Question

Suppose that a large mixing tank initially holds 100 gallons of water in which 50 pounds of salt have been dissolved. Another brine solution is pumped into the tank at a rate of 3 gal/min, and when the solution is well stirred, it is then pumped out at a slower rate of 2 gal/min. If the concentration of the solution entering is 4 lb/gal, determine a differential equation (in lb/min) for the amount of salt A(t) (in lb) in the tank at time t > 0. (Use A for A(t).)

Answer 1

Answer:

dA/dt = 12 - 2A/(100 + t)

Step-by-step explanation:

The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 4 lb/gal

Input rate of brine = 3 gal/min

Thus;

R_in = 4 × 3 = 12 lb/min

Due to the fact that solution is pumped out at a slower rate, thus it is accumulating at the rate of (3 - 2)gal/min = 1 gal/min

So, after t minutes, there will be (100 + t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(100 + t)]lb/gal × 2 gal/min

R_out = 2A(t)/(100 + t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 12 - 2A(t)/(100 + t)

Since we are to use A foe A(t), thus the Differential equation is now;

dA/dt = 12 - 2A/(100 + t)