A Salty Tank

Question

A very large tank initially contains 15 gallons of saltwater containing 6 pounds of salt. Saltwater containing 1 pound of salt per gallon is pumped into the top of the tank at a rate of 2 gallons per minute, while a well mixed solution leaves the bottom of the tank at a rate of 1 gallon per minute.

How much salt is in the tank after 15 minutes?

 

(a) the following is a general rule of thumb for setting up rate of change equation for situations like this where there is an input and an output:
rate of change= rate of change in - rate of change out

In this problem, we are interested in using a differential equation to predict future amounts of Salt S in the tank. Using the above rule of thumb, figure out a rate of change equation for this situation.

(b)If the tank has a maximum capacity of 50 gallons, use your rate of change equation to figure out how much salt is in the tank when the tank is full.

 

Answer 1

rate of Salt coming into tank 2 pounds per minute

rate of water entering tank is 2 gallon per minute

rate of Salt leaving tank is   1*C 

rate of water leaving tank is  1 * 1

 

Initial salt 6 pounds

Initial water 15 gallons

 

Find an expression for the concentration of salt in the tank at any time T

Let S be the amount of salt in the water at any time  S/(15+T) is concentration of salt in the tank and the concentration of salt leaving the tank

 

rate of change is salt in minus salt out.

 

dS/dT  =  2 - S/(15+T)

 

 

To Solve

 

S'  +  S/(15+T)  = 2

 

(15 + T)S' +  S  = 2 (15+T)

 

Integrating both sides yields

 

(15 + T)S  =  30T + T² + c   

 

Using initial conditions.  At time t = 0, S = 6

 

6(15) = c or c = 90

 

at time 15, the tank is full and 

 

30S = 30*15 + 30² + 90 = 1440

 

S = 1440/30 = 48 pound of salt