Differential equations - mixing problem

In summary, the room containing 1000 cubic feet of air was originally free of carbon monoxide. However, at time t=0, cigarette smoke containing 4 percent carbon monoxide was blown into the room at a rate of 0.1 ft^3/min. The well-circulated mixture leaves the room at the same rate. Using the equation C(t) = (4x10^-7)(1-e^(-1.0x10^-4 t)) / 1000, the time when the concentration of carbon monoxide reaches 0.012 percent is approximately 30.05 minutes.
  • #1
braindead101
162
0
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 well-circulated 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
 
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  • #2
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
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
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 (1-e^(1.0x10^-4)t)
so pluggin in c(t)
1.2 x 10^-4 = 4x 10^-10 (1-e^(1.0x10^-4)t)
solving for t
300000 = 1-e^(1.0x10^-4)t
299999 = -e^(1.0x10^-4)t

so.. once again. i am stuck.
 
  • #5
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)(1-e^(1x10^-4 t))

c(t) = s(t)/1000
c(t) = (4x10^-10)(1-e^(1x10^-4 t))

find t when c(t) = 1.2 x 10^-4
1.2x10^-4 = (4x10^-10)(1-e^(1x10^-4 t))
300000 = 1-e^(1x10^-4 t)
299999 = -e^(1x10^-4 t)

stuck. any suggestions or any wrong steps?
 
  • #6
Use natural logarithm.
 
  • #7
i end up ln-ing a negative..
 
  • #8
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
braindead101 said:
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.

Re-check your steps. I already told you the idea

[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]
 
  • #10
braindead101 said:
here is all my work.



(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)

The mistake lies in these steps. You forgot the negative.
 
Last edited:
  • #11
i'm working it out your way now, but the negative i 4got to type in, but i had it on paper and it doesn't really make a difference as when i am plugging in s(0)=0, c is still the same regardless of that.
 
  • #12
braindead101 said:
i'm working it out your way now, but the negative i 4got to type in, but i had it on paper and it doesn't really make a difference as when i am plugging in s(0)=0, c is still the same regardless of that.

It DOES make a difference for the logarithm.
 
  • #13
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)(1-e^(-1.0x10^-4 t)) / 1000
pluggin in c(t) = 1.2x10^-4
1.2x10^-4 = 4x10^-10(1-e^(-1.0x10^-4 t))
300000 = 1-e^(-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
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]
 
Last edited:
  • #15
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
i got the same answer as you, thanks very much.
 

What is a differential equation?

A differential equation is a mathematical equation that relates a function with its derivatives. It describes how a quantity changes over time and is commonly used in mathematical modeling.

What is a mixing problem?

A mixing problem is a type of differential equation that models the rate of change of a substance in a mixture. It is commonly used to solve problems involving mixing two or more substances together.

How do you solve a mixing problem?

To solve a mixing problem, you need to set up a differential equation that describes the rate of change of the substance in the mixture. Then, you can use techniques like separation of variables or integrating factors to solve the equation and find the solution.

What are the key concepts in a mixing problem?

The key concepts in a mixing problem are the initial conditions, which describe the initial amounts of each substance in the mixture; the rate of change, which describes how the substances are being mixed together; and the final solution, which represents the final amounts of each substance in the mixture.

What are some real-life applications of mixing problems?

Mixing problems have many real-life applications, such as in chemical reactions, pharmaceutical manufacturing, environmental engineering, and food processing. They are also used in various fields of science, such as physics, biology, and economics.

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