Question

Air containing 0.06% carbon dioxide is pumped into a room whose volume is 12,000 ft3. The air is pumped in at a rate of 3,000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.3% carbon dioxide, determine the subsequent amount A(t), in ft3, in the room at time t.

Answer 1

Answer:

$A(t)=1.8+34.2*e^{-t}$

Explanation:

The concentration of CO2 in the room will be the amount of CO2 in the room at time t, divided by the volume of the room.

Let A(t) be the amount of CO2 in the room, in ft3 CO2.

The air entering the room is 3000 ft3/min with 0.06% concentrarion of CO2. That can be expressed as (3000*0.06/100)=1.8 ft3 CO2/min.

The mixture leaves at 3000 ft3/min but with concentration A(t)/V. We can express the amount of CO2 leaving the room at any time is A(t).

We can write this as a differential equation

$dA/dt=v_i-v_o=1.8-A$

We can rearrange and integrate

$dA/dt=v_i-v_o=1.8-A\\\\dA/(A-1.8)=-dt\\\\\int(dA/(A-1.8) = -\int dt\\\\ln(A-1.8)=-t+C\\\\A-1.8=e^{-t}* e^{C}=C*e^{-t}\\\\A=1.8+C*e^{-t}$

We also know that A(0) = 12000*(0.3/100)=36 ft3 CO2.

$A(0)=1.8+C*e^{-0}\\36=1.8+C*1\\C=34.2$

Then we have the amount A(t) as

$A(t)=1.8+34.2*e^{-t}$