# Convection in the General Property Balance

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```					Convection in the General
Property Balance

Development of the full equations of motion
Control volume analysis:
Based on application of the balance:

Input + Generation =
Output + Accumulation
For a conserved property y and corresponding flux Y
Consider a control volume in Cartesian coordinates:
Y zA ) 2
(
dV = dx dy dz
dy                        Y yA ) 2
(

Y xA ) 1
(                                          Y xA ) 2
(
dz
z
Y yA ) 1
(                     dx                              y

x
Y zA ) 1
(

Property transport entering or leaving each face of the form, Y A
where A is an area element
Y zA ) 2
(
Input + Generation =
Y yA ) 2
(
Output + Accumulation                                dy

Y xA ) 1
(                                    Y xA ) 2
(
dz
Generation:                                                                 z
Y yA ) 1             dx
y G dV  y G dx dy dz
(                                               y

x
Y zA ) 1
(

 y        y 
Accumulation:          dV       dx dy dz
 t        t 

Input:      Yx 1 dy dz  Yy 1 dx dz  Yz 1 dx dy

Output:     Yx 2 dy dz  Yy 2 dx dz  Yz 2 dx dy
Rearrange the balance:

Accumulation = Output – Input + Generation

Next …

Focus on terms for [Output – Input]
Y zA ) 2
(
In the x-direction we can write:
dy                   Y yA ) 2
(

Y xA ) 1
(                                    Y xA ) 2
(
dz
z
Y yA ) 1
(                    dx                         y

x
Y zA ) 1
(

[Output – Input] =       Yx 2 - Yx 1  dy dz  - Yx 1 - Yx 2  dy dz

Yx              Y              Y
       x dy dz  -    dx dy dz  -    dV
x              x              x
[Output – Input] summary:

Y
x - direction:    -      dV
x

Y
y – direction:    -    dV
y

Y
z – direction:    -    dV
z
Accumulation = [Output – Input] + Generation

 y    Y  Y  Y 
dV          x   y   z  dV  y G dV
-                 
 t                    

Cancel out the dV terms:

y   Y Y Y
t     x   y   z   yG
-                 
                 
Recall that the flux, Y, is a vector:

Y i Yx  jYy  k Yz

Short-hand notation … the divergence relation:

 Yx  Yy  Yz
Y          
x   y   z
A final form for our property balance:

y
 -   Y   y G
t

To solve this equation, we need to know Y in terms of y
In engineering practice, we do this by
splitting the flux up into two components:

Y  Yconv  Ydiff
Yconv is a convective component, and

Ydiff is a diffusive component

Yconv  y U               Ydiff  -  y
- where U is the local convective velocity
y
 -   Y   y G
t
We need the divergence (derivative) of Y

  Y    YDiff  YConv 

  Y  -    y   y U 

  Y  -    y   y   U   U   y
The general property balance,

y
 -   Y   y G
t
with

  Y  -    y   y   U   U   y

becomes

y
 U   y  y G     y  - y   U 
t
Accumulation   Convection 1    Generation   Diffusion   Convection 2
y
 U   y  y G     y  - y   U 
t
Some examples:
Heat transfer, y =  Cp T and we obtain
 CPT 
 U   CPT   y G      CPT   - CPT   U 
t
Mass transfer, y = A or CA (mass or moles respectively) and we obtain

 A
 U    A  y G     D A  - A   U 
t
 CA
 U    CA  y G     D CA  - CA   U 
t
Momentum transfer, y = U and we obtain

 ρU x 
 U   ρU x   ψG    ρU x  - ρU x   U 
t

 ρU y     U   ρU   ψ    ρU  - ρU   U 
y     G           y        y
t

 ρU z 
 U   ρU z   ψG    ρU z  - ρU z   U 
t

Components for each coordinate direction
An important special case for the general balance:
y
 U   y  y G     y  - y   U 
t

Assume generation and diffusion are zero:

ψ
   ψ U   0
t
If conserved property is total mass per unit volume, ,


+    U = 0
t
With constant , /t = 0 and,

   U = U    +    U =    U

Hence the property balance for this case becomes,
 U = 0

or   U = 0
And in this case (constant ), our original property balance

y
 U   y  y G     y  -y   U 
t

becomes:

y
 U   y  y G     y 
t

Divergence of the velocity field is zero
Cases with constant  lead to

  2y  2y  2y 
    y      y    
  x2   y2   z 2 

                     

The dot product, , operating on a scalar is given the
symbol 2 and is called the Laplacian operator

e.g. the steady state conduction equation
2 T 2 T 2 T
            0
x 2
y 2
z 2

describes the temperature field, T(x, y, z), given boundary
conditions at specified edges of a Cartesian “box”
V

x

U
i i
L qu d
i
F a lln g
Fm il
i
V e r t ca l
l
P a te

y

Explain developing region for this problem!

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