Answer:
a) C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)
b) The concentration of salt in the tank attains the value of 0.01 kg/L at time, t = 0.0713 min = 4.28s
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
Then the component balance for the salt
Rate of accumulation = rate of flow into the tank - rate of flow out of the tank
dC/dt = FᵢCᵢ - FC
Fᵢ = 9 L/min, Cᵢ = 0.02 kg/L, F = 9 L/min
dC/dt = 0.18 - 9C
dC/(0.18 - 9C) = dt
∫ dC/(0.18 - 9C) = ∫ dt
(-1/9) In (0.18 - 9C) = t + k
In (0.18 - 9C) = -9t - 9k
-9k = K
In (0.18 - 9C) = K - 9t
At t = 0, C = 0.1/100 = 0.001 kg/L
In (0.18 - 9(0.001)) = K
In 0.171 = K
K = - 1.766
So, the equation describing concentration of salt at anytime in the tank is
In (0.18 - 9C) = -1.766 - 9t
In (0.18 - 9C) = - (9t + 1.766)
0.18 - 9C = e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾
9C = 0.18 - (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)
C = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)
b) when C = 0.01 kg/L
0.01 = 0.02 - ((e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)/9)
0.09 = (e ⁻ ⁽⁹ᵗ ⁺ ¹•⁷⁶⁶⁾)
- (9t + 1.766) = In 0.09
- (9t + 1.766) = -2.408
(9t + 1.766) = 2.408
9t = 2.408 - 1.766 = 0.642
t = 0.642/9 = 0.0713 min = 4.28s
