Lecture 16 Air Pollution (3)
CVEN 301 Introduction to Environmental EngineeringFall 2007
Dr. Qi YingDepartment of Civil Engineering
Atmospheric Dispersion Modeling (2)
Plume Rise
A plume of hot gas emitted vertically rises due to its Momentum Buoyancy
Plume Rise
V0
V
Loses momentum due to entrainment
Ts>>T
Δh
H
Ts”~Ta
Ts’>T
Gradually loses buoyancy and bends over
Parameters affect plume rise Plume rise depends on both plume
and ambient parameters Plume and stack parameters Exit velocity Stack diameter Gas temperature Gas molecular weight
Ambient air parameters Stability Wind speed Temperature
Holland’s Simple Equation Includes stack and plume parameters Does not take atmospheric stability into
consideration
21.5 2.68 10s s s aa s
s
v d T Th p d
u T
vs = stack exit velocity (m/s)ds = stack diameter (m)u = wind velocity (m/s)pa = atmospheric pressureTs = stack temperature (K)Ta = ambient temperature
Holland’s Simple Equation
For large power plants, the heat emission rate (QH) is usually reported instead of stack temperature
1.5 9.6s s Hv d Qh
u u
vs = stack exit velocity (m/s)ds = stack diameter (m)u = wind velocity (m/s)QH = heat emission rate (MW)
Briggs Plume Rise Equations
It is the current EPA recommend method for plume rise calculation
It has better performance for thermally dominated plume (buoyancy >> momentum)
Plume rise can be estimated as a function of downwind distance
http://www.air-dispersion.com/briggs.html
2
4s a
s ss
T TF gv d
T
Buoyancy factor (F):
Stability parameter (s):
0.02 for stability class E
0.035 for stability class F
a
a
gS
T
gS
T
Δh
h
xf
Example – Plume Rise
For Class D stability, calculate the final plume rise using Briggs equations from a power plant stack, given the following information
vs = 20 m/sds = 5 mU = 6 m/sTs = 400 KTa = 280 K
Example – Plume Rise
Calculate Buoyancy Factor
F>55, calculate downwind distance where maximum plume rise happens
Final plume rise
2 2 400 2809.81 20 5 525.5
4 4 280s a
s ss
T TF gv d
T
0.4 0.4119 119 525.5 1458fx F m
1/3 2/31/3 2/3 1 1.6 525.5 1458
1.6 2776fh F x U m
Wind Speed as a Function of Height
The wind speed (u2) at stack height (z2) can be estimated using surface wind measurement(u1 @ z1):
22 1
1
pz
u uz
Stability urban ruralA 0.15 0.07B 0.15 0.07C 0.2 0.1D 0.25 0.15E 0.3 0.35F 0.3 0.35
Dependence of p as a function of stability and surface roughness
Wind speed example
Calculation wind speed at 477m if the wind speed at 10m above surface is 2 m/s. Assume neutral condition in urban area.
U477=U10*(477/10)0.25
=2*2.62=5.3 m/s
Maximum Ground Surface Concentration
The surface concentration can be derived by setting z=0 in the equation:
The maximum ground concentration must occur at y=0
2 2
2 2( , ,0) exp exp
2 2y z y z
E y HC x y
S S U S S
(H=h+Δh)
2
2( ,0,0) exp
2y z z
E HC x
S S U S
Maximum Ground Concentration (Neutral)
Stack
2 2
2 2( , ,0) exp exp
2 2y z y z
E y HC x y
S S U S S
H=25mStability class = DE=1g/sU=1m/s
Maximum Ground Concentration (Unstable)
2 2
2 2( , ,0) exp exp
2 2y z y z
E y HC x y
S S U S S
H=25mStability class = AE=1g/sU=1m/s
Stack
Summarize – Gaussian Dispersion Problem
Determine stability class Calculate plume rise Calculate wind speed Calculate Sy, Sz Calculate pollutant concentration
Example Determine the pollutant surface concentration
at 2 meter above surface, 400 meters directly downwind of the stack. Assume stability class D, wind speed 2m/s at effective stack height, pollutant emission rate 1g/s and an effective stack height of 20 m. Also assume that the pollutant is perfectly reflected when it hits the ground.
Example
Solution:u=2m/s, H=20m, Stability Class=DPosition to calculation concentration
(400,0,2) 2 22
2 2 2( , , ) exp exp exp
2 2 2 2y z y z z
z H z HE yC x y z
S S U S S S
2 22
2 2 2
2 2
2 2
2 20(400,0,2) exp exp exp
2 2 2 2
2 2 exp exp
2 2 2
y z y z z
y z z z
H HEC
S S U S S S
H HE
S S U S S
Example
Calculate Sy, Sz:
Sy = a*x0.894 Sz = c*xd + fx<1km x>1km
Stability a c d f c d fD 68 33.2 0.725 -1.7 44.5 0.516 -13E 50.5 22.8 0.678 -1.3 55.4 0.305 -34F 34 14.35 0.74 -0.35 62.6 0.18 -48.6
0.894
0.678
400 1
68 (400 /1000) 30
22.8 (400 /1000) 1.3 11
y
z
x m km
S m
S m
Example
Calculate concentration
2 2
2 2
2 2
2 2
6 3 3
2 2(400,0,2) exp exp
2 2 2
2 20 2 201 exp exp
2 30 11 2 2 11 2 11
8.56 10 / 8.56 /
y z z z
H HEC
S S U S S
g m g m
Puff Release Sometime we need to
determine pollutant concentrations downwind due to an instantaneous release
The plume is advected downwind as a “puff”
Puff concept
x
y
Mass=mt1=U/x1
Mass=mt2=U/x2
Mass=mt3=U/x3
-Pollutant concentration decreases due to dispersion in all directions.-The total mass in the puff remains unchanged.
Puff concentration
The concentration of pollutant (C) at ground surface (x,y) at any given time (t) can be calculated by
22 2
1.5
1( , ,0, ) exp
22( ) x y zx y z
m x Ut y HC x y t
S S SS S S
Sx, Sy Sz = dispersion parameters (m)U = wind speed at plume release point (m/s)t = time after plume release (s)m = amount of pollutant released (kg)H = height where the puff is released (m)
Dose
The amount of pollutant received during pollutant exposure (grams.second/m3)
At ground level
0( , , ) ( , , , )D x y z C x y z t dt
2 21
( , ,0) exp2y z y z
m y HD x y
S S U S S
Coefficient Stability a b Unstable 0.14 0.92
Sy Neutral 0.06 0.92 Stable 0.02 0.89 Unstable 0.53 0.73
Sz Neutral 0.15 0.7 Stable 0.05 0.61
b by zS ax S ax
Sy and Sz are different from plume dispersion equations
(x in meters)
Ground Level Dose
Neutral Condition M=1kgU=1m/sH=2m
Highway Air Pollution
Emissions from highway account for majority of the CO, NOx and VOC in urban areas
Inappropriate arrangement of highways lead to local “hotspot” in air quality
Gaussian dispersion model can be applied to highway segments to predict pollutant concentrations downwind
Finite Length Line Source (FLLS)
A highway section with uniform emission rate can be modeled as a finite line source
U
Wind directionx axis
y axis
Plume from a differential length y2
y1
y
Sy=f(x)
dy
Receptor at(x,y0)
y0
y0-y
Calculate FLLS concentration
Steady State q = Emission factor (kg/s.m)
Emission rate from a differential length = q.dy (kg/s)
Concentration at receptor due tothe differential emission
Integrate over the entire length:
2 20
0 2 2( , ,0) exp exp
2 2y z y z
y yqdy HdC x y
US S S S
Ux axis
y axisy2
y1
ydy
Receptor at(x,y0)
y0y0-y
2
10 0( , ,0) ( , ,0)
y
yC x y dC x y