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Equation of Continuity
Equation of Continuity
differential control volume:
Differential Mass Balance
Rate of Rate of Rate of
accumulation mass in mass out
mass balance:
Rate of
mass in
x y zx zyv y z v x z v x y
Rate of
mass out
x y zx x z zy yv y z v x z v x y
Rate of mass
accumulationx y z
t
Differential Equation of Continuity
yx zvv v
t x y z
v
divergence of mass velocity vector (v)
Partial differentiation:
yx zx y z
vv vv v v
t x y z x y z
Differential Equation of Continuity
Rearranging:
yx zx y z
vv vv v v
t x y z x y z
substantial time derivative
yx zvv vD
Dt x y zv
If fluid is incompressible: 0 v
Equation of Continuity
π· ππ·π‘
=βπ (π» βπ )ππππ‘
=β(π» β Ο π) or
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
( πππ‘π πππππ π π΄ππ)=ππ΄π₯β¨ π₯β π¦ β π§ ( πππ‘πππ
πππ π π΄ππ’π‘)=ππ΄π₯β¨ π₯+β π₯β π¦ β π§
( πππ‘ππππππππππ‘πππππ πππ π π΄)=π π΄ βπ₯ β π¦ β π§
Equation of Continuity
ππ΄π₯β¨π₯β π¦ β π§βππ΄π₯β¨π₯ +β π₯β π¦ β π§+π π΄β π₯β π¦ β π§=π π π΄
ππ‘βπ₯ β π¦ β π§
Dividing by and letting approach zero,
ππ π΄
ππ‘+( πππ΄π₯
π π₯+
πππ΄ π¦
π π¦+
πππ΄π§
π π§ )=π π΄
Equation of Continuity
ππ π΄
ππ‘+( πππ΄π₯
π π₯+
πππ΄ π¦
π π¦+
πππ΄π§
π π§ )=π π΄
ππ π΄
ππ‘+( π» βππ΄ )=π π΄
In vector notation,
But form the Table 7.5-1 (Geankoplis)
ππ΄= ππ΄+ππ΄ π£ ππ΄=β π π·π΄π΅ π π€π΄/ ππ§and
Equation of Continuity
ππ π΄
ππ‘+( π» βππ΄ )=π π΄
ππ΄= ππ΄+ππ΄ π£ ππ΄=β π π·π΄π΅ π π€π΄/ ππ§
Substituting
ππ π΄
ππ‘+( π» βπ π΄ π£ )β ( π» βπ π· π΄π΅ π»π€ π΄ )=π π΄
Equation of Continuity
ππ π΄
ππ‘+( π» βππ΄ )=π π΄
ππ π΄
ππ‘+( π ππ΄π₯
ππ₯+
π π π΄ π¦
π π¦+
π π π΄ π§
π π§ )=π π΄
Dividing both sides by MWA
Equation of Continuity
Recall: 1. Fickβs Law
π½ π΄β=βπ·π΄π΅
π ππ΄
ππ§2. Total molar flux of A
π π΄= π½ π΄β +ππ΄ π£π
π π΄=βππ·π΄π΅
π π₯π΄
ππ§+π₯π΄(π π΄+ππ΅)
Equation of Continuity
ππ π΄
ππ‘+( π ππ΄π₯
ππ₯+
π π π΄ π¦
π π¦+
π π π΄ π§
π π§ )=π π΄
Substituting NA and Fickβs lawand writing for all 3 directions,
ππ π΄
ππ‘+( π» βπ π΄ π£π )β ( π» βπ π· π΄π΅ π» π₯π΄ )=π π΄
Equation of Continuity
ππ π΄
ππ‘+( π» βπ π΄ π£π )β ( π» βπ π· π΄π΅ π» π₯π΄ )=π π΄
ππ π΄
ππ‘+( π» βπ π΄ π£ )β ( π» βπ π· π΄π΅ π»π€ π΄ )=π π΄
(πππ‘πππ
πππππππ ππππππππ ππ π΄πππ
π’πππ‘ π£πππ’ππ)(
πππ‘ πππ‘ππππππππ‘πππ
πππππππ πππ΄πππ π’πππ‘π£πππ’ππππ¦ππππ£πππ‘πππ
)(πππ‘ πππ‘ππππππππ‘πππ
πππππππ πππ΄πππ π’πππ‘π£πππ’ππππ¦πππππ’π πππ
)(πππ‘πππ
πππππ’ππ‘πππππ πππππ πππ΄πππ π’πππ‘π£πππ’ππ
ππ¦ πππππ‘πππ)
Two equivalent forms of equation of continuity
Equation of Continuity
Equation of Continuity
ππ π΄
ππ‘+( π» βπ π΄ π£π )β ( π» βπ π· π΄π΅ π» π₯π΄ )=π π΄
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
Equation of Continuity
ππ π΄
ππ‘+( π» βπ π΄ π£π )β ( π» βπ π· π΄π΅ π» π₯π΄ )=π π΄
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
Fickβs 2nd Law of diffusion
Equation of Continuity
ππ π΄
ππ‘+( π» βπ π΄ π£π )β ( π» βπ π· π΄π΅ π» π₯π΄ )=π π΄
Special cases of the equation of continuity
3. For constant Ο and DAB (liquids),
ππ π΄
ππ‘+( π» βππ΄ )=π π΄
Starting with the vector notation of the mass balance
We substitute and
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
Expanding the equation and taking c inside the space derivative,
β ( π·π΄π΅ π»2 π π΄ )=π π΄
Assuming steady-state,
ππ π΄
ππ‘β ( π·π΄π΅ π»2 π π΄ )=π π΄
Assuming concentration of A is small, then total c is almost constant and
ππ π΄
ππ‘+( π» βπ π΄ π£π )β ( π» βπ π· π΄π΅ π» π₯π΄ )=π π΄
Equation of Continuity
Assuming that diffusion is along the z-direction only,
β(π· π΄π΅
π2ππ΄
π π§2 )=π π΄
β ( π·π΄π΅ π»2 π π΄ )=π π΄
We can write that since A is disappearing by an irreversible, 1st order reaction
β(π· π΄π΅
π2π π΄
π π§2 )=βπ1β² β² β²π π΄
Equation of Continuity
π· π΄π΅
π2 ππ΄
π π§2 βπ1β² β² β²ππ΄=0
Rearranging,
Looks familiar?How to solve this ODE?
π π΄
π π΄0
=cosh [β πβ² β² β² πΏ2
π· π΄π΅(1β π§
πΏ )]cosh (β πβ² β² β² πΏ2
π· π΄π΅
)
Equation of Continuity
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?
Example 2.