• Support PF! Buy your school textbooks, materials and every day products Here!

Diff. Eqs. Mixing Problem

  • Thread starter zero1207
  • Start date
  • #1
6
0
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?
 

Answers and Replies

  • #2
6
0
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)
 
  • #3
Pyrrhus
Homework Helper
2,178
1
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.
 
Last edited:
  • #4
6
0
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.
 
  • #5
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 untill 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:
  • #6
HallsofIvy
Science Advisor
Homework Helper
41,806
932
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?
 

Related Threads on Diff. Eqs. Mixing Problem

  • Last Post
Replies
3
Views
1K
Replies
5
Views
1K
Replies
1
Views
3K
  • Last Post
Replies
3
Views
1K
Replies
2
Views
318
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
4
Views
5K
Replies
2
Views
4K
  • Last Post
Replies
2
Views
8K
Replies
15
Views
9K
Top