E4014 - Construction Surveying HYDRAULICS. Introduction surveyors –usually not be directly involved in the design of hydraulics systems –most certainly

E4014 - Construction Surveying

HYDRAULICS

Introduction

• surveyors – usually not be directly involved in the

design of hydraulics systems – most certainly will be responsible for the

setting out of such systems.

• surveyor may need to alter the design of a system ‘on-site’ – the principles involved in this field must

be understood.

Introduction

• It is with this in mind and the desire to make you better able to understand the requirements of the engineer or design draftsman responsible for the project that this module is included.

Definitions & DiscussionsHydrostatics & Hydrodynamics

• Hydrostatics– The study of the properties of fluids which

are at rest or in equilibrium (e. g. storage systems such as dams and reservoirs).

• Hydrodynamics– The study of the properties of fluids which

are in motion in pipes and channels (e.g. water or sewerage reticulation and drainage systems).

Definitions & DiscussionsFluid

– can offer no permanent resistance to any force causing change of shape.

– flow under their own weight

– take the shape of any solid body with which they are in contact.

Definitions & DiscussionsFluid cont

– change of shape is caused by shearing forces; therefore if shearing forces are acting in a fluid it will flow.

– Conversely, if a fluid is at rest there can be no shearing forces in it, and all forces are perpendicular (normal) to the planes on which they act.


• Fluids are divided into liquids and gases.

• Liquid – difficult to compress; – a given mass occupies a fixed volume

irrespective of the size of the container holding it

– a ‘free surface’ is formed as a boundary between the liquid and the air above it.


• Fluids are divided into liquids and gases.

• Gas – easily compressed; – expands to fill any vessel in which it is

contained – does not form a free surface.

Definitions & DiscussionsPressure & Intensity of Pressure

• Pressure– force exerted by a fluid on the surfaces with

which it is in contact, or by one part of a fluid on the adjoining part.

• intensity of pressure at any point – force exerted on the unit area at that point

• measured in newton’s per square metre in SI units

• An alternative metric unit is the bar, which is 105 N/m2.

Definitions & DiscussionsPressure - Example

•

Water

50 cm²

25 kg

A mass of 25 kg acts on a piston of area 50 square centimetres. What is the intensity of pressure on the water in contact with the underside of the piston if

the piston is in equilibrium.

Force acting on the piston = mg

= *

= =

25 kg

245.25 N

9.81 m/sec²

245.25 kg m/sec²

= *

=

0.01 m

0.0001 m²

1 cm² 0.01 m

Area of piston =

= *

=

( 50

0.0050 m²

50 cm²

0.0001 )m²

=

Intensity of Pressure = 0.0050 m²245.25 N

49050.0 N/m²

ForceArea

=

Definitions & DiscussionsPressure Scales

• Atmospheric Pressure– The earth is surrounded by an atmosphere many miles high. – The pressure due to this atmosphere at the surface of the earth

depends upon the head of air above the surface. – Atmospheric pressure at sea level is about 101.325 kN/m2,

• equivalent to a head of 10.35m of water • or 760 mm of mercury approximately, and decreases with altitude


• Vacuum– A perfect vacuum is a completely empty

space in which, therefore, the pressure is zero

• Gauge Pressure – the intensity of pressure measured

above or below atmospheric pressure


• Absolute Pressure– the intensity of pressure measured above

the absolute zero, which is a perfect vacuum


– absolute pressure = atmospheric pressure + gauge pressure, or

– p(abs) = p(atm) + p(g)


– absolute pressure = atmospheric pressure + gauge pressure, or

– p(abs) = p(atm) + p(g)

Definitions & DiscussionsDimensions & Units

• Dimensions– various physical properties that all

matter possess– describe the state of things

• Units– internationally agreed measurable quantities of the

various dimensions– SI ( System International )

• kilogram, metre, cubic metre, metre per second, kelvin, ampere, and joule

Definitions & DiscussionsDimensions & Units

• Seven base dimensions, each with its own base unit

HydrostaticsIntroduction

HydrostaticsIntroduction

HydrostaticsHydrostatic laws - Horizontal

Pressure Variation• Horizontal Hydrostatic Law

– Pressure has a constant value at a given horizontal level in a continuous fluid mass

HydrostaticsHydrostatic laws - Horizontal

Pressure Variation

•

HydrostaticsHydrostatic laws - Vertical

Pressure Variation


Pressure Variation


Pressure Variation


Pressure Variation

HydrostaticsHydrostatic laws - Example

•

Calculate the pressure at points A, B, C, D and E

Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3

Acceleration due to gravity, g =

Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

Pressure = gh

PA = kg/m3 * *

= = 68670.0 kg m/sec² 1/m²

The pressure at A is due to the column of water above it and the atmospheric pressure. Ignore the atmospheric pressure making the pressure at the water surface above A equal zero.

7.0 m1000 9.81 m/sec²

68670.0 Pa

68670 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

Pascall has noted:Pressure has a constant value at a given horizontal level in a continuous fluid mass.

So, The pressures at A, F and G are all equal to 68670 Pa.

68670 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa

The pressure at F due to the water above it can be calculated.

PF due to water = kg/m3 * *

= = 49050.0 Pa49050.0 kg m/sec² 1/m²

1000 5.0 m9.81 m/sec²

PF due to water = 49050.0 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


PB = PF - PF due to water

= -

=

So the pressure at B due to the air above it must be the difference between the pressure at F and the pressure due to the water above it.

49050.0 Pa68670.0 Pa

19620.0 Pa

19620 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa

The pressure at C due to the water above it can be calculated.

PC due to water = kg/m3 * *

= =

9.81 m/sec²1000

58860.0 kg m/sec² 1/m² 58860.0 Pa

6.0 m

PG due to water = 58860.0 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa

PG due to water = 58860.0 Pa

PC = PG - PG due to water

= -

=

58860.0 Pa68670.0 Pa

So the pressure at C due to the air above it must be the difference between the pressure at G and the pressure due to the water above it.

9810.0 Pa

9810 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa

PG due to water = 58860.0 Pa9810 Pa

PD =

The pressures at C and D must be equal they are both at a horizontal level in a continuous fluid mass.

9810.0 Pa

9810 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa

PG due to water = 58860.0 Pa9810 Pa 9810 Pa

The pressure at H due to the benzine above it can be calculated.

PH due to benzine = kg/m3 * *

= =

9.81 m/sec²

51796.8 Pa

6.0 m880

51796.8 kg m/sec² 1/m²

PH due to benzine = 51796.8 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa



PH = PD + PH due to benzine

= +

=

51796.8 Pa

61606.8 Pa

9810.0 Pa

61606.8 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa



61606.8 Pa

PJ =

The pressures at H and J must be equal they are both at a horizontal level in a continuous fluid mass.

61606.8 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa



61606.8 Pa

The pressure at J due to the fluid above it can be calculated.

PJ due to benzine = kg/m3 * *

= =

9.0 m9.81 m/sec²

77695.2 Pa77695.2 kg m/sec² 1/m²

880

PJ due to benzine = 77695.2 Pa


•


Air E

AirC D

B

WaterA F G H J

Benzene = kg/m3

Water = kg/m3


Benzene

880

1000

RL 9.0Air

RL 7.0RL 6.0

RL 0.0

RL 5.0

9.81 m/sec²

68670 Pa


19620 Pa



61606.8 Pa

PJ due to benzine = 77695.2 Pa

PE = PJ - PJ due to benzine

= -

= The air pressure above E is a partial vacum

77695.2 Pa61606.8 Pa

-16088.4 Pa

So the pressure at E due to the air above it must be the difference between the pressure at J and the pressure due to the water above it.

-16088.4 Pa

Hydrostatics Pressure Units

• Expressed in two ways– pressure head

• the height of the free surface of a particular liquid above the surface in question

– traditional pressure units• Pascals (Pa) = Newtons/square metre

Hydrostatics Pressure Units - Example

•

In a tank as shown with water depth of 0.6 metres what is the pressure exerted on the base in terms of (i) pressure head and (ii) traditional pressure units

0.60 m


•


0.60 m

(i) Pressure head is 0.6 metres of water.

Pressure (p ) = gh

rho = mass density of liquid ( for water = 1000 kg/m3 )

g = accelleration due to gravity =

h = height of water above the surface under consideration ( metres )

9.81 m/sec²


•


0.60 m

(i) Pressure head is 0.6 metres of water.

Pressure (p ) = gh

= kg/m3 * *

= = 5886.0 N/m²

1000 9.81 m/sec² 0.60 m

5886.0 kg m/sec² 1/m²

= ( Pascals ) = 5886.0 Pa 5.886 kPa

pgh

Note: To convert from traditional pressure units to pressure head rearrange the formula for pressure.

h =

Hydrostatics Hydrostatic Pressure on Surfaces• A fluid in contact with a solid surface will exert a force on every small area of the surface. • The total pressure on the surface can be represented by a point force equal to the product

of the pressure on the small element and its area.

AghF *

Hydrostatics Hydrostatic Pressure on Surfaces

• The point at which this force must act is the centroid.

Hydrostatics Hydrostatic Pressure on Horizontal Surfaces -

example

0.60 m

2.00 m

3.00 m

In a tank as shown with water depth of 0.6 metres what is the total pressure or force exerted on the base in terms of traditional pressure units


example

0.60 m

2.00 m

3.00 m


Construct a pressure diagram showing that equal forces act over each partial area of a horizontal surface within a fluid.


examplePressure = 5886.0 N/m2

0.60 m

2.00 m

3.00 m


Pressure = gh

= kg/m3 * *

=

1000 0.60 m9.81 m/sec²

5886.0 N/m²

The base area is a rectangle.

AreaBase = *

=

3.00 m 2.00 m

6.0 m²

AreaBase = 6.0m2


examplePressure = 5886.0 N/m2

0.60 m

2.00 m

3.00 m


AreaBase = 6.0m2

=

= 5886.0 N/m² * 6.0 m²Total Pressure (Force)

on the base35316.00 N

The centre of pressure will act at the centroid of the base.

35316.00 N

Hydrostatics Hydrostatic Pressure on Inclined Surfaces

Construct a pressure diagram of the dam wall and also describe critical components to be used in the calculations.

y CP

y CG

h CG

h CP

CG

Pressure DisgramCP

Hydrostatics Hydrostatic Pressure on

Inclined Surfaces


y CP

y CG

h CG

h CP

CG

Pressure DisgramCP

F = A gh CG where F is the magnitude of the force, A is the surface area of the immersed surface and h CG is the vertical depth to the centre of gravity ( or centroid ) of the immersed surface

F


Inclined Surfaces


y CP

y CG

h CG

h CP

CG

Pressure DisgramCP

F

The formula to calculate the position of the centre of pressure of an inclined surface is:.

y CP = y CG + I CG

yCGA where y CP is the depth to the centre of pressure measured along the inclined surface ( not vertically ) , y CG is the is the depth to the centre of gravity of the surface measured along the inclined surface and I CG is the second moment of inertia of the immersed surface about its horizontal

Ay

Iyy

CG

CGCGCP


Inclined Surfaces

• ICG formula for common shapes


y CP

y CG

h CG

h CP

CG

Pressure DisgramCP

F

Ay

Iyy

CG

CGCGCP


Inclined Surfaces


y CP

y CG

h CG

h CP

CG

Pressure DisgramCP

F

Ay

Iyy

CG

CGCGCP

Note that for vertical surfaces = 90° hence sin = 1 so h CG = y CG and h CP = y CP.

Hydrostatics Hydrostatic Pressure on Inclined Surfaces

- Example

A

B

In a tank as shown with water depth of 0.6 metres what is the total pressure or force exerted on the hatchered side in terms of traditional pressure units and calculate the position of the centre of pressure.

0.60 m

2.00 m

3.00 m


Inclined Surfaces - ExampleA

B


0.60 m

2.00 m

3.00 m

Construct a pressure diagram of the forces acting on the side. They will act perpendicular to the surface and will increase from zero at the surface to a maximum at the base. The pressure at the bottom of the side is equal to the pressure at the base.



B


0.60 m

2.00 m

3.00 m

F = A gh CG

When the surface is not horizontal it can shown that “the hydrostatic force on an immersed plane surface is the product of the surface area and the pressure at its centroid”.

where F is the magnitude of the force, A is the surface area of the immersed surface and h CG is the vertical depth to the centre of gravity ( or centroid ) of the immersed

)( CGghAF



B


0.60 m

2.00 m

3.00 m

)( CGghAF

d2

F = A gh CG

= * * kg/m3 * *

=

9.81 m/sec² 0.30 m

3531.60 N

20.30 m

2.00 m 0.60 m 1000

= h CG

(Rectangle) = =

0.60 m



B


0.60 m

2.00 m

3.00 m

)( CGghAF

d2

F = A gh CG

= * * kg/m3 * *

=

9.81 m/sec² 0.30 m

3531.60 N

20.30 m

2.00 m 0.60 m 1000

= h CG

(Rectangle) = =

0.60 m

mhCG 30.0)(rectangle

NF 60.3531



B


0.60 m

2.00 m

3.00 m

)( CGghAF NF 60.3531

The formula to calculate the position of the centre of pressure of an inclined surface is:.

y CP = y CG + I CG

yCGA where y CP is the depth to the centre of pressure measured along the inclined surface ( not vertically ) , y CG is the is the depth to the centre of gravity of the surface measured along the inclined surface and I CG is the second moment of inertia of the immersed surface about its horizontal

Ay

Iyy

CG

CGCGCP



B


0.60 m

2.00 m

3.00 m

)( CGghAF NF 60.3531

Ay

Iyy

CG

CGCGCP

bd3 * * * 12

Note that for vertical surfaces = 90° hence sin = 1 so h CG = y CG and h CP = y CP.

* *

=

= 0.30 m +

0.03600 m^4

0.30 m 2.00 m 0.60 m0.03600 m^4

y CP = y CG + I CG

yCGA

I CG

(Rectangle) = = 2.00 m = 0.60 m 0.60 m 0.60 m

12

0.40000 m

myCP 4.0


Inclined Surfaces - Example)( CGghAF NF 60.3531

Ay

Iyy

CG

CGCGCP

myCP 4.0A

B

3.00 m

3531.60 N0.40 m

0.30 m0.60 m

2.00 m

y CP

y CG

Hydrostatics Hydrostatic Pressure on Inclined

Surfaces• When the surface is not horizontal it can be

shown that ‘the hydrostatic force on an immersed plane surface is the product of the surface area and the pressure at its centroid’

F = ApG

• where F is the magnitude of the force, A is the surface area of the immersed surface and pG is the pressure at the centroid


Surfaces

• Formula to calculate the pressure at the centroid

gApF

CGg ghp Where hCG = vertical depth to the centre of gravity (or centroid) of the immersed surface

• So... )( CGghAF


Surfaces• the position of the centre of

pressure of an inclined surface.

• where yCP is the depth to the centre of the pressure measured along the inclined surface (not vertically), yCG is the depth to the centre of gravity of the surface measured along the inclined surface and ICG is the second moment of inertia of the immersed surface about its horizontal centroidal axis.

Ay

Iyy

CG

CGCGCP

Hydrostatics Hydrostatic Pressure on Surfaces• A fluid in contact with a solid surface will exert a force on every small area of the surface. • The total pressure on the surface can be represented by a point force equal to the product

of the pressure on the small element and its area.

AghF *

Documents

E4014 - Construction Surveying HYDRAULICS. Introduction surveyors –usually not be directly involved in the design of hydraulics systems –most certainly