1

An Introduction to Statics and

Dynamics of Fluids

2

Hydrostatics – Fluids at Rest

A fluid may be a gas or a liquid however, they have some distinct differences as described below.

Gases – When held within a sealed container they expand to fill it completely; they are also known to

be compressible.

Liquids – These do not expand to fill sealed containers and are considered to be incompressible.

On this course we shall only be considering liquids.

Pressure (P)

If held within a cylinder, sealed by a piston a fluid may be subjected to pressure if force is applied to

the piston.

Pressure is defined as the applied force divided by area, in this example the area of the piston. Therefore,

we have the equation:

P =F

A

It has units of N/m2 and bar where 1 bar is equal to 105 N/m2.

Since pressures may have high values we may also use kN/m2, MN/m2, and GN/m2.

It is important to remember that pressure at a point is the same in all direction while pressure on a

surface will always act at 90o to it.

Other Useful Properties:

Density (ρ)

Density is defined as mass per unit volume and therefore yields the familiar equation:

ρ =m

v

Units include; kg/m3, tonne/m3, kg/litre, and g/millilitre

Since we often work with water it is worth remembering it density which is 1000 kg/m3 = 1 Mg/m3 = 1

tonne/m3 = 1 kg/litre.

We also have relative density, where relative density = density of substance

density of water.

Specific Weight

This is defined as weight per unit volume, therefore we have;

Specific weight = ρg

Where, g is the acceleration due to gravity.

Specific Gravity

The specific gravity of a substance is defined as weight of substance

weight of equal volume of pure water.

3

Since this is in effect the ratio of the masses it follows that Specific Gravity and Relative Density are in

fact the same.

Pressure in a Liquid Due to its Own Weight

P atmos

Surface

h P

Mass dm, cross-sectional area da

P + dP

dh

dw = dmxg = volume x density x g

Balancing forces acting on element

PdA – (P +dP)dA + dmg = 0

Or PdA – (P + dP)dA +dAdhρg = 0

Therefore P –(P + dP) +dhρg = 0

Or P – P – dP + dhρg = 0

Therefore dP = ρgdh

Or ∫ dP = ∫ ρgdh

Provided ρ and g are constant this gives P = ρgh + C

When h = 0, P = Patmos, therefore P = ρgh + Patmos

This is called absolute pressure.

Usually in fluids we tend to work in terms of gauge pressure, i.e. ρgh (See diagram below).

4

Pressure

P = ρgh (gauge pressure)

Patmos

Pvacuum

Pabsolute

Pabsolute

Measuring Pressure

Absolute pressure is usually measured using a barometer while gauge pressure is measured by a

manometer or pressure gauge such as a Bourdon gauge. Since we have already stated that we are

normally only interested in gauge pressure we will have a brief look at some of these measuring devices.

Piezometer Tube

Height h is known as the pressure head and is normally expressed in mm, cm, or m of water. As an

example a 10 mm head of water would represent

P = ρgh = 1000 x 9.81 x 10 x 10-3 = 98 N/m2

If measuring small pressure, a version with a sloping tube may be used.

Depending on the pressure of the

liquid in the pipe the value of h will

change.

Pressure in pipe = ρgh

5

Another devise is a U-Tube Manometer

U-Tube Manometer

There are several other measuring devises and if you are interested you could search the internet.

Fluid F1 has density ρ

Fluid F2 (usually mercury) has

density ρ’

Balancing pressures on a-a

P + ρgh2 = ρ’gh

Therefore, P can be found

6

Tutorial Problems

1. Find the head of water corresponding to an intensity of pressure of 340,000 N/m2. The specific

weight of water is 9.81 x 103 N/m3.

(34.66 m)

2. 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 10,000 N/m3.

(180,000 N/m2)

3. Determine the pressure in N/m2 at a depth of 6 m below the free surface of a body of water and

at a depth of 9 m below the free surface of a body of oil of specific gravity 0.75.

(58.9 kN/m2; 66.2 kN/m2)

4. What depth of oil, specific gravity 0.8 will produce a pressure of 120 kN/m2? What depth of

water will produce the same pressure?

(15.3 m; 12.2 m)

5. At what depth below the free surface of oil having density of 600 kg/m3 is the pressure equal

to 1 bar?

(17 m)

6. What is the gauge and absolute pressure at a point in water 10 m below the free surface? Assume

atmospheric pressure to be 100 kN/m2 and the density of water 1000 kg/m3.

(98.1 kN/m2; 198 kN/m2)

7. Calculate the absolute pressure at a depth of 8 m in oil of specific gravity 0.8 if the atmospheric

pressure is 0.97 bar.

(159.8 kN/m2)

8. What is the pressure in kN/m2 absolute and gauge at a point 3 m below the free surface of a

liquid having specific weight of 15 x 103 N/m3 if the atmospheric pressure is 750 mm of

mercury? Take the specific gravity of mercury as 13.6 and density of water as 1000 kg/m3.

(145 kN/m2; 45 kN/m2)

7

Hydrodynamics – Fluid Dynamics

All real fluids are viscous, that is offer resistance to shear. They are compressible, have surface tension

and vaporise. An idea fluid does not exist and does not have any of the above properties. We use the

concept however to simplify calculations and any discrepancies between predicted results and actual

results are found by experiment. Coefficients are then used to convert the ideal to the real.

Steady Flow

This occurs when conditions at a point in the moving fluid do not change with time.

Unsteady Flow

When this is the case, conditions do change with time.

Streamlines

If the path of the fluid particles in steady flow are traced the result is a series of smooth curves called

streamlines.

i.e.

Streamline flow

1. The tangent to a streamline gives the motion of the particle at that point.

2. There can be no flow across a streamline.

3. Converging streamlines indicate increasing velocity.

Laminar and Turbulent Flow

Laminar flow occurs when the fluid moves in layers which slide smoothly one over the other. Turbulent

flow occurs when small irregular motions of fluid particles are superimposed on the main motion of the

fluid.

Reynolds’ experiment using dye in water shows that at low velocities the dye flow has a thin streamline

down the centre (laminar flow). Increasing the flow velocity causes an initial wavering (transition) and

then diffusion (turbulent). He also found that the fluids density (ρ), the viscosity (μ), and the diameter

of the pipe (d) also affected the point of transition. By repeated experiments Reynolds’ found that

laminar flow ended when the dimensions group ρud

μ equaled 2300. This is known as the Critical

Reynolds Number.

Laminar Transition Turbulent

8

Bernoulli Equation

In order to understand this better we need to consider all the types of energy that may be involved.

These include:

• Pressure energy

• Potential energy

• Kinetic energy

Pressure Energy

Consider the tank shown below:

h Pressure P, Area A

F = PA

x

In moving piston distance x, work is done.

Where WD = Fx = PAx

If we release the piston, both piston and liquid would return to their original states and we therefore see

that this form of energy is recoverable. This represents pressure energy.

When piston is forced in

Volume in = Ax

Mass in = ρAx (since ρ = m

v)

Hence the amount of energy possessed by unit mass of liquid is given by

WD

Mass=

PAx

ρAx=

P

ρ

It will have units of N

m2 ×m3

kg=

Nm

kg=

J

kg

It follows that pressure energy/unit weight is P

ρg and this will have units of

J

N=

Nm

N= m. This is called

pressure head.

So if a liquid is under pressure P such as that shown above, then h is equivalent to the static head of

liquid which produces the same pressure. In other words, the pressure at depth h is equal to ρgh.

(Something we have already seen.)

Potential Energy

We now know that pressures at some depth h is P = ρgh

It follows that gh = P

ρ

9

Since P

ρ represents the pressure energy/ unit mass, it follows that gh must also represent energy, i.e. the

potential energy required to raise unit mass through a certain distance h.

What we should be seeing now is that potential energy and pressure energy may be converted one into

the other.

1. As we force the piston in, pressure energy is converted into potential energy.

2. As piston is released and level of liquid falls, potential energy is converted back into pressure

energy.

We are therefore able to write

Potential energy per unit mass + Pressure energy per unit mass = a constant

That is gh +P

ρ = a constant, per unit mass

However, it is more usual to write this in terms of energy per unit weight

That is h + P

ρg= a constant, per unit weight

Kinetic Energy

B

h

A u

In falling from B to A

Potential energy lost/unit mass = gh

KE gained/unit mass = 1

2u2

PE lost = KE gained

Therefore, gh = u2

2

Or h = u2

2g = KE/unit weight (Velocity head)

If we need exit velocity this is given by u = √2gh

10

Relationship between Pressure and KE

It is not necessary for any particle to have actually fallen from B to A before flowing from the orifice.

If already at level A it will possess pressure energy P

ρg and this in turn may be converted into KE on

escaping Therefore we are able to write pressure energy lost = KE gained

Or P

ρg=

u2

2g (per unit weight)

The Bernoulli equations is really a way of stating the principle of conservation of energy as applied to

a steady flowing fluid system. That is that the total energy in the system at any two points is the same

provided that energy is neither given to nor extracted from the system. Thus, we are able to write

Pressure energy + Kinetic energy + Potential energy = a constant

Or P

ρg +

u2

2g + h = a constant

The equation is valid provided

• There is no loss of energy due to friction

• The flow is steady

Applying this to flow through a pipe (note the change from h to z)

P2, A2, u2

z2

z1

P1, A1, u1

Datum

Applying Bernoulli’s equation

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2

For real fluids which include losses it becomes necessary to add a viscosity term

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2 + hf

Where hf is a head loss due to friction.

Bernoulli’s equation alone is usually insufficient for solving pipe flow problems and it becomes

necessary to make use of another equation known as the Continuity Equation.

11

Equation of Continuity

m1 m2

A1 A2

m = mass flow rate in kg/s

m = ρQ, where Q = volumetric flow rate in m3/s

Q = Au, where A = area and u = velocity

Assumption is that m1 = m2

Therefore ρ1Q1 = ρ2Q2

Hence ρ1u1A1 = ρ2u2A2

If ρ1 = ρ2, which is the case with a flowing liquid

Then u1A1 = u2A2 this is the Continuity Equation

Note that u is the average velocity across the section because u varies as shown below.

Graph of Velocity across Section

Application of Bernoulli’s Equation

Orifice in a Tank

1

h

2

Applying Bernoulli’s equation between 1 and 2

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2

P1 = P2 therefore pressure energy terms disappear; u1 is

negligible; z1 = h; z2 = 0

12

It follows therefore that h =u2

2

2g or u2 = √2gh

However, the actual velocity in a real fluid is less than this where

uactual

utheoretical= Cv = Coefficient of velocity

So that we should write u2 = Cv√2gh

To calculate the volume flow rate Q where Q = Area ×Velocity

Ao Av

The ratio Av

Ao= Cc = Coefficient of contraction

Actual flow rate Qactual = Cv√2gh×CcAo

Or Qactual = CdAo√2gh

Where Cd = Cc x Cv = Coefficient of discharge which is found by experiment to be approximately 0.6.

Venturi Meter

Pressures at 1 and 2 measured by pressure gauge

Q = Area x Velocity

= A1u1 = A2u2

Z2 2 Diffuser

Throat

Z1 1

Applying Bernoulli’s equation between 1 and 2 and assuming an ideal fluid

Ao = Area of orifice

Av = Area of vena contractor

13

P1

ρg+

u12

2g+ z1 =

P2

ρg+

u22

2g+ z2

Rearranging

P1−P2

ρg+ z1 − z2 =

u22−u1

2

2g

But u2 =A1u1

A2 from the continuity equation

P1−P2

ρg+ z1 − z2 =

(A1A2

u1)2

−u12

2g= [(

A1

A2)

2− 1]

u12

2g

Therefore u1 = √([

P1−P2ρg

]+z1−z2)2g

(A1A2

)2

−1

Qtheoretical = A1u1 = A1 √([

P1−P2ρg

]+z1−z2)2g

(A1A2

)2

−1

Qactual = Qtheoretical×Cd

Where Cd = Coefficient of discharge which is found to be approximately 0.97.

Note:

If the venturi is horizontal, then z1 – z2 = 0

Next consider a venture meter when pressures 1 and 2 are measured using manometers.

h

h1 h2

2 z2

y

1 z1

Since P = ρgh it follows that h1 = P1

ρg and h2 =

P2

ρg

Therefore P1−P2

ρg+ z1 − z2 = h1 − h2 − y

But h1 = h + h2 +y

14

Therefore P1−P2

ρg+ z1 − z2 = h + h2 + y − h2 − y = h

Modifying our previous equation for Qactual we end up with

Qactual = CdA1√2gh

(A1A2

)2

−1

15

Tutorial Problems – First 4 are from Applied Mechanics by Hannah & Hillier

1. A tank of oil, of relative density 0.8, discharges through a 12 mm diameter orifice. The head of

oil above the centre line of the orifice is kept constant at 1.2 m and the measured discharge rate

is 16 kg/min. Calculate the coefficient of discharge for the orifice.

(0.61)

2. A large tank contains water to a depth of 0.9 m. Water issues from a sharp –edged orifice of 25

mm diameter and is collected in a circular tank of 0.9 m diameter. The water level in the cylinder

rises 600 mm in 5 min. Calculate the discharge coefficient for the orifice.

(0.62)

3. A venturi meter has an inlet diameter of 100 mm and a throat diameter of 50 mm. What will be

the difference of head in metres of water between inlet and throat if the flow rate is 15 litres/s

of water? If the flow rate is doubled, what would then be the difference in head?

(2.78 m; 11.12m)

4. The measured discharge of water through a venturi meter is 78 Mg/h. The inlet and throat

diameters are 120 mm and 55 mm respectively. The pressure drop between inlet and throat is

42 kN/m2. Find the discharge coefficient for the meter.

(0.975)

5. A horizontal venturi meter measures the flow of oil of specific gravity 0.9 in a 75 mm diameter

pipe line. If the difference of pressure between the full bore and the throat tapping is 34.5 kN/m2

and the area ratio is 4, calculate the rate of flow, assuming a coefficient of discharge of 0.97.

(0.0097 m3/s)

6. Crude oil of density 850 kg/m3 flows upwards a a volume rate of 0.06 m3/s through a vertical

venturi meter which has an inlet diameter of 200 mm and a throat diameter of 100 mm. The

venturi meter coefficient is 0.98. The vertical distance between the tapping points is 300 mm,

and they are connected to two pressure gauges, calibrated in kN/m2 and positioned at the levels

of their respective tapping points.

Working from first principles (but assuming Bernoulli’s equation), determine the difference of

the readings of the two pressure gauges.

(26.7 kN/m2)

16

Blank page for you to make notes