Question

In a 30 m$^3$ room, a stove in operation consumes wood at the rate of 0.25 kg/h. The inflow and outflow rate of air in the room is the same, i.e., 500 m$^3$/h. This stove emits a VOC species at a rate of 0.2 g/kg-wood. The VOC species gets converted to CO$_2$ at a rate of 0.4 per hour. Given: (i) the air in the room is completely mixed, (ii) initial concentration of the VOC species in the room is negligible, and (iii) concentration of the VOC species in the air entering the room is negligible. The concentration of the VOC species due to two hours of stove operation in the room is __________________________________ (in $\mu$g/m$^3$, rounded off to one decimal place).

Hint:

For a well-mixed room with constant source and first-order loss + ventilation, use $C(t)=\dfrac{S/V}{(Q/V)+k}\big(1-e^{-[(Q/V)+k]t}\big)$. If $[(Q/V)+k]t\gg 1$, the exponential term vanishes and $C(t)$ approaches the steady value $S/\big(Q+kV\big)$.

Answer 1

Correct Answer: 95

Step 1: Set up the completely mixed room mass balance.
For a CSTR with a constant internal source $S$ (mass/time), ventilation $Q$ (m$^3$/h), volume $V$ (m$^3$), and first-order decay $k$ (h$^{-1}$), the VOC concentration $C(t)$ obeys \[ \frac{dC}{dt}=\frac{S}{V}-\left(\frac{Q}{V}+k\right)C \text{(since $C_{\text{in}}\approx 0$)}. \]
Step 2: Evaluate parameters.
$V=30$ m$^3$, \; $Q=500$ m$^3$/h $\Rightarrow \dfrac{Q}{V}= \dfrac{500}{30}=16.6667$ h$^{-1}$.
$k=0.4$ h$^{-1}$ $\Rightarrow$ total removal rate $a=\dfrac{Q}{V}+k=17.0667$ h$^{-1}$.
Wood burn rate $=0.25$ kg/h; emission factor $=0.2$ g/kg $\Rightarrow S = 0.25\times 0.2=0.05$ g/h $=5.0\times 10^4$ $\mu$g/h$.
Source strength per volume $s=\dfrac{S}{V}= \dfrac{5.0\times10^4}{30}=1.6667\times 10^3$ $\mu$g/(m$^3 . $h).
Step 3: Solve the ODE with $C(0)=0$.
\[ C(t)=\frac{s}{a}\left(1-e^{-a t}\right). \]
Step 4: Compute $C$ at $t=2$ h.
\[ \frac{s}{a}=\frac{1666.7}{17.0667}\approx 97.66\ \mu\text{g/m}^3,\qquad e^{-a t}=e^{-17.0667\times 2}\approx e^{-34.13}\approx 0. \] \[ \Rightarrow\ C(2~\text{h})\approx 97.66\times(1-0)=97.66\ \mu\text{g/m}^3. \] Final Answer: \[ \boxed{97.7\ \mu\text{g/m}^3} \]