Page 224 - Fundamentals of Water Treatment Unit Processes : Physical, Chemical, and Biological

Flotation 179

8.3.5.2 Mass Balance Calculations by Spreadsheet (2) Recalling F b ¼ C r =r(air), calculate C r from (8.16),
Table CD8.3 shows the sequence of calculations that permit
F b
determination of P(saturator) for various r. The algorithm N b ¼ pd =6 (8:16)
3
b
utilizes the materials balance equation as basis for calculating
3
the required C r . The spreadsheet is set up to permit the 11 bubbles C r =(1:204 kg air=m air)
1:2 10 ¼
3
3
exploration of ‘‘what-if’’ scenarios, with respect to tempera- m water [p(40 10 6 m) =6]=bubble
ture, elevation, N p , B, and r. Note that the best-fit polynomial C r ¼ 4:785 10 3 kg air=m water
3
equations for the effect of temperature on H(O 2 ) and H(N 2 ),
from Table H.5, was applied to approximate H(air, T). 7. Determine C(saturator):
From the mass balance Equation 8.20, with argu-
ments, C r ¼ 4.785 10 3 kg=m 3 and r ¼ 0.10,
Example 8.7 Rational Design C(saturator) is calculated,
The purpose of this example is to illustrate a design algo- r[C(saturator) C a ] (C a C o )
rithm based on theory, as described in the previous sec- C r ¼
tions. A similar algorithm is used in Table CD8.3. 3 (1 þ r)
4:785 10
Given
0:10 [C(saturator) 0:031] (0:031 0:024)
¼
3
Q ¼ 0.0876 m =s (2.0 mgd) Recycle ratio, r ¼ 0.10 (1 þ 0:10)
d P ¼ 50 mm (average) Saturator efficiency, f ¼ 0.9 3
5:26 10 ¼ 0:10 [C(saturator) 0:031]
d b ¼ 40 Elevation ¼ sea level, i.e., Z ¼ 0.00 m
4
N P ¼ 1.2 10 particles=mL T ¼ 208C (0:031 0:024)
0:0526 ¼ [C(saturator) 0:031] 0:07
B ¼ 10 bubbles=particle
0:0526 ¼ C(saturator) 0:101
3
Required C(saturator) ¼ 0:154 kg air=m water
A(basin), P(sat)
8. Note that C a and C o in Equation 8.20 are calculated
Solution as follows:
1. Determine average rise velocity of particle–bubble Assume D(nozzles) ¼ 3.0 m; therefore, C a is cal-
combination: culated by Henry’s law, i.e.,
v pb (10 bubbles=particle, d p ¼ 50 mm) ¼ 13.7 m=h
(Table CD8.2) C a ¼ H(air) [P(atm) þ (D(nozzle)=10:33 m)
2. Determine area of separation zone, i.e., A(basin) (101:325 kPa)]
. Enter Figure 8.10 with arguments, d p ¼ 50 mm, C a ¼ (0:000238 kg air=m water=kPa air)
3
B ¼ 10 bubbles=particle to obtain, v o ¼ 13.7
m=h ¼ 0.0038 m=s [(101:325 2:338) kPa þ (3:00 m=10:33 m)
. Calculate A(basin): (101:325 kPa)]
3
¼ 0:031 kg air=m water
Q
A(basin) Also, calculate C o by Henry’s law, i.e.,
v o ¼
3
0:0876 m =s
A(basin)
0:0038 m=s ¼ C o ¼ H(air) [P(atm) vapor pressure]
3
A(basin) ¼ 23:0m 2 ¼ (0:000238 kg air=m water=kPa air)
[(101:325 2:338) kPa]
3. Proportion length to width; let L=w 10:1, i.e., 3
¼ 0:0236 ¼ 0:3056 kg air=m water (Ex8:7:1)
w(basin) ¼ 2.0 m
L(basin) ¼ 12.0 m 9. The saturator pressure, P(saturator) is also calculated
4. Depth by Henry’s law, i.e.,
From experience, depth is about 3–5m; let D ¼ 3.0 m.
5. Detention time, u: CðsaturatorÞ¼ f H(air) P(saturator)
3
0:154 kg air=m water ¼ 0:9 (0:000238 kg air=m 3
2
V ¼ 3m 24:0m ¼ 72:0m 3 water=kPa air) P(saturator)
3
3
u ¼ V=Q ¼ 72:0m =0:0876m =s ¼ 822s ¼ 14min P(saturator) ¼ 719 kPa absolute
¼ 618 kPa gage
6. Determine C r :
(1) Calculate N b for N b =N p ¼ B ¼ 10,
Comment
The value for v o ¼ 13.7 m=h compares with the range for
N b
¼ 10
1:2 10 particles=mL practice, 0.05 v o 100 m=h (Table 8.2). The value for
4
P(saturator)¼ 618 kPa gage is at the upper end of the range
5
N b ¼ 1:2 10 bubbles=mL for practice, 300 P(sat) 600 kPa gage (Table 8.4).