A 1,900-m3 water tower has been cleaned with a chlorine solution. The vapors of chlorine in the tower exceed allowable concentrations for the work crew to enter and finish repairs. If the chlorine concentration is 15 mg/m3 and the allowable concentration is 0.0015 mg/L, how long must the workers vent the tank with clean air flowing at 2.35 m3/s

t = 1862 s

Explanation:

To do this, we need first to determine the theorical detention time, which can be determined with the following expression:

t₀ = ∀/Q  (1)

Where:

t₀: detention time

∀: Volume of the fluid in the reactor

Q: Flow rate in the reactor

With this time, we must use the following expression to determine the time that the workers will take to vent the tank:

C = C₀ e^(-t/t₀)   (2)

From here, we must solve for time t, and the expression will be:

t = ln(C₀/C) * t₀   (3)

Now that we know the expression to use, let's solve for t. Using (1) to determine the detention time, ∀ is 1900 m³, and Q is 2.35 m³/s so:

t₀ = 1900 / 2.35 = 808.51 s

Now, let's solve for the time t. C will be 0.0015 mg/L (or 1.5 mg/m³ cause in 1 m³ we have 1000 L) and C₀ 15 mg/m³:

t = ln(15/1.5) * 808.51

Hope this helps

you gotta add the following molecules so we can give u an answer

explanation: