Question 1(a) Explain the advantages of sedimentation over centrifugal separation.(b) An experimental filter press having an area of 0.05 m2 is used to filter anaqueous BaCO3 slurry at a constant pressure of 250 kPa. The filtration equationobtained was:tV = 8.3x106 V + 2.4×103where t is in s and V is in m³. [µ=0.0012 Pa.s]If the same slurry and conditions are used in a leaf press having an area of 7.0m?, how long will it take to obtain 2 m3 of filtrate?(c) 30 um particles in a water suspension are to be removed using acentrifugal separator operating at 2500 rpm. Calculate the effective radiusrequired for the separation when the velocity of separation is 0.003 m/s and theparticle density is 1450 kg/m3. [The liquid viscosity is 4.2 mPa.s with adensity of 1050 kg/m³](d) Explain the functions of a surfactant in a dispersed system.

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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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Question-3. A piece of titanium (Ti) sheet in a thickness of 5 mm was exposed to a nitrogen-rich gas atmosphere on one side and a nitrogen-deficient atmosphere on the other side at 850 °C. After having reached steady-state, the metal sheet was quickly cooled to room temperature. The nitrogen concentrations at the two surfaces were determined to be 0.02 and 0.0018 wt%. The densities for nitrogen (N) and Ti are 0.808 g/cm³ and 4.5 g/cm³, respectively. Answer the following questions: (a) Convert the nitrogen concentrations from weight percent to kilograms per meter cubed. (b) Compute the diffusion coefficient if the diffusion flux is 2.58 × 10-6 kg/m²-s.
P3
1.6 Properties: Diffusivity Estimate the molecular diameter and diffusion coefficient for the proteins ribonuclease (MW 13,700 Da), hemoglobin (MW 68,000), and urease (MW 480,000), assuming the molecules are spherical and the density of each protein molecule is 1.3 g/cm³.
In a counter-current absorption tower, a process gas containing 0.04 (mole fraction) ammonia is scrubbed with pure water. The exit gas contains 0.002 (mole fraction) ammonia.The diameter of the tower is 1m, and the packing has a large area per volume of 200 m2m–3. The overall mass transfer coefficient based on the liquid side driving force, KL, is 1.5×10−5ms–1. The total gas and liquid concentrations are 0.6kmolm–3and 20 kmolm–3respectively and the gas flowrateis 0.03kmols–1. The actual liquid to gas ratio is 1.5 times that of the minimum liquid to gas ratio. The equilibrium for ammonia between the air and water is given by y* = 1.04x. Draw a rough sketch of the operating and equilibrium lines and label the slopes. How many transfer units are needed? What is the height of the tower?
3)Gas absorption (scrubbing) is an operation in which gas mixture is contacted with a liquid for the purpose of preferentially dissolving one or more components of the gas mixture and to provide a solution of them in the liquid phase. (b)What are the differences between Physical Absorption and Chemical Absorption?Demonstrate their effects on gas absorptionprocess
28.21 A 1 x 10-2 m spherical pellet is sprayed with a very thin coat of paint. The paint contains a volatile solvent. To dry the pellet, a 300 K and 1.013 × 105 Pa air stream flows around it with a bulk velocity of 1 m/s. The estimated loading of the solvent in the wet paint is 0.12 g solvent/cm³. Physical properties are vapor pressure of the solvent = 1.27 x 104 Pa mass diffusivity of solvent in air = 9.62 x 10-6 m²/s kinematic viscosity of air density of air = 1.569 x 10-5 m²/s = 1.177 kg/m³ thermal conductivity of air thermal diffusivity of air heat capacity of air = 2.624 x 10-2 J/m.s. K = 2.216 x 10-5 m²/s = 1.006 J/g. K = 78 g/g mole molecular weight of the solvent Use the McAdam's¹¹ equation Nu = 0.37(Redp) 0.6 (Pr) 1/3, where Redp dp vx V to evaluate (a) the heat-transfer coefficient, h; (b) the mass-transfer coefficient, ke; (c) the molar flux of the solvent into the air stream.
In a counter-current absorption tower, a process gas containing 0.04 (mole fraction) ammonia is scrubbed with pure water. The exit gas contains 0.002 (mole fraction) ammonia.The diameter of the tower is 1m, and the packing has a large area per volume of 200 m2m–3. The overall mass transfer coefficient based on the liquid side driving force, KL, is 1.5×10−5ms–1. The total gas and liquid concentrations are 0.6kmolm–3and 20 kmolm–3respectively and the gas flowrateis 0.03kmols–1. The actual liquid to gas ratio is 1.5 times that of the minimum liquid to gas ratio. The equilibrium for ammonia between the air and water is given by y* = 1.04x. What is the mol fraction of ammonia in the liquid?
1) A biofilm is a layer-like aggregation of microscopic animals, plant or bacteria attached to a solid surface. Assume such a biofilm where the reactant is consumed Bulk fluid only at the solid surface, maintaining a particular value of surface concentration. Calculate the steady-state reactant flux into a biofilm of thickness 0.01 cm, if the diffusivity of the reactant in the biofilm is 0.8 cm²/day. Assume the reactant concentration in the bulk fluid to be uniform at 3.2 mg/liter. At the surface, reactant concentration is 0.25 mg/liter. Biofilm Surface
A double effect evaporator is used to concentrate an 18% CuSO4 solution. At the end of evaporation, 26,000 kg/h of 50% CuSO4 solution were obtained. If the water evaporated in the second effect is half the water evaporated in the first effect, determine: a) The amount of solution fed into the first effect. b) The amount of water evaporated in the first and second effect. c) The amount of solution fed into the second effect, as well as its composition. Problem diagram 18% CuSO4 82% H₂O M₁ M₂ M3 X CuSO4 Y H₂O A 2 M4 = 0.5 M₂ M5 H₂O steam M6 M2 M6 is the water that condenses 26 000 kg/hr 50% CuSO4 50% H2₂O
Consider a diffusion process in a semi-infinite solid with a source at x=0 mm, the concentration of the diffusing species is 2 nm-3 at x=5 mm and t=1 hr (since the start of the process). At what time the concentration at x=15 mm will reach 3 nm-3 ?
34. calculate the residence time of the methane stream in the porous absorber given the solar power of 2.3193 kW and a felt temperature of 1758 K. the methane flow rate is 12683.916793505834 cm^3/min. the porous carbon felt absorber has a porosity of 0.9 and a thickness of 3.2 mm.
In a counter-current absorption tower, a process gas containing 0.04 (mole fraction) ammonia is scrubbed with pure water. The exit gas contains 0.002 (mole fraction) ammonia.The diameter of the tower is 1m, and the packing has a large area per volume of 200 m2m–3. The overall mass transfer coefficient based on the liquid side driving force, KL, is 1.5×10−5ms–1. The total gas and liquid concentrations are 0.6kmolm–3and 20 kmolm–3respectively and the gas flowrateis 0.03kmols–1. The actual liquid to gas ratio is 1.5 times that of the minimum liquid to gas ratio. The equilibrium for ammonia between the air and water is given by y* = 1.04x. If the operator increases the gas flow by 20% while keeping the liquid flow and the entering gas and liquid compositions constant,what are the new mole fractions of ammonia in the exit gas and the exit liquid? Assume that the overall mass transfer coefficient does not change with the gas flow.

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