Unformatted text preview: From Eq. (3105), the average rate of sublimation per unit area is,
nA,B / A = k cavg cA i − cA o
(1)
From Eqs. (3137) and (3138), the average masstransfer coefficient is given by,
k cavg = 0.664 DA,B
N Re L
L 1/ 2 N Sc 1/ 3 = 0.664 0.94 × 10 −5
5 × 105
1 From Example (3.14), cA i =4.3 x 102 kmkol/m3 and 1/ 2 2.41 1/ 3 = 0.0059 m / s cA o = 0. From Eq. (1), nA,B / A = 0.0059 4.3 × 10 −2 − 0 = 2.54 × 10 −4 kmol / s  m2
(b) From Eqs. (3132) and (3133), using physical properties in Example (3.14), with a
Reynolds number of onehalf of 5 x 105, since x = 0.5 m,
k cx = 0.332 DA,B
N Re x
x 1/ 2 N Sc 1/ 3 = 0.332 0.94 × 10 −5
2.5 × 105
0.5 1/ 2 2.41 1/ 3 = 0.00417 m / s From Eq. (3105), the local rate of sublimation per unit area is,
nA,B / A = kcx ( cA )i − ( cA )o = 0.00417 ( 4.3 × 10−2 − 0 ) = 1.79 × 10 −4 kmol/sm 2 Exercise 3.30
Subject: Sublimation of a straight, circular tube of naphthalene (A) into air (B) at 100oC (373
K) and 1 atm that flows through it in laminar flow.
Given: Tube diameter, D, is 5 cm (0.05 m). Reynolds number = NRe = 1,500. Physical
properties in Example 3.14.
Assumptions: Fully established laminarflow velocity profile for air entering the tube.
Negligible pressure drop. Equilibrium at naphthaleneair interface.
Find: Tube length, x, for average naphthalene mole fraction in exiting air, yA L , to equal 0.005.
Analysis: From Example 3.14, yA i = PAs / P = 10 / 760 = 0.0132
By material balance for sublimation of naphthalene into air, the rate of mass transfer over the
tube length must be:
nA = ux S cA x − cA 0
(1)
This rate of mass transfer is given by (3105), which with a logmean driving force is,
nA = k cavg A cA i − cA
ln 0 − cA i − cA cA i − cA (2) 0 cA i − cA x x For the entering air, cA 0 = 0. Therefore, combining Eqs. (1) and (2), and solving for k cavg ,
k cavg = cA i − cA
ux S
ln
A
cA i − cA 0 (3) x From the tube geometry, S = πD2/4 = (3.14)(0.05)2/4 = 0.00196 m2
A = πDx = (3.14)(0.05)x = 0.157 x m2
For laminar flow, Eq. (3150) applies, which contains the Peclet number, Eq. (3102):
N Pe M = N Re N Sc = (1,500)(2.41) = 3,620 From Eq. (3150), using the definition of the Sherwood number in Eq. (3148), but for average
conditions, Exercise 3.30 (continued)
Analysis: (continued)
k cavg = 0.0668 N Pe M / x / D
DA,B
D
N Sh avg = A,B 3.66 +
D
D
1 + 0.04 N
/ x/D 2/3 = Pe M = 0.000688 + N Re = 0.00227
x 1 + 1.28 / x 2 / 3 0.0668 3620 / x / 0.05
0.94 × 10−5
3.66 +
=
2/3
0.05
1 + 0.04 3620 / x / 0.05 (4) Dux ρ
N µ (1,500)(0.000215)
, therefore, u x = Re =
= 68 cm / s = 0.68 m / s
0.0327(29)
µ
Dρ
(5)
1,000
ln cA i − cA 0 cA i − cA x = ln yA i − yA 0 yA i − yA x = ln 0.0132 − 0.0
= 0.476
0.0132 − 0.005 Combining Eqs. (3) and (4),
0.000688 + 0.00227
(0.68)(0.00196)
0.00404
=
(0.476) =
0.157 x
x
x 1 + 128 / x 2 / 3
. Rearranging Eq. (5), f x = 0.000688 +
Solving the nonlinear Eq. (6),
x = 3.73 m 0.00227
0.00404
−
=0
x
x 1 + 1.28 / x 2 / 3 (6) (5) Exercise 3.3...
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 Spring '11
 Levicky
 The Land

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