Question

Stirred tank reaction A 100-L tank is filled with pure water when an inflow pipe is opened and a sugar solution with a concentration of 20 gm/L flows into the tank at a rate of 0.5 L/min. The solution is thoroughly mixed and flows out of the tank at a rate of 0.5 L/min.

c. At what time does the mass of sugar reach 95% of its steady-state level?

Accepted Answer

Define the variable $$ m(t)  as $$the mass of sugar (in grams) in the tank at time $$ t $$ minutes. Since the tank volume is constant at 100 L, the concentration in the tank at time $$ t  is$$ $$ \frac{m(t)}{100}  gm/L. $$Set up the differential equation for the mass of sugar in the tank. The rate of change of mass $$ \frac{dm}{dt} $$ equals the rate of sugar entering minus the rate of sugar leaving:

$$ \frac{dm}{dt} = (	ext{inflow rate}) 	imes (	ext{inflow concentration}) - (	ext{outflow rate}) 	imes (	ext{tank concentration}) $$

Substitute the given values:

$$ \frac{dm}{dt} = 0.5 	imes 20 - 0.5 	imes \frac{m(t)}{100} $$ Simplify the differential equation to standard linear form:

$$ \frac{dm}{dt} = 10 - \frac{m(t)}{200} $$ Solve this first-order linear differential equation with the initial condition $$ m(0) = 0  ($$since the tank starts with pure water). The solution will be of the form:

$$ m(t) = 	ext{steady-state mass} 	imes \left(1 - e^{-kt}ight) $$

where $$ k  is a $$positive constant related to the outflow rate and volume. Determine the steady-state mass by setting $$ \frac{dm}{dt} = 0 $$ and solve for $$ m . $$Then, find the time $$ t $$ when $$ m(t) $$ reaches 95% of this steady-state value by solving:

$$ m(t) = 0.95 	imes m_{steady-state} $$

Use the expression for $$ m(t)  to $$isolate $$ t .$$

Verified video answer for a similar problem:

Key Concepts

First-Order Linear Differential Equations

Steady-State Concentration

Exponential Approach to Equilibrium