Engineering, 03.03.2020 21:51 alexportillo859

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?

Answers: 2

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In order to test the feasibility of drying a certain foodstuff, drying data were obtained in a tray...