Question

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

Answer 1

Answer:

25 minutes

Explanation:

V = Volume = $90\ \text{m}^3$

$C_0$ = Radon concentration under steady state = 1.5 Bq/L

k = Radon decay rate = $2.09\times 10^{-6}\ \text{s}^{-1}$

$Q$ = Venting rate = $0.14\ \text{m}^3/\text{s}$

$C_f$ = Final concentration of radon = 0.15 Bq/L

Theoretical detention time is given by

$\theta=\dfrac{V}{Q}\\\Rightarrow \theta=\dfrac{90}{0.14}\\\Rightarrow \theta=642.86\ \text{s}$

We have the relation

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

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