Differential equations problem

eoins23456

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

Fremen

Did you drop a minus sign when you integrated dy/(120-y)?

I find it's often best to try to solve everything symbolically for as long as you can, then substitute actual numbers in at the last minute. It means you can keep track of everything better.

eoins23456

oh ya i did make a mistake there so it would become
-ln(120-y)=1/500000(t)+c
120-y=e^(-t/500000)
120-0.012=e^(-t/500000)
ln(119.088)=-t/500000
t=-ln(119.088)*500000
still the same ans just negative so doesnt make sense.prob going wrong somewhere else

Fremen

You didn't set your initial condition right, either. You need to determine A by

120-Ae^(-t/500000)=y

and

y(0) = 0, assuming the air is clean to begin with.

eoins23456

Well A would equal 120 in that case.
120-120(e^(-t/500000)=0.012
-500000(ln(.9924)=t
t=3814?

Fremen

No, (120-.012)/120 is .9999. This is why you should use letters instead of numbers.

eoins23456

right so got y = 4.8(1-e^(.00005t))
0.012% was the concentraion so would you make the differential equation equal to (1200*.012)/100 and solve for t?