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Environmental Physics
Chapter 4:
Solids, Liquids and Gases
Copyright © 2007 by DBS
Concepts
• The physical states of matter – solids, liquids, gases and plasma
- are determined by intermolecular forces, dependent upon
temperature and pressure
• To change state involves exchange of energy and may also
need ‘seeding’
• Strength and elastcity are important characteristics of solids,
which may be applied to rocks
• In a gas, pressure, temperature and volume are closely related,
giving rise to the ‘gas laws’ which are important in describing
energy flows in the atmosphere
• Fluid dynamics, the study of movements of fluids, is applicable
to a wide range of environmental systems
States of Matter
The physical characteristics of matter contained in
environmental systems determine many of the key processes
and flows within that system
States of Matter
States of Matter
fluids
Solids
•
little thermal energy
•
rigidly bonded together
•
regular pattern
•
may be plastic or rigid
Temp.
Liquids
•
particles flow
•
held close together
•
no regular arrangement
Plasma
•
similar to a gas but higher energy,
composed of isolated e- and nuclei rather
than discrete atoms or molecules
http://www.chem.purdue.edu/gchelp/atoms/states.html
Gases
•
individual molecules
•
well separated
•
move at high speeds
Water
Question
vapor
What are the common names for the three states of water?
Spot the
mistake!
Question
Complete the following:
following
State
Shape
Volume
Solid
Definite
Definite
Liquid
Indefinite
Definite
Gas
Indefinite
Indefinite
106 K
Source: http://ds9.ssl.berkeley.edu/themis/mission_sunearth.html
States of Matter
• Aurora borealis – plasma
– interaction of the solar
wind (plasma) with the
Earth’s atmosphere and
magnetic field
http://www.om3rkp.cq.sk/articles.php?lng=en&pg=220
States of Matter
The Structure of the Earth
Not a solid!
P
On cooling lighter elements
moved to surface
•
•
•
•
•
Lithosphere – rigid
Asthenosphere – plastic/molten
Lower mantle – solid
Outer core – liquid
Inner core – solid (under
pressure)
Core
Mantle
Crust
Question
Label the diagram below.
1.
2.
3.
4.
5.
6.
7.
Outer core
Crust
Mantle
Lithosphere
Inner core
Asthenosphere
Lithosphere
States of Matter
Solutions and Other Mixtures
•
Solution
–
•
Suspension
–
–
•
–
–
Suspended mixture of insoluble liquids
(e.g. oil and water)
Brought about under specific chemical/physical
conditions
Settle out over time
Aerosol
–
•
Small solid particles are mixed in a liquid
Can be separated by filtering or centrifuge
Emulsion
–
•
Molecules or ions of one material are evenly spread
throughout a liquid
Liquid water as tiny droplets too small to fall as rain
Saturation point
–
–
–
Maximum amount of material that can be dissolved in
another
Solids - Increases with temperature e.g. sugar
Gases in liquids - Decreases with temperature
States of Matter
Changes of State
•
Change of state
– Due to changes in temperature
and pressure
– Intermolecular bonds
•
•
•
Ice is unusual as it expands
when it freezes
Water crystallizes into an open
hexagonal form containing more
space than the liquid state
Compressing ice tends to melt
the ice (mpt. decreases)
States of Matter
Changes of State
4° density
maximum
Balance between
breakup of icelike clusters (> ρ)
and thermal
expansion (< ρ)
~ 1 kg L-1
Expansion
Controlled by Hbonding
ρ↓
contraction
expansion
2.2 Matter
Has States
Deposition
Add phase
• States
of
matter
change
terms
Melting
Evaporation
Freezing
Condensation
Sublimation
Demo
• Sublimation
States of Matter
Cloud Seeding and Condensation Nuclei
•
Condenstaion nuclei (CN)
– Gases condense more readily
– Liquids solidify more readily
•
Supersaturated
– Cooling a vapor/gas mixture to
below saturation point
•
Supercooled
– Cooling a pure liquid to below
freezing point
contrail
Demo: Smog in a Bottle
•
Large bottle, matches
•
•
•
•
•
Swirl hot water to coat inside walls of bottle
Light a match and add to bottle
Compress sides to increase the pressure
Release
As the temperature inside the bottle reduces water vapor condenses
States of Matter
Molecular Diffusion
•
•
Random thermal motion
In most cases turbulent
mixing and
pressure/convection
currrents predominate
•
Diffusion is slow
http://www.ap.stmarys.ca/demos/content/thermodynamics/brownian_motion/brownian_motion.html
States of Matter
Molecular Diffusion
•
Fick’s Law- Depends on concentration gradient
Over short
distances and
for low C
Q = kD ΔC / Δz
~ kD (C2-C1)/(z2-z1)
Where Q = flux of material (kg m-2 s-1), ΔC / Δ z = concentration
gradient, kD = diffusivity coefficient (m2 s-1)
C(z) = C0exp[-1/2(z/σ)2]
Where C0 = concentration at source, z = distance from source, σ =
typical distance diffused after a certain time
•
Flux declines with distance exponentially
Question
Consider a local concentration of 106 mg per cm3 which drops by
a factor of 10 over a distance of 1 cm. Assume the diffusivity is
0.1 m2 s-1. Calculate the flux.
Q = kD ΔC / Δz
Q = (0.1 m2 s-1)[-(106 mg cm-3)/10] / 1 cm
Q = (0.1 m2 s-1)[-(1012 mg m-3)/10] / 0.01 m
Q = (0.1 m2 s-1)(-1011 mg m-2) / 0.01 m = 1012 mg m-2 s-1
End
• Review
Properties of Solids
•
Solids are characterized by:
–
–
–
•
Shape
Strength – tensile, compressive, shear
Resistance to external stress
Properties determine material use
Properties of Solids
Elasicity, Stress and Strain
•
•
Stress – internal forces reacting to applied load (= force/Area)
– Compressive, tensile, shear
Strain – geometrical expression of deformation caused by stress
(= extension/length)
Young’s Modulus (stiffness):
E = stress / strain
E = f/A
e/l
•
•
•
Also known as modulus of elasticity
For certain materials stress ~ strain
Strain due to a given stress can be calculated
Question
What are the units of Young’s modulus?
Kg m s-2
m2
N m-2
= Pascals
Material
E (GPa)
Rubber
0.1
Polystyrene
3
Wood
11
Aluminum
69
Titanium
105
Steel
200
Diamond
1000
flexible
stiff
Questions
(i) Is there a relationship
between E and ρ?
(ii) Why are ceramics, such as
silicon carbide, used for
machine tool tips?
Properties of Solids
Elasicity, Stress and Strain
Bulk modulus, K = volumetric stress
volumetric strain
(resistance to compression)
Extension of E to 3-D
(should be low for solids)
Shear modulus, μ = shear stress
shear strain
Deformation of shape at constant volume
Properties of Solids
Poisson’s Ratio
•
•
When a solid is stretched in one
direction it tends to flatten and broaden
in the other direction
Poisson’s ratio,
σ = (ΔB/B)
(Δ L/L)
Material
Good for wine
bottles!
σ
Cork
~0
Steel
0.3
Rubber
0.5
Properties of Solids
Elasicity, Stress and Strain
•
•
Bulk modulus, K = bulk stress
bulk strain
K=
E(1 – σ)
3(1 + σ)(1 - 2σ)
Shear modulus, μ = shear stress
shear strain
Only applicable over elastic
deformation range
(material returns to original
shape after stress is
removed)
μ = ½ E/(1 + σ)
(=0 in fluids, no resistance to shape changes)
Properties of Solids
Elasicity, Stress and Strain
•
•
Elasticity - ability of a material to return to it previous shape after
stress is released
Plasticity – opposite of elasticity -past the elastic limit plastic
deformation occurs
– Molecular bonds break
– Energy is released as heat
– Material reaches breaking point
Slope = E
Properties of Solids
Strength
•
•
Strength measures the resistance of a material to failure, given by the
applied stress (or force per unit area)
Depends on shape of curve
Will vary with
temperature and purity,
e.g. C added to steel
Failure due to metal
fatique, exposure to UV
or ionizing radiation
Question
Sketch stress-strain curves for rubber, silly putty and china.
Rubber – elastic over a large range
Silly putty – plastic over a large range
China – Elastic then plastic over very short range (breaks)
Properties of Solids
Stress and Strain in Rocks
If forces exceed
plastic limit rock will
break – fault line
Over long timescales
solid rock can bend
and be uplifted
leads to distorted
strata with
discontinuity
Properties of Solids
Seismic Waves
•
P and S waves
End
• Review
Pressure
•
Pressure in a fluid is due to the weight of fluid above pressing down
P = f /A
Where f = force (N), A = area (m2)
•
Pressure on an area A at depth h is
due to the weight of a column of fluid
above it,
volume = Ah
mass = Ahρ
Pressure = f / A = mg / A = Ahρg /A
=hρg
•
Atmospheric pressure is due to the height of the column of air above us
Pressure
•
Average atmospheric pressure =
1.013 x 105 Pa = 1 atmosphere
(at sea level)
Range: 0.9 x 105 Pa - 1.1 x 105 Pa
•
Decreases exponentially with height
(air gets thinner)
•
Since density decreases the linear
relationship p = hρg does not hold
•
Major difference - Air is compressible,
unlike water
Pressure
Mercury Barometer
•
Since mercury is a fluid
P = hρg
•
Where = density of mercury = 13600 kg m-3
h=p/ρg
h = 1.013 x 105 Pa / (13600 x 9.8 m s-2)
h = 0.76 m
So 1 atm = 760 mm Hg
Question
What is the pressure (in atm.) at 10 m depth of water?
What is the pressure at 20 and 30 m depth?
(ρ = 1000 kg = m-3)
P = hρg = 10 x 1000 x 10 = 1 x 105 Pa = 1 atm
P = hρg = 20 x 1000 x 10 = 2 x 105 Pa = 2 atm
P = hρg = 20 x 1000 x 10 = 3 x 105 Pa = 3 atm
Every 10 m increase in depth increases P by 1 atm
Pressure
Archimedes’ Principle
•
•
When a body is submerged in a fluid, there
is an upward force equal to the weight of
fluid displaced
Archimedes' principle
– The apparent weight of an object
immersed in a fluid will be smaller than
its “true” weight
Wobject in water = Win air - W displaced water
•
Pressure at the lower surface is greater than
up top – results in net upwards force
•
Bouyancy depends on density
Question
An object weighs 36 g in air and has a volume of 8.0 cm3. What
will be its apparent weight when immersed in water?
When immersed in water, the object is buoyed up by the mass of
the water it displaces, the mass of 8 cm3 of water
Taking the density of water as 1 g cm-3, the upward (buoyancy)
force is 8 g
The apparent weight = 36 g – 8 g = 28 g
Pressure
Flotation
• Objects float if upthrust from fluid > objects weight
• i.e. density < density of fluid
• An iceberg has density 0.9 g cm-3 and is floating with
V1 of its volume above and V2 of its volume below
water
• Sketch a diagram showing the iceberg
Pressure
Flotation
V1
Volume of ice below water:
V = V1 + V2
ρice
V2
ρwater
Weight of ice:
W = mg = Vρiceg
By Archimedes principle: W = mg = V2ρwaterg
Balance forces (flotation):
V2ρwaterg = Vρiceg
V2 = V(ρice/ρwater)
9/10 of iceberg is
below water
= V (0.9 g cm-3/1 g cm-3) = 0.9V
Pressure
Flotation
• What happens when the iceberg melts?
• Meltwater has the same density as liquid water so the volume
contracts (ρice < ρwater)
After melting,
Vmelt = W / (ρwaterg)
Substituting for W, the upwards bouyant force: W = V2ρwaterg
Vmelt = V2ρwaterg / (ρwaterg)
Vmelt = V2
Same volume as ice before melting,
top vanishes, effect on water level is zero
Pressure
Flotation
• Any melting of floating ice in the Artic (sea ice) due to global
warming has no net effect on global sea levels
• Only land ice has an effect (glaciers)
– Is this true???
– Something has been overlooked!
Pressure
Flotation
• Is the density of sea water 1 g cm-3 = 1000 kg m-3?
No!
It is actually 1026 kg m-3
Vmelt = V2ρsea waterg / (ρwaterg)
Vmelt = V2(1026/1000) = 1.026 V2
Volume of meltwater is 2.6 % greater than the displaced
seawater, water levels will rise
Ice displaces its own weight of the underlying liquid, it does not
displace the same volume as the meltwater
End
• Review
Gases
•
•
•
Gases are compressible and have lower density than water
Pressure (P), Temperature (T), and Volume (V) are more closely
related than for solids or liquids
Understanding gas behavior is important in climatology and air
pollution studies
Gases
The Gas Laws
•
At constant T (Kelvin), V is halved when P is doubled
Boyle’s Law: P ~ 1/V or PV = constant
•
At constant P, if T increases then V increases
Charles’ Law: V ~ T or V/T = constant
•
Pressure increases,
particles move closer
together
At constant V, if T increases then P increases
Gay-Loussac’s Law: T ~ P or P / T = constant
KE increases, particles
spread out, increased V
KE increases, particles
collide with the container
walls more rapidly,
exerting increased P
Gases
The Gas Laws
Question
The temperature of a balloon at room temperature (27 ºC) is
increased by 1 ºC. What is the new volume?
V1/300 = k
V2/301 = k
V1/300 = V2/301
V2 = 301V1 / 300
V2 = (300/300 + 1/300)V1
V2 = (1 + 1/300)V1 = V1 + 1/300V1
Gases
The Gas Laws
•
Three laws are combined to give the equation of state:
PV = r where r = constant
T
Where V = volume (m3), P = pressure (Pa), T = temperature (K),
r = 4.16 J kg-1 K-1
The ideal gas law is most accurate for monoatomic gases and is
favored at high temperatures and low pressures. It does not factor
in the size of each gas molecule or the effects of intermolecular
attraction
Gases
The Gas Laws
•
Gases are often measured in terms of molar volume.
One mole of any gas occupies the same volume
Avogadro’s hypothesis: V ~ n or V / n = k
Where n = no. moles and 1 mole = 6.022 x 1023 molecules
•
Equation of state now becomes:
PV = nRT
Where n = no. moles and R = molar gas constant = 8.31 J mol-1 K-1
•
1 mole of any gas occupies 22. 4 L at 1 atm. 273 K
Gases
Conversion to standard temperature and pressure
•
Since volume changes with T and P it is useful to state the volume
occupied at STP (standard temperature and pressure)
P1V1 = P2V2 = r
T1
T2
Where subscripts 1 and 2 refer to values before and after
Question
An industrial plant emits 3000 m3 of gas per hour containing
0.5 % NO2 at T = 77 ºC and P = 1 atm.
Find the volume of NO2 emitted.
Convert to STP:
V2 = P1V1T2 / T1P2
V2 = (1)V1T2 / T1(1) = V1T2 / T1 = 3000 x 273 / 350
= 2340 m3 in one hour = 2340 m3 h-1
Volume of NO2 emitted is 0.5% of 2340 = 11.7 m3 h-1
Gases
Work Done
•
When a gas expands it does work using energy supplied
by the change in temperature
e.g. car engine – energy from gasoline heats cylinder and
expands the gas inside: heat → movement
E = P ΔV
Where E = Energy (J), P = Pressure (N m-2) and V =
Volume (m3)
•
Compressed gases can store energy and be used as a
fuel
E=fxd
1Nm=1J
Gases
Work Done
•
Force exerted on piston
F=P.A
•
Piston moves Δx, what is work done
on the gas, W?
W = P . A . Δx
•
What is ΔV?
ΔV = A . Δx
•
Express work done:
W = PΔV
Question
A cylinder is at pressure of 6000 Pa and a piston has an area of
0.1m2. As heat is slowly added the piston is pushed up a distance
of 5 cm.
Calculate the work done on the surroundings assuming constant
pressure.
E = P ΔV
E = 6000 Pa x (0.1 m2 x 0.05 m)
E = 6000 Pa x 5 x 10-3 m3 = 30 N m-2 x m3 = 30 N m = 30 J
Gases
Heat Capacities of Gases
•
When a gas heats up the temperature rise depends on
– Heat capacity (c) (similar to solids and liquids)
But also:
– Expansion (volume change)
– Pressure
•
Not possible to define simple heat capacity for a gas, need two!
Gases
Heat Capacities of Gases
•
Specific heat capacity at constant volume, Cv (J
mol-1
K-1)
All heat
applied goes
into inc. T
Q = nCv ΔT
Energy required to raise the temperature of 1 mole of gas by 1 K in
a closed container (constant V)
•
Specific heat capacity at constant pressure, Cp (J mol-1 K-1)
Heat applied
goes into inc.
T and
expanding
Q = nCp ΔT
Energy required to raise the temperature of 1 mole of gas by 1 K at
constant pressure
Gases
Heat Capacities of Gases
•
According to first law of thermodynamics
ΔU = Q – W
•
Where U = change in internal energy, Q = heat added to system, W = work
done by the system (PΔV)
ΔU + PΔV = nCp ΔT
•
From the ideal gas law (PV = nRT) under constant pressure conditions:
PΔV = nRΔT
ΔU + nR ΔT = nCp ΔT
•
Since Cv = ΔU/ nΔT (same as above but ΔV = 0)
Cp = Cv + R
Cp > Cv
Takes more energy
since energy is used
in expansion
Gases
Adiabatic and Isothermal Expansion
•
Gas expands at constant T = isothermal
PV = constant
Work is being done in expanding the gas so
to keep T constant heat (E = PΔV) must be
supplied, e.g. squeezing a balloon
•
If no external energy input = adiabatic
PVγ = constant
Where γ = Cp/Cv
e.g. release of gas from
pressurized cylinder
cools the cylinder
Gases
Adiabatic Cooling
•
A body of warm, less dense
air rises in the atmosphere
– Pressure decreases
– Body expands
– No source of external
energy
– Body cools as it does
work to equilibrium point
•
Adiabatic lapse rate –
temperatures decrease with
height
– For dry air
DALR = - 9.8 ºC km-1
– For moist air
MALR = - 6.5 ºC km-1
LH release from
condensation decreases
MALR
Question
Question
End
• Review
Fluids and Fluid Flow
• Applications
– Groundwater flow
– Sedimentation
– Water supply
– Air and water pollutant distribution
– Design of wind turbines
– Reduction of drag in vehicles
– Flight of birds and aircraft
Fluids and Fluid Flow
Characterizing flow
• Viscosity
– Measure of how treacly a liquid is
– ‘fluid friction’
– Newtonian fluid - viscosity increases linearly with rate of
change of velocity
• Compressibility
– How much a fluid can be squashed
– Common liquids are virtually incompressible
– Gases are compressible
Fluids and Fluid Flow
Characterizing flow
• Reynolds number
– Ratio of inertial (ρv) to viscous forces (μ/l)
ρvl
μ
Re used in
modelling flow
Where ρ = density, μ = viscosity, v = velocity, l = length
– Useful in physical modeling, at the same Re two fluids
behave in the same way
Question
A scale model has linear dimensions one quarter of full size,
how should the flow velocity be increased to show the same
behavior?
Re = ρvl
μ
Re1 = Re2
ρv1l1 / μ = ρv2(l1/4) / μ
v2 = 4v1
If l is reduced by a factor of 4 v must be increased by a
factor of 4
Fluids and Fluid Flow
Laminar and Turbulent Flow
•
Laminar flow
– Uniform and orderly
– Occurs at low Reynolds
numbers, where viscous
forces are dominant
– Flow rate easy to calculate
•
Turbulent flow
– Random eddies, vortices and
other flow fluctuations
– Occurs at high Reynolds
numbers and is dominated by
inertial forces
– Flow rate difficult to calculate
Fluids and Fluid Flow
Laminar and Turbulent Flow
The transition between laminar
and turbulent flow is often
indicated by a critical Reynolds
number
Flat surface Recritcial = 5 x 10-5
Pipe Recritcial = 2200
Atmosphere
Re ~ 109-1012
http://www.ae.su.oz.au/aero/fprops/pipeflow/node7.html
Rapid dispersal
of pollutants
Fluids and Fluid Flow
Measuring Flow Rates
•
•
•
•
Flow rate = amount of fluid passing one point at any time (m3 s-1)
Flow velocity = speed of flow (m s-1)
Flow rate for a small stream can be measured with a timed bucket fill,
not practical for larger bodies
– Requires conversion from flow velocity (easier to measure) to
flow rate
Volume passing a point in one second equals the area multiplied by
the distance travelled downstream in one second
Q=vA
Where v = velocity (m s-1), Q = flow rate (m3 s-1) and A = crosssectional area (m2)
Question
How would you measure flow velocity?
With a float and time or a current meter
(looks like an anemometer)
Fluids and Fluid Flow
Measuring Flow Rates
• At constant Q,
decreasing cross-sectional area increases velocity
decrease A
v increases
Principle of continuity of flow
– the same amount of water passes each point in one second anywhere down its length
Fluids and Fluid Flow
Laminar Flow in a Pipe
Fluids and Fluid Flow
Laminar Flow in a Pipe
•
•
One of the few types of flow that can be described by simple equation
Flow is modelled assuming thin layer at wall does not move, water
closer to center moves rapidly
Flow rate = p1 – p2 = π pr 4
R
8ηL
Poiseuille’s Law
•
Where r = radius of pipe, η = viscosity, p = pressure difference, L =
length
Not valid for turbulent flow or large velocity and pressures
Question
Suppose the flow rate is 100 cm3 s-1. What are the effects of
doubling each of the parameters?
Double length → 50 cm3 s-1
Double viscosity → 50 cm3 s-1
Double pressure → 200 cm3 s-1
Double radius → 1600 cm3 s-1
Fluids and Fluid Flow
Bernoulli’s Principle
•
Derived from Law Conservation of Energy - Sum of all forms of
energy in a fluid flowing along an enclosed path is always constant
PV + ½ m v2 + mgh = constant
PV + ½ m v2 + mgh = constant
V
V
V
P + ½ ρ v2 + ρgh = constant
•
•
Where P = pressure, ρ = density, v = velocity, g = gravity and h =
height
Law conservation of energy – a reduction in energy due to a change
in either p, KE or PE must be compensated by an increase in another
Fluids and Fluid Flow
Bernoulli’s Principle
• Pressure decreases when velocity is increased
http://www.wfu.edu/physics/demolabs/demos/avimov/bychptr/chptr4_matter.htm
Fluids and Fluid Flow
Venturi Effect
• Reduction of pressure when velocity is increased
Used to
measure flow
Fluids and Fluid Flow
Aerodynamics and Aerofoils
Newtonian explanation
• Lift occurs when a moving flow of gas is turned by a solid object
– Equal and opposite force acts from below
Bernoulli explanation
• As air travels at different velocities around a solid pressure changes,
adding up the pressure changes x area of wing gives lifting force
Cannot be based on
shape of wing:
Planes fly upside down
Paper planes fly
Fluids and Fluid Flow
Atmospheric Transport
• Dilute and disperse – high stacks
• Temporal and spatial scales are large
– Horizontal motion - wind
– Vertical motion - convection and
turbulence
• Pollution that reaches the stratosphere
may not be removed for 100’s of years
Fluids and Fluid Flow
Atmospheric Transport
•
•
•
Gaussian plume model
Assumes pollution drifts with wind,
spreading out at a rate dependent
on turbulent diffusion and
concentration
Depends on height, wind speed,
surface roughness and
atmospheric effects
Fluids and Fluid Flow
Atmospheric Transport
r
H
z
A
B
Q = rate of emission (kg s-1),
u = average wind speed,
P = concentration (kg m-3)
H
P
B = spread out, dec. P
Fluids and Fluid Flow
Atmospheric Transport
• Amount of pollution passing point A = amount emitted
Q = π r2 u P
• Plume hits ground when r = H, when plume hits the ground:
Pmax =
• To minimize P
– Reduce Q
– Increase H (higher stack)
– Increase u (windy sites)
Q
π u H2
More likely to have pollution
in middle of plume
Fluids and Fluid Flow
Plumes
•
Emitted gas tempeature is higher than
surroundings, initial movement is upwards
Pmax =
NQ
u(H + h)2
Where N is typically 0.15
•
•
Higher stacks reach higher level winds
(less turbulence)
Pollutant may travel 100-1000s km before
being washed out
Fluids and Fluid Flow
Plumes
(a) inversion
coning
(b) weak lapse
(c) weak lapse below,
inversion aloft
trapping
(d) lapse below,
inversion above
(e) Inversion below,
lapse aloft
fummigation
lofting
Plume shape depends on
stability of atmosphere
fanning
Fluids and Fluid Flow
Surface Tension
At the surface of the
liquid, the molecules
are pulled inwards
but they are not
attracted as
intensely by the
molecules in the air
Therefore all of the molecules at
the surface are subject to an net
inward force
Results in relatively strong
surface of water
Each molecule is pulled equally
in all directions by neighboring
molecules, resulting in a net
force of zero
Fluids and Fluid Flow
Surface Tension
•
•
Explains roundness of droplets
(smallest surface area)
Shapes meniscus
Force circumference = gravity
Force = 2 π r γ = mg
2 π r γ = π r2hρg
2 γ = rhρg
h = 2 γ / rρg
h ~ 15 / R
End
• Review
Hydrology
•
h
Hydrology and Hydrogeology
Hydrological Processes
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Hydrology and Hydrogeology
Darcy’s Law
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Hydrology and Hydrogeology
Groundwater Flow
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Books
• Jones, A.M. (1997) Environmental Biology. Routledge, London.
p124.
• Price, M. (1985) Introducing Groundwater. Chapman & Hall,
London.
• Seinfeld, J., and Pyros, S.N. (1998) Atmospheric Chemistry and
Physics: From Air Pollution to Climate Change. John Wiley, New
York.
Further Reading
•
Eastlake, C.N. (2002) An Aerodynamicist's View of Lift, Bernoulli, and
Newton. The Physics Teacher, Vol. 40, pp.166 .
• Noerdlinger, P.D. and Brower, K.R. (200?) The Melting of
Floating Ice Raises the Ocean Level. Unpublished article.
• Rosenfeld, D. (2000) Supression of rain and snow by urban and
industrial air pollution. Science, Vol. 287, pp. 1793.