MET 212 Module 2-hydrostatics

Hydrostatics

1- Fluid pressure

Pressure is the force per unit area

Pressure = forcearea

Unit is Newton/square meter bar = 105 N/m2

Pound/square foot (fps)

Example:

A mass m of 50 kg acts on a piston of area 100 cm2 what is the intensity of pressure on the water in contact with the underside of the position if the piston is in equilibrium

Solution.

Force = mg

= 50 x 9.81 = 490.5 N

Area of piston A = 100 cm2 = 0.01 m2

Pressure = for cearea =

490.50.01 N/m2

= 4.905 x 104 N/m2

Dr. Adel Afify 1 MET 212

For liquids or gases at rest the pressure gradient in the vertical direction at any point in a fluid depends only on the specific weight of the fluid at that point

2- Pascal's Law for Pressure at a Point

Triangular prismatic element of fluid

acts perpendicular to surface ABCD,

acts perpendicular to surface ABFE and

acts perpendicular to surface FECD.

And, as the fluid is at rest, in equilibrium, the sum of the forces in any direction is zero.

Pressure at any point is the same in all directions.This is known as Pascal's Law and applies to fluids at rest.


3. Variation of pressure vertically in a fluid under gravity

Vertical elemental cylinder of fluid

The fluid is at rest and in equilibrium so all the forces in the vertical direction sum to zero. We have

Taking upward as positive, in equilibrium we have

Thus in a fluid under gravity, pressure decreases with increase in height .


4- pressure head Free surface (pressure = p0)

Z p2

z2 h = z2 – z1

p1

z1

y

X

p1 - p2 = ρgh

h = pressure head


example:

A diver is working at a depth of 18 m below the surface of the sea. How much greater is the pressure intensity at this depth than at the surface? Specific weight of sea water is 10000 N/m3

Solution:

p1 - p2 = ρgh

Δ p = 10000 x 18

Δ p=180000 N/m2

Example:

Find the head height (h) of water corresponding to an intensity of pressure (p) of 340000 N/m2. The specific weight of water is 9.81 x 103 N/m3

Solution:

p = ρgh

h = pρg h = 340000

9.81x 1000 = 34.7 m

The hydraulic jack

F1 = PA1 F2 = PA2


Example: If the force F1 is 850 N is applied to the smaller cylinder of a hydraulic jack. The area A1 of a small piston is 15 cm2 and the area A2 of the large piston is 150 cm2. What load can be lifted on the larger piston? The specific weight of the liquid in the jack is 9.81 x 103 N/m3

Solution:

F1 = PA1

850 = P (15/10000) P = 5.667 X 105 N/m2

F2 = PA2

F2 = 5.667 X 105 N/m2 X (150/10000) F2 = 8500 N

Mass of lifted = F 2g Mass of lifted =

85009.81 = 866.5 kg


Measurement of pressure

Pressure

1

Gage pressure @ 1

Local atmospheric

Pressure reference

Absolute pressure @ 1

2 Gage pressure @ 2

Gauge pressure is

Absolute pressure is

Example:

A mountain lake has an average temperature of 10 0c and a maximum depth of 40 m. for a barometric pressure of 598 mm Hg. Determine the absolute pressure (in Pascal’s) at the deepest part of the lake.

Solution:


Pressure is designated as either absolute pressure or gage pressure

A barometer is used to measure atmospheric pressure

Absolute pressure = gauge pressure + atmospheric pressure

P = ϪH2O h + p0

p0 is the pressure at the surface

p0 = ϪHg h

p0 = (133 kN/m3) (0.598m) = 79.5 kN/m2

P = ϪH2O h + p0

P = (9.81x1000 N/m3)(40 m) + 79.5 5 kN/m2

= 392. 5 kN/m2 + 79.5 5 kN/m2

= 472 kN/m2 = 472 kpa (abs)

Manometer

1- Piezometer tube

Consists of a vertical tube, open at the top, and attached to the container in which the pressure is desired

P = Ϫ h + p0

A simple piezometer tube manometerDr. Adel Afify 8 MET 212

Pressure will increase as we move downward

And will decrease as we move upward

Example:

A pressure tube is used to measure the pressure of oil (mass density ρ = 640 kg/m3) in a pipeline. If the oil rises to a height of 1.2 m above the center of the pipe, what is the gage pressure in N/m2 at that point?

Solution:

P = Ϫ h

P = ρ g h

P = (640 kg/m3) x (9.81 m/s2) x (1.2 m)

P = 7.55 kN/m2

2- U- Tube manometer

Pressure in a continuous static fluid is the same at any horizontal level so,


For the left hand arm

For the right hand arm

As we are measuring gauge pressure we can subtract giving

Example:1

What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the surface? ρwater = 1000 kg/m3, and p atmosphere = 101kN/m2. a)

b)

Example:2

At what depth below the surface of oil, relative density 0.8, will produce a pressure of 120 kN/m2? What depth of water is this equivalent to?


a)

b)

Example:3

What would the pressure in kN/m2 be if the equivalent head is measured as 400mm of (a) mercury relative density =13.6 (b) water ( c) oil specific weight 7.9 kN/m3?

a) Mercury

b) Water

c) Oil


Example 4

A manometer connected to a pipe indicates a negative gauge pressure of 50mm of mercury. What is the absolute pressure in the pipe in Newton’s per square meter is the atmospheric pressure is 1 bar?

Example 5What height would a water barometer need to be to measure atmospheric pressure?

Measurement of pressure difference by using a "U"-Tube Manometer.


Pressure difference measurement by the "U"-Tube manometerExample In the figure below two pipes containing the same fluid of density ρ= 990 kg/m3 are connected using a u-tube manometer. What is the pressure between the two pipes if the manometer contains fluid of relative density13.6

h1 = 1.5 m , h2 = 0.5 m , h3 = 0.76 m & ρ2 = SG * ρwater


ρ2 = 13.6 * 1000 kg/m3

Ϫ1 = ρ1g = 990 * 9.81 = 9711.9 N/m3

Ϫ2 = ρ2 g = 13.6 * 1000 * 9.81 = 133416 N/m3

pC = pD

pC = pA + g hA

pD = pB + g (hB - h) + man g h

pA - pB = g (hB - hA) + hg(man - )

= 990 x9.81x(0.75-1.5) + 0.5x9.81 x(13.6-0.99) x 103

= -7284 + 61852

= 54 568 N/m2 (or Pa or 0.55 bar)

Forces on Submerged Surfaces in Static Fluids

1. Fluid pressure on a surface

Pressure is defined as force per unit area. If a pressure p acts on a small area then the force exerted on that area will be

Since the fluid is at rest the force will act at right-angles to the surface.

General submerged plane

Consider the plane surface shown in the figure below. The total area is made up of many elemental areas. The force on each elemental area is always normal to the surface but, in general, each force is of different magnitude as the pressure usually varies.


We can find the total or resultant force, R, on the plane by summing up all of the forces on the small elements i.e.

This resultant force will act through the centre of pressure, hence we can say

If the surface is a plane the force can be represented by one single resultant force,acting at right-angles to the plane through the centre of pressure.

Horizontal submerged plane

For a horizontal plane submerged in a liquid (or a plane experiencing uniform pressure over its surface), the pressure, p, will be equal at all points of the surface. Thus the resultant force will be given by

Curved submerged surface

If the surface is curved, each elemental force will be a different magnitude and in different direction but still normal to the surface of that element. The resultant force can be found by resolving all forces into orthogonal co-ordinate directions to obtain its magnitude and

direction. This will always be less than the sum of the individual forces, .

2. Resultant Force and Centre of Pressure on a submerged plane surface in a liquid.

This plane surface is totally submerged in a liquid of density and inclined at an angle of to the horizontal. Taking pressure as zero at the surface and measuring down from the

surface, the pressure on an element , submerged a distance z, is given by


and therefore the force on the element is

The resultant force can be found by summing all of these forces i.e.

(assuming and g as constant).

The term is known as the 1 st Moment of Area of the plane PQ about the free surface. It is equal to i.e.

where A is the area of the plane and is the depth (distance from the free surface) to the

centroid, G. This can also be written in terms of distance from point O ( as )

The resultant force on a plane

This resultant force acts at right angles to the plane through the centre of pressure, C, at a depth D. The moment of R about any point will be equal to the sum of the moments of the

forces on all the elements of the plane about the same point. We use this to find the position of the centre of pressure.

It is convenient to take moments about the point where a projection of the plane passes through the surface, point O in the figure.

We can calculate the force on each elemental area:

And the moment of this force is:


are the same for each element, so the total moment is

We know the resultant force from above , which acts through the centre of pressure at C, so

Equating gives,

Thus the position of the centre of pressure along the plane measure from the point O is:

It look a rather difficult formula to calculate - particularly the summation term. Fortunately this

term is known as the 2nd Moment of Area , , of the plane about the axis through O and it can be easily calculated for many common shapes. So, we know:

And as we have also seen that 1st Moment of area about a line through O,

Thus the position of the centre of pressure along the plane measure from the point O is:

and depth to the centre of pressure is

How do you calculate the 2nd moment of area?

To calculate the 2nd moment of area of a plane about an axis through O, we use the parallel axis theorem together with values of the 2nd moment of area about an axis though the centroid of the shape obtained from tables of geometric properties.

The parallel axis theorem can be written


where is the 2nd moment of area about an axis though the centroid G of the plane.

Using this we get the following expressions for the position of the centre of pressure

The second moment of area of some common shapes.

The table blow given some examples of the 2nd moment of area about a line through the centroid of some common shapes.

Shape Area A2nd moment of area, , about

an axis through the centroid

Rectangle

Triangle

Circle

Semicircle


Example: 1

The 4- m diameter circular gate is located in the inclined wall of a large reservoir containing water (Ϫ = 9.8 kN/m3). The gate is mounted on a shaft along its horizontal diameter. For a water depth of hc = 10 m above the shaft determine: (a) the magnitude and location of the resultant force exerted on the gate by the water. (b) the moment that would have to be applied to the shaft to open the gate.

Solution

F = Ϫ hc A

F = 9.8 X 103 X 10 X (4π)

F = 1230 X 103 N

X= 10sin60

=11.55 m

A = π R2 = π (22) = 12.57 m2

IGG = π4

R4 = 12.567

Sc = 12.57

12.57 X 11.55+11.55 = 11.637 m

D = Sc * sin 60

D = 11.637 * sin60 = 10.0775 m

(b) the moment

M = F * (Sc – X)

M = 1230 X 103 * (11.637 – 11.55)

M = 1.07 X 105 N . m

Example: 2Determine the resultant force due to the water acting on the 1m by 2m rectangular area AB


shown in the diagram below.

The magnitude of the resultant force on a submerged plane is:

R = pressure at centroid area of surface

This acts at right angle to the surface through the centre of pressure.

By the parallel axis theorem (which will be given in an exam), , where IGG is the 2nd moment of area about a line through the centroid and can be found in tables.

For a rectangle

As the wall is vertical, ,


Example: 3Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular area CD shown in the figure above. The apex of the triangle is at C.

For a triangle

Depth to centre of gravity is .

Distance from P is

Distance from P to centre of pressure is

Forces on submerged surfaces

Example: 4

Obtain an expression for the depth of the centre of pressure of a plane surface wholly submerged in a fluid and inclined at an angle to the free surface of the liquid.A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates about a horizontal axis through its centre. Determine the torque which would have to be applied to


the disk spindle to keep the disk closed in a vertical position when there is a 3m head of fresh water above the axis.

Answer:

The question asks what is the moment you have to apply to the spindle to keep the disc vertical i.e. to keep the valve shut?

So you need to know the resultant force exerted on the disc by the water and the distance x of this force from the spindle.

We know that the water in the pipe is under a pressure of 3m head of water (to the spindle)

Diagram of the forces on the disc valve, based on an imaginary water surface.

, the depth to the centroid of the disc

h' = depth to the centre of pressure (or line of action of the force)

Calculate the force:

Calculate the line of action of the force, h'.


By the parallel axis theorem 2nd moment of area about O (in the surface) where IGG is the 2nd moment of area about a line through the centroid of the disc and IGG = r4/4.

So the distance from the spindle to the line of action of the force is

And the moment required to keep the gate shut is


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MET 212 Module 2-hydrostatics