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Pauline M. Doran Bioprocess Engineering Principles. Pages·· MB·4, Downloads. Example 5. 6: Calculation of heat of reaction from heats of. Solution Manual for Bioprocess Engineering Principles – 2nd Edition Author(s): Pauline M. Doran it include chapter 2 t. downloads Views KB Size Bio Process Engineering Principles [Solutions Manual] - P. Doran () WW. Bioprocess Engineering Chap 7 Solutions. Bioprocess engineering. Download Book Bioprocess Engineering Principles, by Pauline M Doran Ph D. Bioprocess.

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As revolutions is a non-dimensional quantity Section 2. Therefore: 1 kcal 4. From Table A. Case 2 Convert to units of kg, m, s. Other sources may also be used. Answer: 0. Answer: 2. Answer: 1. Assuming that Cp changes linearly with temperature between K and K, the value at As fermentation medium is mostly water, it is reasonable to assume that the density of liquid in the fermenter is the same as that of water.

Unless the culture produces a highly viscous extracellular product such as gum, it is also reasonable to assume that the viscosity of liquid in the fermenter is the same as that of water. From Eq.

The parameter values and conversion factors can now be used to calculate the dimensionless groups in the equation for the Sherwood number.

Therefore, these groups are dimensionless. The dimensions of h can be deduced from its units. From Table 2. From Section 2. Answer: ii. Using the same procedure as in a : 3. Therefore, from Section 2. From the definition of specific volume in Section 2.

Therefore, in 1 gmol of air, there are 0. The molecular weight of air is equal to the number of grams in 1 gmol of air: In g solution, there are 30 g water, 25 g ethanol, 15 g methanol, 12 g glycerol, 10 g acetic acid, 8 g benzaldehyde, and no other components. The total number of moles is 1. Add water to dissolve the solid and make up to the ml mark on the measuring cylinder. Therefore: g 1 gmol Therefore: g 1 gmol 4.

The amount of sucrose transferred in 30 min is: 60 s From the atomic weights in Table C. Because the solution being considered contains only very dilute concentrations of components in water, we can assume that the density of the solution is the same as the density of water or 1 g ml—1 Section 2.

Answer: 25 mM. After addition of the water, there are Therefore, the concentration is As 1 gmol l—1 is the same as 1 M Section 2. Therefore, the mass corresponding to 50 gmol of NaCl is: This mass of NaCl is present in l of solution. Therefore, the concentration is 0. Therefore: 1l 0. Therefore, the concentration of NaCl calculated in iii , 5. The mass of NaCl present in g of solution is 5. Answer: ppm, assuming that the density of the solution is equal to the density of water.

For relatively light gases such as oxygen, carbon dioxide, ammonia and nitrogen, we can assume that the ideal gas law is valid over the range of conditions applying to bioreactor operation Section 2. This means that the relative partial volumes of the component gases will not change with temperature and pressure, so that the gas composition will be unaffected.

Answer: The composition is unaffected. If the specific gravity of the solution is 1. Therefore, the concentration of pharmaceutical is Converting units gives: cm3 1 kg 0. Therefore: 6. If the barometric pressure is Let us assume that the ideal gas law applies to compressed air under the prevailing conditions Section 2. Temperature in the ideal gas equation is absolute temperature; therefore, using Eq. The ratio of the amounts of air in the tank before and after the leak can be determined using Eq.

Substituting values into the equation: 1 atm 0. The amount of air provided to the bioreactor between 4 pm Friday and 9 am Monday, i. Using Eq. Applying Eq. Answer: No. The maximum theoretical yield from the stoichiometric equation is 1 gmol of penicillin for every 1. This is equivalent to Therefore, the actual yield of penicillin from glucose is Using a basis of 1 l of medium, if 50 — 5.

Converting to gmol of glucose, this is equivalent to 2. As the amount of glucose 4. Answer: Glucose. In a litre tank, the total mass of glucose consumed for growth is therefore According to the stoichiometry, this produces 1.

Therefore, in a litre tank, 8. Converting to mass, 0. Answer: g. Converting from moles to mass, this is equivalent to 8. As 4 g l—1 phenylacetic acid is provided, 4 — 1. Answer: Yes. From the stoichiometry, as 1 gmol of hexadecane is required to produce 1. From the stoichiometry, The maximum yield is therefore g of cells per Converting to mass terms using the molecular weight of hexadecane, Answer: 3.

The corresponding volume of air is calculated using the ideal gas law. As temperature in the ideal gas equation is absolute temperature, from Eq.

Substituting these values into the equation for V: 3 From the stoichiometry, this requires 0. According to the stoichiometry, complete conversion of 8. Converting to mass units using the molecular weight of the biomass, Therefore, the volume of air required is 0. According to the stoichiometry, the yield is 4 gmol of N2 for every 5 gmol of acetate consumed. From the reaction stoichiometry, 8 mol of nitrate are required for every 5 mol of acetic acid consumed.

Therefore, complete reaction of 4. As only 5. Answer: Nitrate. From c , nitrate is the limiting substrate and 5. Therefore, in l, the total amount of nitrate converted in the reaction is 5. According to the stoichiometry, for each 8 gmol of nitrate reacted, 4 gmol of N2 are formed. Therefore, conversion of Converting to mass units using the molecular weight of N2, the mass of N2 produced is Answer: m2.

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