# Find the quantity of salt in the tank?

• Math10
Q*V*tNow, show us your balance on the concentration of salt in the tank.Chet=Q/(1+qout*Cin)So, the total amount of salt in the tank at any given time is equal to the sum of the rates of input (qin) and output (qout):Chet=Q*V*t+qout*Cin

## 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.

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
Where did I make a mistake?

Math10 said:

## 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.

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
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.

So I used Q(0)=20,
C=30
Q=50-30e^(-t/50)
Now what?

Math10 said:
(Q)lbs/100 gal*(2)gal/min=Qlbs/50 min
This line I don't understand. Why are you dividing by 100 gal?

So what do I do? Can you guys correct me?

DrClaude said:
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.

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.