Unformatted text preview: ter packed with 4 x 6 mesh silica gel with an external void
fraction of 0.5. Gas is benzene vapor in air, which flows through the bed at a benzenefree flow
rate of 25 lb/min.
Find: External gastoparticle masstransfer and heattransfer coefficients at a location in the
bed where the pressure is 1 atm, temperature is 70oF, and benzene bulk mole fraction = 0.005.
Analysis: From Perry's Handbook, 4mesh and 6mesh screens have openings of 0.476 cm and
0.336 cm, respectively, with an average of (0.476 + 0.336)/2 = 0.406 cm. Aasssuming the silica
gel is crushed particles, the sphericity = ψ = 0.65. Therefore, the effective particle diameter =
0.65(0.406) = 0.264 cm = 0.00264 m = Dp. 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) Because the gas is dilute in benzene, use the properties of air in the coefficient correlations.
At 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 benzene in air at 1 atm = 0.077 cm2/s.
From Eq. (336), the diffusivity is proportional to T to the 1.75 power.
Therefore, Di = 0.077[(70 + 460)/(32 + 460)]1.75 = 0.0877 cm2/s = 0.0877 x 104 m2/s.
PM
1013(29)
.
From the ideal gas law, ρ =
=
= 120 kg / m3
.
RT 8.314(530 / 18)
.
CP µ 1090(183 × 10 −5 )
.
=
= 0.779
k
0.0256
µ
183 × 10−5
.
N Sc =
=
= 174
.
ρDi 1.20(0.0877 × 10−4 ) N Pr = Exercise 15.16 (continued) Analysis (continued)
In the Reynolds number, G = mass flow rate/bed crosssectional area
bed crosssectional area = Ab = πD2/4 = 3.14(2)2/4 = 3.14 ft2 or 0.292 m2
Air flow rate = 25 lb/min = 25/60 = 0.417 lb/s or 0.417/29 = 0.0144 lbmol/s
Benzene flow rate = (0.005/0.995)0.0144 = 0.000072 lbmol/s or 0.000072(78.11) = 0.0056 lb/s
Total gas mass flow = 0.417 + 0.0056 = 0.423 lb/s or 0.192 kg/s
Gas mass velocity = G = 0.192/0.292 = 0.658 kg/m2s
NRe = 0.00264(0.658)/1.83 x 105 = 94.9
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,
0.0877 × 10 −4
0.6
1/ 3
2 + 1.1( 94.9 ) (1.74 )
kc =
= 0.0741 m/s
0.00264 h= 0.0256
0.6
1/ 3
= 170.1 J/m 2 sK
2 + 1.1( 94.9 ) ( 0.779 )
0.00264 Exercise 15.17
Estimation of external gastoparticle masstransfer and heattransfer coefficients in a fixedbed
adsorption column.
Given: Fixed bed of 12.06cm inside diameter packed with 3.3mmdiameter Alcoa F200
activated alumina beads with an external void fraction of 0.442. Gas is air containing water
vapor, which flows through the bed at a flow rate of 1.327 kg/min. Find: External gastoparticle masstransfer and heattransfer coefficients at a location in the
bed where the pressure is 653.3 kPa, temperature is 21oC , and the dewpoint temperature =
11.2oC....
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 Spring '11
 Levicky
 The Land

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