Question

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

Answer 1

Here is the full question:

Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes? (Round your answer to three decimal places.

Answer:

0.046 %

Explanation:

The rate-in;

$R_{in}$ $= \frac{0.04}{100}*2000$

$R_{in}$ = 0.8

The rate-out

$R_{out}$ = $\frac{A}{6000}*2000$

$R_{out}$ = $\frac{A}{3}$

We can say that:

$\frac{dA}{dt}=$ $0.8-\frac{A}{3}$

where;

A(0)= 0.2% × 6000

A(0)= 0.002 × 6000

A(0)= 12

$\frac{dA}{dt} +\frac{A}{3} =0.8$

Integration of the above linear equation =

$e^{\int\limits \frac {1}{3}dt } =$ $e^{\frac{1}{3}t$

so we have:

$e^{\frac{1}{3}t}\frac{dA}{dt}} +\frac{1}{3}e^{\frac{1}{3}t}A$ $= 0.8e^{\frac{1}{3}t$

$\frac{d}{dt}[e^{\frac{1}{3}t}A]$ $= 0.8e^{\frac{1}{3}t$

$Ae^{\frac{1}{3}t} =2.4e\frac{1}{3}t +C$

∴ $A(t) = 2.4 +Ce^{-\frac{1}{3}t$

Since A(0) = 12

Then;

$12 =2.4 + Ce^{-\frac{1}{3}}(0)$

$C= 12-2.4$

$C =9.6$

Hence;

$A(t) = 2.4 +9.6e^{-\frac{t}{3}}$

$A(0) = 2.4 +9.6e^{-\frac{10}{3}}$

$A(t) = 2.74$

∴ the concentration at 10 minutes is ;

=  $\frac{2.74}{6000}*100$%

= 0.0456667 %

= 0.046% to three decimal places