Differential calculus mixing problem:
There is a dam with a volume of 49 cubic metres. Water flows out of this into one of the creeks. Salt water with salt content of 30kg/m3 flowed into the dam at 0.68m3/s and mixed with the fresh water. The mixed water flowed out at the same rate. Flag tail in the dam can tolerate salinity up to 14kg/m3. How long after the flow starts can the fish survive in the dam?

Question

Differential calculus mixing problem:
There is a dam with a volume of 49 cubic metres. Water flows out of this into one of the creeks. Salt water with salt content of 30kg/m3 flowed into the dam at 0.68m3/s and mixed with the fresh water. The mixed water flowed out at the same rate. Flag tail in the dam can tolerate salinity up to 14kg/m3. How long after the flow starts can the fish survive in the dam?

anonymous · Answer

Do you know the formula for mixtures?

anonymous · Answer

I can show you my working so far.

anonymous · Answer

$$\frac{ dQ }{ dt }= Rate In-RateOut$$

anonymous · Answer

Rate In = (Concentration in)(Flow Rate In)
Rate Units = lb/min
Concentration Units = lb/gal
Flow Units = gal/min

Other units may be used if consistent throughout.

anonymous · Answer

I know the formula, and i have attempted the question but the answer i got was 0.045 seconds which must be wrong.

anonymous · Answer

input: 20.4Kg/s
Output: 0.01387755x

nincompoop · Answer

don't you need to solve for t?

anonymous · Answer

i cant be sure i have solved the differential equation correctly. 
as : $$\frac{ dx }{ dt}=20.4 - 0.01387755x$$

anonymous · Answer

How did you get your units output and input above?

anonymous · Answer

concentration x flow rate

anonymous · Answer

How would i go about doing that, don't i have to have a completed model to solve for t.

anonymous · Answer

Lets take it step by step,$$Rate In= (30kg/m^3)(0.63m^3/s) = 18.9\frac{ kg }{ s }$$

anonymous · Answer

20.4

anonymous · Answer

0.68,    = 20.4Kg/s

anonymous · Answer

My apology I just noticed.

anonymous · Answer

All good

anonymous · Answer

if this is correct so far,do i just solve the differential equation and find t?

anonymous · Answer

$$RateOut = (\frac{ y(t)kg }{ 49m^3 })(0.68kg/s)$$

nincompoop · Answer

https://www.youtube.com/watch?v=bqVzRBf_NGM

anonymous · Answer

Not sure as of yet since the 14kg/m^3 may be used to determine the value for C that we find to actually solve for the value of t

anonymous · Answer

I cannot access youtube, i am using a school computer atm. thanks though.

anonymous · Answer

Another thing, is that the fish can only tolerate a salinity of 14Kg/m^3. 
Does this not mean that the amount of salt in the dam would actually be 686Kg as this creates a concentration in the pond equal to 14Kg/m^3?
14 x 49 = 686

anonymous · Answer

It actually may be that we use 14kg/m^3 for the concentration out since that is what we are looking for at a time.

anonymous · Answer

Then solving the equation you will end up with a 2 variables T and C. To solve for C you will use the fresh water initially or Y(0) = 49m^3 so the T will go away and you will find a value for C and from there all you would need to do is algebraically solve for t.

anonymous · Answer

Ok, i'll give it s go.

anonymous · Answer

Ninc, do you know if this is correct?

anonymous · Answer

Doesn't look correct no but even so you did not use the intial value Y(o) = 49m^3 so that C will be 49m^3

anonymous · Answer

In most of the examples i found similar the final model contained e

anonymous · Answer

Correct so have I, that is why I am not seeing how this problem is going so far and the units are not helping me either

anonymous · Answer

http://www4.ncsu.edu/~msolufse/MA432ODEs.pdf

anonymous · Answer

On page 10 of this pdf there is a very similar question, you may be able to follow it better than i did.

anonymous · Answer

We initially followed that.
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