BOD: Biochemical Oxygen Demand
BOD5 is the total amount of oxygen consumed by microorganisms during the first five days of biodegradation
BOD5 = (DOi - Dof)/(P) this is the five-day BOD of a diluted sample
Where DOi = the initial dissolved oxygen (DO) of the diluted wastewater
DOf = the final DO of the diluted wastewater, 5 days later
P = the dilution fraction = (volume of wastewater)/(volume of wastewater plus dilution water)
A standard BOD bottle holds 300 mL, so P is just the volume of wastewater divided by 300 mL
Consider a lake with 100 X 10^6 m^2 of surface area for which the only source of phosphorus is the effluent from a wastewater treatment plant. The effluent flow rate is 0.4 m^3/s and its phosphorus concentration is 10.0 mg/L (10.0 g/m^3). The lake is also fed by a stream having 20 m^3/s of flow with no phosphorus. If the phosphorus settling rate is estimated to be 10 m/yr, estimate the average phosphorus concentration in the lake. What level of phosphorus removal at the treatment plant would be required to keep the average lake concentration below 0.010 mg/L?
Although not stated in the problem, I would assume an outflow to keep the lake level unchanged. Then this is essentially a mass balance problem. Draw a diagram showing the input from the wastewater treatment plant and the stream. Fill in each of the numbers given in the problem, and leave blanks for those you're ...
Waste water mass balance problem
Chemical Engineering: Stoichiometry and Mass Balance
Please see the attached file for the fully formatted problems. Just do A3 and A4, please.
A3. Solid calcium fluoride (CaF2) reacts with sulphuric acid to form solid calcium sulphate (CaSO4) and gaseous hydrogen fluoride (HF). The HF is then dissolved in water to form hydrofluoric acid. A source of calcium fluoride is fluorite ore containing 96% w/w CaF2 and 4% w/w SiO2.
In a typical hydrofluoric acid manufacturing process, fluorite ore is reacted with 93% w/w aqueous sulphuric acid, supplied in 15% excess of the stoichiometric amount. Ninety five percent (95%) of the CaF2 reacts with the acid. Some of the HF formed reacts with the dissolved silica (SiO2) in the reaction. The unreacted HF gas exiting from the reactor is subsequently dissolved in enough water to produce 60% w/w aqueous hydro fluoric acid.
(a) Determine the mass of 93%w/w acid in the feed.
(b) Calculate the quantity of fluorite ore needed to produce 2500 kg of 60% w/w acid. Assume all the SiO2 present in the ore reacts with HF. The reaction equations are:
CaF2 (s) + H2SO4(aq) ---> CaSO4(s) + 2 HF(g)
6 HF(g) + SiO2(aq) ---> H2SiF6(s) + 2 H2O(l)
Ca = 40, F = 19, H = 1, S = 32, O = 16, Si = 28
A4. Formaldehyde (CH2O) is manufactured by the catalytic oxidation of methanol using an excess of air. Formic acid (HCOOH) is also formed if conditions are not properly controlled. The reaction equations are as follows:
CH3OH + ½ O2 → CH2O + H2O --------- (1)
CH2O + ½ O2 → HCOOH --------- (2)
The product gases in a test run have the following composition:
Component Mol %
Determine the following:
(a) The molar ratio of feed air to feed methanol
(b) The per cent excess air based on the main reaction
(c) The conversion of methanol
(d) The yield of formaldehyde
(e) The selectivity of formaldehydeView Full Posting Details