HW-5-solutions - Homework solutions PDF

Preview

Summary

CEE 6330 Physicochemical process: SolutionHOMEWORK #5:1 (30 pts). Activated carbon is used to remove a pesticide residual from a waste stream. Pesticide concentration in the waste stream is 100 mg/liter. The process is done in a batch mode in a series of three stirred tanks. The waste is processed i...

Description

CEE 6330 Physicochemical process: Solution HOMEWORK #5:

1 (30 pts). Activated carbon is used to remove a pesticide residual from a waste stream. Pesticide concentration in the waste stream is 100 mg/liter. The process is done in a batch mode in a series of three stirred tanks. The waste is processed in series by allowing the waste to contact carbon in the first tank until the system is at equilibrium. The waste stream is then separated from the exhausted carbon and sent to the second tank where it again is equilibrated. The waste stream is again separated from the exhausted carbon and sent to a third tank where it it allowed to equilibrate with the activated carbon in this tank. If each tank initially contains fresh carbon, determine the final concentration of the pesticide in the third tank. The adsorption isotherm for this pesticide can be approximated by: qe = K ⋅Ce K = 0.08 liters/mg Volume of each tank is 5 m3 (assume this is the waste volume applied to each tank). Tanks 1, 2 and 3 contain 1 kg, 0.75 kg and 0.25 kg of activated carbon, respectively.

Solution:

qe = K ⋅ Ce

qe =

( Ci − Ce ) ⋅ V m

K ⋅ Ce ⋅ m = ( Ci − C e ) ⋅ V or Ce =

Ci ⋅ V K⋅m+ V

So for the first reactor Ci = 100 mg/L, m = 1 kg, V = 5 m3 Ce = 5.88 mg/L This will be the input concentration to the 2nd reactor. So for the second reactor Ci = 5.88 mg/L, m = 0.75 kg, V = 5 m3 Ce = 0.452 mg/L This then becomes the input to the third reactor. So, for the third reactor Ci = 0.452 mg/L, m = 0.25 kg, V = 5 m3 Ce = 0.09 mg/L (This is the effluent concentration for the process).

2 (35 pts). An isotherm study was conducted for the adsorption of an azo-dye on activated carbon. Varying amounts of carbon were added to 500 ml of a 95 mg/L azo-dye solution. The results of this isotherm study are given in the following table. Carbon added Final soluble TCE (g) conc. (mg/L) 0.593 6 0.382 11 0.296 18 0.220 31 0.147 45 0.030 84 a) Determine a Langmuir and a Freundlich isotherm equation for the data. For the Langmuir isotherm determine determine K and Q0a using plots of 1/qe vs. 1/Ce and Ce/qe vs C. Plot the resulting isotherms on a single graph for comparison. b) Based on the R2 for each regression determine which regression gives the best data fit. Solution: For each bottle:

qe =

(C0 − C) ⋅ V m

V = 500 ml C0 = 95 mg/L m = mass of carbon in each bottle. qe (gm/gm) 0.075 0.110 0.130 0.145 0.170 0.183

C (mg/L) 6 11 18 31 45 84

For a Langmuir isotherm regression of 1/qe versus 1/C: Using gm/L as units for C: This regression gives slope equal to 0.0498 L/gm and intercept = 4.905 (gm/gm). R2= 0.993

Q0a =

1 1 gm = = 0.204 intercept 4.905 gm

K=

1 1 = = 98.5 L / gm 0 ⋅ slope Q a 0.0498⋅ 0.204

If C has units of mg/L then: This regression gives slope equal to 49.765 L/mg and intercept = 4.905 (mg/mg). R2= 0.993

Qa = 0

1 1 mg 1 1 L K= = = 0.204 = = 0.0986 0 intercept 4.905 mg slope ⋅ Q a 49.765⋅ 0.204 mg

For a Langmuir isotherm regression of C/qe versus C: Using gm/L as units for C: This regression gives: slope = 4.852 (gm/gm), intercept = 0.0514 gm/L. R2 = 0.9982

Q0 = a

1 1 = = 0.206 slope 4.852

K=

1 = intercept ⋅ Q0a

1 L = 94.39 gm gm ⋅ 0.206 0.0514 L

Using mg/L as units for C: This regression gives: slope = 4.852 (mg/mg), intercept = 52.36 mg/L. R2 = 0.9982

Q0 = a

1 1 = = 0.206 slope 4.852

K=

1 = intercept ⋅ Q0a

1 L = 0.0927 mg mg ⋅ 0.206 52.36 L

For a Freundlich isotherm regression of log q versus log C: For C in units of mg/L: Get intercept = -1.33 Slope = 0.329

K F = 10

int ercept

n=

R2 = 0.938

= 0.0466

1 = 3.041 slope

1 = 0.329 n

For C in units of gm/L Get intercept = -0.345 Slope = 0.329

R2 = 0.938

K F = 10int ercept = 0.451 n=

1 = 3.041 slope

1 = 0.329 n

Plot of isotherms generated from regression analysis:

0.25

qe (g/g)

0.2 0.15

1/q vs 1/C C/q vs C

0.1

Freundlich data

0.05 0 0

0.02

0.04

0.06

0.08

0.1

Ce (g/L)

b) Either Langmuir regression gives good R2

3. (35 pts). For the GAC pilot plant data presented below, determine the specific throughput for DCE for two beds in series with each bed having an EBCT of 5.0 min. The flow rate is 232 L/d, the adsorber density is 0.457 g/cm3, the average DCE influent concentration is 80 µg/L, and the DCE treatment objective is 5 µg/L. The concentration–history profiles for DCE as a function of elapsed time in days are given below. Column data: EBCT, min 5.0

M, g 395.5

10.0

791.1

t, d 27.3 75

Q · t, L 6,334

L/g 16.0

17,400

22.0

Solution Calculate the specific throughput. The contribution from both the first and second columns is given by: Specific throughput =

Qtc = M1

q1 C0 

+

contribution from first column

Q ∫ t0o C 2 dt M1   contribution from second column

a. Calculate the contribution from the amount of adsorbate in the effluent from the second column using the expression shown in Eq. 15-48 for the second column. For two beds in series with 5.0 min EBCT, the treatment objective from the second column is exceeded after 75 days and the average influent concentration is 80 µg/L. The contribution from the amount of adsorbate in the effluent from the pilot study in cycle is given by the expressions: Q M 1 C inf 1 EBCT1 ρ F Cinf

to

∫0

C2 dt =

to

Q V F ρF C inf

∫0





1

to

C2 dt = L

1 EBCT1 ρ F C inf

 L

to

∫0

C2 dt

1

∫0 C2 dt ≈  5.0 min   457 g   80 µ g  2 (75 − 50) d × (1440 min/d)(5 µg/L) = 0.49 L/g

b. Calculate the contribution to the specific throughput due to adsorption onto the first column. The loading on the first column, when the effluent is 5 µg/L from the second column, is computed using Eq. 15-47 as follows: q1 = to

∫0

(

Q C inf t o − ∫0t o C1 dt M1

)=

1 EBCT1 ρ F

 C t − t o C dt   inf o ∫0 1   

C1 dt ≈ 12 (75 − 20) d × (48 µ g/L)× (1440 min/d)= 1.90× 106 µ g⋅ min/L

Cinf t o = (80 µ g/L) (75 d) (1440 min/d)= 8.64× 106 µ g⋅ min/L q1 =

[(8.64 × 106 − 1.90 × 106 ) µ g ⋅ min/L] = 2950 µg/g [5.0 min× (457g/L)]

The contribution to the specific throughput due to adsorption onto the first column is given by the following expression, which is the expression shown in Eq. 15-48 for the first column: q1 (2950 µg/g) L = = 37 C0 g (80 µ g/L)

2. Calculate the specific throughput using Eq. 15-119 and the values computed in step 1: t

Specific throughput =

Qtc q1 Q ∫ 0o C 2 dt = + = 37 + 0.49 = 37.5 L/g M1 C0 M1...