Question

A brine solution of salt flows at a constant rate of 9 ​L/min into a large tank that initially held 100 L of brine solution in which was dissolved 0.1 kg of salt. The solution inside the tank is kept well stirred and flows out of the tank at the same rate. If the concentration of salt in the brine entering the tank is 0.02 ​kg/L, determine the mass of salt in the tank after t min. When will the concentration of salt in the tank reach 0.01 ​kg/L?

Answer 1

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