Tutorial-6-Solution - tute 6 solution PDF

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tute 6 solution...

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HEAT AND MASS TRANSPORT PROCESSES TUTORIAL 6 SOLUTION 1. The problem described replicates the example given in the lecture notes, where Fick's Second Law is solved for the case of a semi-infinite fluid. In this case the semi-infinite fluid is Nitrogen. Initially, the concentration of n-octane in the bulk of the nitrogen phase is zero CA0 = 0 We assume that the concentration of octane in the gas phase at the interface (CAi ) is the equilibrium value. This, in turn, is given by the vapour pressure of n-octane at 20C. y Ai = C=

10.45 = 0.0138 760

n P 101325 = = = 41.6 mol/m3 V RT 8.314 x 293

⇒ C = 0.574 mol/m3 Ai a) Use Equation 5.25 in the Lecture notes y = 0.2 m DAB = 5.09x10-6 m2/s y 2 DAB t

=

0.2 2 5.09 x10 − 6 x 3600

t= 1 hour = 3600 s

= 0.74

By Reference to the Table in the lecture notes: Erf(0.74) = 0.705 Erfc(0.74) = 1 - 0.705 = 0.295 ∴

C A − C A0 CA = = 0.295 C Ai − C A 0 0.574

CA = 0.17 mol/m3

yA = 0.004

b) At the surface y = 0 Equation 5.26 in the lecture notes becomes:

N A = ( C Ai − C A0 )

DAB πt

i)

The initial mass transfer rate is calculated to be infinite (this is consistent with the assumption that the concentration at the interface instantaneously rises to the equilibrium value!)

ii)

At t = 3600 s NA= 1.22 x 10-5 mol/m2s

iii)

At t = 86,400 NA= 2.49 x 10-6 mol/m2s

c)

We integrate the molar transfer rate over time: 24 x 3600

∫ N A dt

=

DAB

c Ai

86400

π

0

∫t

−1 / 2

dt

0

DAB t

= 2c Ai

π

= 0.43 mol/m2 Mass transfer area =

π d2

= 0.0020 m2

4 Therefore the total evaporation = 0.43 x 0.002 = 8.4x10-4 mol = 8.4x10-4 x 114.2 = 0.096 g 2.

Gas Stream:

Superficial fluid velocity,

v 0 = 50

l 1m3 1 min 4 x x x min 1000l 60s π (0.146) 2

= 0.050m / s Referring to the Table of Mass Transfer Coefficients from Cussler in the Tables section of the notes and using the first correlation for gas in a packed tower: k v  = 3.6  0  aD  av 

0. 70

v    D

1/ 3

(ad )−2.0 (Physical property data for pure air used)

0.05 k   = 3.6 -5 −6  364x 2.26 x10  364x 16.8x 10 

0. 70

 16.8 x10 −6    -5   2.26 x10 

1/ 3

(364 x.013)− 2.0

kc = 0.0052 m/s ky = c kc = kG=

ky P

=

P 101325 kc = x0.0052 = 0.213 mol / m 2 s RT 8.314x 298

0.213 6 = 2.10 x10 − mol/m2.Pa.s 101325

Liquid Stream:

Superficial fluid velocity,

l 1m3 1 min 4 x x x min 1000l 60 s π (0.146) 2

v 0 = 1.26

= 1.25 x10 −3

From the tables, at 298K, v =

m/ s

µ 8.95 x10 −4 = = 8.98x10 −7 m 2 / s ρ 996.3

Using the first correlation for liquid in a packed tower on page 6-7: 1/ 3

 1  k    vg 

v  = 0.0051 0   av  1/ 3

1   k  −7  8.98 x10 x9.81 

0..67

D    v

0.5

(ad )0.4

(Physical property data for pure water used)

 1.25 x10− 3   = 0.0051 −7   364 x8.98 x10 

0..67

 1.83x10- 9    −7   8.98x10 

0.5

(364x.013)0.4

kc = 2.17x10-5 m/s Using the second correlation:  kd     D

1/ 3

0. 45

 dv  = 25  0   v 

v    D

0.5

 .013 x1.25 x10 −3  k x 0.013  25 =   -9  −7  1.83x10   8.98x10

0. 45

   

 8.98x 10−7     1.83 x10 -9   

0.5

kc = 2.87x10-4 m/s While the third correlation gives:

k  dv  =α 0  v0  v 

−0.3

 D   v 

0.5

 .013x1.25x10 −3  k   − = 1 −7 1.25 x10 3   8.98 x10

−0.3

0.5

 1.83x10-9    −7   8.98x10 

kc = 2.37x10-5 m/s Using the first correlation as the best available: kL = kc = 2.17x10-5

m/s

Assuming a dilute solution, the concentration can be calculated as that of pure water: kx = ckc =

mol 996.3 x10 3 g m 2 x x2.17 x10 −5 = 1.20 mol/m s 3 18g m s

Whitman Film thicknesses: D D ⇒ δ= kc = δ kc

⇒

δ = 4.35x10-3 or 4.35 mm for the gas phase δ = 8.43x10-5 m or 84 micron for the liquid

b) 1 1 m 1 1.414 = + = + KOy k y k x 0.213 1.20 KOG =

⇒ KOy = 0.170 mol/m2s

K OY 0.170 = = 1.68 x 10-6 mol/m2s.Pa P 101325

1 1 1 1 1 = + = + KOx kx mky 1.20 1.414 x0.213 (Note that K OL =

⇒ KOx = 0.241 mol/m2s

Koy also may be used to calculate Kox)

KOx 18g m3 = x x 0.241 = 4.36x10 − 6m/s c mol 996.3 x10 3 g

The gas phase is dominating the mass transfer process as KOY is similar to ky whereas KOX is somewhat smaller than kx. However, the effect of the liquid phase is still significant and I would be reluctant to use the phrase 'gas phase control'. c) yA = 0.025 NA = 0.001 mol/m2s N A = K Oy ( y A − y Ae ) ⇒

y Ae = y A −

yAe 0.0191 = = 0.0135 m 1.414 y 0.025 = A = = 0.0177 m 1.414

NA 0.001 = 0.0191 = 0.025 − 0.170 K Oy

xA =

x Ae

yA =0.025

Check: K Ox (x Ae − x A ) = 0.241(0.0177 − 0.0135) = 0.001 N A = k y ( y A − y Ai ) = 0.213(0.025 − y Ai )

⇒ yAi = 0.0203 N A = k x (x Ai − x A ) = 1.20(x Ai − 0.0135) ⇒ xAi = 0.0143

yAi =0.0203 xAi =0.0143 xA =0.0135

3. a) Sh Re

Sc

⇒

=

ρvd µ

=

1.146 × 1× 10 × 10−3 1.846 × 10 −5

ρN 2 =

=

621

=

=

2.225

Sh

=

2 + 0.6 (621)1/2 (2.225)1/3

kd D

=

21.52

⇒k

=

21.52 × 7.24 × 10− 6 10 × 10 − 3 0.0156 m/s

is analogous to

 Cp µ     k 

Sh  kd  :    D

 hx     k 

Pr

101325 × 28.02 8.314 × 10 3 × 298

ρDAB 1.846 ×10 −5 1.146 × 7.24 × 10− 6

Nu

pM B RT

µ

=

= b)

0.6

Sc1/3

2

=

+

Re1/2

=

is analogous to

heat or mass transfer rate molecular transport rate

Sc

 µ   :   ρD 

viscous forces molecular transport rate

Sc = 2.225 which is in the range 0.71...