Problem 3. An experiment requires a tank of 100 gallons of water with salt conce

Question

Problem 3. An experiment requires a tank of 100 gallons of water with salt concentration of 0.1 lb/gal. By mistake, 20 lb of salt was added to fresh water. To correct this, we started adding 3 gal/min of fresh water and draining at the same rate while agitating the tank. Assume that the salt is uniformly mixed in the water. a Formulate a differential equation and initial conditions for the amount of salt. b Solve the equation. c) How long will it take for the salt concentration to reach the desired value?

Explanation / Answer

a)

To set up the IVP

Flow rate of the water entering =3 gal/min

Concentration of the salt in the water entering =0 (since its freshwater)

Flow rate of the water leaving =3 gal/min

Concentration of the salt in the water exiting (we don’t have this yet)

So, we first need to determine the concentration of the salt in the water exiting the tank.  

Since we are assuming a uniform concentration of salt in the tank the concentration at any point in the tank and hence in the water exiting is given by,

The amount at any time t is just Q (t).

The volume is 100 gallons

Hence, Concentration of the salt in the water exiting the tank= Q (t) / 100 = 0.01Q (t)

The equation that we’ll be using to model this situation is:

Rate of change of Q (t) = Rate at which Q (t) enters the tank - Rate at which Q (t) exits the tank

Rate of change of Q (t) =

Rate at which Q (t) enters = (flow rate of water entering) x (concentration of salt in water entering)

= 3 x 0 = 0

Rate at which Q (t) exits the tank = (flow rate of water exiting) x (concentration of salt in water exiting)

= 3 x 0.01Q (t) = 0.03 Q (t)

Therefore Q'(t) = 0 - 0.03 Q (t)

Q'(t) = -0.03 Q (t) with initial condition Q (0) = 20lb as initially there was 20lb of salt

b)

In order to solve the above equation

Separating the variables we get

dQ/Q = - 0.03 t

Integrating both sides

ln Q = -0.03t +C where C is the integration constant

or Q = Ce-0.03t

Substituting the initial condition

20 = C e0 = C

Hence Q(t) = 20e-0.03t is the solution to the equation

c) In order for the salt concentration to reach the desired concentration i.e. Q = 10

we have 10 =  20e-0.03t

0.5 =  e-0.03t

-0.03t = ln(0.5)

t = ln(0.5) / -0.03 = 23.105 minutes

Hence it takes approximately 23 minutes to reach the desired concentration   

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