CEE 370

Name      _________________________________

# Final Exam

December 21, 2006

Closed book, 3 sheets of notes allowed.

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.

#### ·        one from the last four questions #8 - #11 (15%)

total is 100%

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# Answer both of the first two questions #1 & #2

## 1.                  Stoichiometry (20 points)

### Aerobic heterotrophic bacteria can facilitate the following unbalanced biochemical transformation of glucose:

C6H12O6 + O2 ®  CO2 + H2O

### A.Which element (and in which chemical species) is being oxidized in this reaction and what is its initial and final oxidation state?

The carbon in glucose is being oxidized to carbon in carbon dioxide.  The oxidation state goes from (0) to (+IV).

### B.Which element (and in which chemical species) is being reduced in this reaction and what is its initial and final oxidation state?

The oxygen in O2 is being reduced to the oxygen in water.  The oxidation state goes from (0) to (-II).

### D.Determine the Theoretical Oxygen Demand (ThOD) of a 25 mg/L solution of glucose.

The balanced stoichiometric equation is:

The ThOD is then:

# Answer one of the next three questions #3 - #5

## 4.      Hazardous Waste (15 points)

And:

### B.How many moles of sodium hydroxide (NaOH) or hydrochloric acid (HCl) would also be needed to neutralize the reaction mixture if you started with 1 mole of cyanide and the requisite amount of sodium hypochlorite?

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 OH-, again a wash.  This means the answer is:

Zero

## 5.      Air Pollution & Hazardous Waste, short answer (15 points)

### A.Describe at least two ways that particulate air contaminants can be removed from gaseous emissions.Be as specific as possible.

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

### B.Describe how acidic gases can be removed from incinerator waste gases.Include a diagram with your description.

By washing with water containing a base such as “caustic” or NaOH

### C.Describe two in-situ cleanup methods for site remediation of soil contaminated with hazardous wastes.

Vacuum Extraction

Bioventing

### D.Describe the characteristics of a hazardous waste, and how a waste might be defined as "hazardous" under the law.

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

# Answer questions #6 and #7

## 6.      Water Quality Modeling (30 points)

### The Intercontinental Paper Co. is discharging its wastewater directly into the Mill River.The discharge flow is 3.8 ft3/s (cfs) the discharge D.O. is 5.5 mg/L and the discharge ultimate BOD (BODult) is 35 mg/L.They obtain half of this water from an intake 0.5 miles upstream of the wastewater outfall, and half from groundwater via a nearby well field.On average, the Mill River water upstream of the IPC outfall has an ultimate BOD (BODult) of 2.5 mg/L and a D.O. of 8.5 mg/L.If the Mill River has a flow of 12 cfs upstream of the IPC intake, and if the state permits a minimum DO of 7.5 mg/L in the Mill River, will the state have to further restrict the BOD in IPC's wastewater (i.e., is the stream out of compliance)?Calculate the minimum stream D.O. expected as a result of the IPC discharge.

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.

# Answer one of the following four questions #8 - #11

## 8.      Chemical Oxidation (15 points)

### A. Write the two half cell reactions and balance each of them.

Oxidation half reaction:

Reduction half reaction:

### B. Write the overall balanced reaction

Overall Reaction:

### C. How much oxygen (in mg/L) is required to oxidize 1.5 mg/L of ferrous iron and how much ferric hydroxide precipitate is formed?

First the oxygen requirement:

And now the hydroxide sludge:

## 10. Biological Growth Kinetics (15 points)

### You have been studying the performance of a CMFR biological reactor.Your wastewater supports a specific bacterial growth rate of 0.25 d-1.Under these conditions, the effluent substrate (waste) concentration is 0.4 mg/L.When you increase your wastewater strength such that the final substrate concentration leaving the CMFR is doubled to 0.8 mg/L, you observe a specific growth rate of 0.35 d-1.From these data estimate the two Monod coefficients, Ks and mmax.

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:

## 11. Reactor Kinetics (15 points)

### a.      What is the % removal of MCAS if the reactor is operated as a plug flow reaction (PRF)?

Thus the % removal is 100-18 = 82%

### b.      What is the % removal of MCAS if the reactor is operated as a completely mixed flow reactor (CMFR)?

Thus the % removal is 100-37 = 63%

## Appendix

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. Valence 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