The document discusses the fundamental equations of fluid dynamics: A) The continuity equation states that mass is conserved as fluid flows. It relates the rate of change of fluid density to the divergence of fluid velocity. B) The Navier-Stokes equations describe how fluid velocity varies over time and space due to forces such as pressure, viscosity, gravity, etc. They are partial differential equations derived from momentum balance on a fluid parcel. C) Bernoulli's equation relates pressure, density, velocity, elevation, and kinetic energy in fluid flows. It applies to steady, incompressible, inviscid flows and is useful for analyzing potential flows.

1. Fluid Flow Equations

2. Fundamentals
• Three pilers of fluid dynamics
• Continuity : Mass is Conserved
• Momentum : Newton’s Second law
• Energy : Energy is conserved
Fluid dynamics is based on the mathematical statements of these three
physical principles.

3. Mass balance in a flowing fluid –
Continuity
– Fluid of density ῤ and velocity u
The equation of mass balance is
(Rate of Mass flow in) – (Rate of Mass flow out) = Rate of
Accumulation

4. Balance Equations
1. Mass flux in x direction at the face x = 𝜌𝑢 𝑥
2. Mass flux in x direction at the face (x+dx) = 𝜌𝑢 𝑥+𝑑𝑥
Flux is defined as the rate of flow of quantity per unit area
Hence
3. Mass flow rate of the fluid entering the fluid element in x – direction =
𝜌𝑢 𝑥∆𝑦∆𝑧
4. Mass flow rate of the fluid leaving the fluid element in x – direction =
𝜌𝑢 𝑥+𝑑𝑥∆𝑦∆𝑧
5. Rate of Accumulation of momentum in the volume element = ∆𝑥 ∆𝑦 ∆𝑧
𝜕𝜌
𝜕𝑡
Similarly taking mass balance in ‘y’ and ‘z’ directions and substituting in general Mass
balance equation we will get

5. 𝜌𝑢𝑥 − 𝜌𝑢𝑥+∆𝑥 ∆𝑦∆𝑧 + 𝜌𝑣𝑦 − 𝜌𝑣𝑦+∆𝑦 ∆𝑥∆𝑧
+ 𝜌𝑤𝑧 − 𝜌𝑤𝑧+∆𝑧 ∆𝑥∆𝑦 = ∆𝑥 ∆𝑦 ∆𝑧
𝜕𝜌
𝜕𝑡
Dividing by Δx Δy Δz gives
𝜌𝑢𝑥 − 𝜌𝑢𝑥+∆𝑥
∆𝑥
+
𝜌𝑣𝑦 − 𝜌𝑦𝑦+∆𝑦
∆𝑦
+
𝜌𝑤𝑧 − 𝜌𝑤𝑧+∆𝑧
∆𝑧
=
𝜕𝜌
𝜕𝑡
Taking the limit of Δx Δy Δz approach zero gives the differential equation for
conservation of mass
A.
𝜕𝜌
𝜕𝑡
= −
𝜕 𝜌𝑢
𝜕𝑥
+
𝜕 𝜌𝑣
𝜕𝑦
+
𝜕 𝜌𝑤
𝜕𝑧
= − 𝛻. 𝜌𝑽
𝜕𝜌
𝜕𝑡
+ 𝑢
𝜕𝜌
𝜕𝑥
+ 𝑣
𝜕𝜌
𝜕𝑦
+ 𝑤
𝜕𝜌
𝜕𝑧
= − 𝜌
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧

6. For incompressible fluids
𝑫𝝆
𝑫𝒕
= − 𝜌
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
= −𝜌 𝛻. 𝑽
𝛻. 𝑽 =
𝜕𝑢
𝜕𝑥
+
𝜕𝑣
𝜕𝑦
+
𝜕𝑤
𝜕𝑧
= 0
Stream Line
• Imaginary line drawn in a force field that tangent drawn at any point indicates
the direction of velocity vector at that point and at that time.
• Properties
• There cannot be any movement of fluid across the streamlines
• Streamlines never intersect nor two of them cross each other
• Converging streamlines indicate the accelerating nature of fluid flow in a
direction.

7. • Tube of large or small cross section and of any convenient shape entirely
bounded by streamlines.
• No net flow through the walls of the stream tube.
• Mass flowrate through differential c/s area dS is denoted by
𝑑𝑚 = 𝜌 𝑢 𝑑𝑆
• For conduit of cross sectional area S
𝑚 = 𝜌
𝑆
𝑢 𝑑𝑆
• Average velocity of the stream (Ṽ) flowing through the c/s area is given by
Ṽ =
𝑚
𝜌𝑆
=
1
𝑆 𝑆
𝑢 𝑑𝑆
Ṽ = volumetric flow rate per unit c. s area of condiut
Ṽ =
𝑞
𝑆
q = volumetric flow rate
Stream tube

8. Mass Velocity
ρṼ =
𝑚
𝑆
= 𝐺 ≡
𝑘𝑔
𝑚2 𝑠𝑒𝑐
• Advantage of using G is that it is independent of T and P at a steady flow (m) through
a constant cross section.
• Mass velocity can also be designated as the mass flux which represents the quantity
of material passing through unit area in unit time.

9. Differential momentum balance: Equations of motion
(Rate of Momentum Entering) – (Rate of Momentum leaving) + ( Sum of forces
acting on the system ) = Rate of Momentum Accumulation
1. Rate at which x component of momentum enters face x = 𝜌𝑢𝑢 𝑥 ∆𝑦∆𝑧
2. Rate at which x component of momentum leaves face x+dx = 𝜌𝑢𝑢 𝑥+∆𝑥 ∆𝑦∆𝑧
3. Rate at which x component of momentum enters face y = 𝜌𝑣𝑢 𝑦 ∆𝑥∆𝑧
Equations 1, 2 , 3 correspond to convective flow

10. 𝜌𝑢𝑢 𝑥 − 𝜌𝑢𝑢 𝑥+∆𝑥 ∆𝑦∆𝑧 + 𝜌𝑣𝑢 𝑦 − 𝜌𝑣𝑢 𝑦+∆𝑦 ∆𝑥∆𝑧
+ 𝜌𝑤𝑢 𝑧 − 𝜌𝑤𝑢 𝑧+∆𝑧 ∆𝑥∆𝑦 + τ𝑥𝑥 𝑥 − τ𝑥𝑥 𝑥+∆𝑥 ∆𝑦∆𝑧
+ τ𝑦𝑥 𝑦
− τ𝑦𝑥 𝑦+∆𝑦
∆𝑥∆𝑧 + τ𝑧𝑥 𝑧 − τ𝑧𝑥 𝑧+∆𝑧 ∆𝑦∆𝑥 + ∆𝑦∆𝑧 𝑝𝑥 − 𝑝𝑥+∆𝑥
+ 𝜌𝑔𝑥∆𝑥∆𝑦∆𝑧
=
𝜕(𝜌𝑢)
𝜕𝑡
∆𝑥∆𝑦∆𝑧
Dividing by Δx Δy Δz gives and taking the corresponding limit of Δx Δy Δz tend to
zero
Writing similar expressions for convective flow and mass transport of x momentum through all the six faces & substituting
in Momentum balance equation
4. Rate at which x component of momentum enters face x by molecular transport
= τ𝑥𝑥 𝑥 ∆𝑦∆𝑧
5. Rate at which x component of momentum leaves face x+dx by molecular transport
= τ𝑥𝑥 𝑥+∆𝑥 ∆𝑦∆𝑧
6. Rate at which x component of momentum enters face y by molecular transport = τ𝑦𝑥 𝑦
∆𝑦∆𝑧

11. A.
𝜕𝜌𝑢
𝜕𝑡
= −
𝜕 𝜌𝑢𝑢
𝜕𝑥
+
𝜕 𝜌𝑣𝑢
𝜕𝑦
+
𝜕 𝜌𝑤𝑢
𝜕𝑧
−
𝜕 τ𝑥𝑥
𝜕𝑥
+
𝜕 τ𝑦𝑥
𝜕𝑦
+
𝜕 τ𝑧𝑥
𝜕𝑧
−
𝜕𝑝
𝜕𝑥
+ 𝜌𝑔𝑥
𝝆
𝑫𝒖
𝑫𝒕
= −
𝜕𝑝
𝜕𝑥
−
𝜕 τ𝑥𝑥
𝜕𝑥
+
𝜕 τ𝑦𝑥
𝜕𝑦
+
𝜕 τ𝑧𝑥
𝜕𝑧
+ 𝜌𝑔𝑥
𝝆
𝑫𝑽
𝑫𝒕
= −𝛻𝑝 − 𝛻. τ + 𝜌g
Writing similar expressions for y and z components and adding them
For fluid of const. density and viscosity
B. 𝜌
𝜕𝑢
𝜕𝑡
+ 𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
+ 𝑤
𝜕𝑢
𝜕𝑧
= μ
𝜕2𝑢
𝜕𝑥2 +
𝜕2𝑢
𝜕𝑦2 +
𝜕2𝑢
𝜕𝑧2 −
𝜕𝑝
𝜕𝑥
+ 𝜌𝑔𝑥
C. 𝜌
𝜕𝑣
𝜕𝑡
+ 𝑢
𝜕𝑣
𝜕𝑥
+ 𝑣
𝜕𝑣
𝜕𝑦
+ 𝑤
𝜕𝑣
𝜕𝑧
= μ
𝜕2
𝑣
𝜕𝑥2 +
𝜕2
𝑣
𝜕𝑦2 +
𝜕2
𝑣
𝜕𝑧2 −
𝜕𝑝
𝜕𝑦
+ 𝜌𝑔𝑦

12. C. 𝜌
𝜕𝑤
𝜕𝑡
+ 𝑢
𝜕𝑤
𝜕𝑥
+ 𝑣
𝜕𝑤
𝜕𝑦
+ 𝑤
𝜕𝑤
𝜕𝑧
= μ
𝜕2𝑤
𝜕𝑥2
+
𝜕2𝑤
𝜕𝑦2
+
𝜕2𝑤
𝜕𝑧2
−
𝜕𝑝
𝜕𝑧
+ 𝜌𝑔𝑧
𝝆
𝑫𝑽
𝑫𝒕
= −𝛻𝑝 + μ 𝛻2𝑽 + 𝜌g
Navier – Stokes Equation
Vector form of Navier – Stokes Equation
Euler’s Equation
𝝆
𝑫𝑽
𝑫𝒕
= −𝛻𝑝 + 𝜌g
Assumptions
• Fluid is incompressible
• Fluid is Inviscid (ideal)
• Fluid has zero viscosity
• Streamline and irrotational
flow
• Flow is steady

13. Macroscopic Momentum Balances
Mass flow rate = ; Velocity = u
Momentum flow rate =
Momentum carried by the fluid through c/s area dS in unit time
Variation in instantaneous velocity (u) along the flow section changes the
momentum flow rate estimations.
Momentum rate, estimated even from average velocity brings errors in
Momentum estimations
Momentum correction factor is being used in momentum balance equation;
estimated from principle of momentum flux.

14. Momentum Correction factor
For entire c/s for a constant density fluid

15. Steady and Unsteady flow
Flow parameters such as velocity, pressure, density does not change with
time in a steady flow.
𝜕𝑉
𝜕𝑡
= 0;
𝜕𝑝
𝜕𝑡
= 0;
𝜕𝜌
𝜕𝑡
= 0 For steady flow
𝜕𝑉
𝜕𝑡
≠ 0;
𝜕𝑝
𝜕𝑡
≠ 0;
𝜕𝜌
𝜕𝑡
≠ 0 For unsteady flow
Uniform and non-uniform flow
Type of flow in which Velocity, pressure, density does not with respect to
spatial co-ordinates.
𝜕𝑉
𝜕𝑠
= 0;
𝜕𝑝
𝜕𝑠
= 0;
𝜕𝜌
𝜕𝑠
= 0 For uniform flow
𝜕𝑉
𝜕𝑠
≠ 0;
𝜕𝑝
𝜕𝑠
≠ 0;
𝜕𝜌
𝜕𝑠
≠ 0 For non-uniform flow
Steady uniform flow Steady non-uniform flow
Unsteady uniform flow
Unsteady non-uniform flow
Ex: Flow through a channel at constant discharge Ex: Constant flow through expanding or diverging
section
Non practicable situation
Flow in a pipe through a valve.

16. Energy Equation for potential flow –
Bernoulli’s equation w/o friction
𝜌
𝜕𝑢
𝜕𝑡
+ 𝑢
𝜕𝑢
𝜕𝑥
+ 𝑣
𝜕𝑢
𝜕𝑦
+ 𝑤
𝜕𝑢
𝜕𝑧
= −
𝜕𝑝
𝜕𝑥
+ 𝜌𝑔𝑥
For unidirectional flow component of velocity in y(v) and z(w) are zero
Multiplying remaining terms with u gives
𝜌𝑢
𝜕𝑢
𝜕𝑡
+ 𝑢
𝜕𝑢
𝜕𝑥
= − 𝑢
𝜕𝑝
𝜕𝑥
+ 𝜌𝑢𝑔𝑥
𝜌
𝜕(
𝑢2
2 )
𝜕𝑡
+ 𝑢
𝜕(
𝑢2
2 )
𝜕𝑥
= − 𝑢
𝜕𝑝
𝜕𝑥
+ 𝜌𝑢𝑔𝑥
Mechanical energy equation for unidirectional potential flow of fluids
of constant density and when flow rate varies with time

17. 𝑔𝑥 = −𝑔 𝑐𝑜𝑠∅
𝑍 = 𝑍𝑎 + 𝑥 𝑐𝑜𝑠∅
𝑑𝑍 = 𝑐𝑜𝑠∅ 𝑑𝑥
cos ∅ =
𝑑𝑍
𝑑𝑥
Steady flow
𝜌
𝑑(
𝑢2
2
)
𝑑𝑥
+ 𝑢
𝑑𝑝
𝑑𝑥
+ 𝜌𝑢𝑔 𝑐𝑜𝑠∅ = 0
𝑑(
𝑢2
2 )
𝑑𝑥
+
1
𝜌
𝑑𝑝
𝑑𝑥
+ 𝑔
𝑑𝑧
𝑑𝑥
= 0
Integrating the equation between two stations ‘a’ and ‘b’

18. 𝑝𝑎
𝜌
+ 𝑔𝑍𝑎 +
𝑢2
𝑎
2
=
𝑝𝑏
𝜌
+ 𝑔𝑍𝑏 +
𝑢2
𝑏
2
Bernoulli’s equation without friction
𝑝
𝜌
=
𝑁
𝑚2 𝑘𝑔
𝑚3
=
𝑁 𝑚
𝑘𝑔
𝑢2
2
=
𝑚2
𝑠2
=
𝑚2
𝑠2
𝑘𝑔
𝑘𝑔
=
𝑁𝑚
𝑘𝑔 𝑔𝑍 =
𝑚 𝑚
𝑠2
=
𝑚2
𝑠2
𝑘𝑔
𝑘𝑔
=
𝑁𝑚
𝑘𝑔
Total energy per unit mass of fluid at every point in a flow is constant
𝑝𝑎
𝜌𝑔
+ 𝑍𝑎 +
𝑢2
𝑎
2𝑔
=
𝑝𝑏
𝜌𝑔
+ 𝑍𝑏 +
𝑢2
𝑏
2𝑔
𝑝
𝜌𝑔
=
𝑁
𝑚2 𝑘𝑔
𝑚3
𝑚
𝑠2
=
𝑁
𝑘𝑔
𝑠2
=
𝑁𝑚
𝑘𝑔𝑚
𝑠2
=
𝑁 𝑚
𝑁
𝑢2
2𝑔
=
𝑚2
𝑠2 𝑚
𝑠2
= 𝑚 =
𝑁𝑚
𝑁
𝑍 = m =
𝑁𝑚
𝑁
Total energy per unit weight of fluid at every point in a flow is constant
𝑝𝑎 + 𝜌𝑔𝑍𝑎 +
𝜌𝑢2
𝑎
2
= 𝑝𝑏 + 𝜌𝑔𝑍𝑏 +
𝜌𝑢2
𝑏
2
𝑝 =
𝑁
𝑚2
=
𝑁 𝑚
𝑚2𝑚
=
𝑁 𝑚
𝑚3
𝜌𝑢2
2
=
𝑘𝑔 𝑚2
𝑚3𝑠2 =
𝑁 𝑚
𝑚3
𝜌𝑔𝑍 =
𝑘𝑔 𝑚 𝑚
𝑚3𝑠2 =
𝑁 𝑚
𝑚3
Total energy per unit volume of fluid at every point in a flow is constant

19. Bernoulli’s equation: Corrections for effects of solid boundaries
• Problems in engineering involve streams which are influenced by solid
boundary
• Flow of fluid through a pipe where entire stream is in boundary layer
flow
• Practical Situation involves correction terms in Bernoulli’s equation
for
1. Kinetic energy
2. Existence of fluid friction

20. Kinetic energy
Element of c/s area ds
Mass flow rate through the c/s = ρu ds
Energy flow rate through c/s area
𝑑Ë𝑘 = 𝜌𝑢 𝑑𝑠 ∗
𝑢2
2
Total rate of flow of Kinetic Energy through c/s S is
Ë𝑘 =
𝜌
2
∗
𝑠
𝑠 𝑢3𝑑𝑆
Term which replaces u2/2 in Bernoulli’s equation is
Ë𝑘
𝑚
= 𝑠
𝑠 𝑢3
𝑑𝑆
2 𝑠
𝑠 𝑢 𝑑𝑆
= 𝑠
𝑠 𝑢3
𝑑𝑆
2 Ṽ𝑆

21. K.E correction factor
∝ Ṽ2
2
=
Ë𝑘
𝑚
= 𝑠
𝑢3
𝑑𝑆
2 Ṽ𝑆
∝ = 𝑠
𝑢3
𝑑𝑆
Ṽ3𝑆
Correction factor for fluid friction
𝑝𝑎
𝜌
+ 𝑔𝑍𝑎 +
𝛼𝑎𝑉2
𝑎
2
=
𝑝𝑏
𝜌
+ 𝑔𝑍𝑏 +
𝛼𝑏𝑉2
𝑏
2
+ ℎ𝑓
• Whenever flow occurs friction is generated and hf denote the
friction generated per unit mass of fluid (conversion of
mechanical energy into heat) that occurs when fluid flows
between two stations)
• hf is positive value; it will be zero for potential flow
• Skin friction – friction generated in unseparated boundary layers
• Form friction – boundary layers separate and form wakes,
additional energy dissipation appears in the wake

22. Pump work in Bernoulli Equation
𝑊
𝑝 − ℎ𝑓𝑝 = η𝑊
𝑝
𝑊
𝑝- work done by pump per unit mass of fluid
ℎ𝑓𝑝- friction generated in the pump per unit mass of fluid
η < 1
𝑝𝑎
𝜌
+ 𝑔𝑍𝑎 +
𝛼𝑎𝑉2
𝑎
2
+ η𝑊
𝑝 =
𝑝𝑏
𝜌
+ 𝑔𝑍𝑏 +
𝛼𝑏𝑉2
𝑏
2
+ ℎ𝑓