Page 302 - Chemical Process Equipment - Selection and Design

266 DRYERS AND COOLING TOWERS

EXAMPLE 9.8 At a tower diameter of 0.6 m,
Sizing a Pneumatic Conveying Dryer
A granular solid has a moisture content of 0.035 kg/kg dry material u=-- Q 25.0m/sec at 450"C,
which is to be reduced to 0.001 kg/kg. The charge is at the rate of 0.3644 - { 13.6 m/sec at 120°C.
9.72 kg/sec, is at 60°C and may not be heated above 90°C. Inlet air
is at 450°C and has a moisture content of 0.013 kg/kg dry air. These velocities are great enough to carry the largest particles with
settling velocity of 10 m/sec.
Equations are developed over intervals in which Wl+ W,,
Tl+ T,, and Ti+ Ti.
The procedure will be:
1. Start with known Wl, T,, and Ti.
2. Specify a moisture content W,.
3. Assume a value & of the solid temperature.
4. Calculate Ti from the heat balance.
5. Check the correctness of T, by noting if the times for heat and
mass transfers in the interval are equal.
Q Q
6 --=
-ha(AT), 0.47(AT),,

rn, =
ma * Heat balance:
T; = 450 C T, = 60 C fis[0.391( T, - TI) + (W1- Wz)( T2 - TI + 563)]
9, = 0.013 kgkg W, = 0.035 kglkg =fila{[0.25 + 0.48(0.001](T; - Ti)
Specific gravity of the solid is 1.77 and its heat capacity is +0.48(W,-Wz)(T;-60)}.
0.39 cal/g "C. The settling velocity of the largest particle present,
2.5mmdia, is 10m/sec. Heat capacity of the air is taken as Substitute fis/iiit, = 9.72/3.46 = 2.81 and solve for T;.
0.25 cal/g "C and the latent heat at 60°C as 563 cal/g. Experimental
data for this system are reported by Nonhebel and Moss (1971, pp. T; =
240ff) and are represented by the expressions: -0.25048T; + 28.8(W1 - W,) + 2.81
Heat transfer coefficient: x [O. 39( T2 - q) + ( Wl - W,)( T2 - T, + 563)]
ha = 0.47 cal/(kg solid)("C).
0.48(W1 - W,) - 0.25048
Vapor pressure:
- (1)
P = exp(13.7419 - 5237.0/T), atm, K.
g, = 0.013 +,". (W, - 0.013) = 0.013 + 2.81(W, - 0.013). (2)
ma
Mass transfer coefficient:
pl= g1 = g1 (partial pressure in air).
kga = exp(-3.1811- 1.7388 In w - 0.2553(1n w)', 18/29 + g, 0.6207 + g, (3)
g, = 0.013 + 2.81(W2 - 0.013). (4)
where w is the moisture content of the solid (kg/kg) in the units kg
water/(kg solid)(atm)(sec). p - g2 (5)
In view of the strong dependence of the mass transfer - 0.6207 + g, .
coefficient on moisture content and the 35-fold range of that Pa, = exp[13.7419 - 5237.9/(T, + 273.2)], vapor pressure. (6)
property, the required residence time and other conditions will be Pa, = exp[13.7419 - 5237.9/(& + 273.2)]. (7)
found by analyzing the performance over small decrements of the
moisture content.
An air rate is selected on the assumption that the exit of the Ti - TI - (Ti- T,)
solid is at 85°C and that of the air is 120°C. These temperatures (AT)"=ln[(T;- T,)/(T;- T,)]' (9)
need not be realized exactly, as long as the moisture content of the
exit air is below saturation and corresponds to a partial pressure less AQ=O.391(T,- T1)+(W1-W,)(T2-T,+563),
than the vapor pressure of the liquid on the solid. The amount of per kg of solid. (10)
heat transferred equals the sum of the sensible heat of the wet solid @ = 0.5(W1 + WJ. (11)
and the latent heat of the lost moisture. The enthalpy balance is kga = exp[-3.1811- 1.7388 In W - 0.2553(1n W)"]. (12)
based on water evaporating at 60°C:
Oh = AQ/ha(AT),, = AQ/0.43(AT),,, heating time. (13)
fis[(0.39 + 0.001)(85 - 60) + (0.035 - 0.001)(85 - 60 + 563)] Om = (W, - W,)/k,a(AP),, mass transfer time. (14)
= Za[(O.25 + 0.480(0.001))(450 - 120) + 0.48(0.034)(120 - 60)], Z = eh - 9, + 0 when the correct value of T,
has been selected. (15)
After the correct value of T, has been found for a particular
interval, make W,+Wl, Tz+Tl, and T;+T;. Specify a