Equation of Continuity
Equation of Continuity
differential control volume:
Differential Mass Balance
 Rate of   Rate of   Rate of 
mass balance: 



accumulation
mass
in
mass
out

 
 

 Rate of 

    vx  x yz    v y  y xz    vz  z xy
 mass in 
 Rate of 

    vx  x x yz    v y  y y xz    vz  z z xy
 mass out 
 Rate of mass  
xyz


 accumulation  t
Differential Equation of
Continuity
    vx     v y     vz  

 


      v 
t
y
z 
 x
divergence of mass velocity vector (v)
Partial differentiation:
 vx v y vz   


 
  


 vy
 vz
   vx

t
y
z 
 x y z   x
Differential Equation of
Continuity
Rearranging:
 vx vy vz 




 vx
 vy
 vz
  



t
x
y
z
 x y z 
substantial time derivative
 vx vy vz 
D
  


      v 
Dt
 x y z 
If fluid is incompressible:  v  0
Equation of Continuity
𝜕𝜌
= −(𝛻 ∙ ρ𝒗)
𝜕𝑡
or
𝐷𝜌
= −𝜌(𝛻 ∙ 𝒗)
𝐷𝑡
Conservation of mass for pure liquid flow
Equation of Continuity
Applying the conservation of mass to the volume element
* May also be expressed in terms of moles
Equation of Continuity
Equation of Continuity
𝑟𝑎𝑡𝑒 𝑜𝑓
= 𝑛𝐴𝑥|𝑥 ∆𝑦∆𝑧
𝑚𝑎𝑠𝑠 𝐴 𝑖𝑛
𝑟𝑎𝑡𝑒 𝑜𝑓
= 𝑛𝐴𝑥|𝑥+∆𝑥 ∆𝑦∆𝑧
𝑚𝑎𝑠𝑠 𝐴 𝑜𝑢𝑡
𝑟𝑎𝑡𝑒 𝑜𝑓
= 𝑟𝐴 ∆𝑥∆𝑦∆𝑧
𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑎𝑠𝑠 𝐴
𝜕𝜌𝐴
𝑟𝑎𝑡𝑒 𝑜𝑓
=
∆𝑥∆𝑦∆𝑧
𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑚𝑎𝑠𝑠 𝐴
𝜕𝑡
Equation of Continuity
𝜕𝜌𝐴
𝑛𝐴𝑥|𝑥 ∆𝑦∆𝑧 − 𝑛𝐴𝑥|𝑥+∆𝑥 ∆𝑦∆𝑧 + 𝑟𝐴 ∆𝑥∆𝑦∆𝑧 =
∆𝑥∆𝑦∆𝑧
𝜕𝑡
Dividing by ∆𝑥∆𝑦∆𝑧 and letting ∆𝑥, ∆𝑦, 𝑎𝑛𝑑 ∆𝑧 approach zero,
𝜕𝜌𝐴
𝜕𝑛𝐴𝑥 𝜕𝑛𝐴𝑦 𝜕𝑛𝐴𝑧
+
+
+
= 𝑟𝐴
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
Equation of Continuity
𝜕𝜌𝐴
𝜕𝑛𝐴𝑥 𝜕𝑛𝐴𝑦 𝜕𝑛𝐴𝑧
+
+
+
= 𝑟𝐴
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
In vector notation,
𝜕𝜌𝐴
+ 𝛻 ∙ 𝑛𝐴 = 𝑟𝐴
𝜕𝑡
But form the Table 7.5-1 (Geankoplis)
𝑛𝐴 = 𝑗𝐴 + 𝜌𝐴 𝑣
and
𝑗𝐴 = −𝜌𝐷𝐴𝐵 𝑑𝑤𝐴 /𝑑𝑧
Equation of Continuity
𝜕𝜌𝐴
+ 𝛻 ∙ 𝑛𝐴 = 𝑟𝐴
𝜕𝑡
𝑛𝐴 = 𝑗𝐴 + 𝜌𝐴 𝑣
𝑗𝐴 = −𝜌𝐷𝐴𝐵 𝑑𝑤𝐴 /𝑑𝑧
Substituting 𝑛𝐴 𝑎𝑛𝑑 𝑗𝐴
𝜕𝜌𝐴
+ 𝛻 ∙ 𝜌𝐴 𝑣 − 𝛻 ∙ 𝜌𝐷𝐴𝐵 𝛻𝑤𝐴 = 𝑟𝐴
𝜕𝑡
Equation of Continuity
𝜕𝜌𝐴
+ 𝛻 ∙ 𝑛𝐴 = 𝑟𝐴
𝜕𝑡
Dividing both sides by MWA
𝜕𝑐𝐴
𝜕𝑁𝐴𝑥 𝜕𝑁𝐴𝑦 𝜕𝑁𝐴𝑧
+
+
+
= 𝑅𝐴
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
Equation of Continuity
Recall: 1. Fick’s Law
𝐽𝐴∗
𝑑𝑐𝐴
= −𝐷𝐴𝐵
𝑑𝑧
2. Total molar flux of A
𝑁𝐴 = 𝐽𝐴∗ + 𝑐𝐴 𝑣𝑀
𝑑𝑥𝐴
𝑁𝐴 = −𝑐𝐷𝐴𝐵
+ 𝑥𝐴 (𝑁𝐴 + 𝑁𝐵 )
𝑑𝑧
Equation of Continuity
𝜕𝑐𝐴
𝜕𝑁𝐴𝑥 𝜕𝑁𝐴𝑦 𝜕𝑁𝐴𝑧
+
+
+
= 𝑅𝐴
𝜕𝑡
𝜕𝑥
𝜕𝑦
𝜕𝑧
Substituting NA and Fick’s law
and writing for all 3 directions,
𝜕𝑐𝐴
+ 𝛻 ∙ 𝑐𝐴 𝑣𝑀 − 𝛻 ∙ 𝑐𝐷𝐴𝐵 𝛻𝑥𝐴 = 𝑅𝐴
𝜕𝑡
Equation of Continuity
Two equivalent forms of equation of continuity
𝜕𝜌𝐴
+ 𝛻 ∙ 𝜌𝐴 𝑣 − 𝛻 ∙ 𝜌𝐷𝐴𝐵 𝛻𝑤𝐴 = 𝑟𝐴
𝜕𝑡
𝜕𝑐𝐴
+ 𝛻 ∙ 𝑐𝐴 𝑣𝑀 − 𝛻 ∙ 𝑐𝐷𝐴𝐵 𝛻𝑥𝐴 = 𝑅𝐴
𝜕𝑡
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒
𝑖𝑛 𝑚𝑜𝑙𝑒𝑠
𝑜𝑓 𝐴 𝑝𝑒𝑟
𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒
𝑛𝑒𝑡 𝑟𝑎𝑡𝑒 𝑜𝑓
𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛
𝑖𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓
𝐴 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
𝑣𝑜𝑙𝑢𝑚𝑒 𝑏𝑦
𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛
𝑛𝑒𝑡 𝑟𝑎𝑡𝑒 𝑜𝑓
𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛
𝑖𝑛 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓
𝐴 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
𝑣𝑜𝑙𝑢𝑚𝑒 𝑏𝑦
𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛
𝑟𝑎𝑡𝑒 𝑜𝑓
𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛
𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓
𝐴 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡
𝑣𝑜𝑙𝑢𝑚𝑒
𝑏𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛
Equation of Continuity
Equation of Continuity
𝜕𝑐𝐴
+ 𝛻 ∙ 𝑐𝐴 𝑣𝑀 − 𝛻 ∙ 𝑐𝐷𝐴𝐵 𝛻𝑥𝐴 = 𝑅𝐴
𝜕𝑡
Special cases of the equation of continuity
1. Equation for constant c and DAB,
At constant P and T, c= P/RT for gases, and substituting 𝛻𝑥𝐴 = 𝛻𝑐𝐴 /𝑐
Equation of Continuity
𝜕𝑐𝐴
+ 𝛻 ∙ 𝑐𝐴 𝑣𝑀 − 𝛻 ∙ 𝑐𝐷𝐴𝐵 𝛻𝑥𝐴 = 𝑅𝐴
𝜕𝑡
Special cases of the equation of continuity
2. Equimolar counterdiffusion of gases,
At constant P , with no reaction, c = constant,
vM = 0, DAB = constant and RA=0
Fick’s 2nd Law of diffusion
Equation of Continuity
𝜕𝑐𝐴
+ 𝛻 ∙ 𝑐𝐴 𝑣𝑀 − 𝛻 ∙ 𝑐𝐷𝐴𝐵 𝛻𝑥𝐴 = 𝑅𝐴
𝜕𝑡
Special cases of the equation of continuity
3. For constant ρ and DAB (liquids),
Starting with the vector notation
of the mass balance
𝜕𝜌𝐴
+ 𝛻 ∙ 𝑛𝐴 = 𝑟𝐴
𝜕𝑡
We substitute 𝑛𝐴 = −𝜌𝐷𝐴𝐵 𝛻𝑤𝐴 + 𝜌𝐴 𝒗and 𝛻𝑤𝐴 = 𝜌𝐴 /𝜌
Equation of Continuity
Example 1.
Estimate the effect of chemical
reaction on the rate of gas
absorption in an agitated tank.
Consider a system in which the
dissolved gas A undergoes an
irreversible first order reaction
with the liquid B; that is A
disappears within the liquid phase
at a rate proportional to the local
concentration
of
A.
What
assumptions can be made?
Equation of Continuity
1. Gas A dissolves in liquid B and
diffuses into the liquid phase
2. An irreversible 1st order
homogeneous reaction takes
place
A + B  AB
Assumption:
AB is negligible in the solution
(pseudobinary assumption)
Equation of Continuity
𝜕𝑐𝐴
+ 𝛻 ∙ 𝑐𝐴 𝑣𝑀 − 𝛻 ∙ 𝑐𝐷𝐴𝐵 𝛻𝑥𝐴 = 𝑅𝐴
𝜕𝑡
Assuming concentration of A is small, then total c is almost constant and
Expanding the equation and taking c inside the space derivative,
𝜕𝑐𝐴
− 𝐷𝐴𝐵 𝛻 2 𝑐𝐴 = 𝑅𝐴
𝜕𝑡
Assuming steady-state,
− 𝐷𝐴𝐵 𝛻 2 𝑐𝐴 = 𝑅𝐴
Equation of Continuity
− 𝐷𝐴𝐵 𝛻 2 𝑐𝐴 = 𝑅𝐴
Assuming that diffusion is along the z-direction only,
𝜕 2 𝑐𝐴
− 𝐷𝐴𝐵
= 𝑅𝐴
2
𝜕𝑧
We can write that 𝑅𝐴 = −𝑘1′′′ 𝑐𝐴
since A is disappearing by an irreversible, 1st order reaction
𝜕 2 𝑐𝐴
′′′
− 𝐷𝐴𝐵
=
−𝑘
1 𝑐𝐴
2
𝜕𝑧
Equation of Continuity
Rearranging,
𝜕 2 𝑐𝐴 ′′′
𝐷𝐴𝐵
−𝑘1 𝑐𝐴 = 0
2
𝜕𝑧
Looks familiar?
How to solve this ODE?
𝑐𝐴
=
𝑐𝐴0
𝑘 ′′′ 𝐿2
𝑧
cosh[
1− ]
𝐷𝐴𝐵
𝐿
𝑘 ′′′ 𝐿2
cosh(
)
𝐷𝐴𝐵
Equation of Continuity
Example 2.
A hollow sphere with permeable
solid walls has its inner and outer
surfaces maintained at a constant
concentration
CA1
and
CA0
respectively.
Develop
the
expression for the concentration
profile for a component A in the
wall at steady-state conditions.
What is the flux at each surface?