Sugar water tank problem (again)

  • Thread starter RossH
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  • #1
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Thank you for any help that you can give me. I am very bad at differential equations and this one is really stumping me.

Homework Statement


A tank contains 2740 L of pure water. A solution that contains 0.05 kg of sugar per liter enters a tank at the rate 3 L/min The solution is mixed and drains from the tank at the same rate.


Homework Equations


I guess I need to solve for ds/dt, which I think would be something along the lines of
ds/dt=.15-3s/2740, or the rate at which the sugar enters minus the rate at which it leaves.

The Attempt at a Solution



So far I've found the equation above and I think that the solution is something like e^[(-3*t)/2740], but my online homework thing keeps saying that it is wrong. Also, that solutino doesn't work for s(0)=0 and subtracting one from the above equation does not work. Does anybody have any suggestions? I think I might be solving the differential equation wrong, but I'm sure that my integrating factor is correct. Thank you very much.
 

Answers and Replies

  • #2
lanedance
Homework Helper
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the rate of change will be the difference between the sugar in (conc of 0.05kg/l at 3L/M per min) & the sugar out (conc of tank)

I am guessing you have used s for the sugar in the tank?

one would expect it would start at zero and increase assymotically to the concentration of the incoming stream

inlet stream (conc c_0=0.05kg/L) sugar in
(ds/dt)_in = c_0*3L/min = 3*(0.05)
(ds/dt)_out = c_tank*3L/min = (s/2740)*3

so the differential equation looks correct
ds/dt=3*(0.05-s/2740) with initial condition s(0) = 0

(2740/3)*ds/dt = 2740*0.05-s

maybe show how you solve this & we can have a look at it

hint you could try the subtitution & variable change
u = s-2740*0.05
 
  • #3
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the rate of change will be the difference between the sugar in (conc of 0.05kg/l at 3L/M per min) & the sugar out (conc of tank)

I am guessing you have used s for the sugar in the tank?

one would expect it would start at zero and increase assymotically to the concentration of the incoming stream

inlet stream (conc c_0=0.05kg/L) sugar in
(ds/dt)_in = c_0*3L/min = 3*(0.05)
(ds/dt)_out = c_tank*3L/min = (s/2740)*3

so the differential equation looks correct
ds/dt=3*(0.05-s/2740) with initial condition s(0) = 0

(2740/3)*ds/dt = 2740*0.05-s

maybe show how you solve this & we can have a look at it

hint you could try the subtitution & variable change
u = s-2740*0.05

Yeah, I chose s for the amount of sugar in the tank.
Anyway, here is how I would try to solve this (probably wrong)
(2740/3)*ds/dt=2740*0.05-s
ds/dt=.15-3s/2740
ds/dt+3s/2740=.15 I(t)=e^(3t/2740)
ds/dt*e^(3t/2740)+(3s/2740)e^(3t/2740)=.15*e^(3t/2740)
integrate both sides:
e^(3t/2740)s=.15e^(3t/2740)*(2740/3)
simplify:
e^(3t/2740)s=137(e^3t/2740)
And this is where I get stuck. Because then I get that s(t)=137, which obviously is wrong. Sorry about the slow post.
 
  • #4
lanedance
Homework Helper
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i haven't used integrating factors for ages, but how about the substitution?

so if you have
(2740/3)*ds/dt = 2740*0.05-s

let u = s-2740*0.05
then
du/dt = ds/dt

do the equation becomes
(2740/3)*du/dt = -u

should be able to separate, integrate & substitute back in for s
 
Last edited:
  • #5
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i haven't used integrating factors for ages, but how about the substitution?

so if you have
(2740/3)*ds/dt = 2740*0.05-s

let u = s-2740*0.05
then
du/dt = dst/dt

do the equation becomes
(2740/3)*du/dt = -u

should be able to separate, integrate & substitute back in for s

I ended up with -e^(3t/2740)+137=s but my homework thing still said that the answer was wrong.

My steps were:
3dt/2740=du/-u integrate both sides:
3t/2740=ln|-u|
e^(3t/2740)=137-s
-e^(3t/2740)+137=s
The problem is that this just doesn't fit the initial condition of 0 kg sugar. And when I changed the constant so that it was (+/-)e^3t/2740) (either way) -1, it still said that I was wrong, even though the initial condition was met.
 
  • #6
lanedance
Homework Helper
3,304
2
I ended up with -e^(3t/2740)+137=s but my homework thing still said that the answer was wrong.

My steps were:
3dt/2740=du/-u integrate both sides:
3t/2740=ln|-u|
e^(3t/2740)=137-s
-e^(3t/2740)+137=s
The problem is that this just doesn't fit the initial condition of 0 kg sugar. And when I changed the constant so that it was (+/-)e^3t/2740) (either way) -1, it still said that I was wrong, even though the initial condition was met.

you're right that its wrong, that answer makes no sense, s will tend toward negative infinity and the initial condition is not met -> s(0) =-e^(0) + 137, which is not 0

i think the problem is with the negative, move it to the t side of the equation before integrating, then keep a constant when you integrate as well, then choose it to satisfy your initial condition
 

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