# Continuity Equation - PowerPoint

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```					Continuity Equation
• Statement of conservation of mass
• Several different forms:
Mass Divergence Form

                   ( u ) ( v )  ( w)
   ( V )                  
t                     x      y        z

“Convergence of Density”

Change of density
with respect to time

We can derive the “Velocity Divergence Form” by using the product rule
and Euler’s Relation
• Move all terms to the left hand side
• Apply product rule to mass flux (density *
wind) terms
• Apply Euler’s relation to simplify
d            
 u v    w
dt t x  y    z
Velocity Divergence Form

1 d        
   V
 dt

This works great for water bodies, since they are incompressible (density
does not change). This makes the left hand side = 0 and the velocity
divergence form becomes:

u v w
     0
x y z

But the atmosphere is compressible. So another approach is to use pressure
(p) as the vertical coordinate, rather than height (z). Then the velocity divergence
form becomes:

u v 
     0
x y p
u v 
     0
x y p

Here, ω is the change in the vertical velocity with respect to pressure (whereas
‘w’ was the vertical velocity in height coordinates

This equation can be used to compute the vertical velocity at various levels
in the atmosphere. This technique is called the kinematic method because
it only requires information about the winds.

     u v              Integrate both sides from
 (  )               p1 to p2
p     x y
           u v
p2             2p

 p
p1
dp    (  )dp
p1
x y
Simplify

u v
p2

p p  (                 )dp      Rearrange
2      1
p1
x y
u v
p2

p         p1   (  )dp
2
p1
x y

Vertical velocity at p2                             Horizontal divergence
between p1 and p2
Vertical velocity at p1

If the horizontal divergence is constant between p1 and p2, we can represent
this equation in finite difference form:

 u v 
 p2   p1  (   )( p1  p2 )
 x y 
Lab Notes
See revised question 1 on website

For question 4,

 u   v 
 p   p  (   )( p1  p2 )
 x y 
2     1

 u v      u v 
p         p1     * p1     * p2 )
 x y  p1  x y  p2
2

```
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 views: 52 posted: 6/29/2012 language: English pages: 9