Differential Equations

Question

A tank contains 70 kg of salt and 2000 L of water. A solution of a concentration 0.0175 kg of salt per liter enters a tank at the rate 7 L/min. The solution is mixed and drains from the tank at the same rate.

 

What is the concentration of our solution in the tank initially?

Concentration = __ kg/L

 

Find the amount of salt in the tank after 33 hours.

Amount= ___ kg

 

Find the concentration of salt in the solution in the tank as time approaches infinity.

Concentration = __ kg/L

Answer 1

Initial salt concentration in tank, C0 = 70 kg/2000 L = 0.035 kg/L

 

Salt Balance Around Tank Solution Volume:    Note: tank volume of solution is constant since volume rate in = rate out

 

in:  (0.0175 kg salt/L)*(7 L/min)                  = 0.1225 kg salt/min

 

out: C*(7 L/min)

 

Accumulation:  V*dC/dt                      Note: dV/dt = 0   and  V = 2000 L

 

V*dC/dt = 0.1225 - 7*C

 

 

2000*dC/dt = 0.1225 - 7*C

 

dC/(0.1225 - 7*C) = (1/2000)*dt                           (integrate both sides)

 

∫dC/(0.1225 - 7*C) = ∫(1/2000)*dt

 

-(1/7)*ln[(0.1225 - 7*C)/(0.1225 - 7*C0)] = (1/2000)*t        

 

ln[(0.1225 - 7*C)/(0.1225 - 7*C0)] = -(7/2000)*t                             (take exponential of each side)

 

[(0.1225 - 7*C)/(0.1225 - 7*C0)] = e^-(7/2000)*t  

 

0.1225 - 7*C = (0.1225 - 7*C0)*e^-(7/2000)*t                        (solve for C)

 

0.1225 - (0.1225 - 7*C0)*e^-(7/2000)*t = 7*C

 

C = 0.0175 - A*e^-(0.0035)*t        Let A = (0.0175 - C0/7) = 0.0175 - 0.035/7 = 0.0125

 

C(t) = 0.0175 - 0.0125*e^-(0.0035)*t 

 

 

Therefore, the amount of salt in the tank as function of time = S(t) = V*C(t) = 2000*C(t)

 

S(t) = 2000*[ 0.0175 - A*e^-(7/2000)*t ]= 35 - 2000*A*e^-(0.0035)*t         Note:  2000*A = 2000*(0.0125) = 25

 

S(t) = 35 - 25*e^-(0.0035)*t

 

 

at t = 33 hr or 33 hr x 60 min/hr = 1980 min       C0 = 0.035 kg/L

 

S(1980) = 35 - 25*e^{-(0.0035)*(1980)}  = 34.976   ≈ 35.98 kg after 33 hr (or 1980 min)

 

 

at t = infinity  

 

C(infinity) = 0.0175 - 0.0125*e^-(infinity)  = 0.0175  kg/L