Page 272 - Chemical Process Equipment - Selection and Design

9.3. CLASSIFICATION AND GENERAL CHARACTERISTICS OF DRYERS 239

EXAMPLE 9.4 dW
Effects of Moist Air Recyde and Increase of Fresh Ai Rate in -loo -
d0
Belt Conveyor Drying
The conditions of Example 9.3 are taken except that recycle of 34.15(R + 1)o.8(H, - H,), 0.58<W<1.16, (6)
moist air is employed and the equilibrium moisture content is =i 60.33(R+ l)0~*(Hs-H,)(W-0.014), W <0.58. (7)
assumed constant at We = 0.014. The material balance in terms of
the recycle ratio R appears on the sketch: When fresh air supply is simply increased by a factor R + 1 and no
recycle is employed, Eq. (1) is replaced by
W=K790R 97.4(R + 1) + 1400(W - 0.1)
W=1581.4R H, =
7790(R + 1)
A = 7790
The solution procedure is:
r 1. Specify the recycle ratio R (lbs recycle/lb fresh air, dry air basis).
S = 1400 2. Take a number of discrete values of W between 1.16 and 0.1.
W = 140 For each of these find the saturation temperature T, and the
drying rates by the following steps.
A = air, W = water, S = dry solid 3. Assume a value of T,.
4. Find H,, Ps, H,, and C from Eqs. (1)-(4).
Humidity of the air at any point is obtained from the water balance
5. Find the value of T, from Eq. (5) and compare with the assumed
value. Apply the Newton-Raphson method with numerical
2581.4R + 97.4 + 1400(W - 0.1)
H, = derivatives to ultimately find the correct value of T, and the
7790(R + 1) corresponding value of H,.
6. Find the rate of drying from Eqs. (6), (7).
The vapor pressure is
7. Find the drying time by integration of the reciprocal rate as in
Example 9.3, with the trapezoidal rule.
ps = exp[11.9176 - 7173.9/(T, + 389.5)] atm. (2) The printout shows saturation temperatures and reciprocal
rates for R = 0, 1, and 5 with recycle; and for R = 1 with only the
The saturation humidity is
fresh air rate increased, using Eq. (8). The residence times for the
four cases are
K = (18/29)pS/(1 -pJ (3)
R = 0, moist air, 0 = 3.667 hrs
The heat capacity is = 1, moist air, =2.841
-5, moist air, = 1.442
C = 0.24 + 0.456-1,. (4)
= 1, fresh air, = 1.699.
With constant air temperature of 170"F, the equation of the Although recycling of moist air does reduce the drying time
adiabatic saturafion line is because of the increased linear velocity, an equivalent amount of
il 900 fresh air is much more effective because of its lower humidity. The
170 - T, = -- (Hs - H,) - (H, - H,). points in favor of moist air recycle, however, are saving in fuel
6: C
when the fresh air is much colder than 170°F and possible avoidance
The drying rate equations above and below the critical moisture of case hardening or other undesirable phenomena resulting from
content of 0.58 are contact with very dry air.

R E 1, fresh air
3

1 1 Rate W 1 I Rate