1. Rehearsal Hydrostatics
p=0 p=0
p=F/A [N/m2]
1

2. Rehearsal Hydrostatics
paverage≠0 paverage≠0 y
p    g  y [N/m2]
1

3. CU06997 Fluid Dynamics
Principles of fluid flow / Conservation laws
2.2 Classification of flows (page 21,22)
2.3 Visualization of flow patterns (page 22,23,24)
2.4 Fundamental equations of fluid dynamics (page 24,25)
2.5 Application of the conservations laws to fluid flows (page 25-32)
2

4. Discharge / Flow rate [debiet]
In hydrology, the discharge or outflow [afvoer] of
a river is the volume of water transported by it in a
certain amount of time. Has to do with the outflow
of a catchment area.
The flow rate [debiet]in fluid dynamics, is the
volume of fluid which passes through a given
surface per unit time.
Q [m3/s]
2

5. Classification of flows
Based on timescale
• Steady (no change Q in time) [Eenparig]
• Unsteady (change …) [Niet eenparig]
Based on scale of distance
• Uniform (no change A along the flow path)
[Uniform]
• Non-uniform (change….) [Niet uniform]
3

6. Classification of flows
1. Steady uniform flow [Eenparig uniform]
example: pipe with constant D and Q
example: channel with constant A and Q
2. Steady non-uniform flow
example: pipe with different D and constant Q
example: channel with different A and constant Q
3. Unsteady uniform flow[Niet eenparig , uni..]
example: pipe with constant D and different Q
example: channel with constant A and different Q
4. Unsteady non-uniform flow
example: pipe with different D and Q
3 example; channel with different A and Q

7. Visualization of flow patterns
A streamline [stroomlijn] is a
line representing the direction
of flow. Streamlines can not
cross
A set of streamlines may
be arranged to form a
imaginary pipe. This is
called a streamtube
[stroombaan]
4

8. Example streamlines
4

9. One-, two- and three dimensional flow
With three dimensional flow the streamlines
changes in x, y and z directions. Calculations
become difficult.
Most of the time it is possible to simplify to
one or two dimensional.
Steady uniform flow in a pipe is considered
to be one dimensional
4

10. Fundamental laws of physics
1. Conservation of matter [Behoud van materie]
(water can’t disappear of appear from nothing)
2. Conservation of energy [Behoud van energie]
(potential and kinetic energy)
3. Conservation of momentum (impuls) [ Behoud van impuls]
A body in motion cannot gain or lose momentum unless
some external force is applied. Has to do with speed and
mass
5

11. 1. Conservation of matter
Mass flow entering = Mass flow leaving
Water is incompressible
Q entering = Q leaving
5

12. Discharge / Flow rate
𝑄= 𝑢∙ 𝐴
𝑄 = Flow rate / Discharge [m3/s]
𝑢 = Mean Fluid Velocity [m/s]
𝐴 = Wetted Area [m2]
𝑢 = v [m/s]
Both are used for velocity
5

13. Equation of Continuity
The continuity equation is a mathematical statement
that, in any steady state process, the rate at which
mass enters a system is equal to the rate at which
mass leaves the system
𝑢1 ∙ 𝐴1 = 𝑢2 ∙ 𝐴2 = 𝑢3 ∙ 𝐴3 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
5

14. 2. Conservation of energy
𝐸 𝑡𝑜𝑡𝑎𝑙 = 𝐸 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 + 𝐸 𝑘𝑖𝑛𝑒𝑡𝑖𝑐
= 𝑚 ∙ 𝑔 ∙ 𝑑 + 1 𝑚 ∙ 𝑢2
2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [𝐽 = 𝑁𝑚]
𝐸 𝑡𝑜𝑡𝑎𝑙 𝑢2 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
= 𝑑+ = [m]
𝑚∙𝑔 2𝑔 𝑚∙𝑔
𝑝
𝑑= 𝑧+ 𝑦= 𝑧+ [𝑚]
𝜌∙ 𝑔
𝑽𝟐 Total Head
𝑯= 𝒚+ 𝒛+ [m]
6 𝟐∙𝒈 [Energiehoogte]

15. Piezometric Head [Piezometrisch nivo]
𝑧1 + 𝑦1 = 𝑧2 + 𝑦2 = 𝑧3 + 𝑦3
Stagnant, not flowing
𝑦3 Piezometric Head
𝑦2 𝑦1 Surface / water level
[druklijn]
Horizontal reference line / datum
𝑝
𝑦= =Pressure Head[drukhoogte] [m]
𝜌∙ 𝑔
6 =Potential Head[plaatshoogte] [m]

16. 3. Conservation of momentum
You need an external force to change
momentum (impuls)
Newton’s Second Law of motion
7

17. Momentum equation
𝐹𝑥 = 𝜌 ∙ 𝑄 𝑉2𝑥 − 𝑉1𝑥
𝐹= Force [N]
𝜌= fluid density [Kg/m3]
𝑄= Flow rate [m3/s]
𝑉1 = Mean velocity before [m/s]
𝑉2 = Mean velocity after [m/s]
Steady flow for a region of uniform velocity
7

18. 2. Conservation of energy
𝐸 𝑡𝑜𝑡𝑎𝑙 = 𝐸 𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 + 𝐸 𝑘𝑖𝑛𝑒𝑡𝑖𝑐
= 𝑚 ∙ 𝑔 ∙ 𝑑 + 1 𝑚 ∙ 𝑢2
2
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [𝐽 = 𝑁𝑚]
𝐸 𝑡𝑜𝑡𝑎𝑙 𝑢2 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
= 𝑑+ = [m]
𝑚∙𝑔 2𝑔 𝑚∙𝑔
𝑝
𝑑= 𝑧+ 𝑦= 𝑧+ [𝑚]
𝜌∙ 𝑔
𝑽𝟐 Total Head
𝑯= 𝒚+ 𝒛+ [m]
8 𝟐∙𝒈 [Energiehoogte]

19. Flowing water and total head
2
u
H1  z1  y1  1
[m ]
2g
Total head H [m]
u12/2g Velocity head [m]
Surface level [m]
y1 h = Pressure head [m]
u1 P1
z1 z = Potential head [m]
Reference /datum [m]
8

20. Bernoulli’s law (without energy losses)
2 2
u u
y1  z1  y2  z2   y 3  z3 
2
 constant
3
2g 2g
Total Head [m]
P1 y1 u22/2g u32/2g Velocity head [m]
z1 y2 Surface Level [m]
P2
u1=0 P3 y3
u2>0 z2 u3>u2 z3
Reference[m]
8

21. Bernoulli’s Equation
2 2
𝑢1 𝑢2
𝑦1 + 𝑧1 + = 𝑦2 + 𝑧2 + = 𝐻 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡[𝑚]
2𝑔 2𝑔
p
y= = Pressure Head[m] [drukhoogte]
ρ∙g
𝑧= Potential Head[m] [plaatshoogte]
V2
=
2g Velocity Head[m] [snelheidshoogte]
8

23. Pipe with no head loss
2 2
u u
y1  1
 y4  4
H Q  u1  A1  u4  A4
2g 1 2g 4
Total Head
[Energiehoogte]
Pressure Head
[Drukhoogte]
Reference / datum

24. Assume no energie loss, calculate
difference in pressure
D=0,15 m D=0,30 m
Q=140 l/s

25. Solution
u2/2g=3,14 m u2/2g=0,20 m
∆p=3,14 – 0,20 = 2,94 m
D=0,15 m D=0,30 m
Q=140 l/s

26. Assume no energie loss, calculate
difference in pressure
D1=0.3m D2=0.15m
Q=0.14 m3/s

27. Solution
u2/2g=0,20 m u2/2g=3,14 m
∆p=0,20-3,14 = -2,94 m
D1=0.3m D2=0.15m
Q=0.14 m3/s

28. Example book