# Diff. Eqs. Mixing Problem

• zero1207
In summary, to lower the salt concentration to 0.01 lb/gal of water in under one hour, you would need to pour water into the tank at a rate of gal/min.

#### zero1207

I haven't found any examples anywhere on how to do a mixing problem where you don't know the rate of volume change, could someone give me a hand just setting up the equation? Here is the problem. . .

1. A tank contains 500 gal of a salt-water solution containing 0.05 lb of salt per gallon of water. Pure water is poured into the tank and a drain at the bottom of the tank is adjusted so as to keep the volume of solution in the tank constant. At what rate (gal/min) should the water be poured into the tank to lower the salt concentration to 0.01 lb/gal of water in under one hour?

Sorry I didn't post what work I've tried:
if x(t) is amount of salt in that tank then
dx/dt=Volume in*Salt Concentration in-Volume out*Salt Contentration out
dx/dt=Volume in*(0)-Volume Out*(x/500)
dx/dt=-(x/500)*(Volume Out)

Hey there, in the beginning before pumping out the salt (x(0)=25 lb), you know the tank will have 25 pounds of salt dissolved, then you want to diminish that to 5 pounds of salt (x(t) = 5 lb). You are given the time too, so t = 60, if the time is in minutes.

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Ok, maybe I'm missing some obvious thing but does the information you gave let me calculate dx/dt? I'm still confused. I was pretty sure I'd use that information after I had calculated the differential equation.

zero1207 said:
Ok, maybe I'm missing some obvious thing but does the information you gave let me calculate dx/dt? I'm still confused. I was pretty sure I'd use that information after I had calculated the differential equation.

You're looking for volume/time at time t where salt concentration =.01, essentially what it's asking you to do is to rearrange the equation so that the unknown is = to the variables you are given, have a play around until you find an equation that relates pounds and time to volume

Once you have that you should be able to integrate to find your answer. at f'(59.59) t=59.59 or just under an hour where salt concentration=.01.

ie with these values of the variables in your equation with the unknowns (x) etc.

EDIT: Sorry I meant integrate obviously.

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zero1207 said:
Sorry I didn't post what work I've tried:
if x(t) is amount of salt in that tank then
dx/dt=Volume in*Salt Concentration in-Volume out*Salt Contentration out
dx/dt=Volume in*(0)-Volume Out*(x/500)
dx/dt=-(x/500)*(Volume Out)
And "Volume Out" is the constant you want to find: call it v.

dx/dt= -(v/500)x

Can you integrate that? What must v be so that x(1)< 0.01?