Question

A tank with a height of 100 feet and a constant cross sectional area of 10 ft has a constant input flow of 15 ft /hour of pulp stock at 1% consistency and has a screen on the exit flow that only allows water to be removed and keeps all of the fiber in the tank. The tank is well mixed and completely full (i.e, an overflow type condition where flow in equals flow out). What is the consistency as a function of time if the tank contents starts off at 0% consistency. Plot it using excel or another spreadsheet tool, see below. The tank will become clogged when the consistency reaches 6%, when will this happen?

Answer 1

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

$\frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*C_{o}$

In this case, Co is 0, because the exit flow is only water and 0% fiber.

$frac{dC}{dt}=Q_{i}*C_{i}-Q_{o}*0=Q_{i}*C_{i}$

Rearranging adn integrating

$dC = (Q_{i}*C_{i})dt\\\int dC = \int (Q_{i}*C_{i})dt\\C=(Q_{i}*C_{i})*t+constant$

At t=0, C=0,

$C=(Q_{i}*C_{i})*t+constant\\0=(Q_{i}*C_{i})*0+constant\\0=constant\\\\C=(Q_{i}*C_{i})*t$

$C=(15*0.01)*t=0.15*t$

(b) The time at when the concentration reaches 6% is 0.4 hours or 24 minutes.

$C=0.15*t\\0.06=0.15*t\\t=0.06/0.15=0.4$

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