Answer:
(a) The consistency as a function of time is C=0.15*t.
(b) The tank will become clogged in 24 minutes.
Explanation:
The rate of accumulation of the pulp stock can be defined as
[tex]\frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*C_{o}[/tex]
In this case, Co is 0, because the exit flow is only water and 0% fiber.
[tex]frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*0=Q_{i}*C_{i}[/tex]
Rearranging adn integrating
[tex]dC = (Q_{i}*C_{i})dt\\\int dC = \int (Q_{i}*C_{i})dt\\C=(Q_{i}*C_{i})*t+constant[/tex]
At t=0, C=0,
[tex]C=(Q_{i}*C_{i})*t+constant\\0=(Q_{i}*C_{i})*0+constant\\0=constant\\\\C=(Q_{i}*C_{i})*t[/tex]
[tex]C=(15*0.01)*t=0.15*t[/tex]
(b) The time at when the concentration reaches 6% is 0.4 hours or 24 minutes.
[tex]C=0.15*t\\0.06=0.15*t\\t=0.06/0.15=0.4[/tex]
