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Simple D.E mixing tank problem

  • Thread starter leehufford
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  • #1
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Hello,
Having trouble with one quirk to this mixing tank problem:

All large mixing tank initially holds 300 gallons of water, with 50 lbs of salt already in it (initial value problem). A brine solution of 2 lb/gal is pumped in at 3 gallons per minute. The water is pumped out at the faster rate of 3.5 gal/min. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0.


I came up with

dA/dt + 7A/600 = 6. A(0) = 50.

Now I understand intuitively that the volume of the tank is decreasing so the 600 in the denominator is changing with time. The answer is

dA/dt + 7A/(600-t) = 6. A(0) = 50

My question is how do they know to subtract exactly t. The tank is losing 0.5 gal/min. Thanks in advance,

Lee.
 

Answers and Replies

  • #2
Dick
Science Advisor
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Hello,
Having trouble with one quirk to this mixing tank problem:

All large mixing tank initially holds 300 gallons of water, with 50 lbs of salt already in it (initial value problem). A brine solution of 2 lb/gal is pumped in at 3 gallons per minute. The water is pumped out at the faster rate of 3.5 gal/min. Determine a differential equation for the amount of salt A(t) in the tank at time t > 0.


I came up with

dA/dt + 7A/600 = 6. A(0) = 50.

Now I understand intuitively that the volume of the tank is decreasing so the 600 in the denominator is changing with time. The answer is

dA/dt + 7A/(600-t) = 6. A(0) = 50

My question is how do they know to subtract exactly t. The tank is losing 0.5 gal/min. Thanks in advance,

Lee.
The volume in V in the tank is V=300-t/2, right? As you said, you are losing 1/2 gal/min. The concentration of salt is A/V. How much salt are you losing per minute if you are pumping 3.5 gal/min out?
 
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