Question

3.4 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 0.186 m². The bone-dry sample weight was 3.765 kg dry solid. At equilibrium after a long period, the wet sample weight was 3.955 kg H2O + solid. Hence, 3.955 - 3.765, or 0.190, kg of equilibrium moisture was present. The following sample weights versus time were obtained in the drying test. Time (h) Weight (kg) Time (b) Weight (kg) (a) Calculate the free moisture content X kg 0.0 4.944 4.2 4.241 H2O/kg dry solid for each data point and plot 0.4 4.885 5.0 4.150 X versus time. 4.808 7.0 4.019 (b) Measure the slopes, calculate the drying rates 4.699 3.978 2.2 R in kg H2O/h.m. and plot R versus X. 4.554 3.955 3.0 (c) Using this drying-rate curve, predict the total 4.404 time to dry the sample from X = 0.20 to X = 0.04. Use graphical integration for the falling-rate period. What is the drying rate RC in the constant-rate period and Xc. 0.8 12.0

please show all calculation step-by-step and sketch

Answer 1

0 In order to test the feasibility of drying a certain food Stuff doying data were obtained in a tvay dryer with air flow over the top exposed Surface having an area OP 0.186 m2. The bone'-dry Sample Weight W3 3.765 kg dry Solid. At equilibrium after a long period, the coot Sample weight coas 8.955 kq HO + solid. Hence 3.955 - 5765 or 0.190. So! The following clata are given A-0.186 m2 Ws = Lg = 3.765 kg dry solid Equilibrium moisture - 0.19 kg H2O @ X. Wr - Wg - equilibrium moisture Ws X = 4.944 - 3.765 -0.19 3.765 1x = 0.26263 kg Hooling doy solid. Time (hors) * kg H20 /kg doy Solid) 0.26268 0.24701 0.22656 lot oot of o 0.19964 0.15940 0.11926 Scanned with CamScanner

b) 3lope R = / 0.24701 -0,21262) 3,765 0:126 TR = 0.79301 R. (8 g H2/162) x C kg H20) kg dooy Solid) 0-993011 0.26262 1.934945 0.24 701 0.946703 0.974462 1. 003065 0- 22656 O. 1976) O. 15910 0.11926 Scanned with LeamScanner

ига лhp 11, 500 0.30 Paroo Most 0 0 14.0 6. 0 O 12.0 6.0 TIME Scanned with CamScanner

t (0.04) = 4..g hury + (0-1) = 0. + b Total t ..4..1 hor Re - 0.999 kg H2O/n.me - 3.765 / 0.20 -0.127 to line 56.186 0,998) It = 1.63 hrs] R VS X chying rate 0.200.250.30 0.05 0.10 0.23 YR 0.966 1.024 1.026 0.20 0.16 0.992 1.369 0.12 0.08 Scann@uobith Cam Saler 1.635 2.840

VR vs x 3000t 2.500 2000 1,500 1.000 0.500 0.000 0.05 0.10 D.IS 7.2 Al ) 1805 X 0.04 = 0.04722 Az = Az = 1,5020 X 0.03 = 0.04 506 1.8175 x 0.01 = 0.01818 A tot = 0.11 te 45 dx - 30365 (0.11) . 2.24 Total t = 2.94 + 1.63 = 3.87 hrs. Scanned with CamScanner

YR V3 X 3.000 2.500 21000 SDD ..500 0.000 0.00 0.05+ 0.100. 150.20025 cs Scanned with CamScanner