# q = 6.74x1 = mg/l x 3.78x10 L/d = 3.4x10 mg / day a) Single CMFR mg/g C Organic Load = Carbon requirement = 6.74 mg 1000 g C inf = 10 mg/l

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1 Example An industrial wastewater contains 10 mg/l chlorophenol, and is going to be treated by carbon adsorption. 90% removal is desired. The wastewater is discharged at a rate of 0.1 MGD. Calculate the carbon requirement for a) a single, mixed contactor (CMFR) b) two mixed (CMFR) contactors in series c) a column contactor. Freundlich isoherm 0.41 q = 6.74xC mg/g C mg/l

2 a) Single CMFR C inf = 10 mg/l Organic Load = Carbon requirement = 0.41 q = 6.74x1 =6.74 mg/g C mg/l x.78x10 L/d =.4x10 mg / day 6 g C 1 kg.4x10 mg/day x x = 505 kg/day 6.74 mg 1000 g C eff = 1 mg/l

3 b) Two CSTRs connected in series 10 mg/l 5 mg/l 1 mg/l We assume it. It is not given in the question. If you change it, you will calculate a different value.

4 Contactor 1 C inf = 10 mg/l C eff = 5 mg/l 0.41 q = 6.74x5 =1.0 mg/g C Organic Load = mg/l x.78x10 L/d = 1.89x10 mg/day Carbon requirement = 6 g C 1 kg 1.89x10 mg/day x x = 145 kg / day 1.0 mg 1000 g

5 Contactor 2 C inf = 5 mg/l C eff = 1 mg/l 0.41 q = 6.74x1 =6.74 Organic Load = Carbon requirement = mg/g C mg/l x.78x10 L/d = 1.51x10 mg/day 6 g C 1 kg 1.51x10 mg/day x x = 224 kg/day 6.74 mg 1000 g

6 qe C e of 1 CMFR and 2 nd Contactor Ce C e of 1 st Contactor C o

7 Total C requirement = =69 kg/day a) a single, mixed contactor (CMFR) b) two mixed (CMFR) contactors in series C requirement decreased because, in the 1 st contactor, we are able to put more on the surface of the carbon.

8

9 Column Height Flow Direction 0 C o Everything happens in the primary adsorption zone (or mass transfer zone, MTZ). This layer is in contact with the solution at its highest concentration level, C o. As time passes, this layer will start saturating. Whatever escapes this zone will than be trapped in the next zones. As the polluted feed water continues to flow into the column, the top layers of carbon become, practically, saturated with solute and less effective for further adsorption. Thus the primary adsorption zone moves downward through the column to regions of fresher adsorbent. Concentration (mg/l) Here you start observing your breakthrough curve when the last layer starts getting saturated.

10 0 C o Flow Direction Last primary adsorption zone. It is called primary because the upper layers are not doing any removal job. They are saturated. When breakthrough occurs there is some amount of carbon in the column still not used. Generally, this is accepted to be 10-15%.

11 Primary adsorption zone Region where the solute is most effectively and rapidly adsorbed. This zone moves downward with a constant velocity as the upper regions become saturated.

12 Active zones at various times during adsorption and the breakthrough curve.. Ref:

13 Column Contactor C inf = 10 mg/l C eff = 1 mg/l 0.41 q = 6.74x10 =17. mg/g C Organic Load = mg/l x.78x10 L/d =.4x10 mg/day Assume that the breakthrough occurs while 10% of the carbon in the column is still not used. Carbon requirement = 6 g C 1 kg 100%.4x10 mg/day x x x = kg/day 17. mg 1000 g 90%

14 Packed Column Design It is not possible to design a column accurately without a test column breakthrough curve for the liquid of interest and the adsorbent solid to be used.

15 Theoretical Breakthrough Curve

16 i. Scale up procedure ii. and Kinetic approach Packed Column Design are available to design adsorption columns. In both of the approaches a breakthrough curve from a test column, either laboratory or pilot scale, is required, and the column should be as large as possible to minimize side wall effects. Neither of the procedures requires the adsorption to be represented by an isotherm such as the Freundlich equation.

17 Packed Column Design Scale up Procedure for Packed Columns Use a pilot test column filled with the carbon to be used in full scale application. Apply a filtration rate and contact time (EBCT) which will be the same for full scale application (to obtain similar mass transfer characteristics). Obtain the breakthrough curve. Work on the curve for scale up.

18 Example An industrial wastewater having a TOC of 200 mg/l will be treated by GAC for a flowrate of 150 m /day. Allowable TOC in the effluent is 10 mg/l. Pilot Plant Data Q = 50 L/hr Column diameter = 9.5 cm Column depth (packed bed) = 175 cm Packed bed carbon density = 400 kg/m V breakthrough = 8400 L V exhaustion = 9500 L

19 Breakthrough Curve of the Pilot Plant

20 a) Filtration rate of the pilot plant FR = Q L cm = 50 x x = 2 A hr π d 1L 2 cm 705 hr.cm 2 d = 9.5 cm The same FR applies to Packed Column.

21 b) Area of the Packed Column FR = Q A Q A = FR 2 6 m 1 h.cm 1 d 10 cm A = 150 x. x x = d 705 cm 24 h 1 m 8865cm 2 4 x 8865 d= = π 106cm

22 c) Empty Bed Contact Time of the Pilot Plant τ = Q 2 2 d 9.5 = A x Height = π x H =.14 x x175 = 12,404 cm = L 12.4 L τ = = hr = L/hr 15 min 15 mins.is the EBCT of the Packed Column.

23 d) Height of the Packed Column Q cm 1 hr τ x = 15 min x 705 x = 2 A hr.cm 60 min 176 cm The same as the height of the Pilot Plant. Because height τ Q/A is set by and, and these are the same for Pilot Plant and the Packed Column.

24 e) Mass of Carbon required in the Packed Column π 4 2 = 1.76 m x x 1.06 = 1.55 m Packed bed carbon density is given by the supplier. density = 400 kg/m kg m 1.55 m x 400 = 621 kg

25 f) Determination of q e Mass of carbon in the pilot column = Volume of the pilot column = 12.4 L x density kg = m x 400 = 4.96 m TOC removed by 5 kg of carbon = 5 kg 200 mg/l x 9500 L = 6 1.9x10 mg 6 1.9x10 mg TOC q e = = 5 kg C mg 80 C g

26 g) Fraction of Capacity Left Unused (Pilot Plant) mg Total capacity = 9500 L x 200 = L 6 1.9x10 mg TOC removed before breakthrough = mg 8400 L x 200 = L x10 mg Fraction of capacity left unused = f = x10 6 x x10 12% This fraction of capacity left unused will apply to the Packed Column also.

27 h) Breakthrough time of the Packed Column Organic Loading = mg m 1000 L 200 x 150 x = L d 1 m 6 0x10 mg/d 6 mg 1 Carbon consumption rate = 0x10 x = d 80 mg/g C 78.9 kg/d Amount of carbon consumed = 621 kg x = kg Breakthrough time = kg 78.9 kg/d 7 days The same as the Packed Column : 1 hr 1 d 8400 L x x = 50 L 24 hr 7 days

28

29 i) Volume Treated Before Breakthrough treated m = 150 x 7 days = d 1050 m

30 Kinetic Approach Packed Column Design This method utilizes the following kinetic equation. C 1 C k o 1 + e Q 1 q M - C V o o

31 where C = effluent solute concentration C o = influent solute concentration k 1 = rate constant q o = maximum solid phase concentration of the sorbed solute, e.g. g/g M = mass of the adsorbent. For example, g V = throughput volume. For example, liters Q = flow rate. For example, liters per hour

32 Kinetic Approach Packed Column Design The principal experimental information required is a breakthrough curve from a test column, either laboratory or pilot scale. One advantage of the kinetic approach is that the breakthrough volume, V, may be selected in the design of a column.

33 Packed Column Design Assuming the left side equals the rigth side, cross multiplying gives k t Q q M - C V o 1 + e = C o Co Rearranging and taking the natural logarithms of both sides yield the design equation

34 Packed Column Design Rearranging and taking the natural logarithms of both sides yield the design equation. C k q M k C V ln o -1 = 1 o - 1 o C Q Q y = b - mx

35 Example A phenolic wastewater having a TOC of 200 mg/l is to be treated by a fixed bed granular carbon adsorption column for a wastewater flow of 150 m /d, and the allowable effluent concentration, C a, is 10 mg/l as TOC. A breakthrough curve has been obtained from an experimental pilot column operated at 1.67 BV/h. Other data concerning the pilot column are as follows: inside diameter = 9.5 cm, length = 1.04 m, mass of carbon = 2.98 kg, liquid flowrate = 12.9 L/h, unit liquid flowrate = L/s.m 2, and the packed carbon density = 400 kg/m. The design column is to have a unit liquid flowrate of 2.04 L/s.m 2, and the allowable breakthrough volume is 1060 m.

36 Example Using the kinetic approach for design, determine : The design reaction constant, k 1, L/s-kg. The design maximum solid - phase concentration, q o, kg/kg. The carbon required for the design column, kg. The diameter and height of the design column, m. The kilograms of carbon required per cubic meter of waste treated.

37 V (L) C(mg/L) C/Co Co/C Co/C-1 ln(co/c-1) 0 0 0,000 78,0 9 0,045 22,222 21,222,06 984,0 11 0,055 18,182 17,182 2,84 124,0 8 0,040 25,000 24,000,18 190,0 9 0,045 22,222 21,222, ,0 0 0,150 6,667 5,667 1,7 2520, ,500 2,000 1,000 0, , ,825 1,212 0,212-1,55 290,0 19 0,965 1,06 0,06 -,2 126, ,000 1,000 0,000

38 C, mg/l V, Liters

39 ln(co/c - 1) V (L) Plot of Complete Data Set

40 ln(co/c - 1) 15 y = -0,0064x + 15, Volume treated (L) Take the linear range only!

41 q k M = 0 1 Q k -1 C L = 1 0 Q (0.0064L-1) (12.9 L) a)k = h = L 1 mg 200 mg h L L 1h 10 mg =0.11 L mg h 600s 1kg kg s -4 6

42 q 0.11 L 2.98kg b)15.787= 0 kg s 12.9 L h 600s 1h L h 600s 1h q = L 2.98kg kg s q =0.166 kg 0 kg

43 k = mg h q = kg kg 4 L Q = 6250 L / h V = L C 0 = 200 mg / L Using Co k1qom k1cov ln -1 = - C Q Q L kg L mg 200 mg h kg mg h L ln -1 = - 10 L L h h M L M= mg =1545 kg

44 M =1545 kg Packet carbon density = 400 kg / m Then, design bed volume is; 1545 kg V.86 m kg 400 m Q = 6250 L / h = 1.76 L / s (given) Unit liquid flowrate = 2.04 L / s m 2 (given) 1.76 L / s Cross section area = 0.85m d = 1.04 m Breakthrough time is; L / s m.86 m m Column height = 1050 m T B = 7 d 150 m / d 4.54 m

45 Scale-up approach: 1. The design bed volume (BV) is found as; 150 m / d 1.67 BV / h = = 6.25 m / h BV =.74 m 24 h/ d 2. The mass of carbon required is; M = BV =.74 m 400 kg / m =1500 kg From the breakthrough curve the volume treated at the allowable breakthrough (10 mg/l TOC) is 2080 L. So, the solution treated per kilogram of carbon is 2080 L/2.98 kg or 698 L/kg (pilot scale). The same applies to the design column; for a flow rate of 150 m /d.. The weight of carbon exhausted per hour (M t ) is 150 m / d kg 1000 L t M = 24 h 698 L m kg/h

46 C, mg/l V, Liters

47 4. The breakthrough time is; 1500 kg T = =168 h = 7 d kg / h 5. The breakthrough volume of the design column is; V = Q T =150 m / d 7 d = 1050 m B Comparing the results of two approaches: Kinetic approach M=1545 kg V = 1050 m B T B = 7 d V =.86 m Design Scale-up approach M = 1500 kg V = 1050 m B T B = 7 d V =.74 m Design

48 Example A phenolic wastewater that has phenol concentration of 400 mg/l as TOC is to be treated by a fixed bed granular carbon adsorption column for a wastewater flow of L/d, and the allowable effluent concentration, C a, is 5 mg/l as TOC. A breakthrough curve has been obtained from an experimental pilot column operated at 1.67 BV/h. Other data concerning the pilot column are as follows: inside diameter = 9.5 cm, length = 1.04 m, mass of carbon = 2.98 kg, liquid flowrate = L/h, unit liquid flowrate = L/s.m 2, and the packed carbon density = 401 kg/m. The design column is to have a unit liquid flowrate of 2.8 L/s.m 2, and the allowable breakthrough volume is 850 m.

49 V (L) C (mg/l) C/Co Co/C Co/C - 1 ln(co/c - 1)

50 C, mg/l V, Liters

51 ln(co/c - 1) V (L)

52 ln(co/c - 1) 2 1 y = -0,0146x + 18,657 R² = 0, V (L)

53 q k M = 0 1 Q k C L -1= 1 0 Q ( L -1) (17.42 L) h L L mg mg h kg s L k = = L h 600s 1h q = L 2.98kg kg s q =0.171 kg 0 kg

54 k = mg h q = kg kg 4 L Q = L / h V = L C 0 = 400 mg / L Using Co k1qom k1cov ln -1 = - C Q Q L kg L mg 400 mg h kg mg h L ln -1 = - 5 L L h h M L M=2190 kg

55 M =2190 kg Packet carbon density = 401 kg / m Then, design bed volume is; 2190 kg V 5.46 m kg 401 m Q = L / h = 2.6 L / s (given) Unit liquid flowrate = 2.8 L / s m 2 (given) 5.46 L / s Cross section area = 2.29m d = 1.71 m Breakthrough time is; L / s m 5.46 m m Column height = T = 850 m.74 d m / d B 2.8 m

56 Scale-up approach: The design bed volume (BV) is found as; L/ d 1.67 BV / h = = L / h 24 h/ d BV = L = 5.67 m The mass of carbon required is; M = BV = 5.67 m 401 kg / m = 2272 kg From the breakthrough curve the volume treated at the allowable breakthrough (5 mg/l TOC) is 1110 L. So, the solution treated per kilogram of carbon is 1110 L/2.98 kg or 72.5 L/kg (pilot scale). The same applies to the design column; for a flow rate of L/d, the weight of carbon exhausted per hour (M t ) is M = t L / d kg 24 h /d 72.5 L 25.4 kg/h

57 C, mg/l V, Liters

58 The breakthrough time is; 2272 kg T = = 89.5 h =.7 d 25.4 kg / h The breakthrough volume of the design column is; V = Q T = m / d.7 d = m B Comparing the results of two approaches: Kinetic approach Scale-up approach M=2190 kg M = 2272 kg V = 850 m B T B =.74 d V = 5.46 m Design V = m B T B =.7 d V = 5.67 m Design

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