝐿
→ 𝜂 𝑚𝑎𝑥 =
𝐻 − ℎ 𝐿
𝐻
=
3 ⋅ ℎ 𝐿 − ℎ 𝐿
3 ⋅ ℎ 𝐿
=
2
3
𝜂 𝑚𝑎𝑥 = 0.667
HYDROSTATIC FORCES
Hydrostatic forces on Plane surfaces
Inclined Surfaces
Taking a small elemental area dA, we can calculate force on the small element and the total force can be
calculated by integrating.
𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑡𝑜𝑡𝑎𝑙 ℎ𝑦𝑑𝑟𝑜𝑠𝑡𝑎𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒 𝑖𝑠 𝐹 = 𝓌 ⋅ 𝐴 ⋅ 𝑥̅,
𝐴 → 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑠𝑢𝑟𝑓𝑎𝑐𝑒, 𝑥̅ → 𝑣𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐶𝑒𝑛𝑡𝑟𝑒 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 𝑓𝑟𝑜𝑚 𝑓𝑟𝑒𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒.
𝑃 = 𝜌 ⋅ 𝑔 ⋅ 𝑥̅ = 𝓌 ⋅ 𝑥̅ → 𝐹 = 𝑃 ⋅ 𝐴 = 𝓌 ⋅ 𝑥̅ ⋅ 𝐴

Centre of Pressure
It’s the point through which total hydro static force is supposed to be acting.
Case-1 (Inclined Surface)
From the principle of Moments, the centre of pressure can be calculated,
𝑥 𝑐⋅𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
⋅ 𝑠𝑖𝑛2
𝜃
𝐼 𝐺𝐺 =
𝑏 ⋅ 𝑑3
12
, 𝐴 = 𝑏 ⋅ 𝑑
IGG is the moment of Inertia about centroidal axis, which is parallel to OO’.
θ is the angle made by surface with respect to free surface.
The centre of pressure is below C.G because the pressure increases with depth.
Case-2 (Plane Vertical surface)
Put θ=90ᵒ in case 1
𝐹 = 𝓌 ⋅ 𝐴 ⋅ 𝑥̅
𝑥 𝑐.𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
⋅ 𝑠𝑖𝑛2
𝜃 → 𝑥 𝑐.𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
⋅ 𝑠𝑖𝑛2
90ᵒ
→ 𝑥 𝑐.𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
Case-3 (Plane Horizontal surface)
Put θ=0ᵒ in case 1
𝑥 𝑐.𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
⋅ 𝑠𝑖𝑛2
𝜃 → 𝑥 𝑐.𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
⋅ 𝑠𝑖𝑛2
0ᵒ → 𝑥 𝑐.𝑝 = 𝑥̅
Case Force Centre point
Inclined 𝓌Ax̅
𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
⋅ 𝑠𝑖𝑛2
𝜃
Vertical 𝓌Ax̅
𝑥 𝑐.𝑝 = 𝑥̅ +
𝐼 𝐺𝐺
𝐴 ⋅ 𝑥̅
Horizontal 𝓌Ax̅ 𝑥 𝑐.𝑝 = 𝑥̅

Hydrostatic force on Curved surfaces
𝑑𝐹 = 𝑃 ⋅ 𝑑𝐴 → 𝑑𝐹 = 𝜌 ⋅ 𝑔 ⋅ 𝑥 ⋅ 𝑑𝐴
𝑑𝐹 𝐻 = 𝑑𝐹 ⋅ 𝑠𝑖𝑛 𝜃 = 𝜌 ⋅ 𝑔 ⋅ 𝑥 ⋅ 𝑑𝐴 ⋅ 𝑠𝑖𝑛 𝜃
The horizontal component of force on curved surface is equal to hydrostatic force on vertical projection area, and
this force acts at the centre of pressure of corresponding area.
𝑑𝐹𝑣 = 𝑑𝐹 ⋅ 𝑐𝑜𝑠 𝜃
𝑑𝐹𝑣 = 𝜌 ⋅ 𝑔 ⋅ 𝑥 ⋅ 𝑑𝐴 ⋅ 𝑐𝑜𝑠 𝜃 = 𝜌 ⋅ 𝑔 ⋅ (𝑥 ⋅ 𝑑𝐴 𝑐𝑜𝑠 𝜃) = 𝜌 ⋅ 𝑔 ⋅ 𝑉 = 𝑚𝑔
𝑑𝐹𝑣 = 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
The vertical component of force on the surface is equal to weight of the liquid contained by curved surface taken
up to free surface and this weight will act from the centre of gravity of corresponding weight.
Special Cases

Turbulent flow
In turbulent flow, as there is a continuous mixing of fluid particles, velocity fluctuates continuously. Hence no
turbulent flow is purely steady.
In turbulent flow the shear stress is due to fluctuation of velocity in flow as well as in the transverse direction.
Head loss in turbulent is proportional to (v1.75 to v2), where as in laminar flow head loss is proportional to v.
Boussinesq equation
𝜏 = 𝜇
𝑑𝑢
𝑑𝑦
+ 𝜂
𝑑𝑢
𝑑𝑦
(𝜂 = 𝑒𝑑𝑑𝑦 𝑣𝑖𝑠𝑐𝑜𝑠𝑖𝑡𝑦)
It’s very difficult to find eddy viscosity. Hence this equation is not used.
Reynold developed an equation for turbulent shear stress, τ = ρ⋅ u’⋅ v’, where u’ & v’ are fluctuating component of
velocity in x & y directions respectively.
Prandtl’s mixing theory
Mixing length is that length in transverse direction, where in fluid particles after colliding lose excess momentum
and reach momentum of new region. According to Prandtl mixing length ‘l’ is equal to 0.4y, where y is distance
measured from pipe wall.
At the pipe wall mixing length is zero.
𝑢′
= 𝑣′
= 𝑙 ⋅
𝑑𝑢
𝑑𝑦
𝜏 = 𝜌 ⋅ 𝑢′
⋅ 𝑣′
= 𝜌 ⋅ (𝑙 ⋅
𝑑𝑢
𝑑𝑦
) ⋅ (𝑙 ⋅
𝑑𝑢
𝑑𝑦
) = 𝜌 ⋅ 𝑙2
⋅ (
𝑑𝑢
𝑑𝑦
)
2
𝜏
𝜌
= 𝑙2
⋅ (
𝑑𝑢
𝑑𝑦
)
2
→ √
𝜏
𝜌
= 𝑙 ⋅ (
𝑑𝑢
𝑑𝑦
)
𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡 √
𝜏
𝜌
= 𝑉∗ (𝑠ℎ𝑒𝑎𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦)
𝑉∗
= 𝑙 ⋅ (
𝑑𝑢
𝑑𝑦
) = 0.4𝑦 ⋅ (
𝑑𝑢
𝑑𝑦
) (𝑙 = 4𝑦)
𝑉∗
0.4
×
𝑑𝑦
𝑦
= 𝑑𝑢
Integrating,
∫
𝑉∗
0.4
×
𝑑𝑦
𝑦
= ∫ 𝑑𝑢
𝑢 = (
5
2
𝑉∗
⋅ 𝑙𝑛 𝑦) + 𝐶
The velocity distribution in turbulent flow is logarithmic nature.

𝑇ℎ𝑒 𝑙𝑎𝑚𝑖𝑛𝑎𝑟 𝑠𝑢𝑏 − 𝑙𝑎𝑦𝑒𝑟 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 ‘𝛿’ 𝑖𝑠
11.6 ⋅ 𝑣
𝑉∗
(𝑝𝑖𝑝𝑒𝑠)𝑎𝑛𝑑 𝛿′
=
5 ⋅ 𝑣
𝑉∗
(𝑝𝑙𝑎𝑡𝑒𝑠)
𝐼𝑓 𝑘 𝑖𝑠 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 ℎ𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑟𝑜𝑢𝑔ℎ𝑛𝑒𝑠𝑠, 𝑡ℎ𝑒𝑛
𝑘
𝛿′
< 0.25, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑎𝑡 𝑝𝑖𝑝𝑒 𝑖𝑠 𝑠𝑚𝑜𝑜𝑡ℎ 𝑝𝑖𝑝𝑒.
𝐼𝑓
𝑘
𝛿′
> 6, 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝑝𝑖𝑝𝑒 𝑖𝑠 𝑟𝑜𝑢𝑔ℎ 𝑝𝑖𝑝𝑒.
𝐼𝑓 0.25 <
𝑘
𝛿
< 6, 𝑖𝑡 𝑖𝑠 𝑡𝑟𝑎𝑛𝑠𝑖𝑡𝑖𝑜𝑛 𝑝𝑖𝑝𝑒.
𝑢 − 𝑣
𝑉∗
= 5.75 ⋅ 𝑙𝑜𝑔10 (
𝑦
𝑅
) + 3.75
u→ local velocity, v→ average velocity, V*→ Shear velocity, y→ displacement from pipe walls, R→ pipe radius.
Equation valid for both rough & smooth pipes.
Note
𝐸𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝜏 𝑜 =
𝜌 ⋅ 𝑓 ⋅ 𝑣2
8
𝑖𝑠 𝑣𝑎𝑙𝑖𝑑 𝑓𝑜𝑟 𝑡𝑢𝑟𝑏𝑢𝑙𝑒𝑛𝑡 𝑓𝑙𝑜𝑤 𝑎𝑙𝑠𝑜.
At the centre y=R, u=umax.
Substituting in equation,
𝑢 𝑚𝑎𝑥 − 𝑣
𝑉∗
= 5.75 ⋅ 𝑙𝑜𝑔10 (
𝑅
𝑅
) + 3.75 →
𝑢 𝑚𝑎𝑥 − 𝑣
𝑉∗
= 0 + 3.75 → 𝑢 𝑚𝑎𝑥 − 𝑣 = 3.75 ⋅ 𝑉∗
𝜏 𝑜 =
𝜌 ⋅ 𝑓 ⋅ 𝑣2
8
→ √
𝜏 𝑜
𝜌
= √
𝑓 ⋅ 𝑣2
8
= √
𝑓
8
⋅ 𝑣 → √
𝜏 𝑜
𝜌
= 𝑉∗
= √
𝑓
8
⋅ 𝑣
Substituting V* in umax equation,
𝑢 𝑚𝑎𝑥 − 𝑣 = 3.75 ⋅ √
𝑓
8
⋅ 𝑣
𝑢 𝑚𝑎𝑥
𝑣
= 1 + 1.326√𝑓
DIMENSIONAL ANALYSIS
There are 2 methods of grouping, i.e., Rayleigh’s method and Buckingham’s pi theorem.
Rayleigh’s method
For a laminar flow in a pipe, the pressure drop ΔP is a function of pipe length ‘l’ & velocity ‘v’ and viscosity ‘μ’
using Rayleigh’s method, develop an expression for ΔP.
𝛥𝑃 = 𝑓(𝐿, 𝐷, 𝑣, 𝜇)
𝛥𝑃 = 𝐾 ⋅ 𝐿 𝑎
⋅ 𝐷 𝑏
⋅ 𝑣 𝑐
⋅ 𝜇 𝑑
→ 𝑀1
𝐿−1
𝑇−2
= 𝑀0
𝐿0
𝑇0
⋅ 𝐿 𝑎
⋅ 𝐿 𝑏
⋅ (𝐿𝑇−1) 𝑐
⋅ (𝑀𝐿−1
𝑇−1) 𝑑
→ 𝑀1
𝐿−1
𝑇−2
= 𝑀0+𝑑
⋅ 𝐿0+𝑎+𝑏+𝑐−𝑑
⋅ 𝑇0−𝑐−𝑑
𝑑 = 1 , 𝑎 + 𝑏 + 𝑐 − 𝑑 = −1, − 𝑐 − 𝑑 = −2
𝑐 = 1 𝑎 + 𝑏 = −1 𝑏 = −𝑎 − 1
Substituting,
𝛥𝑃 = 𝐾 ⋅ 𝐿 𝑎
⋅ 𝐷−1−𝑎
⋅ 𝑣1
⋅ 𝜇1
= 𝐾 ⋅ 𝜇 ⋅ 𝑣 ⋅
𝐿 𝑎
𝐷1+𝑎
𝛥𝑃 =
𝐾 ⋅ 𝜇 ⋅ 𝑣
𝐷
⋅ (
𝐿
𝐷
)
𝑎

Buckingham’s pi theorem
If there are n no. of total variables, and m no. of fundamental quantities, then given systems can be grouped into
n-m pi terms.
The resistance force F of a ship is a function of length ‘l’, velocity ‘v’, acceleration due to gravity ‘g’ and fluid
properties like density ‘ρ’, viscosity ‘μ’, and write the relationship in dimensionless form using buckingham’s pi
theorem.
𝐹 = 𝜙(𝐿, 𝑣, 𝑔, 𝜌, 𝜇)
𝑛 = 6, 𝑚 = 3
[𝐹 → 𝑀𝐿𝑇−2
, 𝑔 → 𝐿𝑇−2
, 𝜌 → 𝐿−3
, 𝜇 → 𝑀𝐿−1
𝑇−1
, 𝑣 → 𝐿𝑇−1]
𝑇ℎ𝑒 𝑔𝑖𝑣𝑒𝑛 𝑠𝑦𝑠𝑡𝑒𝑚 𝑣𝑎𝑛 𝑏𝑒 𝑔𝑟𝑜𝑢𝑝𝑒𝑑 𝑖𝑛𝑡𝑜 6 − 3 = 3, 3 𝜋 𝑡𝑒𝑟𝑚𝑠
Selection of repeated variables
1. Repeated variables must be selected from independent variables.
2. Number of repeated variables is equal to number of fundamental quantities.
3. Each repeated variable must have its own dimension.
4. Repeated variable group must contain all fundamental quantities.
5. Most fundamental quantity must be selected as repeated variable.
→ 𝐹 = 𝜙(𝐿, 𝑣, 𝑔, 𝜌, 𝜇)
→ 𝜋1 = 𝐹 ⋅ 𝐿 𝑎1
⋅ 𝑣 𝑏1
⋅ 𝜌 𝑐1
→ 𝜋2 = 𝑔 ⋅ 𝐿 𝑎2
⋅ 𝑣 𝑏2
⋅ 𝜌 𝑐2
→ 𝜋3 = 𝜇 ⋅ 𝐿 𝑎3
⋅ 𝑣 𝑏3
⋅ 𝜌 𝑐3
𝝅 𝟏 →
𝑀0
𝐿0
𝑇0
= 𝑀𝐿𝑇−2
⋅ 𝐿 𝑎1
⋅ (𝐿𝑇−1) 𝑏1
⋅ (𝑀𝐿−3) 𝑐1
𝑀0
𝐿0
𝑇0
= 𝑀1+𝑐1 ⋅ 𝐿1+𝑎1+𝑏1−3𝑐1 ⋅ 𝑇−2−𝑏1
𝑐1 = −1 𝑏1 = −2 𝑎1 = −2
𝜋1 =
𝐹
𝐿2 𝑣2 𝜌
𝑆𝑖𝑚𝑖𝑙𝑎𝑟𝑙𝑦, 𝜋2 =
𝑔𝐿
𝑣2
𝜋3 =
𝜇
𝜌𝑣𝐿
→ 𝐹 = 𝜙(𝐿, 𝑣, 𝑔, 𝜌, 𝜇)
→ 𝜋1 = 𝜙(𝜋2, 𝜋3) →
𝐹
𝐿2 𝑣2 𝜌
= 𝜙 (
𝑔𝐿
𝑣2
,
𝜇
𝜌𝑣𝐿
) → 𝐹 = 𝜌𝐿2
𝑣2
⋅ 𝜙 (
𝑔𝐿
𝑣2
,
𝜇
𝜌𝑣𝐿
)
Various forces in fluid mechanics
Inertia force
𝐹𝑖 = 𝑚 ⋅ 𝑎
[𝜌 =
𝑚
𝑉3
=
𝑚
𝐿3
→ 𝑚 = 𝜌 ⋅ 𝐿3
, 𝑎 =
𝑣
𝑇
]
𝐹𝑖 = 𝜌 ⋅ 𝐿3
⋅
𝑣
𝑇
= 𝜌 ⋅ 𝐿2
⋅
𝐿
𝑇
⋅ 𝑣 → 𝐹𝑖 = 𝜌 ⋅ 𝐿2
⋅ 𝑣2
Pressure Force
𝑃 =
𝐹𝑃
𝐴
→ 𝐹𝑃 = 𝑃 ⋅ 𝐴
𝐹𝑃 = 𝑃 ⋅ 𝐿2

Gravity force
𝐹𝑔 = 𝑚𝑔
𝐹𝑔 = 𝜌 ⋅ 𝐿3
⋅ 𝑔 [𝑚 = 𝜌 ⋅ 𝐿3]
Surface tension force
𝜎 =
𝐹𝑠
𝐿
𝐹𝑠 = 𝜎 ⋅ 𝐿
Viscous force
𝐹𝑣 =
𝜇 ⋅ 𝐴 ⋅ 𝑣
𝑦
𝐹𝑣 =
𝜇 ⋅ 𝐿2
⋅ 𝑣
𝐿
𝐹𝑣 = 𝜇 ⋅ 𝐿 ⋅ 𝑣
Elastic force
When a fluid is compresses, there is a rise in pressure, this rise in pressure is proportional to bulk modulus and this
rise in pressure gives rise to a force known as elastic force.
𝐹𝑒 = 𝐾 ⋅ 𝐿2
Various dimensionless numbers in fluid mechanics
Reynold’s number
It’s defined as the ratio of inertia force to viscous force.
→ 𝑅𝑒 =
𝐹𝑖
𝐹𝑣
=
𝜌 ⋅ 𝐿2
⋅ 𝑣2
𝜇 ⋅ 𝐿 ⋅ 𝑣
=
𝜌 ⋅ 𝐿 ⋅ 𝑣
𝜇
→ 𝑅𝑒 =
𝜌 ⋅ 𝐿 ⋅ 𝑣
𝜇
Euler number
𝐸𝑢 =
𝐹𝑖
𝐹𝑃
=
𝜌 ⋅ 𝐿2
⋅ 𝑣2
𝑃 ⋅ 𝐿2
𝐸𝑢 =
𝜌 ⋅ 𝑣2
𝑃
Froude number
𝐹𝑟 =
𝐹𝑖
𝐹𝑔
=
𝜌 ⋅ 𝐿2
⋅ 𝑣2
𝜌 ⋅ 𝐿3 ⋅ 𝑔
𝐹𝑟 =
𝑣2
𝑔 ⋅ 𝐿
Weber number
It’s the ratio of inertia force to surface tension force.
𝑊𝑒 =
𝐹𝑖
𝐹𝑠
=
𝜌 ⋅ 𝐿 ⋅ 𝑣2
𝜎 ⋅ 𝐿
𝑊𝑒 =
𝜌 ⋅ 𝑣2
𝜎

Mach number
𝑀 = √
𝐹𝑖
𝐹𝑒
= √
𝜌 ⋅ 𝑣2 ⋅ 𝐿2
𝐾 ⋅ 𝐿2
= √
𝜌 ⋅ 𝑣2
𝐾
=
𝑣
√
𝐾
𝜌
=
𝑣
𝐶
𝑀 =
𝑣
𝐶
𝑅𝑒 → 𝑉𝑖𝑠𝑐𝑜𝑢𝑠 𝐸𝑢 → 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝐹𝑟 → 𝐺𝑟𝑎𝑣𝑖𝑡𝑦 𝑊𝑒 → 𝑆𝑢𝑟𝑓𝑎𝑐𝑒 𝑡𝑒𝑛𝑠𝑖𝑜𝑛 𝑀 → 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑓𝑜𝑟𝑐𝑒
Similitude and Modelling
Geometric similarity
Model and prototype are said to be in geometric similarity, if the ratio if corresponding dimensions of model and
prototype are same.
𝑙 𝑚
𝑙 𝑝
=
𝑏 𝑚
𝑏 𝑝
=
𝑡 𝑚
𝑡 𝑝
= 𝐿 𝑟 (𝑆𝑐𝑎𝑙𝑒 𝑟𝑎𝑡𝑖𝑜)
𝐴 𝑚
𝐴 𝑝
=
𝑙 𝑚 ⋅ 𝑏 𝑚
𝑙 𝑝 ⋅ 𝑏 𝑝
= 𝐿 𝑟
2
𝑉𝑜𝑙 𝑚
𝑉𝑜𝑙 𝑝
=
𝑙 𝑚 ⋅ 𝑏 𝑚 ⋅ 𝑡 𝑚
𝑙 𝑝 ⋅ 𝑏 𝑝 ⋅ 𝑡 𝑝
= 𝐿 𝑟
3
Kinematic Similarity
Model and prototype are said to be in kinematic similarity, if the ratio of velocity & acceleration at corresponding
points in model and prototype are same.
Note
For kinematic similarity, geometric similarity I must.
Dynamic Similarity
Model and prototype are said to be in dynamic similarity, if the ratio of forces at corresponding points in model
and prototype are same.
Note
For achieving dynamic similarity, various dimensionless numbers are equated in model and prototype.
𝐹𝑖 1𝑚
𝐹𝑖 1𝑝
=
𝐹𝑣 1𝑚
𝐹𝑣 1𝑝
⇒
𝐹𝑖 1𝑚
𝐹𝑣 1𝑚
=
𝐹𝑖 1𝑝
𝐹𝑣 1𝑝
𝑅𝑒 𝑚 = 𝑅𝑒 𝑝

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Fluid mechanics notes for gate