Respuesta :
Answer:
a) [tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]
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:
[tex]\frac{dx}{dt} =[/tex] ( 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:
[tex]conc = \frac{x(t)}{1000}\frac{kg}{L}[/tex]
- The amount of salt leaving the tank per unit time can be determined from:
salt flow-out = conc * V( flow-out )
salt flow-out = [tex]\frac{x(t)}{5000}\frac{kg}{L}*\frac{50 L}{min}\\[/tex]
salt flow-out = [tex]\frac{x(t)}{100}\frac{kg}{min}[/tex]
- The ODE becomes:
[tex]\frac{dx}{dt} = 0 - \frac{x}{100}[/tex]
- Separate the variables and integrate both sides:
[tex]\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}^)[/tex]
- 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:
[tex]13 = C*e^0 = C[/tex]
- The solution to the ODE becomes:
[tex]x(t) = 13*e^(^-^\frac{t}{100}^)[/tex]
- We will use the derived solution of the ODE to determine the amount amount of salt in the tank after t = 20 mins:
[tex]x(20) = 13*e^(^-^\frac{20}{100}^)\\\\x(20) = 13*e^(^-^\frac{1}{5}^)\\\\x(20) = 10.643 kg[/tex]
- The amount of salt left in the tank after t = 20 mins is x = 10.643 kg
