Question 2.15: A hold tank is installed in an aqueous effluent-treatment pr...

Basis: increment of time Δt
To illustrate the general solution to this type of problem, the balance will be set up in
terms of symbols for all the quantities and then actual values for this example substituted.
Let, Material in the tank = M,
Flow-rate = F,
Initial concentration in the tank = C_{0} ,
Concentration at time t after the feed concentration is increased = C,
Concentration in the effluent feed = C_{1} ,
Change in concentration over time increment Δt D ΔC,
Average concentration in the tank during the time increment = C_{av} .
Then, as there is no generation in the system, the general material balance (Section 2.3)
becomes: Input – Output = Accumulation

Material balance on acetone.
Note: as the tank is considered to be perfectly mixed the outlet concentration will be
the same as the concentration in the tank.
Acetone in – Acetone out = Acetone accumulated in the tank
FC_{1}\Delta t-FC_{av} \Delta t=M(C+\Delta C)-MC
F\left(C_{1}-C_{av} \right) =M\frac{\Delta C}{\Delta t}
Taking the limit, as \Delta t\longrightarrow 0
\frac{\Delta C}{\Delta t}=\frac{dC}{dt} , C_{av}=C

Integrating

Substituting the values for the example, noting that the maximum outlet concentration
will occur at the end of the half-hour period of high inlet concentration.

t = 0.5 h
C_{1} = 1000 ppm
C_{0} = 100 ppm (normal value)
M = 500 m^{3} = 500,000 kg
F = 45,000 kg/h

0.5=-\frac{500,000}{45,000}ln\left[\frac{1000-C}{1000-100} \right]
0.045=-ln\left[\frac{1000-C}{900} \right]

C= \underline{\underline{140ppm} }

So the maximum allowable concentration will not be exceeded.