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In a furnace, 2 chemical reactions are happening – 1 mole of solid carbon reacts with 1 mole of oxygen gas to generate 1 mole of carbon dioxide gas; 1 mole of solid carbon reacts with 0.5 mole of oxygen gas to generate 1 mole of carbon monoxide gas. In a given process, 100 kmol of carbon is burned in a furnace. (a) Calculate the theoretical oxygen gas needed by assuming that all the carbon is burned completely to carbon dioxide gas. (b) Calculate the theoretical air needed by assuming that all the carbon is burned completely to carbon dioxide gas and there is only 21 % of oxygen gas. (c) Determine the amount of air required (in kmol) if 50 % excess oxygen gas must be satisfied for (a) and (b). (d) It has latter been found that 20 % of the carbon undergoes incomplete combustion resulting to carbon monoxide gas production. The rest of the carbon undergoes complete combustion. Calculate the total oxygen gas required stoichiometrically based on the actual process.
REACTION ENGINEERING - EXAMPLE 13.2 : A batch reactor is designed for the system of the irreversible, elementary liquid-phase hydration of butylene oxide that produces butylene glycol. At the reaction temperature T = 323 K, the reaction rate constant is k = 0.00083 L / (mol - min). The initial concentration of butylene oxide is 0.25 mol / L = Ca. The reaction is conducted using water as the solvent, so that water is in large excess. (a) Let the molecular weight of water is 18 g / mol and the mass of 1 kg in 1 L of water, calculate the molar density of water, Cb in the unit of mol / L. (b) Determine the final conversion, X of butylene oxide in the batch reactor after t = 45 min of reaction time. Use the formula X = 1 - 1 / exp [ kt (Cb) ] derived from material balance. (c) Find the equation of t as a function of X.
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Question 68 – Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
What is the most common cause of solid size segregation in bulk solid systems?
what is the role of a chemical engineer on a cement plant?
Explain what type of flow measurement devices is best for slurries?
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