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 (water plus solid). Hence, 3.955-3.765 or 0.190 kg of equilibrium moisture was present. The following table provides the sample weights versus time during the drying test:

Time (hr)
Weight (kg)
Time (hr)
Weight (kg)
Time (hr)
Weight (kg)

0
4.944
2.2
4.554

7.0
4.019
0.4
4.885

3.0
4.404
9.0
3.978

0.8
4.808
4.2
4.241

12.0
3.955
1.4
4.699

5.0
4.150

A) Calculate the free moisture content, X (kg water per kg dry solid) for each data point and plot X versus t;
B) Using the slope, calculate the drying rate, R in kg water per hour per square meter and plot R versus X;
C) Using this drying rate curve, predict the total time to dry the sample from X = 0.20 to X = 0.04. Use numerical integration in the falling rate period. What is the drying rate in the constant rate period and what is X in the constant rate period?

Answer / Explanation:

(A)  To start, we calculate the free moisture content  multiplies by Kg H₂O / Kg Dry solid for each data point and plot X versus Time

Therefore,

X = Wt - Ws - Equilibrium Moisture / Ws

= 4.944 - 3.765 - 0.190 / 3.785

 = 0.26268 Kg H₂0 / Kg dry solid

(a) To also calculate the free moisture content X Kg H₂O / Kg Dry solid for each data point and plot X versus Time.

Kindly refer to the plotted graph below of X against or versus Time

(B) To measure the slope, we calculate the drying rate in Kg H₂0 /H.M²

Now, recalling the equation for calculating slope:

we have Slope = ΔX /ΔT

Therefore, R =Ls ΔX / A ΔT

Hence, R = 3.765  ( 0.24701 - 0.26268) / 0.186 ( 0.4 - 0)

= 0.79301

(b) To measure the slope, we calculate the drying rate in Kg H₂0 /H.M²

Kindly refer to the plotted graph below of R against or versus X

(C) Since t ( 0.4) = 4.8 hours

                t (0.2 ) = 0.7 hours

total time (t)  = 4.1 hours

This can also be represented in the graph below:

(c) USING THIS DRYING RATE CURVE, PREDICT THE TOTAL TIME TO DRY THE SAMPLE FROM X=0.20 TO X=0.04. USE GRAPHICAL INTERGRATION FOR THE FALLING-RATE PERIOD. WHAT IS THE DRYING RATE R  IN THE CONSTANT-RATE PERIOD AND X

Moving forward,

Rc = 0.998 Kg H₂0 /hm²

= t = Ls dx/ A R = 3.765 / 0.186 ( 0.20 - 0.12 / 0.998)

= 1.63

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