Ps is equal to 760 mm of Hg, one atm of pressure, 101,330 N/m , etc. depending on the system of units used.

As the pressurized flow from the air saturation tank is released into the flotation unit, the pressure reduces to atmospheric, Pa. It would be accurate to assume that the condition at this point in the flotation unit is saturation at the prevailing temperature and pressure. Let C0SA,SP represent the saturation dissolved oxygen concentration at standard pressure at the prevailing ambient temperature of the flotation tank. Thus, after pressure release, the remaining dissolved air Ao in the recycled portion of the flow is

Note that f is no longer in this equation, because enough time should be available for saturation to occur. The air utilized for flotation is Aj - Ao - Aused. Or,

The solids in the influent is Qi[Xt]. The air used to solids ratio A/S is then

4.29 RQiP

P (fCos ,t,spP — C os,a,spP a ) 4 29 R R( fC P — C P )

Similar derivation may be undertaken for operation without recycle. If this is done, the following equation is obtained.

5.3.1 Laboratory Determination of Design Parameters

To design a flotation unit, the overflow area, the pressure in the air saturation tank, and the recirculation ratio must be determined. The overflow velocity needed to estimate the overflow area may be determined in the laboratory. The right-hand side of the top drawing of Figure 5.15 shows a laboratory flotation device. A sample of the sludge is put in the pressure tank and pressurized. The tank is then shaken to ensure the sludge is saturated with the air. The pressure and temperature readings in the tank are noted. A valve leading to the flotation cylinder is then opened to allow the pressurized sludge to flow into the cylinder. The rate of rise of the sludge

Flotation cylinder

Pressure tank

Flotation cylinder

Pressure tank

interface in the flotation cylinder is followed with respect to time; this gives the rise velocity. This is equated to the overflow velocity to compute the overflow area.

The value of the air saturation tank pressure and the recirculation ratio may be designed depending upon the A/S ratio to be employed in the operation of the plant. This A/S ratio, in turn, is determined by the laboratory flotation experiment just described. A sample of subnatant is taken from the bottom of the flotation cylinder and the clarity or turbidity determined. Thus, by performing several runs for different values of A/S, corresponding values of clarity will be obtained, thereby producing a relationship between A/S and clarity. The A/S corresponding to the desired clarity is then chosen for the design calculations. Alternatively, the solids content of the float may be analyzed to obtain relationships between A/S and solids content. The A/S corresponding to the desired solids content may then be chosen for design.

In performing the laboratory float experiment, no recirculation is used. Thus, the formula to be used to compute the A/S is Equation 5.51. Ensuring, during the experiment, that the pressure tank is fully saturated, f can be considered unity.

Example 5.13 A laboratory experiment is performed to obtain the air-to-solids ratio A/S to be used in the design of a flotation unit. The pressure gage reads 276 kN/m and the temperature of the sludge and the subnatant in the flotation cylinder is 20°C. The prevailing barometric pressure is 100.6 kN/m . The total solids in the sludge is 10,000 mg/L and ¡3 was originally determined to be 0.95. Determine the A/S ratio.

Solution:

Standard barometric pressure = 101.33 kN/m

4.29(0.95)[i1X9)2X276)000 + 100,600) - 9.2(100,600)] 101,330(10,000)

Example 5.14 It is desired to thicken an activated sludge liquor from 3,000 mg/L to 4% using a flotation thickener. A laboratory study indicated an A/S ratio of 0.010 is optimal for this design. The subnatant flow rate was determined to be 8 L/m -min. The barometric pressure is assumed to be the standard of 101.33 kN/m and the design temperature is to be 20°C. Assume f = 0.5; ß = 0.95. The sludge flow rate is 400 m3/d. Design the thickener with and without recycle.

Solution:

(a) Without recycle:

= 4.29(0.95)[(0.5)(9.2)P-9.2(101,330)] ' 101,330(3,000)

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