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

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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 𝜌𝐷𝑡 =−𝜌 (𝛻 ∙𝒗 )

𝜕𝜌𝜕𝑡 =−(𝛻 ∙ 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

( 𝑟𝑎𝑡𝑒 𝑜𝑓𝑚𝑎𝑠 𝑠 𝐴𝑖𝑛)=𝑛𝐴𝑥∨𝑥∆ 𝑦 ∆ 𝑧 ( 𝑟𝑎𝑡𝑒𝑜𝑓

𝑚𝑎𝑠 𝑠 𝐴𝑜𝑢𝑡 )=𝑛𝐴𝑥∨𝑥+∆ 𝑥∆ 𝑦 ∆ 𝑧

( 𝑟𝑎𝑡𝑒𝑜𝑓𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑖𝑜𝑛𝑜𝑓 𝑚𝑎𝑠𝑠 𝐴)=𝑟 𝐴 ∆𝑥 ∆ 𝑦 ∆ 𝑧

Equation of Continuity

𝑛𝐴𝑥∨𝑥∆ 𝑦 ∆ 𝑧−𝑛𝐴𝑥∨𝑥+∆ 𝑥∆ 𝑦 ∆ 𝑧+𝑟 𝐴∆ 𝑥∆ 𝑦 ∆ 𝑧=𝜕 𝜌 𝐴

𝜕𝑡 ∆𝑥 ∆ 𝑦 ∆ 𝑧

Dividing by and letting approach zero,

𝜕𝜌 𝐴

𝜕𝑡 +(𝜕𝑛𝐴𝑥

𝜕 𝑥 +𝜕𝑛𝐴 𝑦

𝜕 𝑦 +𝜕𝑛𝐴𝑧

𝜕 𝑧 )=𝑟 𝐴

Equation of Continuity

𝜕𝜌 𝐴

𝜕𝑡 +(𝜕𝑛𝐴𝑥

𝜕 𝑥 +𝜕𝑛𝐴 𝑦

𝜕 𝑦 +𝜕𝑛𝐴𝑧

𝜕 𝑧 )=𝑟 𝐴

𝜕𝜌 𝐴

𝜕𝑡 +(𝛻 ∙𝑛𝐴 )=𝑟 𝐴

In vector notation,

But form the Table 7.5-1 (Geankoplis)

𝑛𝐴= 𝑗𝐴+𝜌𝐴 𝑣 𝑗 𝐴=− 𝜌 𝐷𝐴𝐵 𝑑 𝑤𝐴/𝑑𝑧and

Equation of Continuity

𝜕𝜌 𝐴

𝜕𝑡 +(𝛻 ∙𝑛𝐴 )=𝑟 𝐴

𝑛𝐴= 𝑗𝐴+𝜌𝐴 𝑣 𝑗 𝐴=− 𝜌 𝐷𝐴𝐵 𝑑 𝑤𝐴/𝑑𝑧

Substituting

𝜕𝜌 𝐴

𝜕𝑡 +(𝛻 ∙𝜌 𝐴 𝑣 )− (𝛻 ∙𝜌 𝐷 𝐴𝐵 𝛻𝑤 𝐴 )=𝑟 𝐴

Equation of Continuity

𝜕𝜌 𝐴

𝜕𝑡 +(𝛻 ∙𝑛𝐴 )=𝑟 𝐴

𝜕𝑐 𝐴

𝜕𝑡 +( 𝜕 𝑁 𝐴𝑥

𝜕𝑥 +𝜕 𝑁 𝐴 𝑦

𝜕 𝑦 +𝜕 𝑁 𝐴 𝑧

𝜕 𝑧 )=𝑅𝐴

Dividing both sides by MWA

Equation of Continuity

Recall: 1. Fick’s Law

𝐽 𝐴∗=−𝐷𝐴𝐵

𝑑 𝑐 𝐴

𝑑𝑧2. Total molar flux of A

𝑁 𝐴= 𝐽 𝐴∗ +𝑐𝐴 𝑣𝑀

𝑁 𝐴=−𝑐𝐷𝐴𝐵𝑑 𝑥𝐴

𝑑𝑧 +𝑥𝐴(𝑁 𝐴+𝑁 𝐵)

Equation of Continuity

𝜕𝑐 𝐴

𝜕𝑡 +( 𝜕 𝑁 𝐴𝑥

𝜕𝑥 +𝜕 𝑁 𝐴 𝑦

𝜕 𝑦 +𝜕 𝑁 𝐴 𝑧

𝜕 𝑧 )=𝑅 𝐴

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

𝜕𝑐 𝐴

𝜕𝑡 +(𝛻 ∙𝑐 𝐴 𝑣𝑀 )− (𝛻 ∙𝑐 𝐷 𝐴𝐵 𝛻 𝑥𝐴 )=𝑅𝐴

Equation of Continuity

𝜕𝑐 𝐴

𝜕𝑡 +(𝛻 ∙𝑐 𝐴 𝑣𝑀 )− (𝛻 ∙𝑐 𝐷 𝐴𝐵 𝛻 𝑥𝐴 )=𝑅𝐴

𝜕𝜌 𝐴

𝜕𝑡 +(𝛻 ∙𝜌 𝐴 𝑣 )− (𝛻 ∙𝜌 𝐷 𝐴𝐵 𝛻𝑤 𝐴 )=𝑟 𝐴

(𝑟𝑎𝑡𝑒𝑜𝑓

𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑖𝑛𝑚𝑜𝑙𝑒𝑠𝑜𝑓 𝐴𝑝𝑒𝑟

𝑢𝑛𝑖𝑡 𝑣𝑜𝑙𝑢𝑚𝑒)(

𝑛𝑒𝑡 𝑟𝑎𝑡𝑒𝑜𝑓𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛

𝑖𝑛𝑚𝑜𝑙𝑒𝑠𝑜𝑓𝐴𝑝𝑒𝑟 𝑢𝑛𝑖𝑡𝑣𝑜𝑙𝑢𝑚𝑒 𝑏𝑦𝑐𝑜𝑛𝑣𝑒𝑐𝑡𝑖𝑜𝑛

)(𝑛𝑒𝑡 𝑟𝑎𝑡𝑒𝑜𝑓𝑎𝑑𝑑𝑖𝑡𝑖𝑜𝑛

𝑖𝑛𝑚𝑜𝑙𝑒𝑠𝑜𝑓𝐴𝑝𝑒𝑟 𝑢𝑛𝑖𝑡𝑣𝑜𝑙𝑢𝑚𝑒𝑏𝑦𝑑𝑖𝑓𝑓𝑢𝑠𝑖𝑜𝑛

)(𝑟𝑎𝑡𝑒𝑜𝑓

𝑝𝑟𝑜𝑑𝑢𝑐𝑡𝑖𝑜𝑛𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓𝐴𝑝𝑒𝑟 𝑢𝑛𝑖𝑡𝑣𝑜𝑙𝑢𝑚𝑒

𝑏𝑦 𝑟𝑒𝑎𝑐𝑡𝑖𝑜𝑛)

Two equivalent forms of equation of continuity

Equation of Continuity

Equation of Continuity

𝜕𝑐 𝐴

𝜕𝑡 +(𝛻 ∙𝑐 𝐴 𝑣𝑀 )− (𝛻 ∙𝑐 𝐷 𝐴𝐵 𝛻 𝑥𝐴 )=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

𝜕𝑐 𝐴

𝜕𝑡 +(𝛻 ∙𝑐 𝐴 𝑣𝑀 )− (𝛻 ∙𝑐 𝐷 𝐴𝐵 𝛻 𝑥𝐴 )=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

𝜕𝑐 𝐴

𝜕𝑡 +(𝛻 ∙𝑐 𝐴 𝑣𝑀 )− (𝛻 ∙𝑐 𝐷 𝐴𝐵 𝛻 𝑥𝐴 )=𝑅𝐴

Special cases of the equation of continuity

3. For constant ρ and DAB (liquids),

𝜕𝜌 𝐴

𝜕𝑡 +(𝛻 ∙𝑛𝐴 )=𝑟 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,

− ( 𝐷𝐴𝐵 𝛻2 𝑐 𝐴 )=𝑅𝐴

Assuming steady-state,

𝜕𝑐 𝐴

𝜕𝑡 − ( 𝐷𝐴𝐵𝛻2 𝑐𝐴 )=𝑅𝐴

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

𝜕𝑐 𝐴

𝜕𝑡 +(𝛻 ∙𝑐 𝐴 𝑣𝑀 )− (𝛻 ∙𝑐 𝐷 𝐴𝐵 𝛻 𝑥𝐴 )=𝑅𝐴

Equation of Continuity

Assuming that diffusion is along the z-direction only,

−(𝐷 𝐴𝐵𝜕2𝑐 𝐴

𝜕 𝑧2 )=𝑅𝐴

− ( 𝐷𝐴𝐵 𝛻2 𝑐 𝐴 )=𝑅𝐴

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

−(𝐷 𝐴𝐵𝜕2𝑐 𝐴

𝜕 𝑧2 )=−𝑘1′ ′ ′𝑐 𝐴

Equation of Continuity

𝐷 𝐴𝐵𝜕2 𝑐 𝐴

𝜕 𝑧2 −𝑘1′ ′ ′𝑐 𝐴=0

Rearranging,

Looks familiar?How to solve this ODE?

𝑐 𝐴

𝑐 𝐴0=

cosh [√𝑘′ ′ ′ 𝐿2

𝐷 𝐴𝐵(1− 𝑧

𝐿 )]cosh (√𝑘′ ′ ′ 𝐿2

𝐷 𝐴𝐵)

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.

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