Help – How to solve questions 11-139 on solid drying in the Tianjin University version of the Fundamentals of Chemical Engineering textbook? SOS
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Could everyone please help me figure out how to solve question 11-139? When drying a certain wet material under constant drying conditions, the processing rate of the wet material is 5 kg/m2, the moisture content on a dry basis is 0.5, the critical moisture content of the material is 0.15 kg of water per kg of absolutely dry material, and the equilibrium moisture content is 0.0215 kg of water per kg of absolutely dry material. The drying rate at a constant speed is 1.8 Kg/m2·h; the drying rate during the deceleration phase is linearly related to the moisture content on a dry basis. Then, what is the time required to reduce the moisture content of the material to 0.05? ? Why is the answer C? Is 0.05 the moisture content on a dry basis?
Reply 1# woshiwang*aona Under constant conditions, it is divided into constant-speed drying and decelerating drying； Constant-rate drying refers to the process of removing non-bound water, and it denotes the stage in which the wet material is dried to its critical moisture content ； The speed-reduced drying stage removes some of the bound water and unbound water, referring to the period from the critical stage to the target stage ； The formula is τ=τ1+τ2=G’(X1-XC)/UCS+G’(XC-X*)/UCS*LN(XC-X*)/(X2-X*)
Reply to 2# zjwmcl: By substituting the values given in the problem, the result is 1.509 hours. Here, 0.05 refers to the moisture content on a dry basis.
Does the moisture content of the wet material need to be converted before it can be used in the formula?
Reply to 4# woshiwang*aona: Yes, the moisture content on a wet basis is given by w=x/(1+x), where x represents the moisture content on a dry basis. It’s best to carefully study the section on drying; there aren’t many useful formulas in the exam handbook for the Chemical Engineering Principles part of the exam, so you need to memorize them yourself
Hello, may I ask you why this doesn’t calculate to 1.5 hours? ? ? Could you please list the process in detail for me? Thank you! ! !
As given in the problem, G = 5 kg/m2, /S = 5 kg/m2, X1 = 0.5, XC = 0.15, X* = 0.02, Uc = 1.8 kg/m2, and X2 = 0.05. The drying time during the constant-speed phase is calculated as follows: τ1 = G’(X1 – XC) / UCS = 5 / 1.8 * (0.5 – 0.15) = 0.972. The drying time during the decelerating drying phase is calculated as: τ2 = G’(XC – X*) / UCS * ln(XC – X*) / (X2 – X*) = 5 * (0.15 – 0.02) / 1.8 * ln(0.15 – 0.02) / (0.05 – 0.02) = 0.530. Therefore, τ = τ1 + τ2 = 1.502
Calculation of drying time in the reduced-speed drying stage: τ2 = G’(XC – X*)/UCS* ln(XC – X*)/(X2 – X*) = 5*(0.15 – 0.02)*1.8*ln(0.15 – 0.02)/(0.05 – 0.02) = 0.530; τ = τ1 + τ2 = 1.502