Practice problems in ADSORPTION and ION EXCHANGE – Solutions Solutions
Q1. Adsorption isotherms
We considered three forms of isotherms: Linear:
q Kc
- gives straight line on plot of q vs. c [1]
Freundlich:
q Kc n
- gives straight line on log-log plot of q vs. c [2]
Langmuir:
q
q0 c K c
- gives straight line of plot of 1/q 1/q vs. 1/c 1/c [3]
The data are closest to a straight line on plot [3], indicating the Langmuir isotherm is the best fit. The linearized form of the Langmuir equation is: 1 q
K 1
1
q0 c
q0
Using the line of best fit for the data in plot [3]: Intercept q0
1 q0
6.78
1 6.78
0.147
g glucose g alumina
and Slope
K q0
0.124
K 0.124q0 0.1240.147 0.0182
g glucose m 3 solution
Q2. Batch adsorption
Given: M = 3 kg granular activated carbon q F = 0 kg phenol / kg carbon 3
S = 2.5 m solution 3 c F = 0.25 kg phenol / m solution
Material balance on adsorbate (phenol), assuming solution v olume S is constant: q F M c F S qM cS
03 0.25 2.5 q 3 c 2.5 Operating line equation : q 0.208 0.833c
Plot the equilibrium data and the operating line. At equilibrium, q and c are in equilibrium, at the intersection of the two lines:
The equilibrium data are best fit by a Freundlich-type isotherm, as shown on the graph. Solving analytically (or graphically) for the intersection of the two lines: ceq 0.109 qeq 0.118
kg phenol m 3 solution kg phenol kg carbon
Percent phenol extracted: - Assuming solution volume S is constant c F ceq c F
100%
0.25 0.109 0.25
100% 56.4%
Q3. Fixed bed adsorption
From Example 2 in class (slide 13): Height of bed: H T = 14 cm Diameter of bed: D = 4 cm 3 Flow rate = 754 cm /s At the breakpoint: c/c0 = 0.01; t b = 3.65 h Total capacity of bed: t t = 5.16 h Usable capacity of bed: t u = 3.65 h Length used to break point: H B = 9.9 cm Length of unused bed: H UNB = 4.1 cm On graph: A1 = 3.65 h; A2 = 1.51 h
(a) For a constant flowrate, a change in the break point time (to t b = 8.5 h) will only change the length of the column, not the diameter ( D = 4 cm). H UNB is constant; new H B is proportional to t b: 8.5h H B (new) 9.9cm 23.1cm 3.65h Height of new column: HT (new) HUNB H B (new) 4.1 23.1 27.2 cm H T
Fraction of capacity used up to the break point: - On a plot of c/c0 vs t , A1 (~t b) will change with the length of the column, but A2 will be constant (same mass transfer zone)
c 0 1 c0 dt A1 A2 t b,new A2 8.5 1.51 10.01h t b,new 8.5h
t t ,new t u ,new t u ,new t t ,new
8.5 10.01
0.849
(b) When the flowrate changes, the diameter need s to change to keep similar flow (velocity) and mass transfer characteristics. The length of the bed and the fraction of capacity used will change with the break point as calculated in (a): H T = 27.2 cm; t u/t t = 0.849. For a constant flow velocity, the cross-sectional area will chan ge proportionally to the volumetric flow rate: Anew
4
2 Dnew
Dnew
Flow new Flowold
2000 754 4 2000 754
Aold
2 Dold
4cm2 6.5cm
Q4. Ion-exchange column
Given: H T = 30.5 cm D = 2.59 cm Mass of ion-exchange resin: 99.3 g 3
Feed flow rate: 1.37 cm /s Feed concentration: c0 = 0.18 M (0.18 mol CuSO4 / L) At the break point (t b), c/c0 = 0.010
Plotting the breakthrough data:
A1
A2
t b
t s
Break point time: from the graph, when c/c0 = 0.010, t b = 460 s
Fraction of total capacity used up to the break point: Usable capacity of the bed: -
= ∫ 1 − =
Because the break point concentration is small,
≅ = 460
Total capacity of the bed: -
= ∫∞ 1 − = +
Can be solved by integrating the area above the curve graphically or numerically Alternatively, it can be shown that ( A1+ A2) = t t ≈ t s, the time when c/c0 = 0.5 From the graph,
≅ = 645 The fraction of total bed capacity used up to break point is then:
= 460 = 0.713 645 Length of unused bed:
= ⇒ = 460 645 ×30.5 = 21.75
= − ⇒ = 30.5 − 21.75 = 8.75 Saturation loading capacity of the ion-exchange resin: Amount of Cu being fed into the 99.3 g resin is equal to:
1 × 645 ×0.18 ×63.54 = 10.106 1.37 × 1000 Saturation loading capacity of the solid:
= 10.106 = 0.102 99.3
Q5. Scale-up of ion-exchange column
Given: Column I:
H T = 0.4 m 3 Flow rate = 0.2 m /h t b = 8.0 min t u/t t = 0.65
Column II:
t b = 13.0 min
In mass transfer zone, length of unused bed ( H UNB) is:
= (1 − ) ( ) ⇒ = 1 − 0.65 × 0.4 = 0.14
Length of used bed ( H B)is:
= ⇒ = 0.65 × 0.4 = 0.26
For Column II, the following relation can be applied to determine the length of used bed:
, = , ⇒ = , × , , , , , ⇒ , = 0.268× 13 = 0.4225
Total length for the column II, assuming that the length of unused bed ( H UNB) is unchanged:
, = , + ⇒ , = 0.4225 + 0.14 = 0.5625
