$9\mathrm{.}7-2.$ Drying Tests with a Foodstuff. In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray dryer with air flow over the top exposed surface having an area of $0\mathrm{.}186$ m$.$ The bone$-$dry sample weight was $3\mathrm{.}765$ kg dry solid. At equilibrium after a long period, the wet sample weight was $3\mathrm{.}955$ kg HO $+$ solid. Hence, $3\mathrm{.}955-3\mathrm{.}765,$ or $0\mathrm{.}190,$ kg of equilibrium moisture was present. The following sample weights versus time were obtained in the drying test: Time $($h$)$ Weight $($kg$)$ Time $($h$)$ Weight $($kg$)$ Time $($h$)$ Weight $($kg$)04\mathrm{.}9442\mathrm{.}24\mathrm{.}5547\mathrm{.}04\mathrm{.}0190\mathrm{.}44\mathrm{.}8853\mathrm{.}04\mathrm{.}4049\mathrm{.}03\mathrm{.}9780\mathrm{.}84\mathrm{.}8084\mathrm{.}24\mathrm{.}24112\mathrm{.}03\mathrm{.}9551\mathrm{.}44\mathrm{.}6995\mathrm{.}04\mathrm{.}150($a$)$ Calculate the free moisture content X kg HO$/$kg dry solid for each data point and plot X versus time. $($Hint: For $0$ h$,4\mathrm{.}944-0\mathrm{.}190-3\mathrm{.}765=0\mathrm{.}989$ kg free moisture in $3\mathrm{.}765$ kg dry solid. Hence, X $=0\mathrm{.}989/3\mathrm{.}765.)($b$)$ Measure the slopes, calculate the drying rates R in kg HO$/$h$.$ m$,$ and plot R versus X$.($c$)$ Using this drying$-$rate curve, predict the total time to dry the sample from X $=0\mathrm{.}20$ to X $=0\mathrm{.}04.$ Use numerical integration for the falling$-$rate period. What are the drying rate Rc in the constant$-$rate period and Xc$?$ Ans. $($c$)$ Rc $0\mathrm{.}996$ kg HO$/$h m$,$ Xc $=0\mathrm{.}12,$ t $=4\mathrm{.}1$ h $($total$)=$