chapter 5 . air pollution meteorology
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Chapter 5 . Air Pollution Meteorology. Selami DEMİR Asst. Prof. Outline. Introduction Solar Radiation Atmospheric Pressure Lapse rate & Potential Temperature Atmospheric Stability Coriolis Force & Gravitational Force Pressure Gradient Force Overall Atmospheric Motion - PowerPoint PPT PresentationTRANSCRIPT
Chapter 5. Air Pollution Meteorology
Selami DEMİRAsst. Prof.
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Outline Introduction Solar Radiation Atmospheric Pressure Lapse rate & Potential Temperature Atmospheric Stability Coriolis Force & Gravitational Force Pressure Gradient Force Overall Atmospheric Motion Equations of Motion Wind Speed Profile
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Introduction (1/2)
Air pollutant cycle Emission Transport, diffusion, and transformation Deposition Re-insertion
In large urban areas, there are several concentrated pollutant sources All sources contribute to pollution at any specific site Determined by mainly meteorological conditions Dispersion patterns must be established Need for mathematical models and meteorological input data
for models
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Introduction (2/2)
Three dominant dispersion mechanisms General mean air motion that transport pollutants downwind Turbulent velocity fluctuations that disperse pollutants in all
directions Diffusion due to concentration gradients
This chapter is devoted to meteorological fundamentals for air pollution modelling
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Solar Radiation (1/6)
Solar constant 8.16 J/cm2.min 0.4-0.8 µ visible range, maximum intensity
Ref: http://www.globalwarmingart.com/images/4/4c/Solar_Spectrum.png
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Solar Radiation (2/6)
Distribution of solar energy on earth
Ref: OpenLearn Web Site, http://openlearn.open.ac.uk/file.php/1697/t206b1c01f26.jpg
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Solar Radiation (3/6)
At right angle on June, 21 Tropic of cancer At right angle on December, 21 Tropic of capricorn At right angle on March, 21 and september, 21 Equator
http://upload.wikimedia.org/wikipedia/commons/8/84/Earth-lighting-equinox_EN.png
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Solar Radiation (4/6)
Example: What is the Sun’s angle over Istanbul on June, 21? Note that Istanbul is located on 40° N latitude.
Solution: Sunlight reaches Tropic of Cancer (23° 27′) at right angle on June, 21.
Where θ = Sun’s angle at the given latitudeL2 = Latitude of given region
L1 = Latitude of region where sunlight reaches surface at right angle
1290 LL
'' 277327234090
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Solar Radiation (5/6)
Example: What is the Sun’s angle over a city located on 39° N latitude when the sunlight reaches surface at right angle on 21° S latitude?
Solution:
30213990
90 12
LL
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Solar Radiation (6/6)
Homework (due 18.04.2008) Make a brief research on Stefan-Boltzman Law and write a one
page report for your research. Comment on what would happen if earth’s inclination were 24°
instead of 23°27′. What determines the seasons? Why some regions of earth get
warmer than other regions. Calculate the sunlight angle over Istanbul
on March, 21 on June, 21 on September, 21 on December, 21
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Atmospheric Pressure (1/4)
Force on earth surface due to the weight of the atmosphere Defined as force exerted per unit surface area Units of measurement Pascal (Pa), atmospheric pressure
unit (apu, atm), newtons per meter-squared (N/m2), water column (m H2O), etc. 1 atm = 101325 Pa 1 atm = 10.33 m H2O 1 atm = 760 mm Hg 1 Pa = 1 N/m2
Atmospheric pressure at sea level is 1 atm
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Atmospheric Pressure (2/4)
Consider a stationary air parcel as shown Force balance (assuming no horizontal
pressure gradient)
gdhdP
ghPg
hP
hgPhgAhAgVgmgPA
mgPAAPP
GAPPAPFF
hh
Net
00limlim
0
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Atmospheric Pressure (3/4)
Integrating from h = z0 to h = z produces
00 exp
;
00
zzgRTMzPzP
gdhRTM
PdP
gdhRTM
PdPg
RTPM
dhdP
RTPMg
dhdP
A
z
z
A
zP
zP
AA
A
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Atmospheric Pressure (4/4)
Homework (due 18.04.2008) Make a research about pressure measurement devices and
prepare a one-page report for your research. Give brief explanations for each type.
Calculate the atmospheric pressure on top of Everest if it is 1013 mb at sea level.
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Lapse Rate & Potential Temperature (1/5)
Adiabatic no heat exchange with surroundings
Consider an air parcel moving upward so rapidly that it experiences no heat exchange with surrounding atmosphere
Enthalpy change:where
H1 = initial enthalpy of air parcelH2 = final enthalpy of air parcel
U1 = initial internal energyU2 = final internal energy
V1 = initial volume
V2 = final volume
12 HHH
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Enthalpy change is a function of only temperature when pressure is constant
Substituting differential pressure as follows:
Since the process is adiabatic, no heat exchange occurs
Lapse Rate & Potential Temperature (2/5)
By enthalpy’s definition
In infinitesimal expression
Internal energy substitution
By internal energy definition
pVUH
pVUpVUH
12
pdVVdpdUpVddUdH
pdVVdpWQddH
VdpdQpdVVdppdVdQdH
dTCVdpdQdH p
dTCgVdhdQ p
mC
CgV
dhdT
dTCgVdh
p
p
10098.0
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Lapse Rate & Potential Temperature (3/5)
This approximation assumed there is no phase change in the air parcel called Dry Adiabatic Lapse Rate (DALR)
If any phase change takes place during the motion, the temperature change will be far more different from DALR Called Saturated (Wet) Adiabatic Lapse Rate (SALR, WALR) Variable, must be calculated for each case Also significant in some cases; this course does not focus on it
For standardization purposes, Standard Lapse Rate (SLR), also known as Normal Lapse Rate (NLR), has been defined On average, in middle latitude, temperature changes from 1°C
to -56.7°C SLR = -0.66°C/100 m
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Lapse Rate & Potential Temperature (4/5)
Lapse rate measurements are taken by a device called Radiosonde Results of measurements are plotted to obtain Environmental
Lapse Rate (ELR) ELR is real atmospheric lapse rate Another significant concept is Potential Temperature
Defined as possible ground level temperature of an air parcel at a given altitude
HDALRTTp * whereθ = Tp = potential temperature of air parcelT = Temperature of air parcelH = Height of air parcel from groundDALR = Dry adiabatic lapse rate
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Lapse Rate & Potential Temperature (5/5)
Homework (due 18.04.2008) Calculate potential temperature for given data
Calculate the atmospheric temperature at 800 m from the ground if the atmosphere shows adiabatic characteristic and the ground level temperature is 12°C.
Height, m Temperature, °C
350 8
750 2
1200 14
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Atmospheric Stability (1/8)
If ELR < DALR Then Superadiabatic meaning unstable
ElseIf ELR = DALR Then Neutral
ElseIf DALR < ELR < 0 Then Subadiabatic meaning stable (weakly stable)
ElseIf DALR < 0 < ELR Then Inversion meaning strongly stable
EndIf
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Atmospheric Stability (2/8)
Superadiabatic
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Atmospheric Stability (3/8)
Neutral
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Atmospheric Stability (4/8)
Subadiabatic
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Atmospheric Stability (5/8)
Inversion
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Atmospheric Stability (6/8)
If dθ/dz < 0 Then Superadiabatic
ElseIf dθ/dz = 0 Then Neutral
ElseIf dθ/dz > 0 Then Subadiabatic
EndIf
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Atmospheric Stability (7/8)
Example: Calculate vertical temperature gradient and comment on atmospheric stability condition if the atmospheric temperature at 835 m is 12 °C when the ground temperature is 25 °C.
Solution:
The atmosphere is said to be unstable since ELR < DALR m
CmC
mC
zTT
dzdTELR groundaloft
10056.10156.0
08352512
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Atmospheric Stability (8/8)
Homework (due 25.04.2008) Following measurements are taken over Istanbul at different
times. Determine atmospheric stability condition for each case.
Briefly explain stable air, unstable air, neutral air and inversion. Make a brief research about the role of atmospheric stability in
dispersion of pollutants in the atmosphere and prepare a-one-page report for your research.
What is conditional stability? Explain.
Height, m
Temperature, °C
Case 1 Case 2 Case 3 Case 4
0 14 22 17 4
1000 8 8 7 6
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Coriolis Force
“The Coriolis effect is an apparent deflection of moving objects from a straight path when they are viewed from a rotating frame of reference. Coriolis effect is caused by the Coriolis force, which appears in the equation of motion of an object in a rotating frame of reference.” (Wikipedia Web Site, http://en.wikipedia.org/wiki/Coriolis_Force)
xvm
Ff
xvmF
Corioliscoriolis
Coriolis
2
2
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Gravitational Force (1/3)
The force exerted by the earth on an object in earth’s attraction range
Caused by attraction forces between two masses
m1 being the mass of earth (M) and m2 is that of an object near earth surface
321
rrmmFA
FA = attraction forceγ = 6.668*10-11 Nm2/kg2
m1,m2 = objects’ massesr = distance bw masses
kgNf
rrM
mFf
rrMmF
G
GGG
81.9
33
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Gravitational Force (2/3)
Example: Determine the acceleration of an object near the Erath’s surface due to gravitational attraction force
Solution:
2
3
242
211
3
sec85.9
11000*6360*
11000*6360
10*9736.5*10*668.6
mf
kmmkm
kmmkm
kgkgNm
rrMf
G
G
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Gravitational Force (3/3)
Homework (due 25.04.2008) Determine the acceleration of an object near the Martian surface
due to gravitational attraction force Determine the acceleration of an object near the Moon’s surface
due to gravitational attraction force
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Pressure Gradient Force
Consider an air parcel accelerating in a horizontal direction
In three dimensional representation,
xPf
xP
VPA
mPA
mF
f
PAF
APPPAFF
X
X
X
X
P
PP
P
Net
1
zPf
yPf
Z
Y
P
P
1
1
PfP
1
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Overall Atmospheric Motion (1/7)
Consider an air parsel accelerating around the Earth Overall acceleration
fxdtfd
dtfdNote
ar
rMPa
aaaa
rotatingfixed
fNet
fGPGNet
:
13
rxxVxdtVd
dtVda
rotatingfixedfixedNet
2
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Overall Atmospheric Motion (2/7)
Neglecting vertical terms and re-arranging, we get
SinfxP
fv
yP
fu
jxP
fi
yP
fV
2
1
1
11
u = velocity of atmospheric motion in east-west direction
v = velocity of atmospheric motion in north-south direction
Ω = rotational speed of earth = 7.29*10-5 r/s
Φ = latitude on which the motion occurs
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Overall Atmospheric Motion (3/7)
Example: Briefly explain the mechanisms that forced radioactive pollutants towards Turkey’s coasts after Chernobyl. Tell about the meteorological conditions then. Show the pressure centers and wind patterns on the day of accident and two day after the accident on a brief map. Consider the aspects of geostrophic winds.
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Overall Atmospheric Motion (4/7)
Solution
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Overall Atmospheric Motion (5/7)
Example: Isobars are shown in the figure below, for 40° latitude in the Northern Hemisphere, at an altitude of 5600 meters. Determine the geostrophic wind speed in km/hour
Temperature at 5600 m : -28°C Coriolis force: 2 Ω V sin β Ω= 7.3 x 10-5 radians/s; β= Latitude degrees ; V= geostrophic
wind speed 1 mb = 100 N/m3
180 km
500 mb
504 mb
N
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Overall Atmospheric Motion (6/7)
Example:Suppose a nuclear accident occurs at a place of 3,000 km west of Istanbul. Radioactive pollutants are pumped above the planetary boundary layer (PBL) with the power of explosion. On the day of nuclear accident, the radiosonde data taken at different places of Europe shows that atmospheric pressure is decreasing towards north at a rate of 0.0015 N/m3 and this pattern is valid for the whole Europe. Will the radioactivity affect Istanbul? If yes, when? Note that Istanbul is located on 40° northern latitude and world’s angular speed of rotation is 7.3 * 10-5 radians/sec. You may assume the density of air at the level where geostrophic wind equations apply as 0.70 kg/m3.
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Overall Atmospheric Motion (7/7)
Solution
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Equations of Motion (1/3)
Eularian Approach The observer stays stationary and observes the change in the
value of a function f (concentration, atmospheric parameters, etc.)
The coordinate system (reference frame) is stationary The objective is moving
Lagregian Approach The observer moves with the moving objective and observes
the change in the value of a function f The coordinate system is moving with the objective at the same
speed and direction
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Equations of Motion (2/3)
Lagregian Approach (cont’d)
wzfv
yfu
xf
tf
dtdf
dtdy
zf
dtdy
yf
dtdy
xf
tf
dtdf
fVdttfdf
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Equations of Motion (3/3)
Examples will be given later…
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Wind Speed Profile (1/2)
Due to friction near surface, wind speed increases with height exponentially
Wind speed is measured by a device called anemometer 10 m should be chosen for anemometer height
pzmuzu
10
10 Stability Class P
A 0.15
B 0.15
C 0.20
D 0.25
E 0.40
F 0.60
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Wind Speed Profile (2/2)
Homework (due 25.08.2008) Calculate wind speeds for Class B stability at 20, 30, 50, 100,
200, and 500 m if it is 1.2 m/sec. Plot the results. Comment on how the wind speed would change with altitude if
the stability class were Class E.