Differential equations problem rate in/rate out problem

[lolcaust]

A bar opens at 6 and is quickly filled with customer,the majority of whom are cigarette smokers.The bar has ventilators which exchange the smoke-air mixture with for fresh air.Cigarette smoke contains 4% carbon monoxide anda prolonged exposure to a concentration of more then 0.012% can be fatal.The bar has a floor area of 20, by 15m and a height of 4m.It is estimated that smoke enters the room at a constant rate of 0.006m^3/min and that the ventilators remove the smoke-air mixture at 10 times the rate at which smoke is produced.
The question is when is it advisable to leave the bar?or in other words when does the concentration of carbon monoxide reach the lethal limit?

Formulate the differential equation for the changing concentration of carbon
monoxide in the bar over time.
(b) By solving the differential equation, establish the time at which the lethal
limit will be reached.




attempted solution


20*15*4=1200m^3=volume of room
smoke enter room at 0.006m^3/min
smoke leaves room at 0.06m^3/min

rate in = 0.04*0.006=0.00024
rate out = 0.06(y)/1200=0.00005(y)


dy/dt= 0.00024-0.00005y
dy/dt-1/500000(120-y)

dy/(120-y)=1/500000(dt)
ln(120-y)=1(t)/500000+c
120-y=Ae^(t/500000)
120-Ae^(t/500000)=y




second part 120-Ae^(t/500000)=0.012
119.088=e^(t/500000)
ln(119.088)(500000)=t
2.39*10^6=t

answer is slighty absurd so can anyone spot any mistakes?
thanks in advance

 

[jvicens]

Your formulation of the differential equation is correct. However, there are a few mistakes in your calculations and solution for the second part.

Firstly, the rate at which smoke enters the room should be 0.04 * 0.006 = 0.00024 m^3/min, as the concentration of carbon monoxide in cigarette smoke is 4% (0.04) and the volume entering the room is 0.006 m^3/min.

Secondly, the rate at which smoke leaves the room should be 0.06 * (120-y) / 1200 = 0.00005(y). This is because the rate at which smoke leaves the room is dependent on the concentration of carbon monoxide in the room, which is represented by (120-y), and the total volume of the room, which is 1200m^3.

Thirdly, when solving the differential equation, it should be ln(120-y) = t/500000 + c, not ln(120-y) = 1(t)/500000 + c.

Lastly, when solving for the time at which the lethal limit will be reached, you should use the initial condition y(0) = 0 (no carbon monoxide in the room initially), instead of y(0) = 120 (which would mean the room is already filled with carbon monoxide at the beginning). This will give you a more reasonable answer.

Correct solution:

dy/dt = 0.00024 - 0.00005y
dy/(120-y) = 1/500000(dt)
ln(120-y) = t/500000 + c
120-y = Ae^(t/500000)
y = 120 - Ae^(t/500000)

Initial condition: y(0) = 0
0 = 120 - A * e^(0/500000)
A = 120

Substituting A = 120 into the equation:
y = 120 - 120 * e^(t/500000)

To find the time at which the lethal limit is reached (y = 0.012):
0.012 = 120 - 120 * e^(t/500000)
e^(t/500000) = 99.66
t/500000 = ln(99.66)
t = 2.39 * 10^6 seconds

Therefore, it is advisable to leave the bar after