1. E. coli is grown in a chemostat with a working volume of 1.0 litre. A medium containing 4 g litre¹ ofglucose as growth limiting substrate is fed to the fermenter at a constant flow rate of 0.5 litre h¹¹. Theglucose concentration in the outlet stream is 0.2 g litre-1. The cell yield (Yxs) is 0.42 g g¹. Determinethe cell concentration in the output stream and the specific growth rate.[Ans: x= 1.6 g L¹; μ = 0.5h¹]

Given data: Volume of chemostat = 1 L Initial concentration of substrate = 4 g/L Final concentration…

Answered: E. coli is grown in a chemostat with a…

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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An evaporation-crystallization process (as shown below) is used to obtain solid potassium sulfate from an aqueous salt solution. The feed contains 18.7 wt% K2SO4. The wet filter cake contains solid sulfate crystals and a 40.0 wt% salt solution. The filtrate is recycled to mix with the feed. The evaporator has a capacity of 175 kg of water evaporated per second. The filtrate has 40.0 wt% salt solution, as well, and from the water entering the evaporator, 45.0% is evaporated. In the wet filter cake, there is 10 kg of the solid crystals for every 1 kg of the salt solution. a.) Calculate the fresh feed rate. (Include units) b.) Determine the production rate of crystal. (Include units) c.) Calculate the recycle ratio. (kg recycle/kg feed) Include units
Write down the total and species (virus) mass balances for the air in the plane. Derive an equation for the concentration of virus particles in the air in the plane as a function of time.
6..... 70 m^3 of air flows in an intake airway towards a working section. Before entering the working section 10 m^3/s is drawn off and passed through a filter leaving with a dust concentration of 120p/ml. 50 m^3 is passed through a bulk air cooler leaving with a dust concentration of 100 p/ml. After the remixing of the air streams and before the air enters the working section,the dust count of the mixture is 130p/ml. Calculate the dust concentration of the original 70 m^3/s.
An evaporation-crystallization process (as shown below) is used to obtain solid potassium sulfate from an aqueous salt solution. The feed contains 18.7 wt% K2SO4. The wet filter cake contains solid sulfate crystals and a 40.0 wt% salt solution. The filtrate is recycled to mix with the feed. The evaporator has a capacity of 175 kg of water evaporated per second. The filtrate has 40.0 wt% salt solution, as well, and from the water entering the evaporator, 45.0% is evaporated. In the wet filter cake, there is 10 kg of the solid crystals for every 1 kg of the salt solution. a.) Calculate the fresh feed rate. (Include units) b.) Determine the production rate of crystal. (Include units) c.) Calculate the recycle ratio. (kg recycle/kg feed) Include units
Salt can be made from seawater by solar evaporation. In the first stage of the process, all of the CaCO3 and CaSO4 are precipitated and then the solution is sent to a crystallization pond. In the crystallization pond, water is evaporated and crystalline NaCl is recovered. The pond outlets are a stream containing 91% of the inlet NaCl (pure NaCl crystals) and a stream of mother liquor (the remaining solution), which has an sg of 1.26. The total salt content of the inlet sea water is 3.7 wt% with an sg of 1.03. The composition of the inlet sea water on a dry basis is 78 wt% NaCl, 11 wt% MgCl2, 4.5 wt% MgSO4, 3.5 wt% CaSO4, 2.5 wt% K2SO4, 0.3 wt% CaCO3 and 0.2 wt% MgBr2. Assume that the losses of NaCl, MgSO4, MgCl2, MgBr2, and K2SO4 are negligible. a) Calculate the water (ton/hr) evaporated for an inlet sea water flowrate of 1 ton/hr. b) Calculate the concentration (wt% total salt) of the mother liquor.

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