Question

A tank of 5 mº methanol (CH3OH) was dumped into a small lake with a volume of 1000 m². The water temperature in the lake is 20°C. Assume that methanol is mixed into the water. Please specify: a) Enstimate the uBOD and BODs of the tank after the waste is shaken (know that: specific weight of methanol was 791.8 kg/m²). The reaction as follow: CH3OH + (3/2) O2 = CO2 + (2) H20 b) The time needed to reduce the BODs concentration to 1% of the initial value? (the temperature of the lake is stable at 20°C). c) If the temperature in the tank is 10°C, estimate the time needed to reduce the BOD5 concentration in the tank to 1%? Assuming the methanol process in the lake is treated based on the stable process of the lake, the decomposition mechanism occurs in the first order reaction, with k = 0.3 minutes

Answer 1

Notation used are standard notation.
Date Page No For methanol Kal= 0.3 minutes Volume of methanol Vm = 5m3 Volume at leke Vi= 1000 m². Sepecific weight of methanol = 791.8 kg/m3. CH₂ OH + 302 > CO₂ + 2H2O 48 12 Molecular weight of CH ₂ OH = 32 Weight of methana) present. 791.885 3959 kg 32 then 3959 kg expe methanol requires 48 kg oxy Oxygen methanol requires 5938.5 8.5kg B.O.D =) 48 x 3959 Ultimate 3.0.0 5938.5 1ooo = 5.939 kg/m² ka=0.3 min 0.3 x 2,4860 432 B.O.DE B.O.Du (1-16 kelt) -432X5 5.939 C T *5) lo 5.93 kg/m3 B.O.Ds = 5.93 Kg/m3