Connection between outlet concentration, degree of re-use and effectiveness of the water treatment system

Connection between outlet concentration, degree of re-use and effectiveness of the water treatment system

When starting up a re-use system the concentration of substances in the system will gradually increase until it is stabilized at a given level.

Example

A simple re-use system uses a degree of re-use of 50%; a filter with 50% efficiency is installed in the re-use circuit (Fig. 10.5). Show how many times the water must circulate in the re-use system before the system is in balance (this condition is assumed as ideal) regarding the concentration level of metabolic products, presented as parts of M.

As shown, the system will be stabilized when the water has completed four circuits.

An equation to determine the concentration in the tank outlet in a re-use system (C) compared to the outlet concentration in a flow-through tank has been developed. This is based on the degree of re-use (R) and the removal efficiency (re) of the filter system in the circuit and is as follows:

C =1/(1− R +(R _re))

Example

A system has a degree of re-use of 96%, while the removal efficiency is 50%. Calculate the concentra-tion in the outlet of the system compared to a tradi-tional flow-through system.

C =1/(1−0.96+(0.96×0.5)) = 1.92

This means that the concentration of substances is 1.92 times those from a flow-through system. For instance, if the flow-through system had an SS con-centration of 20 mg/l in the outlet, the concentration in the re-use system will be 20 × 1.92 = 38.4 mg/l. If the maximum concentration that the fish tolerate without growth reduction is 25 mg/l, the re-use system is not useful because the SS concentration is too high. Either a better filter must be installed or the degree of re-use must be lowered, which means greater dilution by adding more new water.

Based on this, it is possible to calculate how much new water has to be added to have a system that functions. First the maximum allowed SS concen-tration C_max, is found:

C_max=25/20=1.25 mg/l

Then this value substituted in the formula and in the equation solved for the degree of re-use (R):

C =1/(1− R + R re)

1.25 = 1/(1 −R+ 0.5R)

R =0.4

This means that the maximum degree of re-use that can be used is 40% and 60% of new water must be added. In practice, however, a better filter unit will be installed instead of adding so much new water.

The general formula is as follows:

C =(1/(1− R + R re))M_f/Q_out

Where M_f/Q_out represents the outlet concentration in a flow-through system. By rearranging this equation it can be used to find the necessary efficiency of a filter system, based on acceptable outlet concentrations and degree of re-use. The equation may also be rearranged and solved to find the accept-able degree of re-use when a filter system has been chosen.

If the inlet water contains the substances in question, their concentrations must also be added to the equation. The mass of substances in the new inlet water (M_i) must be calculated using the mass balance equation and be added to M_f. This can be expressed as C_inQ_i, which gives the following equation:

C =(1/(1− R + R re))(M_f+(C_inQ_i))/Q_out