Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN: 9781259696527
Author: J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher: McGraw-Hill Education
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Leaching of Oil from Soybeans in a Single Stage. Repeat Example 31.2-1 for single-stage leaching of oil from soybeans. The 100 kg of soybeans contains 22 wt % oil and the solvent feed is 80 kg of solvent containing 3 wt % soybean oil. 

EXAMPLE 31.2-1. Single-Stage Leaching of Flaked Soybeans In a single-stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybeans containing 20 wt % oil is leached with 100 kg of fresh hexane solvent. The value of N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow V1 and the underflow slurry L1 leaving the stage. Solution: The process flow diagram is the same as given in Fig. 31.2- 2a. The known process variables are as follows: for the entering solvent flow, V2 = 100 kg, x42 = 0, XC2 = 1.0; for the entering slurry stream, B = 100(1.0 – 0.2) = 80 kg insoluble solid, Lo = 100(1.0 – %3D 0.8) = 20 kg A, No = 80/20 = 4.0 kg solid/kg solution, yA0 = 1.0. To calculate the location of M, substituting into Eqs. (31.2-4), (31.2-5), and (31.2-6) and solving, L, + V, = 20 + 100 = 120 kg = M %3D LY A0 + V,*42 = 20(1.0) + 100(0) = 120x AM
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Transcribed Image Text:EXAMPLE 31.2-1. Single-Stage Leaching of Flaked Soybeans In a single-stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybeans containing 20 wt % oil is leached with 100 kg of fresh hexane solvent. The value of N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained. Calculate the amounts and compositions of the overflow V1 and the underflow slurry L1 leaving the stage. Solution: The process flow diagram is the same as given in Fig. 31.2- 2a. The known process variables are as follows: for the entering solvent flow, V2 = 100 kg, x42 = 0, XC2 = 1.0; for the entering slurry stream, B = 100(1.0 – 0.2) = 80 kg insoluble solid, Lo = 100(1.0 – %3D 0.8) = 20 kg A, No = 80/20 = 4.0 kg solid/kg solution, yA0 = 1.0. To calculate the location of M, substituting into Eqs. (31.2-4), (31.2-5), and (31.2-6) and solving, L, + V, = 20 + 100 = 120 kg = M %3D LY A0 + V,*42 = 20(1.0) + 100(0) = 120x AM
Lo 3- N N vs. YA V2- M N vs. XA 0.5 1.0 YA, XA Figure 31.2-3. Graphical solution of single-stage leaching for Example 31.2- 1. Hence, XAM = 0.167. B = N,L, = 4.0(20) = 80 = N,(120) N = 0.667 The point M is plotted in Fig. 31.2-3 along with V2 and Lo The vertical tie line is drawn locating L1 and Vị in equilibrium with each other. Then, N1 = 1.5, yA1 = 0.167, xA1 = 0.167. Substituting into Eqs. (31.2-4) and (31.2-6), and solving or using the lever-arm rule, L1 = 53.3 kg and V = 66.7 kg.
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Transcribed Image Text:Lo 3- N N vs. YA V2- M N vs. XA 0.5 1.0 YA, XA Figure 31.2-3. Graphical solution of single-stage leaching for Example 31.2- 1. Hence, XAM = 0.167. B = N,L, = 4.0(20) = 80 = N,(120) N = 0.667 The point M is plotted in Fig. 31.2-3 along with V2 and Lo The vertical tie line is drawn locating L1 and Vị in equilibrium with each other. Then, N1 = 1.5, yA1 = 0.167, xA1 = 0.167. Substituting into Eqs. (31.2-4) and (31.2-6), and solving or using the lever-arm rule, L1 = 53.3 kg and V = 66.7 kg.
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