Find the quantity of salt in the tank?

  • Thread starter Math10
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  • #1
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Homework Statement


A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it's about to overflow.

Homework Equations


None.

The Attempt at a Solution


(1/4)lbs/gal*(4)gal/min=1lbs/min
(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
dQ/dt=1-Q/50=(50-Q)/50
50/(50-Q)dQ=dt
-50ln abs(50-Q)=t+C
ln abs(50-Q)= -t/50+C
50-Q=Ce^(-t/50)
Q=50-Ce^(-t/50)
Q(0)=0
C=50
Q(t)=50(1-e^(-t/50))
Q(50)=31.606
But the answer is Q(50)=47.5
Where did I make a mistake?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


A 200 gallon tank initially contains 100 gallons of water with 20 pounds of salt. A salt solution with 1/4 pound of salt per gallon is added to the tank at 4 gal/min, and the resulting mixture is drained out at 2 gal/min. Find the quantity of salt in the tank as it's about to overflow.

Homework Equations


None.

The Attempt at a Solution


(1/4)lbs/gal*(4)gal/min=1lbs/min
(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
dQ/dt=1-Q/50=(50-Q)/50
50/(50-Q)dQ=dt
-50ln abs(50-Q)=t+C
ln abs(50-Q)= -t/50+C
50-Q=Ce^(-t/50)
Q=50-Ce^(-t/50)
Q(0)=0
C=50
Q(t)=50(1-e^(-t/50))
Q(50)=31.606
But the answer is Q(50)=47.5
Where did I make a mistake?

I didn't check all your work, but I notice you used Q(0)=0. But the problem says the tank started with 20lbs of salt.
 
  • #3
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So I used Q(0)=20,
C=30
Q=50-30e^(-t/50)
Now what?
 
  • #4
DrClaude
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(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
This line I don't understand. Why are you dividing by 100 gal?
 
  • #5
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So what do I do? Can you guys correct me?
 
  • #6
haruspex
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This line I don't understand. Why are you dividing by 100 gal?
It's attempting to calculate the instantaneous rate at which salt is leaving the tank (in lb/min).
Math10, you've overlooked that fact that the volume is increasing too. It's only 100 gal at the start.
We shouldn't have to guess what you are doing. Explain as you go along.
 
  • #7
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You need to do 2 material balances: (1) The volume of water in the tank and (2) the amount of salt in the tank.

Let V (gal) represent the volume of water in the tank at time t, and let C (lb/gal) represent the concentration of salt in the tank at time t. You need to solve for both of these. The balances should be of the form

(rate of input) - (rate of output) = (rate of accumulation).

Let qin represent the rate of water addition to the tank, and let qout represent the rate of water output from the tank. Let Cin represent the concentration of salt in the inlet stream. Assume that the tank is always well-mixed, so that the concentration of salt in the tank is always uniform.

To start with, show us your balance on the volume of water in the tank.

Chet
 

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