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Air Pollution-Introductionfor CVL212-Environmental Engineering
(Second Semester 2017-18)
Dr. Arun Kumar Civil Engineering (IIT Delhi)
Courtesy: Dr. Irene Xagoraraki (U.S.A.)
May 3, 2018 [email protected] 6
Point-Source Gaussian Plume Model
• Model Structure and Assumptions
– pollutants released from a “virtual point source”
– advective transport by wind
– dispersive transport (spreading) follows normal (Gaussian)distribution away from trajectory
– constant emission rate
– wind speed constant with time and elevation
– pollutant is conservative (no reaction)
– terrain is flat and unobstructed
– uniform atmospheric stability
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Effective Stack Height
Where:
H = Effective stack height (m)
h = height of physical stack (m)
∆H = plume rise (m)
HhH ∆+=
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Effective Stack Height (Holland’s formula) for
neutral conditions
where vs = stack velocity (m/s)
d = stack diameter (m)
u = wind speed (m)
P = pressure (kPa)
Ts = stack temperature (ºK)
Ta = air temperature (ºK)
( )
−×+=∆
−d
T
TTP
u
vH
a
ass 21068.25.1
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• How much will be % error in C(x,0,0) if one uses Heffective(unstable) for stability class? Think qualitatively.
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Wind Speed Correction
• Unless the wind speed at the virtual stack height is known, it must be estimated from the ground wind speed
Where: ux = wind speed at elevation zx
p = empirical constant
p
z
zuu
=
1
212
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Example 2
• A stack in an urban area is emitting 80 g/s of NO. It
has an effective stack height of 100 m. The wind speed is 4 m/s at 10 m. It is a clear summer day with the sun nearly overhead.
• Estimate the ground level concentration at: a) 2 km downwind on the centerline and b) 2 km downwind, 0.1 km off the centerline.
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1. Determine stability class
Assume wind speed is 4 km at ground surface. Description suggests strong solar radiation.
Stability class B
Example 2
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3. Estimate the wind speed at the effective stack height
Note: effective stack height given – no need to
calculate using Holland’s formula
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4. Determine concentration
a. x = 2000, y = 0
−
−=
22
220
100
2
1exp
290
0
2
1exp
)6.5)(220)(290(
80)0,2000(
πC
33 µg/m g/m 3.641043.6)0,2000( 5=×=
−C
Example 2
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b. x = 2000, y = 0.1 km = 100 m
−
−=
22
220
100
2
1exp
290
100
2
1exp
)6.5)(220)(290(
80)100,2000(
πC
33 µg/m g/m 6.601006.6)0,2000( 5=×=
−C
Example 2
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Example 3
• If in example #2, there is another stack (downwind distance from 1st stack =500m) with physical height (203m). Now, calculate overall ground level concentration at 2 km downwind on the center line? This 2nd stack is also emitting NO at same 80 g/s rate (all other conditions remain constant) (for stack #2: inside diameter =1.07m; air temp:13degC; barometric pressure =1000 milibars; stack gas velocity=9.14m/s; stack gas temp: 149degC)
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Example 3 hints
• From stack #1, we know conc (C1)
• For stack #2, first calculate effective stack height using Holland’s formula� then calculate conc. at given
distance using approach given in Example 2 (apply correction for x= distance of receptor from stack #2)�say we get conc. C2
• Now total conc. at receptor =Ctotal=C1+C2
• Now see if this is less than Callowable
• If not, then we need to control stack heights or source strength
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Example 4
• Question: Suppose an anemometer at a height of 10 m above ground measure wind velocity =2.5m/s. estimate the wind speed at an elevation of 300 m in rough terrain if atmosphere is unstable (i.e., k=0.2)?
• Answer:
• U300/u10=(300/10)(0.2)
• Wind velocity at 300m=(2.5)*(30)(0.2)=4.9m/s
p
z
zuu
=
1
212
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CPCB minimum guideline for stack
based on SO2 emission
• CPCB minimum stack height =30m
• So Choose maximum (30m; hSO2)
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Example 5
• A 40% efficient 1000MW coal fired power plant emitts SO2 at rate =6.47*108 microgram/s. the stack has effective height =20m (CPCB recommended minimum height =30m). An anemometer on a 10-m pole measures 2.5m/s of wind and atmospheric class is C.
• Predict the ground-level concentration of SO2 4 km directly downwind?
• What would be this concentration if stack height is changed to 30 m?
• What is the recommended stack height based on SO2 emission rate?
• Which stack height would you choose?
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Example 6
• Repeat Example 5 for stability classes : B,C and D for calculating C(x,0,0) where X=0-100m with 4 m gap. Now plot C(x,0,0) versus distance or for different stability classes. Use effective height obtained from Example 6.
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Dry Adiabatic Lapse Rate
Temperature, T (oC)
Altitu
de
, z (
km
)
Adiabatic lapse rate
1
2
= (T2-T1)/(z2-z1)
When any parcel of air moves up or down, it’s
temperature will change according to the adiabatic
lapse rate
For this parcel of air the
change in temperature with
altitude was:
T1T2
z1
z2= (10-20)oC/(2000-1000)m
= -1 oC/100m
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Stability
• Dry adiabatic lapse rate: temperature decreases with increased altitude
• Atmospheric (actual) lapse rate
< Г (temperature falls faster) unstable (super-adiabatic)
> Г (temperature falls slower) stable (sub-adiabatic)
= Г (same rate) neutral
ft 1000F mC/100 /4500.1 °=°−=−=Γ .- dz
dT
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Example 1
Z(m) T(ºC)
10 5.11
202 1.09
C/m °−=−
−=
−
−=
∆
∆0209.0
10202
11.509.1
12
12
zz
TT
z
T
m C/100 °−= 09.2
Since lapse rate is more negative than Г, (-1.00 ºC/100 m)=> atmosphereis unstable
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Unstable Conditions Rapid vertical mixing
takes place.
-1.25 oC/100 m < -1 oC/100m Unstable air encourages the dispersion and dilution of pollutants.
actual temperature falls faster than Г
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Stable Conditions Air at a certain altitude remains
at the same elevation.
-0.5 oC/100 m > -1 oC/100m
Stable air discourages
the dispersion and dilution of pollutants.
actual temperature falls slower than Г
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Neutral Conditions Air at a certain altitude remains
at the same elevation.
Neutrally stable air discourages the dispersion
and dilution of pollutants.
-1 oC/100 m = -1 oC/100m