M E 433
Professor John M. Cimbala
Lecture 15
Today, we will:
•
Continue derivation of the Gaussian plume model – equation and solution
•
Modify the solution for a buoyant plume, and do some examples/applications
y
U
Top view
x
m j ,s
z
Side view
U
x
hs
From the previous lecture, we derived the differential equation for a non-buoyant Gaussian
plume, assuming gradient diffusion, constant U, and constant diffusion coefficients:
∂c j
∂c  ∂ 
∂c  ∂ 
∂c 
∂
∂
=
− (Uc j ) +  Daj ,x j  +  Daj ,y j  +  Daj ,z j 
(0)
∂t
∂x
∂x 
∂x  ∂y 
∂y  ∂z 
∂z 
Terms
1
2
3
4
5
Assumptions and Approximations (to simplify this as much as possible):
Simplified equation for a steady Gaussian plume without buoyancy:
∂c j
∂ 2c j
∂ 2c j
=
U
Daj ,y 2 + Daj ,z 2
∂x
∂y
∂z
Now we apply boundary conditions (BCs) and solve (1) for cj(x,y,z) in this plume.
(1)
Dispersion coefficients: Tables scanned from Cooper, C. D. and Alley, F. C. Air Pollution Control: A
Design Approach, Edition 4, Waveland Press, Inc., Long Grove, IL, 2011, pp. 662-663.