Answer:
A differential equation to model the rate of change of the amount of new chemical y in the tank at time t is given as
(dy/dt) = 15 - 0.01y
Step-by-step explanation:
First of, we take the overall balance for the system,
Let V = volume of solution in the tank at any time
The rate of change of the volume of solution in the tank = (Rate of flow into the tank) - (Rate of flow out of the tank)
The rate of change of the volume of solution = (dV/dt)
- Rate of flow into the tank = Fᵢ = 3 L/min
- Rate of flow out of the tank = F = 3 L/min
(dV/dt) = Fᵢ - F
(dV/dt) = (Fᵢ - F) = 3 - 3 = 0
dV = 0 dt
∫ dV = 0 ∫ dt
V = 0 + c = c
where c = constant of integration.
At t = 0, V = 300 L
With the flow rate in and out of the tank the same, then it means the volume of the tank is unchanged.
Component balance for the amount of chemical in the tank.
- Let the initial amount of chemical in the tank be y₀ = 0 mg
- Let the rate of flow of the amount of chemical coming into the tank = yᵢ = 5 mg/L × 3 L/min = 15 mg/min
- If the amount of chemical in the tank, at any time = y mg
- Concentration of chemical in the tank at any time = (y/V) mg/L = (y/300) mg/L
- Rate at which that amount of chemical is leaving the tank = 3 L/min × (y/300) mg/L = (y/100) mg/L = (0.01y) mg/L
Rate of Change in the amount of chemical in the tank = (Rate of flow of chemical into the tank) - (Rate of flow of chemical out of the tank)
(dy/dt) = 15 - 0.01y
Hope this Helps!!!
