Differential Equations

Question

A tank is filled with 1000 liters of pure water. Brine containing 0.07 kg of salt per liter enters the tank at 5 liters per minute. Another brine solution containing 0.05 kg of salt per liter enters the tank at 7 liters per minute. The contents of the tank are kept thoroughly mixed and the drains from the tank at 12 liters per minute.

 

Determine the differential equation which describes this system. Let S(t)S(t) denote the number of kg of salt in the tank after tt minutes.Then dS/dt=

 

Solve the differential equation for S(t). S(t)=

Answer 1

The volume of salt solution or water in the tank is constant at V = 1000 L since the volumetric rate in = rate out.

Need to perform a mass salt balance around the tank as a salt concentration times volume of tank solution.

Since tank initially contains pure water (at t = 0) then C0 = 0 kg salt/L.

 

In:       (0.07 kg salt/L)*(5 L/min) + (0.05 kg salt/L)*(7 L/min)             = 0.70 kg salt/min

 

Out:     C*(12 L/min)

 

Accum: V*dC/dt                     Note:  V = constant = V0 = 1000 L

 

100*dC/dt = 0.70 - 12*C        (separate variables)

 

dC/(0.70 - 12*C) = (0.01)*dt

 

∫dC/(0.70 - 12*C)  = (0.01)∫dt

 

-(1/12)*ln[(0.70 - 12*C)/(0.70 - 12*C0)] = (0.01)*t                (C0 = 0)

 

ln[(0.70 - 12*C)/(0.70)]  = -(0.12)*t                                       (take exponential both sides)

 

[(0.70 - 12*C)/(0.70)]  = e^-(0.12)*t

 

1 - (12/0.7)*C = e^-(0.12)*t                                 (solve for C)

 

1 -  e^-(0.12)*t  = (12/0.7)*C

 

C(t) = (0.7/12)*(1 - e^-(0.12)*t)                            ≡ kg salt/L in tank at t

 

S(t) = V*C(t)                                              V = 1000 L

 

S(t) = 1000*[(0.7/12)*(1 - e^-(0.12)*t]  

 

S(t) = 58.33*(1 - e^-(0.12)*t)                                 ≡ kg salt in tank at t