CEE 370 |
|
Name _________________________________
December 21, 2006
Closed book, 3 sheets of notes allowed.
Show all work. Be neat, and box-in your answer.
If you cannot finish a problem (e.g., due to lack of time, or a missing formula), explain on your exam paper how you would have completed it.
total is 100%
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C6H12O6 + O2 ® CO2 + H2O
The carbon in glucose is being
oxidized to carbon in carbon dioxide.
The oxidation state goes from (0) to (+IV).
The oxygen in O2 is being reduced to
the oxygen in water. The oxidation state
goes from (0) to (-II).
The balanced stoichiometric equation
is:
The ThOD is then:
And:
First calculate the
stoichiometry: for every one mole of CN-
converted to bicarbonate and nitrogen gas, you need 2.5 moles of sodium
hypochlorite (1 to convert to CNCl and 3/2 to convert CNO- to N2
and HCO3-). This
means:
Again you need to calculate the
stoichiometry: for every one mole of CN-
converted to bicarbonate and nitrogen gas, you need exactly zero moles
of sodium hypochlorite. This is because
the first reaction produces two hydroxides and the second reaction consumes
them. The third reaction forms 3 H+
but also 3
Zero
n
Baghouse Filters
n
Ray: Figure 13.8,
Masters: Fig. 7.33, M&D: Figure 11-28
n
heat-resistant porous
fabric
n
cleaned by vibration
n
Electrostatic
Precipitators (ESP)
n
Masters: Fig. 7.32,
M&D: Fig 11-30 &31
n
use high voltage
electric field; particles are ionized (by negative plate) and collected on the
positive plate
n
Cyclones
By washing with water containing a
base such as “caustic” or NaOH
Vacuum Extraction
Bioventing
n
Ignitability
n
Corrosivity
n
pH 2 and below; or pH
12.5 and above
n
corrodes steel 0.25 in/y
at 55 C
n
Reactivity
n
reacts violently with or
without water
n
generates toxic gases
n
Toxicity
n
Toxicity Characteristic
Leaching Potential (TCLP) and toxicity characteristics
Additional assumptions: BOD deoxygenation rate: k1(or kd) = 0.23 d-1
reaeration rate constant (k2 or kr) of 0.82 d-1
D.O saturation concentration = 9.5 mg/L
River flow velocity = 0.5 ft/s
Perform a mass balance at the
outfall
Qupstream = 12-0.5*(3.8)
= 10.1 cfs
BODupstream = 2.5 mg/L
DOupstream = 8.5 mg/L
QWW = 3.8 cfs
BODww = 35 mg/L
DOww = 5.5 mg/L
And immediately after the point of
mixing:
Qoutfall = 10.1 cfs + 3.8
cfs = 13.9 cfs
BODoutfall = ((2.5 mg/L *
10.1 cfs) + (35 mg/L * 3.8 cfs))/13.9 cfs = 11.38 mg/L
DOoutfall = ((8.5 mg/L *
10.1 cfs) + (5.5 mg/L * 3.8 cfs))/13.9 cfs = 7.68 mg/L
Doutfall = 9.5 mg/L –
7.68 mg/L = 1.82 mg/L
And now:
The minimum DO is slightly below the
7.5 mg/L standard, so the state will have to further restrict BOD discharges.
Oxidation half reaction:
Reduction half reaction:
Overall Reaction:
First the oxygen requirement:
And now the hydroxide sludge:
Initial case:
High-strength case:
Solve the two equations for the two unknowns. Rearrange the first equation:
And substitute in to the 2nd equation
Now take the above equation and solve for umax:
Thus the % removal is 100-18 = 82%
Thus the % removal is 100-37 = 63%
Some physical constants of Water:
Temp., oC |
Density, kg/m3 |
Viscosity, N-s/m2 |
Kinematic Viscosity, m2/s |
0 |
999.8 |
1.781x10-3 |
1.785x10-6 |
5 |
1000.0 |
1.518 x10-3 |
1.519x10-6 |
10 |
999.7 |
1.307 x10-3 |
1.306 x10-6 |
15 |
999.1 |
1.139 x10-3 |
1.139 x10-6 |
20 |
998.2 |
1.002 x10-3 |
1.003 x10-6 |
25 |
997.0 |
0.890 x10-3 |
0.893 x10-6 |
30 |
995.7 |
0.798 x10-3 |
0.800 x10-6 |
35 |
994.0 |
0.725 x10-3 |
0.729 x10-6 |
40 |
992.2 |
0.653 x10-3 |
0.658 x10-6 |
Selected Chemical Constants
Element |
Symbol |
Atomic # |
Atomic Wt. |
|
Electronegativity |
|
Aluminum |
Al |
13 |
26.98 |
3 |
1.47 |
|
Boron |
B |
5 |
10.81 |
3 |
2.01 |
|
Calcium |
Ca |
20 |
40.08 |
2 |
1.04 |
|
Carbon |
C |
6 |
12.01 |
2,4 |
2.50 |
|
Cerium |
Ce |
58 |
140.12 |
3,4 |
1.06 |
|
Chlorine |
Cl |
17 |
35.453 |
1 |
|
|
Holmiuum |
Ho |
67 |
164.93 |
3 |
1.10 |
|
Hydrogen |
H |
1 |
1.01 |
1 |
2.20 |
|
Iron |
Fe |
26 |
55.85 |
2,3 |
|
|
Magnesium |
Mg |
12 |
24.31 |
2 |
1.23 |
|
Manganese |
Mn |
25 |
54.94 |
2,3,4,6,7 |
1.60 |
|
Nitrogen |
N |
7 |
14.0067 |
3 |
|
|
Osmium |
Os |
76 |
190.2 |
2,3,4,8 |
1.52 |
|
Oxygen |
O |
8 |
16.00 |
2 |
3.50 |
|
Potassium |
K |
19 |
39.10 |
1 |
0.91 |
|
Sodium |
Na |
11 |
22.99 |
1 |
1.01 |
|
Sulfur |
S |
16 |
32.06 |
2,4,6 |
2.44 |
|
Useful conversion factors
1 ft = 0.305 m