Respuesta :
Answer:
Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L
Explanation:
The lagoon can be modelled as a Mixed flow reactor.
From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.
The performance equation of a Mixed flow reactor is given from the material and component balance thus:
(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)
V = volume of the reactor (The lagoon) = ?
C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³
F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s
C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L
For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³
(-r) = kC (Since we know this decay process is a first order reaction)
This makes the performance equation to be:
(kVC₀/F₀) = (C₀ - C)/C
V = F₀(C₀ - C)/(kC₀C)
k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s
V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)
V = 86580 m³
Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.
The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)
Hope this helps!
The volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".
lagoon relation:
The volume for the lagoon relation:
[tex]\to Q_1C_1 = Q_2C_2+KC_2V\\\\[/tex]
[tex]\to Q_1=0.10\\\\\to C_1=30\\\\\to Q_2=0.2\\\\\to C_2=10\\\\\to K=0.10[/tex]
Putting the value into the formula and calculating the volume:
[tex]\to (0.10 \times 30)=(0.2\times 10)+(0.10\times 10\times V \times (\frac{1}{24 \ hrs}) \times (\frac{1 \ hr}{3,600 \ sec})) \\\\\to (3)=(2)+(1\times V \times \frac{1}{86400}) \\\\\to 3-2=(1\times V \times \frac{1}{86400}) \\\\\to 1=(1\times V \times \frac{1}{86400}) \\\\\to 1 \times 86400 = V\\\\\to V= 8.64 \times 10^4 \ m^3\\\\[/tex]
Therefore, the calculated volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".
Find out more information about the volume here:
brainly.com/question/1578538
