Air Pollution 3

CVEN 301 Introduction to Environmental Engineering Fall 2007

Lecture 16 Air Pollution (3) Atmospheric Dispersion Modeling (2)

Dr. Qi Ying Department of Civil Engineering

Plume Rise

Plume Rise  A plume of hot gas emitted vertically rises due to its  Momentum  Buoyancy Ts’>T Ts>>T

Ts”~Ta

V

Δh

V0

H Loses momentum due to entrainment

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 vd h  s s u



 Ts  Ta 1.5  2.68  10  p   a   Ts 









2

vs = stack exit velocity (m/s) ds = stack diameter (m) u = wind velocity (m/s) pa = atmospheric pressure Ts = stack temperature (K) Ta = ambient temperature

d s 

Holland’s Simple Equation  For large power plants, the heat emission rate (QH) is usually reported instead of stack temperature vs d s QH h  1.5  9.6 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

Buoyancy factor (F):  Ts  Ta   4 T s  

F  gvs d s2 

Stability parameter (s): S  0.02

g for stability class E Ta

S  0.035

Δh h xf http://www.air-dispersion.com/briggs.html

g for stability class F Ta

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/s ds = 5 m U = 6 m/s Ts = 400 K Ta = 280 K

Example – Plume Rise  Calculate Buoyancy Factor  Ts  Ta  4Ts

F  gvs d s2 



 400  280    525.5  4  280 

2   9.81 20  5  



 F>55, calculate downwind distance where maximum plume rise happens x f  119  F 0.4  119  525.50.4  1458m

 Final plume1.6 rise 525.5 h  1.6 F x f U 1/3

2/3

1



1/3

6

14582/3

 277 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): p

 z  u2  u1  2   z1 

Dependence of p as a function of stability and surface roughness

Stability urban A 0.15

rural 0.07

B C D

0.15 0.2 0.25

0.07 0.1 0.15

E F

0.3 0.3

0.35 0.35

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:   E y2  H2  C ( x, y, 0)  exp   exp   2   2S y 2    S y S zU 2 S  z   

(H=h+Δh)

 The maximum ground concentration must occur at y=0  E H2  C ( x, 0, 0)  exp   2   S y S zU  2S z 

Maximum Ground Concentration (Neutral)   E y2  H2  C ( x, y, 0)  exp   exp   2   2S y 2    S y S zU 2 S  z    Stack

H=25m Stability class = D E=1g/s U=1m/s

Maximum Ground Concentration (Unstable)   E y2  H2  C ( x, y, 0)  exp   exp   2   2S y 2    S y S zU 2 S  z    Stack

H=25m Stability class = A E=1g/s U=1m/s

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=D Position to calculation concentration (400,0,2) C ( x, y , z ) 

2 2        z  H z  H     y    exp     exp   2  2 2    2S y   2S z  2S z      



E exp   2 S y S zU 

2

2 2        2  H 2  H     E 0  C (400, 0, 2)  exp    exp     exp   2  2 2     2 S y S zU 2S z  2S z   2 S y          2H2    2  H  2   E   exp     exp   2 2    2 S y S zU  2 S 2 S z z      



2

Example  Calculate Sy, Sz: Sy = a*x0.894 Sz = c*xd + f x<1km

x>1km

Stabilit y D E

a 68 50.5

c 33.2 22.8

d 0.725 0.678

f -1.7 -1.3

c 44.5 55.4

d 0.516 0.305

f -13 -34

F

34

14.35

0.74

-0.35

62.6

0.18

-48.6

Q x  400m  1km  S y  68  (400 /1000) 0.894  30m  S z  22.8  (400 /1000) 0.678  1.3  11m

Example  Calculate concentration E C (400, 0, 2)  2 S y S zU







 exp   

 2H  2S z 2





2

 exp    

   2  20  1   exp   2 2  30  11 2  2  11    8.56 106 g / m3  8.56  g / m3



 2 H  

2



2S z 2



 exp    

2

     

 2  20   2 112

2

     

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 y

x Mass=m t1=U/x1

Mass=m t2=U/x2

Mass=m t3=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 C ( x, y, 0, t ) 

m  exp  2( S x S y S z ) 1.5 

 

2 2 2      H  1  x  Ut  y             2   Sx   S z     Sy  

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) 

D ( x, y, z )   C ( x, y, z , t )dt 0

 At ground level D ( x, y, 0) 

m  exp   S y S zU 



2 2   H    1  y          2   Sy   S z    

Sy and Sz are different from plume dispersion equations Coeffici Stabili ent ty a b Unstab   le 0.14 0.92 Neutra Sy l 0.06 0.92   Stable 0.02 0.89 Unstab   le 0.53 0.73 Neutra Sz l 0.15 0.7   Stable 0.05 0.61

S y  ax b

S z  ax b (x in meters)

Ground Level Dose Neutral Condition M=1kg U=1m/s H=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 y2

y axis Plume from a differential length

U

Sy=f(x)

x axis

Wind direction dy

y

y0 y0-y Receptor at (x,y0)

y1

Calculate FLLS concentration y2

 Steady State

U

q = Emission factor (kg/s.m)

x axis

Emission rate from a differential length = q.dy (kg/s)

dy

Concentration at receptor due to the differential emission

y1

  y0  y  qdy dC ( x, y0 , 0)  exp   2   US y S z 2 S y 

2



H2  exp   2    2S z  

Integrate over the entire length: y2

C ( x, y0 , 0)   dC ( x, y0 , 0) y1

y axis



y y0-y

y0 Receptor at (x,y0)