Help with differential equations problem!

Question

A container initially contains 10L of a salt solution. Water flows into the container at a rate of 3 L/min, and the well-stirred mixture flows out at a rate of 2L/min. After 5 min the concentration of salt in the container is 0.2 g/L. Find: a. the amount of salt initially in the container. b. the volume of solution in the container when the concentration of salt is 0.1 g/L.

Answer 1

Let   V0 = initial volume of a salt solution = 10 L sol'n

 

Find the solution volume in the container (V) at a function of time, t (min):    Note: in - out = accumulation

 

in:   3 L/min

 

out: 2 L/min

 

Accum: dV/dt

 

dV/dt = 3 L/min - 2 L/min = 1 L/min      (integrate both sides)

 

∫dV/dt = (1 L/min) ∫dt

 

V - V0 = t   or   V(t) = V0 + (1 L/min)* t           where: V0 = 10 L salt solution

                                                                                  t = time water flows into container (min)

 

Now need mass salt balance:  S(t) = C(t)*V(t)    where: S(t) = mass of salt in container at t  (g salt)

                                                                                   C(t) = salt concentration in container at t  (g salt/L sol'n)

                                                                                   V(t) = V0 + 1*t     (solved above)  (L sol'n)

 

Let: C = salt concentration in container at t, (g salt/L sol'n)

       C0 = initial salt concentration in container at t = 0  (g salt/L sol'n)    - need to solve for C0

 

in:                   0       (no salt flows into container - only pure water)

 

out:                C*(2 L/min)

 

accumulation:  d(V*C)/dt     use chain rule:  V*dC/dt + C*dV/dt

 

V*dC/dt + C*dV/dt = 0 - 2*C          note: dV/dt = 1  (from earlier balance)  and V = V0 + 1*t  (see above)

 

(V0 + 1*t)*dC/dt + 1*C = - 2*C          (separate variables and solve for C)

 

dC/C = - 3*dt/(V0 + 1*t)                  (integrate both sides)

 

ln(C) = -3*ln(V0 + 1*t)                    (from C = C0 to C and t = 0 to t)

 

ln(C/C0)  = -3*ln[(V0 + 1*t)/V0]         (take exponential both sides)

 

C/C0 = [(V0 + 1*t)/V0]^-3       (rearrange)

 

C0/C = [(V0 + 1*t)/V0]³       (solve for C0)    Note: C = 0.2 g/L   at t = 5 min

 

C0 = C*[(V0 + 1*t)/V0]³      (insert given values)

 

C0 = (0.2 g/L)*{[(10 L + (1 L/min)*(5 min)]/10 L}³

 

C0 = (0.2 g/L)*3.375 = 0.675 g/L        (initial salt concentration in container)

 

a) S(t = 0) = C(0)*V(0) = C0*V0 = (0.675 g/L)*(10 L) = 6.75 g salt initially in container

 

For part b) question, you need the time when the salt concentration, C(t=?), is equal to 0.1 g/L  then plug that time, t. into volume equation, V(t) = V0 + 1*t, to solve for solution volume in the container at time, t:

 

C/C0 = [(V0 + 1*t)/V0]^-3      (start with this equation from above and solve for t)    

 

[(V0 + 1*t)/V0]³ = C0/C

 

 [(V0 + 1*t)/V0] = (C0/C)^1/3 

 

V0 + 1*t = V0*(C0/C)^1/3 

 

t =  [V0*(C0/C)^1/3 - V0]/(1 L/min)  = [10 L*(0.675/0.1)^1/3 - 10 L]/(1 L/min)  = 8.8988 = 8.9 min

 

Therefore, insert t = 8.9 min into V(t=8.9 min)

 

V (8.9 min) = V0 + 1* t = 10 L + (1 L/min)*(8.9 min) = 18.9 L salt solution in container when C = 0.1 g/L 

 

b) V = 18.9 L