Differential Equations

Question

1. A tank contains 1000L of pure water. Brine that contains 0.02kg of salt per liter enters the tank at a rate of 5L/min. Also, brine that contains 0.09kg of salt per liter enters the tank at a rate of 10L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15L/min. Answer the following questions.

 

How much salt is in the tank after t minutes?

Answer (in kilograms): S(t)=

 

How much salt is in the tank after 4 hours?
Answer (in kilograms):

 

 

2. A tank contains 100L of water. A solution with a salt concentration of 0.2kg/L is added at a rate of 6L/min. The solution is kept thoroughly mixed and is drained from the tank at a rate of 4L/min. Answer the following questions.

 

If y(t) is the amount of salt (in kilograms) after t minutes, what is the differential equation for which y is satisfied? Use the variable y for y(t).
Answer (in kilograms per minute): dy/dt=

 

How much salt is in the tank after 50 minutes?
Answer (in kilograms)=

Answer 1

Provide overall volumetric flow balance around the tank:   Flow In - Flow Out = Accumulation

 

In:   5 L salt/min + 10 L salt/min

Out: 15 L salt/min

Accum:  0    since In flow = Out Flow  

 

therefore, the water or salt solution in the tank is kept at a constant volume, V = 100 L (given)

 

Since the salt concentration  in the tank is changing with time until steady state is reached, you want to

solve for the salt concentration in the tank with time. Then you can multiply the salt concentration by the

constant tank volume of water or salt solution to determine the salt mass as a function of time, S(t).

 

Therefore, provide a salt mass balance around the tank in terms of salt concentration (since it is changing with time):

 

Let C = salt concentration (kg salt/L)  in tank at time, t  (well-mixed)

      C0 = 0  kg/L  (no salt in tank at t = 0  because it is initially fresh water in tank)

 

In:        (0.02 kg salt/L)*(5 L/min) + (0.09 kg salt/L)*(10 L/min) 

 

Out:      C*(15 L/min)

 

Accum: V*dC/dt             (remember V is constant at 100 L)

 

Therefore:   V*dC/dt = (0.10 kg salt/min + 0.9 kg salt/min) - C*(15 L/min)

 

or   100*dC/dt = 1.0 - 15*C             (separate variables)

               

dC/(1 - 15*C)  = 0.01*dt                  (integrate both sides)

 

∫dC/(1 - 15*C) = 0.01∫dt                 

 

-(1/15)[ln(1 - 15*C)/(1 - 15*C0)] = t             ( C0 = 0    and multiply both sides by -15 )

 

[ln(1 - 15*C)]  = -15*t

 

take exponential of both sides and solve for C:

 

(1 - 15*C)) = e^-15*t      

 

(1 - 15*C) = e^-15*t

 

1 -  e^-15*t  = 15*C

 

 

C = (1 - e^-15*t)/15               ≡ kg salt/L

 

therefore,  S(t) =  V*C(t) =  (100 L)*C(t) =   100*[1 - e^-15*t]/15 = 6.67*[1 - e^-15*t]

 

1a)   S(t) = 6.67*[1 - e^-15*t]

 

1b)   for 4 hrs, t = 4*60 min = 240 min

      S(240) = 6.67*[1 - e^-15*240]  =  6.67 kg salt  (it has long reached steady-state)

 

2)  Since the In volume rate > Out volume rate then the tank volume, V, will change, along with the salt concentration;

therefore, you need to solve for both the V(t) and C(t) to combine to obtain the mass of salt in the tank at time, t

 

 

y(t) = V(t)*C(t)

 

First solve for the volume of brine or water in tank at time, t

 

dV/dt = 6 L/min - 4 L/min = 2 L/min

 

∫dV = 2∫dt

 

V - V0 = 2*t

 

V(t) = V0 + 2*t        ≡ L brine solution or water

 

C0 = 0 kg salt/L   (only water in tank at t = 0)

 

Now perform a salt mass balance around tank volume:

 

In: (0.2 kg salt/L)*(6 L/min)    ≡ kg salt/min

Out: C*(4 L/min)                     ≡ kg salt/min

Accum: d(V*C)/dt  use chain rule:  = V*dC/dt + C*dV/dt           ≡ kg salt/min

 

V*dC/dt + C*dV/dt = 1.2 - 4*C

 

Note: dV/dt = 2 L/min    and V =V0 + 2*t  = 100 + 2*t

 

(100 + 2*t)*dC/dt + 2*C = 1.2 - 4*C      (combine like terms and solve for C)

 

(100 + 2*t)dC/dt = 1.2 - 6*C           (divide all terms by 2)

 

(50 + t)*dC/dt = 0.6 - 3*C             (separate variables)

 

dC/(0.6 - 3*C) = dt/(50 + t)           (integrate both sides)

 

∫dC/(0.6 - 3*C)  = ∫dt/(50 + t)

 

-(1/3)*ln[(0.6 - 3*C)/(0.6 - 3*C0)] = ln[(50 + t)/(50 + t0)]           Note:  C0 = 0   t0 = 0

 

ln[(0.6 - 3*C)/(0.6)] = -3*ln[(50 + t)/(50)]

 

ln(1 - 5*C)  = -3*ln(1 + 0.02*t)                (take exponential both sides)             

 

1 - 5*C  =  (1 + 0.02*t)^-3

 

1 - (1+ 0.02*t)^-3 = 5*C

 

 

5*C = 1 - (1 + 0.02*t)^-3

 

 

 

C =   0.2*[1 - (1 + 0.02*t)^-3]  ≡ kg salt/L

 

Therefore,

 

y(t) = V(t)*C(t) = 0.4*(50 + t)*[1 - (1 + 0.02*t)^-3]

 

2 a)   y(t) = 0.4*(50 + t)*[1 - (1 + 0.02*t)^-3]                ≡ kg salt 

 

 at t = 50 min 

 

y(50) = 0.4*(50 + 50)*[1 - (1 + 0.02*50)^-3]  = 35  kg salt