Unformatted text preview: Analysis: Dp = 3.3 mm = 0.0033 m. Provided that the dimensionless groups are in the specified
ranges of the correlations, Eqs. (1565) and (1566) can be used to estimate the coefficients: Dp G
D
k c = i 2 + 11
.
µ
Dp
Dp G
D
h = i 2 + 11
.
µ
Dp 0 .6 0.6 µ
ρDi 1/ 3 CP µ
k 1/ 3 (1) (2) At the dewpoint temperature, the vapor pressure of water = 0.192 psi or 1.32 kPa. Therefore, at
653.3 kPa, the mole fraction of water in the air is only 1.32/653.3 = 0.002. Because the gas is so
dilute in water vapor, use the properties of air in the coefficient correlations.
At 21oC = 70oF,
µ = 183 micropoise = 1.83 x 105 kg/ms
k = 0.0256 J/msK
CP = 1.09 kJ/kgK = 1090 J/kgK
From Perry's Handbook, at 32oF, the diffusivity of water vapor in air at 1 atm = 0.22 cm2/s.
From Eq. (336), the diffusivity is proportional to T to the 1.75 power and inversely proportional
to the pressure. Therefore,
Di = 0.22(101.3/653.3)[(70 + 460)/(32 + 460)]1.75 = 0.0388 cm2/s = 0.0388x104 m2/s.
At the high pressure of 653.3 kPa, the compressibility factor of air = 0.998. From the generalized
PM
653.3(29)
gas law, ρ =
=
= 7.75 kg / m3
ZRT 0.998(8.314)(530 / 18)
.
CP µ 1090(183 × 10 −5 )
.
=
= 0.779
k
0.0256
µ
183 × 10 −5
.
N Sc =
=
= 0.609
ρDi 7.75(0.0388 × 10 −4 ) N Pr = Exercise 15.17 (continued)
Analysis (continued)
In the Reynolds number, G = mass flow rate/bed crosssectional area
Bed crosssectional area = Ab = πD2/4 = 3.14(0.1206)2/4 = 0.0114 m2
Gas flow rate = 1.327 kg/min = 1.327/60 = 0.0221 kg/s
Gas mass velocity = G = 0.0221/0.0114 = 1.94 kg/m2s
NRe = 0.0033(1.94)/1.83 x 105 = 350
All dimensionless groups and the particle diameter are in the specified ranges of applicability of
the correlations of Eqs (1) and (2). Therefore, using those two equations, kc = h= 0.0388 × 10−4
0.6
1/ 3
2 + 1.1( 350 ) ( 0.609 )
= 0.0392 m/s
0.0033 0.0256
0.6
1/ 3
= 280 J/m 2 sK
2 + 1.1( 350 ) ( 0.779 )
0.0033 Exercise 15.18
Subject: Effective diffusivity of acetone in the pores of activated carbon. Given: Activated carbon with ρp = 0.85 g/cm3, εp = 0.48, dp = 25 Angstroms = 25 x 108 cm,
and tortuosity = τ = 2.75. Diffusion of acetone vapor in nitrogen, where from Example 15.7,
temperature = 297 K, pressure = 136 kPa, and molecular (bulk) diffusivity = Di = 0.085 x 104
m2/s or 0.085 cm2/s.
Assumptions: Negligible surface diffusion.
Find: Effective diffusivity of acetone vapor through nitrogen in the pores, accounting for bulk
(molecular) diffusion and Knudsen diffusion.
Analysis: From a modification of Eq. (1575), omitting surface diffusion, De = εp 1 τ 1
1
+
Di DK = 0.48
2.75 1
1
1
+
0.085 DK = 0.175
1
11.76 +
DK (1) From Eq. (1421), using 58 for the molecular weight of acetone,
DK = 4850 dp(T/Mi)1/2 = 4850(25 x 108)(297/58)1/2 = 0.00274 cm2/s
Substituting this value into Eq. (1) gives:
0.175
0.175
De =
=
= 4.66 × 10−4 cm 2 /s
1
11.76 + 364
11.76 +
0.00274
It is noted that Knudsen diffusion is very important and largely controls diffusion. Exercise 15.19
Subject...
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 Spring '11
 Levicky
 The Land

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