Unformatted text preview: Use the ChiltonColburn jfactor method to obtain kc . Combining Eqs. (3165) and (3166), 0.023G N Re
kc =
ρ
N Sc
N Re = −0 . 2 2/3 0.023ux ρ N Re
=
ρ
N Sc −0.2 2/3 = 0.023ux N Re −0 . 2 N Sc 2/3 (3) DG Dux ρ (5.08)(152)(10)
.
=
=
= 86,800 (confirms turbulent flow) (4)
µ
µ
0.0089
N Sc = 0.0089
µ
=
= 970
ρDAB (10)(9.18 × 10−6 )
. (5) Exercise 3.32 (continued)
Analysis: (continued)
Substitution of the results of Eqs. (4) and (5) into (3) gives,
k c = 0.023(152) 86,800
970 −0 . 2 2/3 = 0.00367 cm/s From Example 3.15, equilibrium exists at the inside wall of the tube such that the concentration
of benzoic acid in the water is at its equilibrium value of cA wall = 0.0034 g/cm3
From Eq. (2), cAout =0.0034 1 − exp − (0.00367)(4,870)
(152)(20.25) = 0.000020 g/cm3 Therefore, the concentration of benzoic acid in the exiting water is way below the solubility
value. Exercise 3.33
Subject: Sublimation of a long, circular cylinder of naphthalene (A) to air (B) flowing normal
to it.
Given: Long, circular cylinder of 1inch (0.0254 m) diameter, D. Air at 100oC (373 K) and 1
atm flows at a Reynolds number, NRe , of 50,000. Physical properties given in Example 3.14.
Assumptions: Negligible bulk flow effect.
Find: Average sublimation flux in kmol/sm2.
Analysis: Only masstransfer resistance is in the air. From Eq. (3105), the flux is,
nA / A = N A = k c cA s − cA ∞ (1) Using the ChiltonColburn analogy, Eq. (3169) applies, which combined with Eq. (3165) gives, 0.0266G N Re
kc =
ρ
N Sc −0.195 (2) 2/3 From the Reynolds number, mass velocity, G, is, using µ = 0.0215 cP (2.15 x 105 kg/sm2),
G= N Re µ (50,000)(2.15 × 10−5 )
=
= 42.3 kg / s  m2
D
0.0254 From Example 3.14, gas density = 0.0327(29) = 0.948 kg/m3 and NSc = 2.41. 0.0266G N Re
From Eq. (2), k c =
ρ
N Sc −0.195 2/3 −0.195 0.0266(42.3) 50,000
=
2/3
0.948
2.41 = 0.080 m/s Using properties in Example 3.14,
cA s PAs
10
= yA s c =
c=
(0.0327) = 4.3 × 10−4 kmol / m3 and cA ∞ = 0
P
760 ( ) From Eq. (1), N A = kc cA s − cA∞ = (0.080)(4.3 × 10−4 − 0) = 3.44 ×10 −5 kmol/sm2 Exercise 3.34
Subject: Sublimation of a sphere of naphthalene (A) into air (B) flowing past it.
Given: Sphere of 1inch (0.0254 m) diameter, D. Air at 100oC (373 K) and 1 atm flows at a
Reynolds number, NRe , of 50,000. Physical properties given in Example 3.14.
Assumptions: Negligible bulk flow effect.
Find: Initial average sublimation flux in kmol/sm2 and comparison to that in a bed packed
with the same spheres at a void fraction of 0.5.
Analysis: Only masstransfer resistance is in the air. From Eq. (3105), the flux is,
nA / A = N A = k c cA s − cA ∞ (1) Using the ChiltonColburn analogy, Eq. (3170) applies, which combined with Eq. (3165) gives, 0.37G N Re
kc =
ρ
N Sc −0 . 4 2/3 (2) From the Reynolds number, mass velocity, G, is, using µ = 0.0215 cP (2.15 x 105 kg/sm2),
G= N Re µ (50,000)(2.15 × 10−5 )
=
= 42.3 kg / s  m2
D
0.0254 From Example 3.14, gas density = 0.0327(29) = 0.948 kg/m3 and NSc = 2.41.
−0 . 4 0.3...
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 Spring '11
 Levicky
 The Land

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