Rehearsal Hydrostatics

p=F/A [N/m2]

p=0 p=0

1

[N/m2] ygp

paverage≠0 paverage≠0 y

Rehearsal Hydrostatics

1

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

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]

Discharge / Flow rate [debiet]

2

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

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

example; channel with different A and Q 3

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

Example streamlines

4

One-, two- and three dimensional flow

Steady uniform flow in a pipe is considered

to be one dimensional

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.

4

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

1. Conservation of matter

Mass flow entering = Mass flow leaving

Water is incompressible

Q entering = Q leaving

5

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

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

2. Conservation of energy

𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐸𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 + 𝐸𝑘𝑖𝑛𝑒𝑡𝑖𝑐

= 𝑚 ∙ 𝑔 ∙ 𝑑 + 12𝑚 ∙ 𝑢2

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [𝐽 = 𝑁𝑚]

𝐸𝑡𝑜𝑡𝑎𝑙

𝑚∙𝑔= 𝑑 +

𝑢2

2𝑔=

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑚∙𝑔 [m]

𝑑 = 𝑧 + 𝑦 = 𝑧 +𝑝

𝜌 ∙ 𝑔 [𝑚]

𝑯 = 𝒚 + 𝒛 +𝑽𝟐

𝟐∙𝒈 [m] Total Head

[Energiehoogte] 6

Piezometric Head [Piezometrisch nivo]

Stagnant, not flowing

Piezometric Head

Surface / water level

[druklijn]

=Pressure Head[drukhoogte] [m]

=Potential Head[plaatshoogte] [m]

Horizontal reference line / datum

𝑦 =𝑝

𝜌 ∙ 𝑔

𝑧1 + 𝑦1 = 𝑧2 + 𝑦2 = 𝑧3 + 𝑦3

𝑦3 𝑦2 𝑦1

6

3. Conservation of momentum

You need an external force to change

momentum (impuls)

Newton’s Second Law of motion

7

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

2. Conservation of energy

𝐸𝑡𝑜𝑡𝑎𝑙 = 𝐸𝑝𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 + 𝐸𝑘𝑖𝑛𝑒𝑡𝑖𝑐

= 𝑚 ∙ 𝑔 ∙ 𝑑 + 12𝑚 ∙ 𝑢2

= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 [𝐽 = 𝑁𝑚]

𝐸𝑡𝑜𝑡𝑎𝑙

𝑚∙𝑔= 𝑑 +

𝑢2

2𝑔=

𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

𝑚∙𝑔 [m]

𝑑 = 𝑧 + 𝑦 = 𝑧 +𝑝

𝜌 ∙ 𝑔 [𝑚]

𝑯 = 𝒚 + 𝒛 +𝑽𝟐

𝟐∙𝒈 [m] Total Head

[Energiehoogte] 8

Flowing water and total head

][2

2

1111 m

g

uyzH

u1

Reference /datum [m]

Surface level [m]

Total head H [m]

P1

z1

y1

u12/2g Velocity head [m]

h = Pressure head [m]

z = Potential head [m]

8

Bernoulli’s law (without energy losses)

u2>0

Reference[m]

Surface Level [m]

Total Head [m]

P2

z2

y2

u22/2g Velocity head [m]

constant22

2

333

2

22211

g

uzy

g

uzyzy

P3 z3

y3

u32/2g

u3>u2 u1=0

P1

z1

y1

8

Bernoulli’s Equation

Pressure Head[m] [drukhoogte]

Potential Head[m] [plaatshoogte]

Velocity Head[m] [snelheidshoogte]

𝑦1 + 𝑧1 +𝑢1

2

2𝑔= 𝑦2 + 𝑧2 +

𝑢22

2𝑔= 𝐻 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡[𝑚]

y =p

ρ∙g=

𝑧 =

V2

2g=

8

Pipe with no head loss

Hg

uy

g

uy

4

2

44

1

2

11

22 4411 AuAuQ

Total Head

[Energiehoogte]

Pressure Head

[Drukhoogte]

Reference / datum

Assume no energie loss, calculate

difference in pressure

D=0,15 m D=0,30 m

Q=140 l/s

Solution

D=0,15 m D=0,30 m

Q=140 l/s

u2/2g=3,14 m u2/2g=0,20 m

∆p=3,14 – 0,20 = 2,94 m

D1=0.3m D2=0.15m

Q=0.14 m3/s

Assume no energie loss, calculate

difference in pressure

D1=0.3m D2=0.15m

Q=0.14 m3/s

Solution

u2/2g=0,20 m u2/2g=3,14 m

∆p=0,20-3,14 = -2,94 m

Example book