Question

A tank contains 5,000 L of brine with 13 kg of dissolved salt. Pure water enters the tank at a rate of 50 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate.
Required:
a. How much salt is in the tank after t minutes?
b. How much salt is in the tank after 20 minutes?

Answer 1

Answer:

a) $x(t) = 13*e^(^-^\frac{t}{100}^)$

b) 10.643 kg

Step-by-step explanation:

Solution:-

- We will first denote the amount of salt in the solution as x ( t ) at any time t.

- We are given that the Pure water enters the tank ( contains zero salt ).

- The volumetric rate of flow in and out of tank is V(flow) = 50 L / min  

- The rate of change of salt in the tank at time ( t ) can be expressed as a ODE considering the ( inflow ) and ( outflow ) of salt from the tank.

- The ODE is mathematically expressed as:

                            $\frac{dx}{dt} =$ ( salt flow in ) - ( salt flow out )

- Since the fresh water ( with zero salt ) flows in then ( salt flow in ) = 0

- The concentration of salt within the tank changes with time ( t ). The amount of salt in the tank at time ( t ) is denoted by x ( t ).

- The volume of water in the tank remains constant ( steady state conditions ). I.e 10 L volume leaves and 10 L is added at every second; hence, the total volume of solution in tank remains 5,000 L.

- So any time ( t ) the concentration of salt in the 5,000 L is:

                             $conc = \frac{x(t)}{1000}\frac{kg}{L}$

- The amount of salt leaving the tank per unit time can be determined from:

                         salt flow-out = conc * V( flow-out )  

                         salt flow-out = $\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}$

                         salt flow-out = $\frac{x(t)}{100}\frac{kg}{min}$

- The ODE becomes:

                               $\frac{dx}{dt} = 0 - \frac{x}{100}$

- Separate the variables and integrate both sides:

                       $\int {\frac{1}{x} } \, dx = -\int\limits^t_0 {\frac{1}{100} } \, dt + c\\\\Ln( x ) = -\frac{t}{100} + c\\\\x = C*e^(^-^\frac{t}{100}^)$

- We were given the initial conditions for the amount of salt in tank at time t = 0 as x ( 0 ) = 13 kg. Use the initial conditions to evaluate the constant of integration:

                              $13 = C*e^0 = C$

- The solution to the ODE becomes:

                           $x(t) = 13*e^(^-^\frac{t}{100}^)$

- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:

                           $x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg$

- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg