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Given the problem and formulas used. Please give a detailed solution and explanation how it was solved.

Given the problem and formulas used. Please give a detailed solution and explanation how it was solved.

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:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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Given the problem and formulas used. Please give a detailed solution and explanation how it was solved. 

4. A filter cake 610 mm square and 50 mm thick supported on a screen is dried from both sides with air at 70°C dry bulb and 27°C wet bulb. The air flows parallel with the faces of the cake at a velocity of 2.5 m/s. The dry density of the cake is 1920 kg/m³. The equilibrium moisture is zero and the critical moisture is 9% on a dry basis. Assume an equivalent diameter of 150 mm. a) What is the drying rate at the constant rate period? b) What is the MTC, k' c) How long would it take to dry from 20% to 8% moisture (dry basis)? Given: W=0.09 = Xc X*= 0 A = 610*610 mm² Ax = 50 mm Pdry = 1920 kg/m³ Dv = 150 mm t = 70°C tw = 27°C Air flows parallel: v = 2.5 m/s Required: a) Rc c) 0 (20% b) k' → 8% d.b.)
Transcribed Image Text:4. A filter cake 610 mm square and 50 mm thick supported on a screen is dried from both sides with air at 70°C dry bulb and 27°C wet bulb. The air flows parallel with the faces of the cake at a velocity of 2.5 m/s. The dry density of the cake is 1920 kg/m³. The equilibrium moisture is zero and the critical moisture is 9% on a dry basis. Assume an equivalent diameter of 150 mm. a) What is the drying rate at the constant rate period? b) What is the MTC, k' c) How long would it take to dry from 20% to 8% moisture (dry basis)? Given: W=0.09 = Xc X*= 0 A = 610*610 mm² Ax = 50 mm Pdry = 1920 kg/m³ Dv = 150 mm t = 70°C tw = 27°C Air flows parallel: v = 2.5 m/s Required: a) Rc c) 0 (20% b) k' → 8% d.b.)
BATCH DRYING EXAMPLES Given: W₁ = 0.09 = X₁ X*= 0 A = 610*610 mm² Ax = 50 mm t = 70°C t₁ = 27°C Required: a) Rc Constant Rate Equation in terms of HTE: h = Required: b) k' Pary Air flows parallel: v = 2.5 m/s Constant Rate Equation in terms of MTE: 0 = Rc = 8.8G0-8 D02 h= 27.361 W/m²K A₂ = 2433.16 KJ/kg (based on ts/tw → Table 2-150, 8th /2-69, 9th ) The heat of vaporization AH, is calculated by C11-7)+CZ+C#T=CMT? AH, where T, = T/T, T, is the critical temperature from Table 2-141, AH, is in J/kmol, and T is in K. h in W/m².K G- air mass velocity in kg/s-m² D, -equivalent diameter of airflow channel in m Q ARC h(t - ts) 2s c) e (20%→ 8% d.b.) BATCH DRYING EXAMPLES Given: W₁ = 0.09 = X₂ X*= 0 A = 610*610 mm² Ax = 50 mm Pdry = 1920 kg/m³ Dv = 150 mm t = 70°C t = 27°C Air flows parallel: v = 2.5 m/s Combined CRP and FRP: X₁ = 0.2 đại - xe 0 = 3.325 hrs 1920 kg/m³ Dv = 150 mm X₂ (X₁Xc) + Xc In In Rc = 1.741 kg/hr-m² G = UGPG 1 + H Vμ PG Rc = k'(Hs - H) k'= 93.797 kg/hr-m²-AH G= 2.569 kg/s-m² p= 1.028 kg/m³ X₂ = 0.08 X = 0.09 Q = Pary * Vol = 35.726 kg A = 2(0.61m)² = 0.7442 m²
Transcribed Image Text:BATCH DRYING EXAMPLES Given: W₁ = 0.09 = X₁ X*= 0 A = 610*610 mm² Ax = 50 mm t = 70°C t₁ = 27°C Required: a) Rc Constant Rate Equation in terms of HTE: h = Required: b) k' Pary Air flows parallel: v = 2.5 m/s Constant Rate Equation in terms of MTE: 0 = Rc = 8.8G0-8 D02 h= 27.361 W/m²K A₂ = 2433.16 KJ/kg (based on ts/tw → Table 2-150, 8th /2-69, 9th ) The heat of vaporization AH, is calculated by C11-7)+CZ+C#T=CMT? AH, where T, = T/T, T, is the critical temperature from Table 2-141, AH, is in J/kmol, and T is in K. h in W/m².K G- air mass velocity in kg/s-m² D, -equivalent diameter of airflow channel in m Q ARC h(t - ts) 2s c) e (20%→ 8% d.b.) BATCH DRYING EXAMPLES Given: W₁ = 0.09 = X₂ X*= 0 A = 610*610 mm² Ax = 50 mm Pdry = 1920 kg/m³ Dv = 150 mm t = 70°C t = 27°C Air flows parallel: v = 2.5 m/s Combined CRP and FRP: X₁ = 0.2 đại - xe 0 = 3.325 hrs 1920 kg/m³ Dv = 150 mm X₂ (X₁Xc) + Xc In In Rc = 1.741 kg/hr-m² G = UGPG 1 + H Vμ PG Rc = k'(Hs - H) k'= 93.797 kg/hr-m²-AH G= 2.569 kg/s-m² p= 1.028 kg/m³ X₂ = 0.08 X = 0.09 Q = Pary * Vol = 35.726 kg A = 2(0.61m)² = 0.7442 m²
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