2 Fluids First Part

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Fluids

Content of the Lecture1. The four phases of matter

2. Hydrodynamics versus Hydrodynamics3. Variables of State (intensive and extensive)4. Units and Dimensions (and energy per unit weight)5. Pressure increase and approximate relation for small kikes and dams6. Gauges and Barometers7. Steady flow8. Turbulent flow and Reynolds number9. The Continuity equation and conservation of mass

10. Bernoulli equation and the conservation of energy

11. Applications (Torricelli reservoir theorem, constricted section in ahorizontal tube, Venturei velocity meter using constricted tube andmanometer, Pitot tube velocity meter, vacuum water-pump, atomizer spraycan)

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There are four phases

Solid: small intermolecular space, definite shape & volume.

Fluid, including:

Liquid: intermediate binding forces and movement, takes theshape of the container.

Gas (adj. Gaseous): large intermolecular distance, nobinding, fills any available space.

Plasma: charged gas, large intermolecular distance, nobinding, fills any available space.

1. Phases of Matter

Fixed shape Fixed

volume

Fluidity

Fluids Gas No No YesFluids Liquid No Yes Yes

ChargedFluid

Plasma No No Yes

Solid Yes Yes No

Fluids are liquids and gases. They are generally compressible,

however, liquids are generally considered as non-compressible asan approximation.

Hydrodynamics vs hydrostatics.

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3.1 hydrostatics

3.1.1 Variables of State

- independent (basic) variables of state:

to describe the components of a system, we need to know thevalues of the following intensive and extensive variables(parameters):

- (by intensive we mean that the value of the parameter does notchange with the change of the amount of the material, whereasby extensive we mean that the value of the parameter doeschange with the change of the amount of the material).

- intensive

- Temperature: the criterion for equilibrium with respect tothermal energy transfer (K).

- Pressure: the criterion for mechanical equilibrium (N m-2known as Pascal, Pa)

- Chemical Potential: the criterion for equilibrium with respect totransfer of matter, it determines the relative tendency of thecomponent to be transformed from one phase to another (or tobe transformed into a different compound) (J kg-1 or J mol-1).

- extensive

- Entropy: the principal criterion for thermodynamic equilibrium(J K-1). The change of entropy S is defined as the transferred

heat q divided by temperature T:S =

dq

T

- Volume: it describes the spatial extent of a component (m3).

- Mass: the amount of material (kg, or the number of moles for apure component).

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Derived variables:example:density (ML-3)

= =mV

kgm

, ][ 3

pressure (M L T-2 L-2 = M T-2 L-1)

Observations show that every surface in a fluid at rest experiencesa force directed normally to it and proportional to its area.- pressure (N m-2) is the force that acts perpendicular to the unit

area of a given surface:

P FA

resure orcerea

=

We use the SI unit systemPressure is measured in N m-2

1 N.m-2 is known as 1 PascalForce = mass x acceleration

= kg x m s-2

Force is measured in Newton (N)1 N = 1 kg . m s-2

= 105 g . cm s-2

= 105 dynes(1 dyne = 1 g . cm s-2)

Pressure is reported in Pascal (N.m-2)1 N.m-2 = 1 kg m s-2 m-2

= 1 kg m-1 s-2

Pressure= Force / area= (kg m s-2) / m2

= kg m-1 s-2

= Pascal

[ ] =N

m

kg m

m s

kg

m s= pascal = Pa

2 2 2 2P =

=

also Pressure can be expressed using several units:4


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- 1 N.m-2 i.e. 1 Pascal= 10 dynes.cm-2

= 9.869 x 10-6 atm

= 10-5

bar= 10-2 mbar= 7.501 x 10-4 cm Hg= 1.0198 x 10-2 cm H2O

1 dyne.cm-2 = 0.1 Pascal1 atm = 1.013 x 106 dynes.cm-2

= 1.013 x 105 N.m-2(Pascal)= 101.3 kPa (kilo Pascal)= 1013 hPa (hundred Pascal)

1 atm approx. = 105 Pa= 0.1013 MPa= 10.13 N.cm-2

= 1.013 bar = 1013 mbar

= 76 cm Hg= 1033.3 cm H2O= 1 kg air cm-2

1.013 MPs = 10 atm1 MPs approx.= 10 atm1 bar = 106 dyne cm-2

= 105 N m-2 (Pascal) (Pa)= 103 hPa (hundred Pascal)= 102 kPa (kilo Pascal)= 0.9869 atm= 75.01 cm Hg= 1019.8 cm H2O= 10.198 m H2O

mbar = 103 dyne cm-2

= 102 N m-2 (Pascal)

= 9.869 x 10-4 atm= 0.07501 cm Hg

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= 1.0198 cm H2O= 0.010198 m H2O

1 cm H2O(at 4oC)= 980.6 dynes.cm-2

= 98.06 N m-2

(Pascal)= 0.9677 x 10-3 atm= 0.9806 x 10-3 bar= 0.9806 mbar = 0.0735 cm Hg

1 cm Hg (at 4oC) = 1.333 x 104 dynes cm2

= 1333 N m2

= 0.01316 atm= 0.01333 bar = 13.33 mbar

1 mm Hg (at 4oC)i.e. 1 torrcillie= 1333 dynes cm2

= 133.3 N m2 (Pascal)

= 0.001316 atm= 0.001333 bar = 1.333 mbar

Avoid the misuse of units forpressure and energy:

Force = mass * acceleration;Acceleration = velocity / time

= m s-2

Energy = Force * distance;(= work done)The unit used to express energy (and work) is the Joule:

= kg . m s-2 * m = kg m2 s-2 = Joule (J)= N * m

In soil water potential calculation, we need to express energy

relative to the mass, volume or weight of moisture contentcompared to free water in a dish at the soil surface:

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Energyper unitmass (kg)= (kg m2 s-2 ) / kg= J / kg

= m2 s-2

Energyper unitvolume (m3)= (kg m2 s-2) / m3

= kg m-1 s-2

= Pascal

Energyper unitweight(kg m s-2)= (kg m2 s-2) / (kg m s-2)= m

To convert between these units, all you need to know is thedensity of water, and the acceleration of gravity:assume waterdensity 1 Mg/m3 and gravity = 9.8 m s-2. Two examples:1 J / kg = 1 k Pa= 0.001MPa = 0.1 m (= 0.01 bar).100 J / kg = 100k Pa= 0.1 MPa = 10 m (= 1 bar).

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increase of pressure (over atmospheric) at a pointinside a liquid depends on:

- liquid density- depth of the point

How to calculate the value of this increase of pressure?it is equal to the weight (mg) of the liquid column acting over across section area (force divided by area):

Pressure increase= force / area= mass * acceleration / area= [(volume) * density] * g / area= [(area x height)* density) * g / area= A * h * * g / A

= h g

Or to remember easily g hThis is pressure of incompressible (uniform) fluid.

Another way to report this is to consider a tank filled with liquid,depth = h, cross-section = A. The weight of liquid is balanced bythe upward force from the bottom of the tank supporting theliquid (Newtons third law):

weight of liquid = reaction

A * h * * g = P * Ah * * g = Pg h = P

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Approximate relation for small dikes and dams:

the average of pressure at the water column half depth= g h

so the total force on the exposed area of the dam body= g h * area

Example 1:Find pressure on the ocean floor at a depth of 3 km.Solution:

P g h

m

s

kg

m m

kg

m s

= =

=

=

9 80 100 10 3 0 10

2 94 10 29 4 300

2

3

3

3

7

2

. . .

. . MPa atm

Example 2:Estimate the height of the Earths atmosphere,(use the approximation, 1 atm = 105 Pa).Solution: We use the barometric formula in a crude

approximation that the air density is uniform vertically.

Air density at STP (atmospheric pressure and normal temperature)is about 1 kg.m-3, and atmospheric pressure ~105 Pa=105 kg.m-1s-2,so:

P gh h

h

=

1kg

m9.80

m

sm

kg

m s

m = 10 km

3 2 210

10

5

4

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Gauges and the Barometer

Absolute pressure (measured above vacuum pressure)P:P Po Pg= +

Gauge pressure (measured above or below atmospheric pressure):Pg P Po=

P0 is the atmospheric pressure

Pg is an excess (gauge) pressure additional to the atmosphericpressure (it could have a negative sign, as it is the case when weuse a vacuum pump).

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2 Fluids First Part