Answer:
a) dx/dt = 600 - 6x
b) x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)
c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s
Explanation:
Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out
Let the concentration of salt in the tank at anytime be C
Let the concentration of salt entering the tank be Cᵢ
Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)
Let the flowrate in be represented by Fᵢ
Let the flowrate out = F₀ = F
Fᵢ = F₀ = F = 6 L/min
a) Then the component balance for the salt
Rate of accumulation = rate of flow into the tank - rate of flow out of the tank
dx/dt = Fᵢxᵢ - Fx
Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min
dC/dt = 24 - 6C
dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25
(1/25)(dx/dt) = 24 - (6/25)x
dx/dt = 600 - 6x
b) dC/dt = 24 - 6C
dC/(24 - 6C) = dt
∫ dC/(24 - 6C) = ∫ dt
(-1/6) In (24 - C) = t + k (k = constant of integration)
In (24 - 6C) = -6t - 6k
-6k = K
In (24 - 6C) = K - 6t
At t = 0, C = 15 g/25 L = 0.6 g/L
In (24 - 6(0.6)) = K
In 20.4 = K
K = 3.02
So, the equation describing concentration of salt at anytime in the tank is
In (24 - 6C) = K - 6t
In (24 - 6C) = 3.02 - 6t
24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾
6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)
C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6
C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)
x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)
x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)
c) when x = 20 g
20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)
80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)
- (6t - 3.02) = In 80
- (6t - 3.02) = 4.382
(6t - 3.02) = -4.382
6t = -4.382 + 3.02
t = - 1.362/6 = 0.227 min = 13.62 s
