Question 2, Morning Session, 2014 Professional Case Study
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Dry material with a wet bulk density of 650 Kg/m3 and 30 wt% water content in a rotary vacuum batch dryer. The dryer has a working volume of 5.5 m3, a wetted surface area of 19.1 m2, an actual drying rate of 4.9 Kg of water/(h·m2), and an operational peak rate of 7.4 Kg of water/(h·m2). If the water content in the dried product can be disregarded, what are the values for the amount of dried product produced per batch (in Kg) and the average production rate (in Kg/h)? (A) 2503, 243.9 (B) 2503, 218.4 (C) 3520, 243.9 (D) 3520, 218.4
I found this question quite awkward when it was tested last year. The amount of dried product is G = G’ * ((1 – x0) / (1 – x1)) = 650 * 5.5 * ((1 – 0.3) / (1 – 0)) = 2502.5 kg per batch. For the second question, no information regarding the non-drying time was provided, nor was it suggested that this time should be ignored. One can only consider it to be negligible; Q=2503/(650*0.3*5.5/(4.9*19.1))=218.4 kg of product per hour
The example on rotary drum filtration in the Tianjin University edition of Chemical Engineering Principles is slightly simpler than the examples in the book, with different data; the solution method is basically the same as that of the examples. It is recommended to take a look at the Tianjin University version of Chemical Engineering Principles.