Equation of Continuity

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Equation of Continuity. Equation of Continuity. differential control volume:. Differential Mass Balance. mass balance:. Differential Equation of Continuity. divergence of mass velocity vector (  v ). Partial differentiation:. Differential Equation of Continuity. Rearranging:. - PowerPoint PPT Presentation

Transcript of Equation of Continuity

Equation of Continuity

Equation of Continuity

differential control volume:

Differential Mass Balance

Rate of Rate of Rate ofaccumulation mass in mass out

mass balance:

Rate ofmass in

x y zx zyv y z v x z v x y

Rate ofmass out

x y zx x z zy yv y z v x z v x y

Rate of mass accumulation

x y zt

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 v v v vt x y z x y z

Differential Equation of Continuity

Rearranging:

yx zx y z

vv vv v vt 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

𝐷 πœŒπ·π‘‘ =βˆ’πœŒ (𝛻 βˆ™π’— )

πœ•πœŒπœ•π‘‘ =βˆ’(𝛻 βˆ™ ρ 𝒗) 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

( π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“π‘šπ‘Žπ‘  𝑠 𝐴𝑖𝑛)=𝑛𝐴π‘₯∨π‘₯βˆ† 𝑦 βˆ† 𝑧 ( π‘Ÿπ‘Žπ‘‘π‘’π‘œπ‘“

π‘šπ‘Žπ‘  𝑠 π΄π‘œπ‘’π‘‘ )=𝑛𝐴π‘₯∨π‘₯+βˆ† π‘₯βˆ† 𝑦 βˆ† 𝑧

( π‘Ÿπ‘Žπ‘‘π‘’π‘œπ‘“π‘”π‘’π‘›π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›π‘œπ‘“ π‘šπ‘Žπ‘ π‘  𝐴)=π‘Ÿ 𝐴 βˆ†π‘₯ βˆ† 𝑦 βˆ† 𝑧

Equation of Continuity

𝑛𝐴π‘₯∨π‘₯βˆ† 𝑦 βˆ† π‘§βˆ’π‘›π΄π‘₯∨π‘₯+βˆ† π‘₯βˆ† 𝑦 βˆ† 𝑧+π‘Ÿ π΄βˆ† π‘₯βˆ† 𝑦 βˆ† 𝑧=πœ• 𝜌 𝐴

πœ•π‘‘ βˆ†π‘₯ βˆ† 𝑦 βˆ† 𝑧

Dividing by and letting approach zero,

πœ•πœŒ 𝐴

πœ•π‘‘ +(πœ•π‘›π΄π‘₯

πœ• π‘₯ +πœ•π‘›π΄ 𝑦

πœ• 𝑦 +πœ•π‘›π΄π‘§

πœ• 𝑧 )=π‘Ÿ 𝐴

Equation of Continuity

πœ•πœŒ 𝐴

πœ•π‘‘ +(πœ•π‘›π΄π‘₯

πœ• π‘₯ +πœ•π‘›π΄ 𝑦

πœ• 𝑦 +πœ•π‘›π΄π‘§

πœ• 𝑧 )=π‘Ÿ 𝐴

πœ•πœŒ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘›π΄ )=π‘Ÿ 𝐴

In vector notation,

But form the Table 7.5-1 (Geankoplis)

𝑛𝐴= 𝑗𝐴+𝜌𝐴 𝑣 𝑗 𝐴=βˆ’ 𝜌 𝐷𝐴𝐡 𝑑 𝑀𝐴/𝑑𝑧and

Equation of Continuity

πœ•πœŒ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘›π΄ )=π‘Ÿ 𝐴

𝑛𝐴= 𝑗𝐴+𝜌𝐴 𝑣 𝑗 𝐴=βˆ’ 𝜌 𝐷𝐴𝐡 𝑑 𝑀𝐴/𝑑𝑧

Substituting

πœ•πœŒ 𝐴

πœ•π‘‘ +(𝛻 βˆ™πœŒ 𝐴 𝑣 )βˆ’ (𝛻 βˆ™πœŒ 𝐷 𝐴𝐡 𝛻𝑀 𝐴 )=π‘Ÿ 𝐴

Equation of Continuity

πœ•πœŒ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘›π΄ )=π‘Ÿ 𝐴

πœ•π‘ 𝐴

πœ•π‘‘ +( πœ• 𝑁 𝐴π‘₯

πœ•π‘₯ +πœ• 𝑁 𝐴 𝑦

πœ• 𝑦 +πœ• 𝑁 𝐴 𝑧

πœ• 𝑧 )=𝑅𝐴

Dividing both sides by MWA

Equation of Continuity

Recall: 1. Fick’s Law

𝐽 π΄βˆ—=βˆ’π·π΄π΅

𝑑 𝑐 𝐴

𝑑𝑧2. Total molar flux of A

𝑁 𝐴= 𝐽 π΄βˆ— +𝑐𝐴 𝑣𝑀

𝑁 𝐴=βˆ’π‘π·π΄π΅π‘‘ π‘₯𝐴

𝑑𝑧 +π‘₯𝐴(𝑁 𝐴+𝑁 𝐡)

Equation of Continuity

πœ•π‘ 𝐴

πœ•π‘‘ +( πœ• 𝑁 𝐴π‘₯

πœ•π‘₯ +πœ• 𝑁 𝐴 𝑦

πœ• 𝑦 +πœ• 𝑁 𝐴 𝑧

πœ• 𝑧 )=𝑅 𝐴

Substituting NA and Fick’s lawand writing for all 3 directions,

πœ•π‘ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘ 𝐴 𝑣𝑀 )βˆ’ (𝛻 βˆ™π‘ 𝐷 𝐴𝐡 𝛻 π‘₯𝐴 )=𝑅𝐴

Equation of Continuity

πœ•π‘ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘ 𝐴 𝑣𝑀 )βˆ’ (𝛻 βˆ™π‘ 𝐷 𝐴𝐡 𝛻 π‘₯𝐴 )=𝑅𝐴

πœ•πœŒ 𝐴

πœ•π‘‘ +(𝛻 βˆ™πœŒ 𝐴 𝑣 )βˆ’ (𝛻 βˆ™πœŒ 𝐷 𝐴𝐡 𝛻𝑀 𝐴 )=π‘Ÿ 𝐴

(π‘Ÿπ‘Žπ‘‘π‘’π‘œπ‘“

π‘–π‘›π‘π‘Ÿπ‘’π‘Žπ‘ π‘’π‘–π‘›π‘šπ‘œπ‘™π‘’π‘ π‘œπ‘“ π΄π‘π‘’π‘Ÿ

𝑒𝑛𝑖𝑑 π‘£π‘œπ‘™π‘’π‘šπ‘’)(

𝑛𝑒𝑑 π‘Ÿπ‘Žπ‘‘π‘’π‘œπ‘“π‘Žπ‘‘π‘‘π‘–π‘‘π‘–π‘œπ‘›

π‘–π‘›π‘šπ‘œπ‘™π‘’π‘ π‘œπ‘“π΄π‘π‘’π‘Ÿ π‘’π‘›π‘–π‘‘π‘£π‘œπ‘™π‘’π‘šπ‘’ π‘π‘¦π‘π‘œπ‘›π‘£π‘’π‘π‘‘π‘–π‘œπ‘›

)(𝑛𝑒𝑑 π‘Ÿπ‘Žπ‘‘π‘’π‘œπ‘“π‘Žπ‘‘π‘‘π‘–π‘‘π‘–π‘œπ‘›

π‘–π‘›π‘šπ‘œπ‘™π‘’π‘ π‘œπ‘“π΄π‘π‘’π‘Ÿ π‘’π‘›π‘–π‘‘π‘£π‘œπ‘™π‘’π‘šπ‘’π‘π‘¦π‘‘π‘–π‘“π‘“π‘’π‘ π‘–π‘œπ‘›

)(π‘Ÿπ‘Žπ‘‘π‘’π‘œπ‘“

π‘π‘Ÿπ‘œπ‘‘π‘’π‘π‘‘π‘–π‘œπ‘›π‘œπ‘“ π‘šπ‘œπ‘™π‘’π‘  π‘œπ‘“π΄π‘π‘’π‘Ÿ π‘’π‘›π‘–π‘‘π‘£π‘œπ‘™π‘’π‘šπ‘’

𝑏𝑦 π‘Ÿπ‘’π‘Žπ‘π‘‘π‘–π‘œπ‘›)

Two equivalent forms of equation of continuity

Equation of Continuity

Equation of Continuity

πœ•π‘ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘ 𝐴 𝑣𝑀 )βˆ’ (𝛻 βˆ™π‘ 𝐷 𝐴𝐡 𝛻 π‘₯𝐴 )=𝑅𝐴

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

πœ•π‘ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘ 𝐴 𝑣𝑀 )βˆ’ (𝛻 βˆ™π‘ 𝐷 𝐴𝐡 𝛻 π‘₯𝐴 )=𝑅𝐴

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

πœ•π‘ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘ 𝐴 𝑣𝑀 )βˆ’ (𝛻 βˆ™π‘ 𝐷 𝐴𝐡 𝛻 π‘₯𝐴 )=𝑅𝐴

Special cases of the equation of continuity

3. For constant ρ and DAB (liquids),

πœ•πœŒ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘›π΄ )=π‘Ÿ 𝐴Starting 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,

βˆ’ ( 𝐷𝐴𝐡 𝛻2 𝑐 𝐴 )=𝑅𝐴

Assuming steady-state,

πœ•π‘ 𝐴

πœ•π‘‘ βˆ’ ( 𝐷𝐴𝐡𝛻2 𝑐𝐴 )=𝑅𝐴

Assuming concentration of A is small, then total c is almost constant and

πœ•π‘ 𝐴

πœ•π‘‘ +(𝛻 βˆ™π‘ 𝐴 𝑣𝑀 )βˆ’ (𝛻 βˆ™π‘ 𝐷 𝐴𝐡 𝛻 π‘₯𝐴 )=𝑅𝐴

Equation of Continuity

Assuming that diffusion is along the z-direction only,

βˆ’(𝐷 π΄π΅πœ•2𝑐 𝐴

πœ• 𝑧2 )=𝑅𝐴

βˆ’ ( 𝐷𝐴𝐡 𝛻2 𝑐 𝐴 )=𝑅𝐴

We can write that since A is disappearing by an irreversible, 1st order reaction

βˆ’(𝐷 π΄π΅πœ•2𝑐 𝐴

πœ• 𝑧2 )=βˆ’π‘˜1β€² β€² ′𝑐 𝐴

Equation of Continuity

𝐷 π΄π΅πœ•2 𝑐 𝐴

πœ• 𝑧2 βˆ’π‘˜1β€² β€² ′𝑐 𝐴=0

Rearranging,

Looks familiar?How to solve this ODE?

𝑐 𝐴

𝑐 𝐴0=

cosh [βˆšπ‘˜β€² β€² β€² 𝐿2

𝐷 𝐴𝐡(1βˆ’ 𝑧

𝐿 )]cosh (βˆšπ‘˜β€² β€² β€² 𝐿2

𝐷 𝐴𝐡)

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.