How can I solve a mixing problem without knowing the rate of volume change?

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The discussion focuses on solving a mixing problem involving a salt-water solution in a tank. The initial conditions include a 500-gallon tank with a salt concentration of 0.05 lb/gal, and the goal is to reduce the concentration to 0.01 lb/gal within one hour by adding pure water while maintaining constant volume. The differential equation established is dx/dt = -(v/500)x, where v represents the volume outflow rate, which needs to be determined to achieve the desired salt concentration. Participants emphasize the importance of rearranging the equation to isolate the unknown and integrating to find the solution.

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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?
 
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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.
 
Last edited:
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?
 

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