Problem $33\mathrm{.}10-5$ of the text $($pg $1101)$ Continuous Drying of a Moist Solid. $(70$ pts$)$

Solve this problem as shown below from the problem statement in the textbook, except for the

following changes. Base the calculations on a higher feed rate of $Ls=1,000lbm$ dry solid$/$h

compared to $700$ in the original problem statement, and with the same drying area per unit mass

$(\frac{A}{L}S)$ as in the original problem statement. The entering air is at a higher flow rate of $17,500$

$lbm$ dry air$/$h and at a lower humidity $(H_2)$ of $0\mathrm{.}026lb$ water vapor $/lb$ dry air compared to the

original problem statement. The moist solids enter the process at the wet bulb temperature of the

inlet air stream and remain at this $T_W$ within the dryer.

Drying in a Continuous Tunnel Dryer: A rate of feed of $700l{b}_n$ dry solid$/$h contain$-$

$x_2=0\mathrm{.}0374lb{H}_2\frac{O}{1}b$ dry solid in a continuous$-$counterflow tunnel dryer. A flow of

$13280l{b}_m$ dry air$/$h enters at $203F$ with an $H_2=0\mathrm{.}0562lbH{H}_2\frac{O}{l}b$ dry air. The stock

enters at the wet bulb temperature of $119F$ and remains essentially constant in tem$-$

perature in the dryer. The saturation humidity at $119F$ from the humidity chart is

$H_W=0\mathrm{.}0786lb{H}_2\frac{O}{l}b$ dry air. The surface area available for drying is $(\frac{A}{L_S})=$

$0\mathrm{.}30f\frac{t^2}{l}{b}_m$ dry solid.

A small$-$batch experiment was performed using approximately the same

constant$-$drying conditions, air velocity, and temperature of the solid as in the con$-$

tinuous dryer. The equilibrium critical moisture content was found to be

$x_C=0\mathrm{.}0959lbH{H}_2\frac{O}{l}b$ dry solid, and the experimental value of $k_y{M}_B$ was found as

$30\mathrm{.}15l{b}_{m}ai\frac{r}{h}*f{t}^2./$ In the falling$-$rate period, the drying rate was directly propor$-$

tional to $x.$

For the continuous dryer, calculate the time in the dryer in the constant$-$rate

zone and the falling$-$rate zone.

Ans. $H_C=0\mathrm{.}0593lb{H}_2\frac{O}{l}b$ dry air, $H_1=0\mathrm{.}0760\frac{lbH2O}{lb\text{d}\text{r}\text{y}\text{a}\text{i}\text{r}},t=4\mathrm{.}20h$ in the

constant$-$rate zone; $t=0\mathrm{.}47h$ in the falling$-$rate zone

Note, that these answers are valid for the original problem statement, not the modified

statement here.