Consider the air over a city to be a box 100km on a side that reaches up to an altitude of 1.0 km. THe air is blowing into the box along one of its sides with a speed of 4 m/s. Suppose an air pollutant with a reaction rate of K = 0.20/hr is emitted into the box at a total mass loading rate of 10 kg/s. Find the steady state concentration if the air in the box is assumed to be perfectly mixed.
V= 100*100*1 = 10000 km^3
Q = 4/1000*1*100 = 0.4 km^3/s
m (loading) = 10kg/s, k = 0.20/hr
1 km^3 = 10^12 L (if u want kg/L as concentration)
Cin = 10kg/s / 0.4 km^3/s = 25 kg/km^3
for CSTR: Cout = Cin/(1+k*t), t = V/Q = 25000 s or 6.94 hr
Cout = 25/(1+0.20*6.94) = 10.47 kg/km^3
Well, well, well, looks like we have a pollution problem blowing in! Let's crunch some funny numbers to find the steady state concentration of this air pollutant, shall we?
First, we need to find the volume of our box. With sides of 100km and an altitude of 1.0km, the volume is 100 km * 100 km * 1.0 km. But hold your breath, we need to convert that into cubic meters! 100 km * 1000 m/km * 1000 m/km * 1.0 km = 10^12 m^3. Phew, that's a big volume!
Now, let's consider the mass loading rate of this pollutant. They're spewing it in at 10 kg/s. Don't worry, it's just a number, not the smell! So, in 1 hour, that would be 10 kg/s * 3600 s/hr = 36,000 kg/hr. Time to convert that into grams, because why not? 36,000 kg/hr * 1000 g/kg = 36,000,000 g/hr. We're getting heavy here!
Since the air is perfectly mixed, the steady state concentration is the mass loading rate divided by the volume. So, that would be 36,000,000 g/hr / 10^12 m^3. Let's put that on a weight loss program and convert g/hr into g/m^3. 36,000,000 g/hr / 10^12 m^3 = 3.6 x 10^-5 g/m^3. Ta-daaa!
So, the steady state concentration of this air pollutant is 3.6 x 10^-5 g/m^3. It's not a laughing matter, though. Let's make sure we take pollution seriously and do our part to keep the air clean and fresh!
To find the steady-state concentration of the air pollutant in the box, we need to calculate the rate at which the pollutant is being removed from the box and then use that information to find the concentration.
1. Calculate the volume of air in the box:
The volume of a rectangular box is given by V = length x width x height. In this case, the length, width, and height of the box are all 100 km (since it is a cube). Convert 100 km to meters: 100 km = 100,000 meters.
So, V = (100,000) x (100,000) x (1000) = 1 x 10^15 m^3.
2. Calculate the rate at which the pollutant is being removed from the box:
The reaction rate of the pollutant, K, is given as 0.20/hr. We need to convert this to a numerical value in units of 1/s (since the rate of removal should have the same units as the rate of emission).
0.20/hr = 0.20/3600 s = 5.6x10^-5 1/s.
This means that for every molecule of pollutant present, 5.6x10^-5 molecules will be removed per second.
3. Calculate the total rate of removal:
The rate of removal is directly proportional to the concentration of the pollutant. Since the air in the box is assumed to be perfectly mixed, the concentration will be uniform throughout the entire volume of the box.
Therefore, the total rate of removal can be calculated as:
Total rate of removal = Concentration x rate at which the pollutant is removed per molecule.
Since we want to find the steady-state concentration, the total rate of removal should be equal to the rate at which the pollutant is emitted into the box.
4. Calculate the steady-state concentration:
Total rate of removal = Rate at which the pollutant is emitted into the box.
Concentration x rate of removal per molecule = Mass loading rate.
Concentration x (5.6x10^-5 1/s) = 10 kg/s.
To find the concentration, divide the mass loading rate by the rate of removal per molecule:
Concentration = Mass loading rate / (rate of removal per molecule).
Concentration = 10 kg/s / (5.6x10^-5 1/s) = 1.79x10^5 kg/m^3.
Therefore, the steady-state concentration of the air pollutant in the box is 1.79x10^5 kg/m^3.