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A 90-m3 basement in a residence is found to be contaminated with radon coming from the ground through the floor drains. The concentration of radon in the room is 1.5 Bq/L under steady-state conditions. The room behaves as a CSTR and the decay of radon is a first-order reaction with a decay rate constant of 2.09 3 1026 s21 . If the source of radon is closed off and the room is vented with radon-free air at a rate of 0.14 m3 /s, how long will it take to lower the radon concentration to an acceptable level of 0.15 Bq/L

Sagot :

Boffeemadrididn

Answer:

25 minutes

Explanation:

V = Volume = [tex]90\ \text{m}^3[/tex]

[tex]C_0[/tex] = Radon concentration under steady state = 1.5 Bq/L

k = Radon decay rate = [tex]2.09\times 10^{-6}\ \text{s}^{-1}[/tex]

[tex]Q[/tex] = Venting rate = [tex]0.14\ \text{m}^3/\text{s}[/tex]

[tex]C_f[/tex] = Final concentration of radon = 0.15 Bq/L

Theoretical detention time is given by

[tex]\theta=\dfrac{V}{Q}\\\Rightarrow \theta=\dfrac{90}{0.14}\\\Rightarrow \theta=642.86\ \text{s}[/tex]

We have the relation

[tex]C_f=C_0e^{-(\dfrac{1}{\theta}+k)t}\\\Rightarrow t=\dfrac{\ln\dfrac{C_f}{C_0}}{-(\dfrac{1}{\theta}+k)}\\\Rightarrow t=\dfrac{\ln\dfrac{0.15}{1.5}}{-(\dfrac{1}{642.86}+2.09\times 10^{-6})}\\\Rightarrow t=1478.25\ \text{s}=\dfrac{1478.25}{60}=24.63\approx 25\text{minutes}[/tex]

The time taken to reach the acceptable level of concentration is 25 minutes.

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