Question

In a sewage treatment plant, a large concrete tank initially contains 440,000 liters of liquid and 10,000 kg of fine suspended solids. To flush this material out of the tank, water is pumped into the vessel at a rate of 40,000 l/h. Liquid containing solids leaves at the same rate. Estimate the concentration of suspended solids in the tank at the end of 4 h. You can assume that the initial concentration of solids in the tank

Answer 1

Answer:

$Concentration = 8.26kg/m^3$

Step-by-step explanation:

Given

  $V = 440000L$ --- volume of tank

$m = 10,000 kg$ --- solid mass

$r = 40000L/hr$ --- outflow rate  

Required

Determine the concentration at the end of 4 hours

First, calculate the amount of liquid that has been replaced at the end of the 4 hours.

$Amount = r * Time$

$Amount = 40000L/hr * 4hr$

$Amount = 40000L * 4$

$Amount = 160000L$

This implies that, over the 4 hours; The tank has 160000 liters of liquid out of 440000 liters were replaced

Calculate the ratio of the liquid replaced.

$Ratio = \frac{Amount}{Volume}$

$Ratio = \frac{160000L}{440000L}$

$Ratio = \frac{16}{44}$

$Ratio = \frac{4}{11}$

Next, calculate the amount of solid left.

$Amount (Solid)= Ratio * m$

$Amount (Solid)= \frac{4}{11} * 10000kg$

$Amount (Solid)= \frac{40000}{11}kg$

$Amount (Solid)= 3636kg$

Lastly, the concentration is calculated as:

$Concentration = \frac{Amount (Solid)}{Volume}$

$Concentration = \frac{3636kg}{440000L}$

Convert L to cubic meters

$Concentration = \frac{3636kg}{440000* 0.001m^3}$

$Concentration = \frac{3636kg}{440m^3}$

$Concentration = 8.26kg/m^3$