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Differential equations  mixing problem 
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#1
Feb1107, 10:31 PM

P: 162

A room containing 1000 cubic feet of air is originally free of carbon monoxide. Beginning at time t=0 cigarette smoke containing 4 percent carbon monoxide is blown into the room at 0.1 ft^3/min, and the wellcirculated mixture leaves the room at hte same rate. Find the time when the concentration of carbon monoxide in the room reaches 0.012 percent.
rate = rate in  rate out ? the 4% carbon monoxide part is really throwing me off. i don't exactly know what to do with this number. but other than that, i should be alright. CO enters : 0.04 x 0.1 OR 0.1, i'm not sure whether to factor in the 4% CO CO leaves : 0.04 x 0.1 S(t)/1000 OR 0.1 s(t)/1000 i need to figure that part out.. after that, everything i can handle. can someone please help me with the 4% carbon monoxide part 


#2
Feb1107, 10:53 PM

P: 162

okay so i tried working it out with the 0.04 factored in as shown above... and at the end i get a ln #, so i guess that is wrong.
so now i am doing it without the 0.04. if i do it without the 0.04, then i do not understand the concentration part. it says to find the time when conc is 0.012 %, so set c(t)=0.00012 OR shoudl i be factoring in 4% CO somewhere still. i am really confused 


#3
Feb1107, 11:29 PM

HW Helper
P: 2,279

Here's a hint: make M(t) be the quantity of carbon monoxide in the
room at any time and then the concentration is given by C(t) = M(t)/1000. 


#4
Feb1107, 11:38 PM

P: 162

Differential equations  mixing problem
i have used, s(t) = amount of CO
c(t)=s(t)/1000 so i have rate in as: 0.04 x 0.1 rate out: 0.1 s(t)/1000 i have modified my rate in and rate out from when i first posted the eq'n as i don't believe i have to add the 4% CO factor twice. tell me what you think now. i have solved this ALL the way to plugging in C(t) i plugged in c(t) as 1.2 x 10^4 my eq'n is c(t) = 4x 10^10 (1e^(1.0x10^4)t) so pluggin in c(t) 1.2 x 10^4 = 4x 10^10 (1e^(1.0x10^4)t) solving for t 300000 = 1e^(1.0x10^4)t 299999 = e^(1.0x10^4)t so.. once again. i am stuck. 


#5
Feb1107, 11:47 PM

P: 162

here is all my work.
rate in: 0.04 x 0.01 rate out: 0.1 x s(t)/1000 s(0) = 0 s'(t) = 0.004  1.0x10^4 s(t) s'(t) + 1.0x10^4 s(t) = 0.004 solving diff eq'n: a(t) = 1.0x10^4, b(t)=0.004 using formula: u(t) = exp(integ(a(t)dt)) u(t) = exp(integ(1.0x10^4 dt)) u(t)=exp(1.0x10^4 t) using formula: d/dt (u(t) s(t) ) = u(t)b(t) d/dt ( (e^(1.0x10^4 t )) s(t) ) = (e^(1.0x10^4 t)) x 0.004 (e^(1.0x10^4 t )) s(t) = integ (0.004(e^(1.0x10^4 t))dt) (e^(1.0x10^4 t )) s(t) = 0.004(1.0x10^4)(e^(1.0x10^4 t) + C s(t) = [(4x10^7) (e^(1.0x10^4 t)) + C]/(e^(1.0x10^4 t))] s(t) = 4x10^7 + Ce^(1.0x10^4 t) sub s(0)=0 0 = 4x10^7 + C C= 4x10^7 so, s(t)=(4x10^7)(1e^(1x10^4 t)) c(t) = s(t)/1000 c(t) = (4x10^10)(1e^(1x10^4 t)) find t when c(t) = 1.2 x 10^4 1.2x10^4 = (4x10^10)(1e^(1x10^4 t)) 300000 = 1e^(1x10^4 t) 299999 = e^(1x10^4 t) stuck. any suggestions or any wrong steps??? 


#7
Feb1207, 02:52 AM

P: 162

i end up lning a negative..



#8
Feb1207, 05:56 PM

P: 162

can anyone find the mistake?
i have some help from the instructor as he said the percentage is percentage of the volume, so i guess that is the part that is wrong??? i'm not sure what he means by that. 


#9
Feb1207, 09:07 PM

HW Helper
P: 2,279

[tex] \frac{dm(t)}{dt} = 0.1 * 0.04  0.1 \frac{m(t)}{1000} [/tex] Maybe you will have less mistakes in your calculations if you solved it like this. [tex] c(t) = \frac{m(t)}{1000} [/tex] [tex] 1000\frac{dc(t)}{dt} = 0.1 * 0.04  0.1 \frac{1000c(t)}{1000} [/tex] 


#11
Feb1207, 10:51 PM

P: 162

i'm working it out your way now, but the negative i 4got to type in, but i had it on paper and it doesnt really make a difference as when i am plugging in s(0)=0, c is still the same regardless of that.



#13
Feb1307, 12:49 AM

P: 162

okay
so here is my work: starting from pluggin in the concentration: s(t) = (4x10^7)  (4x10^7)e^(1.0x10^4 t) C(t) = s(t)/1000 c(t) = (4x10^7)(1e^(1.0x10^4 t)) / 1000 pluggin in c(t) = 1.2x10^4 1.2x10^4 = 4x10^10(1e^(1.0x10^4 t)) 300000 = 1e^(1.0x10^4 t) 299999 = e^(1.0x10^4 t) ln 299999 = ln e^(1.0x10^4 t) i am stuck agian.. at the same place 


#14
Feb1307, 10:49 AM

HW Helper
P: 2,279

Well i decided to do the problem to see why you weren't getting it right, anyway, i see another mistake, your integration is wrong. Btw, i get 30.05 as the answer, is t in minutes?, looks awfully fast if it was in seconds.
[tex] \int e^{kx} dx = \frac{1}{k} e^{kx} + C [/tex] 


#15
Feb1307, 05:28 PM

P: 162

oh my god , thank you so much.
I have checked over my work so many timse and still did not catch that! reworking my solution as we speak and hopefully i get the same answer as you. 


#16
Feb1307, 05:36 PM

P: 162

i got the same answer as you, thanks very much.



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