Total thrust and centre of pressure (vertical circular surface)

A circular opening in a vertical dam face is closed by a gate mounted on trunnions on its horizontal centre line. The gate has a diameter of 4.2 m and its horizontal centre line is 5.0 m below the water level in the dam.

Determine the magnitude of the pulling force required to be applied at a point 2.0 m above the centroid of the gate to just keep the gate closed against the hydrostatic thrust of the water.

Total force acting on circular opening, say F1 from equation:

F = ρ gA y

= 1000 x 9.81 x

π x 4.24 ² x 5

∴ F₁ = 679559 N

Centre of pressure for force F1 lies at a distance GC below trunnion centre line.

⇒ GC = Equilibrium conditions for opening are:

∑ C.W. MOMENTS = A.C.W. MOMENTS

Pulling force F₂ x 2.0 = 679559 x .2205

∴ MAGNITUDE OF PULLING FORCE = 74.9 kN

PULLING FORCE

Head of a liquid

Whilst deriving Bernoulli’s equation in Outcome 4, the total specific energy of a liquid stream was stated as:

Potential energy + kinetic energy + flow energy

⇒ gZ +

 + p 2 C²

(specific energy IN J kg^-1)

Dividing these terms by gravitational acceleration, g, each term can then be re-stated

as:-⇒ Z +

J + p g 2 C²

(head of liquid in metres, m)

i.e. POTENTIAL HEAD + KINETIC HEAD + PRESSURE HEAD The energy quantity p

is also referred to as flow work or pressure energy and is the energy which must be continuously expended by a pump (or its equivalent) in forcing a liquid along a pipeline in the presence of hydrostatic pressure.

Although in common use, the term pressure energy tends to imply that a liquid may have its energy increased by pressurization. This is not the case since most liquids are practically incompressible. The term flow work may be considered more appropriate.

Likewise, the three separate types of energy listed above for a liquid stream are often expressed more conveniently as heads of liquid.

Static pressure head and hydrostatic pressure

The hydrostatic pressure, p, at a point in a liquid stream, may be imagined to be due to its being at a depth, h, below the free surface of the same liquid. Thus, if a small hole be drilled into a pipeline carrying a flowing liquid and then fitted with a piezometer tube, the liquid would rise up the tube due to the pressure in the system and settle at a height when equilibrium was reached.

FIGURE 5G

Reference above figure, the height, h, to which the liquid rises in the tube provides a means whereby the hydrostatic pressure at that point in the pipeline can be

determined. This height, h, is called the static pressure head of the liquid in the pipeline.

Hence,

Static pressure head, h =

g FLOW WORK

J =

p (m)

and hydrostatic pressure, p = ρ gh (N m^-2)

The static pressure head is easily measured and may be regarded as a head of liquid equivalent to the flow work.

PIE Z OM ET ER T U BE

PR ESSU R E

p

D E N SIT Y

S E C T IO N O F P IP E LIN E

Pressure measurement

Manometry may be defined as the science of utilising vertical columns (heads) of liquid to measure fluid pressures.

Many different types of pressure measuring devices exist but our studies will be limited to the barometer, piezometer tube and U-tube manometers.

Before considering these devices, however, it is appropriate to recap on definitions of pressure quantities already covered in this unit.

Vacuum A perfect vacuum is a completely empty space and has zero pressure.

Atmospheric pressure, patm The planet earth is surrounded by an atmosphere. The pressure due to this atmosphere depends upon the head of air above the earth’s surface. At sea level atmospheric pressure is normally taken as 1.013 bar (101.3 kN m^-2), equivalent to a head of 10.35 m of water or 760 mm of mercury, and decreases with altitude.

Gauge pressure, p_g is the intensity of pressure measured above or below atmospheric pressure. When a gauge shows a zero reading it means the pressure is atmospheric.

Absolute pressure, p is the intensity of pressure measured above absolute zero, which is a perfect vacuum.

i.e. ABSOLUTE PRESSURE p = GAUGE PRESSURE pg + ATMOSPH. PRESS. patm

Barometer

The figure alongside shows a very basic mercury barometer, a device suitable for measuring the pressure of the earth’s atmosphere. The device consists of a small diameter glass tube about 1 m long. The tube is filled with mercury and inverted with its open end in a dish of mercury. A vacuum is created at the top of the tube and the atmospheric pressure acting on the surface of the mercury in the dish supports a column of mercury in the tube of height h.

FIGURE 5H

If B is a point in the tube at the same level as the free surface of the mercury in the dish, then the pressure pB acting upwards at B will be equal to the atmospheric pressure p_atm acting downwards, since, in a fluid at rest, the pressure is the same at all points at the same level.

The column of mercury in the tube is in equilibrium due to the action of the force at B acting upwards against the force (m x g) of the column of mercury of height h acting vertically downwards.

For equilibrium and summing forces in the vertical direction:-F = O = pB x A - ρ ghA

Hence, pB = ρ gh and since pB = patm

∴ ATMOSPHERIC PRESSURE patm = ρHggh (kg m^-3 x m s^-2 x m = N m^-2) When the height, h, of the liquid column is 760 mm and density of mercury taken at 13600 kg m^-3.

Atmospheric pressure p_atm = 13600 x 9.81 x .76

= 101.396 kN m^-2

Piezometer tube

Pipelines and vessels carrying liquid under pressure can have their pressure measured by manometers or pressure gauges employing liquid columns.

The simplest type of manometer is the piezometer tube, which is a single vertical transparent open top tube fitted into the pipeline or vessel carrying pressurized liquid whose pressure is to be measured. Two types are shown below.

FIGURE 5J

Due to hydrostatic pressure in the system, the free surface of the liquid in the open tube will rise and stabilise at a height, h, above the centre line of the pipeline.

The height, h, is the pressure head and allows gauge pressure to be calculated, viz.

At position/level A.

Gauge pressure, p_g = ρ gh (kg m^-3 x m s^-2 x m = N m^-2) For example, a pressure head of 40 mm of water converts to a pressure:

Gauge pressure, p_g = ρ gh = 1000 x 9.81 x 3

10 40

∴ p_g = 392.4 N m^-2

The length of vertical tube which can be conveniently used limits the piezometer to measuring pressures in the lower ranges. For higher liquid pressures, U-tube manometers are often more appropriate.

A A

LIQUID DENSITY

PIEZOMETER TUBES

The U-tube manometer

In order to measure higher pressure levels in pipelines, U-tube manometers employing mercury liquid columns are in widespread use.

FIGURE 5K

In above figure, the level of mercury in the LH limb of the U-tube is at section x-x and distance y below the axis of the pipeline. In the open limb, the mercury column is in equilibrium at a point C, height h above section x-x due to the pressure in the pipeline. B is a point in the LH limb on the same horizontal level as the pipe axis A.

Working on the basis of gauge pressure then, from figure:-Pressure at C, pc = 0 (i.e. atmospheric)

and pressure at x-x in RH limb = ρHg g h

likewise pressure at x-x in LH limb =ρHgg h (same horizontal level) Thus, p_B, pressure at B in LH limb = ρHg g h - ρ g y

∴ pressure in pipeline, pA = pB = g (ρ^Hg h - ρ y) ***

A

MERCURY DENSITY Hg ATER DENSITY

B

C

X X

y

U-TUBE MANOMETER (POSITIVE GAUGE PRESSURE)

Pressure difference between two pipes

The pressure difference between two levels in a fluid can be measured directly by using a differential manometer.

FIGURE 5L

The U-tube differential manometer shown in figure above is to measure the pressure difference between levels A and B in pipelines carrying a liquid of specific weight w₁. The U-tube contains mercury of specific weight w2.

Since C and D are at the same level in a liquid at

rest:-pressure at C = pressure at D

⇒ p_c = p_D

For the LH limb pc = pA + w1X

For the RH limb p_D = p_B + w₁ (Y – h) + w₂h

∴ p_A + w₁X = p_B + w₁Y – w₁h + w₂h Hence, pressure difference p_A - p_B = w1Y – w1h + w2h – w1X

= w1 (Y – X) + h (w2– w1) Pressure difference between the two pipes can be deduced from the formula:

*** pA– pB = w1 (Y – X) + h (w2– w1)

C A

B

D X

Y h

W¹ W1

Density to specific weight conversion

When working through problems such as in manometry pressure calculations it is often convenient to convert one fluid quantity into another.

Density

This quantity has earlier been defined as mass per unit volume.

⇒ Density (RHO) ρ =

V = m E UNIT VOLUM

MASS (UNIT kg m^-3)

Specific weight

This is defined as weight per unit volume.

⇒ Specific weight, w =

E UNIT VOLUM

WEIGHT

Since weight = mass x gravitational acceleration = mg (N)

⇒ Specific weight, w =

V

mg (kg x m s^-2 x m^-3 = N m^-3)

∴ w = ρ g (kg m^-3 x m s^-2 = N m^-3)

Thus multiplying density by gravitational acceleration yields specific weight (symbol w, unit N m^-3).

Example

Oil of density 800 kg m^-3 has a specific weight of 800 x 9.81 = 7848 N m^-3