Respuesta :
Answer:
[tex]A(t) = 4*e^(^-^\frac{t}{60}^)[/tex]
Step-by-step explanation:
Solution:-
- The amount of salt in the solution is ( A ) at any time t.
- Pure water enters the tank ( no salt ( A = 0 ) ).
- The volumetric rate of flow in and out of tank is V(flow) = 5 L / min
- The rate of change of salt in the tank at time ( t ) can be expressed as a first order ordinary differential equation for the salt solution that flows in and out of the tank
- The first order ordinary differential equation is expressed as:
[tex]\frac{dA}{dt}[/tex] = ( salt flow in ) - ( salt flow out )
- Fresh water with zero salt content flows in then ( salt flow in ) = 0
- The concentration of salt within the tank changes with time ( t ).
- The volume of water in the tank remains constant ( steady state conditions ). I.e 5 Liters volume leaves and 5 Liters is added; hence, the total volume of solution in tank remains 300 Liters.
- So at any instant the concentration of salt in the 300 Liter tank is:
[tex]conc = \frac{A(t)}{300}*\frac{kg}{L}[/tex]
- The amount of salt-solution flowing out of the tank per unit time would be:
[tex]flow-out = conc * V( flow-out )[/tex]
[tex]flow-out = \frac{A(t)}{300}\frac{kg}{L} * 5\frac{L}{min}[/tex]
[tex]flow-out = \frac{A(t)}{60} \frac{kg}{min}[/tex]
- The differential equation becomes:
[tex]\frac{dA}{dt} = 0 - \frac{A}{60}[/tex]
- Separate the variables and integrate both sides:
[tex]\int {\frac{1}{A} } \, dA = - \int {\frac{1}{60} } \, dt + c \\\\Ln ( A ) = -\frac{t}{60} + c\\\\A = C*e^(^-^\frac{t}{60}^)[/tex]
- Initial conditions: A ( 0 ) = 4 grams. Use the initial conditions to evaluate the constant of integration:
[tex]4 = C*e^0 = C[/tex]
- The solution to the differential equation becomes::
[tex]A(t) = 4*e^(^-^\frac{t}{60}^)[/tex]
