Page 136 - Modelling in Transport Phenomena A Conceptual Approach

116 CHAPTER 4. EVALUATION OF TRANSFER COEFFICIENTS

Solution
Assumption
1. The system is isothermal.

Analysis
The superficial velocity through the packed bed is
0.65
21, = = 0.677 m/ s
(1.2) (0.8)
Substitution of the values into Eqs. (4.6-1) and (4.6-2) gives the friction factor
and the Reynolds numbes. as a function of porosity in the form
-E~ Dp(AP(
f@= l--E pvZL
)
=- e3 [(6 x 10-3)(3200)] = 1.164 ( -
c3
1 - (1.2)(0.677)2(30) 1-€

- [ (6 x 10-3)(0.677)(1.2)] -- - 270.8 ('>
-
1
1.8 x 10-5 1-€ 1--E
Substitution of Eqs. (1) and (2) into Eq. (4.6-5) gives

e3 - 0.476 Z + 2.455 E - 1.979 = 0 (3)
a) Equation (3) can be solved andytically by using the procedure described in Sec-
tion A.7.1.2 in Appendix A. In order to calculate the discriminant, the terms M
and N must be calculated from Eqs. (A.7-5) and (A.7-6)) respectively:
(3)(2.455) - (0.476)2
M= = 0.793
9
- (9)(0.476)(2.455) + (27)(1.979) + (2)(0.476)3
N= = 0.799
54
Therefore, the discriminant is
A = M3+ N2
= (0.793)3 + (0.799)2 = 1.137
Since A > 0, Eq. (3) has only one real root as given by Eq. (A.7-7). The tern 5'
and T in this equation are calculated as
S = (N +

= (0.799 + = 1.231