Question

A salt solution flows at a constant rate into a mixing tank; and the concentration of salt coming into the tank is always Cᵢₙ = 5g/L. Also the flow rate -in , Fᵢₙequal the flow-rate-out,Fₒᵤₜ. Fᵢₙ =3.0L/s and Fₒᵤₜ =3.0L/s Intially the mixing the tank (t= 0s) held 3.0L of pure water ( no salt ) Let A = A(t) = amount of salt in the tank at time t,t≥0 a) Write down an IVP that has for its solution, A(t). b)Solve the IVP in part (a). c)Find the value of t for which the amount salt in the tank is 10kg.

Answer 1

a) The IVP for the amount of salt in the tank is A'(t) = (Cᵢₙ * Fᵢₙ) - (A(t)/Vᵢₙ * Fₒᵤₜ) , A(0) = 0. b) The solution to the IVP is A(t) = (Cᵢₙ * Fᵢₙ * Vᵢₙ / Fₒᵤₜ) * (1 - e^(-Fₒᵤₜ * t / Vᵢₙ)). c) The value of t for which the amount salt in the tank is 10kg is t = (Vᵢₙ / Fₒᵤₜ) * ln((Cᵢₙ * Fᵢₙ * Vᵢₙ) / (10kg * Fₒᵤₜ)).

Step-by-step explanation:

a) The initial value problem (IVP) for the amount of salt in the tank is:

A'(t) = (Cᵢₙ * Fᵢₙ) - (A(t)/Vᵢₙ * Fₒᵤₜ) , A(0) = 0

b) The solution to the IVP is:

A(t) = (Cᵢₙ * Fᵢₙ * Vᵢₙ / Fₒᵤₜ) * (1 - e^(-Fₒᵤₜ * t / Vᵢₙ))

c) To find the value of t for which the amount of salt in the tank is 10kg, we can substitute A(t) = 10kg into the solution from part (b), and solve for t. This gives us:

t = (Vᵢₙ / Fₒᵤₜ) * ln((Cᵢₙ * Fᵢₙ * Vᵢₙ) / (10kg * Fₒᵤₜ))