Respuesta :
Answer:
A) (dA/dt) = 5 - 0.05A
B) A(t) = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾
C) At t = 10 mins, A = 45.42 g
D) When A = 199 lb, t = 62.1 minute
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 = 100 gallons (constant because flowrate in and out is the same)
Rate of flow into the tank = Fᵢ = 5 gallons/min
Rate of flow out of the tank = F = 5 gallons/min
A) Component balance for the concentration.
Let the initial amount of salt in the tank be A₀
The rate of flow of salt coming into the tank be 1 lb/gallon × 5 gallons/min = 5 lb/min
Amount of salt in the tank, at any time = A
Rate of flow of salt out of the tank = (A × 5 m³/min)/V = (5A/V) g/min
But V = 100 gallons
Rate of flow of salt out of the tank = 0.05A lb/min
The balance,
Rate of Change of the amount of salt in the tank = (rate of flow of salt into the tank) - (rate of flow of salt out of the tank)
(dA/dt) = 5 - 0.05A
B) dA/(5 - 0.05A) = dt
∫ dA/(5 - 0.05A) = ∫ dt
Integrating the left hand side from A₀ to A and the right hand side from 0 to t
- 20 In [(5 - 0.05A)/(5 - 0.05A₀)] = t
In (5 - 0.05A) - In (5 - 0.05A₀) = - 0.05t
In (5 - 0.05A) = In (5 - 0.05A₀) - 0.05t
A₀ = 10 lb
In (5 - 0.05A) = (In 4.5) - 0.05t
In (5 - 0.05A) = 1.504 - 0.05t
(5 - 0.05A) = e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾
A(t) = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾
C) When t = 10 min
A = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾
1.504 - 0.05 (10) = 1.004
A = 100 - 20e⁽¹•⁰⁰⁴⁾
A = 100 - 54.58
A = 45.42 g
D) When A = 199
A = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾
199 = 100 - 20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾
20e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾ = 99
e⁽¹•⁵⁰⁴ ⁻ ⁰•⁰⁵ᵗ⁾ = 4.95
-(1.504 - 0.05t) = In 4.95 = 1.599
0.05t = 1.599 + 1.504 = 3.103
t = 62.1 minute
