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Gel 4335 K. Bower
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Gel 4335: Environmental Geology
Problem Solving Manual
Worked Problem Examples
K. Bower, Eastern Illinois University
January 11, 2010
Units – SI (International System of Units) and USCS (U.S. Customary system) 10 -12^ pico p 10 -6^ micro 10 -3^ milli m 10 -2^ centi c 103 kilo k 106 mega M
Significant Digits and Rounding Numbers
To round numbers: If the digits to the right of the digit you are rounding are greater than five, round up. If the digits to the right of the digit you are rounding are less than five, round down. If the digits to the right of the digit you are rounding are exactly equal to five, round up or down so that the digit you are rounding to is an even number. (reference: R.C. Brinker, Elementary Surveying, International Textbook Company, 1969)
Examples: Round 234.5075 to the thousandths place 234. Round 0.0467 to the hundredths place 0.
To determine the number of significant digits: Define the following terms: leading zero – zero to left of first non-zero digit in a number trailing zero - zero to right of last non-zero digit in a number trapped zero – zero between two significant digits (zero or non-zero) Rules of significant digits:
- all non-zero digits are significant digits
- all trailing zeros to right of decimal point are significant digits
- all trapped zeros are significant digits
- all other zeros are not significant digits
Examples: 34608 – 5 significant digits 2110 - 3 significant digits 000456 - 3 significant digits 0.000456 - 3 significant digits 0.456000 - 6 significant digits 4500.000 - 7 significant digits
Hints for problem solving
- Write down on a single page every equation that is likely to be helpful for this problem set.
- Write down on same page every constant parameter that is likely to be helpful for this problem set. Include conversion factors.
- Read each problem several times until you understand what is being given and what is being asked. Sketch a figure if possible. Write down what you know and identify equations that will probably be useful. a. Some people like to work backward; what do you need to determine the answer and how do you find it.
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b. Some people like to work forward; if you know this, what more can you find out and does that get you closer to the final answer. c. Do not work backward from the final answer. In real life the final answer is unknown. Use the final answer only to check your work when you are finished with the problem.
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b. What is the molecular weight of methane gas?
12.01 g + 4 * 1.01 g = 16.0 g
c. What is the concentration of the gas in mg/m^3?
Assume 14 L of methane in 10^6 L total volume
n = PV/RT = 1 atm * 14 L / (0.082056 L atm/K moles * 298.15 K) = 0.572 moles * 16.0 g/mole = 9.16 g
9.16 g/ 10^6 L * 10^3 mg/g * 10^3 L/m^3 = 9.16 mg/ m^3
- A proposed air quality standard for ozone (O 3 ) is 0.08 ppmv. At the elevation of Denver, the pressure is about 0.82 atm. Express the ozone standard in g /m^3 at that pressure and at a temperature of 15oC.
0.08 ppmv * [(1L/10^6 L)/1ppmv] = 8 x 10-8^ Lozone / 1 L (^) total volume
n = PV/RT = 0.82 atm * 8 x 10-8^ Lozone / (0.082056 L atm/K mole *288.15K) = 2.77 x 10-9^ moles
C = (2.77 x 10-9^ moles ozone/ 1 L total volume) * (10^3 L/ 1 m^3 ) * (48 gozone/ 1 moleozone) * (10^6 g / 1 g)
= 1.33 x 10^2 g /m^3
- The federal Air Quality Standard for carbon monoxide is 9.0 ppmv. Express this as a percent by volume as well as in mg/m^3 at 1 atm and 25 C. In a million volume of air there are 9.0 volumes of CO. Therefore 9.0 m^3 = .0009 % by volume 1 x 10^6 m^3 The atomic weights of C and O are 12 and 16 respectively. The molecular weight of CO is then 28 g. This means that there is 28 g of CO in 1 mole of CO. There are 9.0 ppmv – this means than there are 9 m^3 of CO in 10^6 m^3 of air. (Note: 1 m^3 = 10^3 liters and thus we have 9 *10^3 liters of CO alone). From the ideal gas law, the number of moles is:
n=PV/RT = 1atm * 9.0 x 10^3 L/ (.082056 L. atm. K -1. mol -1^ * (25+273.15) K) = 367.87 moles
Thus there are 367.87 moles in 10^6 m^3 of air. To convert to mg/ m^3
367.87 moles /10^6 m^3 *28 g/mole * 10^3 mg/g = 10.3 mg/m^3
- A wastewater treatment plant receives 10 MGD (million gallons per day) of flow. This wastewater has a solids concentration of 192 mg/L. How many kilograms of solids enter the plant every day? (from Vesilind and Morgan, 2004)
192 mg/L *10 x 10^6 gal/day * (1 kg/ 10^6 mg) * (3.785 L/ 1 gal) = 7.27x10^3 kg/ day; 7.27 x 10^3 kg enter plant daily
- Two stream, each with their own chloride concentration, combine. What is the concentration of chlorides out of a stream? Assume steady state, conservative conditions.
Input rate = output rate
C 1 *Q 1 + C 2 *Q 2 = C 3 * Q 3
Solving for Cm
C 3 = C 1 *Q 1 + C 2 *Q 2 = C 1 *Q 1 + C 2 *Q 2
Q 3 Q 1 + Q 2
C 3 = (20.0 mg/L15.0 m^3 /s + 50.0 mg/L5.0 m^3 /s)/ 20.0 m^3 /s = 27.50 mg/ L
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- A stream carries water at a rate of 3000 L/min into a pond. A second stream adds an additional flow of 1500 L/min. What is the flow out of the pond?
- Stream A and Stream B meet to form the Matt River. Stream A has a flow of 3000 L/sec with a salt load of 150 mg/L. Stream B has a flow of 1200 L/s and carries a salt load of 175 mg/L.
a. What is the water flow in the Matt River? 3000 L/s + 1200 L/s = 4200 L/s
b. What is the salt concentration of the Matt River? Assume steady state, conservative conditions
3000 L/s * 150 mg/L + 1200 L/s * 175 mg/L
= (3000 L/s + 1200 L/s) * CO
CO = 6.60 x 10^5 mg/s / 4200 L/s = 1.57 x 10^2 mg/L
- Influent pollutant is added to clean stream. How much clean water must be added to outlet water to bring stream water up to EPA standard of 150 ppm? Assume conservative, no accumulation.
Divide into two problems:
1 st^ part - Find C 3 and Q 3
Input Rate = Output Rate
C 1 Q 1 + C 2 Q 2 = C 3 Q 3 --------21.0 m^3 /s0 + 4.0 m^3 /s1500 ppm = (21.0+4.0) m^3 /s*C 3
C 3 = 240 ppm
2 nd^ part
C 3 Q 3 +C 4 Q 4 = C 5 Q 5 ------------------ 240 ppm25.0 m^3 /s + 0Q 4 = 150ppm*(25.0 +Q 4 ) m^3 /s
Q 4 = 15 m^3 /s
- A river with 400 ppm of salts (a conservative substance) and an upstream flow of 25.0 m^3 /s receives an agricultural discharge of 5.0 m^3 /s carrying 2000 mg/L of salts. The salts quickly mix in the river. A municipality just downstream withdraws water and mixes it with enough pure water from another source to deliver water having no more than 500 ppm salts to its customers. What should be the ratio of pure water to river water, F = Q 5 /Q 4?
1500 L/min
3000 L/min (^) Q 3
Q 1 + Q 2 = Q 3
3000 L/min + 1500 L/min = 4500 L/min
QA = 3000 L/sec CA = 150 mg/L
QB = 1200 L/sec CB = 175 mg/L
Matt River, Qo CO
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Input rate = output rate + decay
Break problem into two parts
1 st^ part: assume no accumulation Q 1 C 1 = Q 2 C 2 + KCV 1 25.0 m^3 /sec15.0 mg/L= 25.0 m^3 /sec C 2 +0.15/day (1day/24hrs)(1hr/3600 sec) C 2 1x10^6 m^3 C 2 = 14.025 mg/L 2 nd^ part: 14.025 mg/L25.0 m^3 /sec =25.0m^3 /secC 3 +0.15/day (1day/24hrs)(1hr/3600 sec)* C 3 2x10^6 m^3 C 3 = 12.315 mg/L Convert this to ppm 12.315 mg/L * (1L/10^3 g)(1g/10^3 mg)* 1ppm/(1g/10^6 g)= 12.315 ppm
- A polluted stream has decay and a sewage outfall. Assume the pollution is completely mixed in the lake AND assume no evaporation or other water losses or gains. Find the steady-state concentration. What is the OUTLET concentration?
Input rate = output rate + KCV
Q 1 *C 1 + Q 2 *C 2 = Q 3 *C 3 +KCV
Q 1 *C 1 + Q 2 C 2 = (Q 1 +Q 2 )C 3 +KC 3 V
(5.0 m^3 /s10.0 mg/L +0.5 m^3 /s100.0 mg/L) * 10^3 L/m^3 =
(5.0m^3 /s + 0.5 m^3 /s) C10^3 L/m^3 + 0.20/day C*10.0 x10^6 m^3 10^3 L/m^3 /(24 hr/day3600s/hr)
C 3 = 3.50 mg/L
- Suppose the condition of the lake (in the above problem) is deemed unacceptable and it is decided to suddenly divert the sewage outfall completely around the lake, eliminating it as a source of pollution. The incoming stream still has Q 1 = 5. m^3 /s and C 1 =10.0 mg/L. With the sewage outfall removed, the outgoing flow Q 3 = 5.0 m^3 /s. Assuming complete mix conditions, find the concentration of pollution in the lake one week after the diversion and find the new final steady- state concentration
Need two equations: C = Q 1 C 1 / (Q 3 + KV) C 3 (t) = C + (C 0 - C ) e – (K+ Q3/V)*t
C = Q 1 C 1 / (Q 3 + KV) = 5.0 m^3 /s 10.0 mg/L/(5.0 m^3 /s + 0.20/day10 x 10^6 m3) = 1.8 mg/L C 3 (t) = C + (C 0 - C ) e – (K+ Q3/V)t C 3 (7 days)= 1.8 mg/L + (3.5-1.8) e – (0.2/day+5.0 m3/sec /1 0 x 106 m3)7days = 2.1 mg/L recall that C 0 = 3.5 mg.L
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- Find Q 3 and C 3 0.2 days after the smaller inflow (Q 2 = 25 m^3 /s) suddenly stops flowing. Assume a non-conservative pollutant with a decay rate of K=0.3/day
Initial conditions 200 m^3 /sec * 12 mg/ m^3 + 25 m^3 /sec * 350 mg/ m^3 = 225 m^3 /sec * C 3 + 0.3/day * 8 x 10^5 m^3 * C 3 * (1 day/24 hr * 1 hr/ 3600sec) Co = C 3 = 4.90 x 10^1 mg/ m^3
Transient conditions C = Q 1 C 1 /(Q3,∞ +KV) = (200 m^3 /sec * 12 mg/m^3 ) / [200 m^3 /sec + 0.3/day * 8 x 10^5 m^3 *(1 day/24 hr * 1 hr/ 3600 sec)] = 11.84 mg/ m^3 C(t) = C + (Co-C ) e – (K + Q3 /V) t C(0.2 days) = 11.84 mg/m^3 + (48.95 mg/m^3 – 11.84 mg/m^3 ) e – (0.3/day * 1 day/24 hr *1 hr/3600 sec + 200 m3/sec / 8 x 105 m3) * 0.2 day * 24 hr /1 d * 3600 sec/hr)
= 1.23 x 10^1 mg/ m^3
Q 1 = 200 m^3 /s C 1 = 12 mg/m^3
Q 2 = 25 m^3 /s C 2 = 350 mg/m^3
Q 3
V = 8 x 10 5 m^3 C 3
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pH = -log (3.5x10-6) = 5.
- The mineral fluorite is CaF 2. What is the equilibrium concentration of the ion, fluoride, when fluorite is dissolved in water? Assume there is an unlimited supply of the solid.
CaF 2 Ca2+^ + 2 F-^ Ksp = 3x10-
The equilibrium equation is: [Ca2+][ F-]^2 = 3x10- Now, no OH or H is involved in this reaction (though water is necessary) so we don’t use the equation for the dissolution of the water molecule. In the stochiometric equation, for every mole of calcium, there are two moles of fluoride. Let s = [Ca2+] then 2s = [ F-] Then s(2s)^2 = 3x10-11^ and s= 2x10-4^ mol/L [ F-] = 2s = 4x10-4^ mol/L [ F-] = 2s = 4x10-4^ mol/L * 19 g/mol1000 mg/g = 7.6 mg/L
- Carbonic acid, H 2 CO 3 , completely ionizes when dissolved in water. Calculate the pH of a solution containing 52 mg/L of carbonic acid.
H 2 CO 3 2H+^ + CO 3 -
52 mg / L * (1 g/ 10^3 mg) * (1 mole (^) H2CO3 / 62.0g) * (2 mole (^) H+ / 1 mole (^) H2CO3) = 1.677 x 10-3^ moles/ L of H+ pH = -log(1.677 x 10-3) = 2.
- The volume of oxygen in air is about 21% by volume. Find the equilibrium concentration of O 2 in water at 25 C and 1 atm. Recalculate it for Los Alamos with an elevation of 7500 ft above sea level.
Kh = .0012630 mol/L atm at 25 C Pg = fraction * pressure P = Po – 1.15x10-4^ * h The partial pressure of oxygen in air is Pg = fraction * pressure = 0.21 * 1 atm = 0.21 atm [gas] = KhPg [O 2 ] = .0012630 mol/L atm * 0.21 atm = 2.65x10-4^ mol/L Now at 7500 ft above sea level: P = Po – 1.15x10-4h P = 1 atm – 1.15x10-4(7500 ft * 0.3048 m/1 ft) = 0.73711 atm Pg = fraction * pressure = 0.21 * 0.73711 atm = 0.155 atm [O 2 ] = .0012630 mol/L atm * 0.155 atm = 1.96 x10-4^ mol/L
- Anaerobic digestion of an industrial waste, largely acetic acid, produces carbon dioxide and methane gas. Calculate the volume of CO 2 produced daily at 20 C if there is an average daily waste production of 500 kg of CH 3 COOH. [there is no oxygen gas or water involved in this reaction]
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Assume sea level so that P= 1 atm
CH 3 COOH CO 2 + CH 4 -balanced equation 500 kg * [1 mole (^) CH3COOH / (212.01 g +4 1.01 g +2*16.00 g)] * (10^3 g/ 1 kg) (1 mole (^) CH3COOH /1 mole (^) CO2) = 8.33 x 103 moles (^) CO V = nRT/P = 8.33 x 10^3 moles (^) CO2 * 0.082056 L atm/(K mole) *(20 +273.15)K / 1 atm = 2.00 x10^5 L
- Nitrogen in a wastewater treatment plant is in form of ammonia and ammonium. Find the fraction of nitrogen in form of ammonia (strippable) as function of pH at 25 C. The value of K is 1.82x10-5.
NH 3 + H 2 O NH 4 +^ + OH- The fraction of NH 3 is the fraction of the ammonia to the total nitrogen in solution or fraction is: f = [ NH 3 ]/ ([ NH 3 ]+ [NH 4 +]) = 1/(1+ [NH 4 +]/[ NH 3 ]) the equilibrium equation for reaction is: [NH 4 +][ OH-]/[ NH 3 ] = K (^) NH3 = 1.82x10- The equation for water is:
H+^ OH-^ = Kw = 1x10- Rearranging [NH 4 +]/[ NH 3 ]= K (^) NH3//[ OH-]= K (^) NH3 / (Kw / H+^ ) Putting this in the fraction for ammonia f = 1/(1+^ [NH 4 +]/[ NH 3 ]) = 1/(1+ K (^) NH3 / (Kw / H+^ )) =1/(1+ 1.82x10-5/ (1x10-14/10-pH)) = 1/(1 + 1.82x10(9-pH))
- Carbon dioxide dissolves in water to form carbonic acid which then disassociates into bicarbonate and hydrogen ions as follows:
CO 2 + H 2 O H+^ + HCO 3 -^ K = 4.47 x 10-
The amount of [HCO 3 - ] depends on the pH. Find the fraction of carbon ions that is bicarbonate as a function of pH. What would be the bicarbonate fraction for pH = 9.2? Fraction: f = [HCO 3 - ] / ( [CO 2 ] + [HCO 3 - ] ) =1 / ([CO 2 ]/ [HCO 3 - ] + 1)
from the equilibrium equation: [HCO 3 - ] * [H+] / [CO 2 ] = 4.47 x 10-
[HCO 3 - ]/ [CO 2 ] = 4.47 x 10-7^ / [H+] = 4.47 x 10-7/ 10-pH
[CO 2 ]/ [HCO 3 - ] = 10-pH^ / 4.47 x 10-7^ = 2.2371 x 106- pH
f = 1 / (2.24 x 106-pH^ + 1)
f(9.2) = 0.
- Calculate the equilibrium concentration of dissolved carbon dioxide in water at 2300 m elevation above sea level and at 20C. Carbon dioxide is 0.03% of the atmosphere
at 2300 m, Pg = 0 .0003 * (1 atm - 1.15 x 10-4^ * 2300m) = 0.00022065 atm
[CO 2 ] = 0.00022065atm * 0.039172 moles / L atm * (44 g/ 1 mole) = 3.80 x 10-4^ mg/L
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Chapter 4 - Risk Analysis
- There is some risk to eating peanut butter from the aflatoxins that cause cancer. The Food and Drug Administration restricts aflatoxin in peanut products to 20 ppb. At this level, eating 4 tablespoons a day is estimated to cause 0.8 cancer deaths /year per 100,000 people who eat peanut butter. Assuming there are 260,000,000 people in the U.S. who eat 4 tablespoons of peanut butter every day, how many people will die of cancer each year?
(0.8 deaths / 100,000 persons)*260,000,000 persons = 2080 deaths/year
What percentage of the yearly U.S. death rate from cancer would this be?
521,000 deaths /year from cancer in U.S. - increase = 2080 / 521,000 = 0.40%
- For every hour spent in a coal mine, a worker increases his mortality risk of dying by Black Lung Disease by 1 in 1,000,000. Suppose 600 coal miners work at the United Carbon Coal Mine for 40 hours each week, 50 weeks per year and 25 years. What percent of the coal miners will die from Black Lung Disease from working in the mine?
((1 / (10^6 * 1 hr)) * (40 hr/1 wk * 50 wk/1 yr * 25 yr) *100 = 5%
- Suppose a 70 kg person drinks 2 L of water every single day for 70 years with chloroform concentration of 0.10 mg/L (drinking water standard). Chloroform has potency factor = 6.1x10-3^ (mg/kg-day)-1. What is upper bound lifetime cancer risk for this person?
PF = 6.1 x10-3^ (mg/kg-day)- CDI (mg/kg-day) = average daily dose/body weight = 0.10 mg/L* 2 L/day / 70 kg = .00286 mg/kg-day Risklifetime = CDI * Potency factor = 0.00286(mg/kg-day) * 6.1x10-3^ (mg/kg-day)-1^ = 17.4x10-6^ - lifetime risk to 1 person
- Suppose drinking water contains 1.0 mg/L of toluene and 0.01 mg/L of tetrachloroethylene (C 2 Cl 4 ). An adult drinks this water for 10 years. What is the hazard index? Tetrachloroethylene is probably carcinogenic. What is the lifetime risk to one person? Is this water safe?
Toluene – assume not carcinogenic: ADD = C * dose/body mass = 1.0 mg/L * 2 L/d / 70 kg = 0.028571 mg/ kg-day Hazard Quotient = 0.028571 mg/ kg-day / 0.200 mg/kg-day = 0. Tetrachloroethylene: ADD = C * dose/body mass = 0.01 mg/L * 2 L/d / 70 kg = 2.857E-4 mg/ kg-day Hazard Quotient = 2.857E-4 mg/ kg-day / 0.01 mg/kg-day = 0. Hazard Index = 0.143 + 0.0286 = 0.172 <1 which suggest the water is not a toxin Carcinogen – Tetrachloroethylene Risk = 5.1 x 10-2^ kg-day/mg * 0.01mg/L * 2L/d * (350 days/yr/365days/yr) * (10yrs/70 yrs)/70 kg = 2.00E-6 twice what is considered safe
- Suppose Marion’s water supply has 52 ppb of carbon tetrachloride in it. Using the PCB oral potency factor (Table 4.9 of your text) and the EPA recommended exposure factors (Table 4.10).
a. What is the CDI over a lifetime? CDI = 52 x 10-3^ mg/L * 2 L/d * 350 d/yr * 30 yr / (70 kg * 365 d/yr * 70 yr) = 6.11 x 10-4^ mg/ kg-d
b. What is the individual lifetime cancer risk for an adult residential consumer? Risk = CDI * PF = 6.1057 x 10-4^ mg/ kg-d * 0.13 kg-d/mg = 7.94 x 10-
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c. If 20,000 people live in Marion, estimate the number of extra cancers per year caused by the carbon tetrachloride in the water supply. Extra cancers/yr = 20000 people * 7.94 x 10-5^ cancers/person-life / 70 yr/life = 0.0227 cancers/yr d. The average cancer death rate in the U.S. is 193 per 100,000 persons per year. How many cancer deaths would be expected in Marion? Do you think the additional cancers caused by the carbon tetrachloride in the drinking water would be detectable? Expected cancer rate = 193 deaths /yr * 20000 people/ 100,000 people = 38.6 deaths/yr
Adding 0.0227 cancers per year would be an increase of 0.0588% - not detectable
- Find the DWEL for drinking water with chloroform that would result in a 10-6^ risk. Chloroform has potency factor = 6.1x10-3^ (mg/kg-day) -1.
a 70 kg adult consumes 2 L/day of water 350 days /year for 30 years CDI = risk/ PF = 10-6^ / 6.1x10-3^ kg-day/mg = 1.64x10-4^ mg/kg-day CDI = Concentration * dose / weight Concentration = C = CDI / weight/dose = 1.64x10-4^ mg/kg-day * 70kg / [(2L/day) *(30 years/70 years) * (350 days/365 days)] = .0140 mg/L DWEL = 14.0 g/L would result in an upper bound risk of 10-
- Compute the drinking water equivalent level (DWEL) for methylene chloride based on a 10-6^ risk for lifetime consumption of 2 L of water per day for an adult individual.
10 -6^ = PF * CDI 10 -6^ = 7.5 x 10-3^ kg-d/mg * C * 2 L/d * (350 d / 365 d) * (30 yr / 70 yr) * 1/70 kg DWEL = C = 1.14 x 10-2^ mg/L
- The cancer risk caused by exposure to radiation is thought to be approximately 1 fatal cancer per 8000 person-rems of exposure (e.g. 1 cancer death if approximately 8000 people are exposed to 1 rem each). Living in a home with 1.5 pCi/L of radon is thought to cause cancer risk equivalent to that caused by about 400 mrem/yr of radiation.
a. Estimate the annual cancers in the U.S. caused by radon gas in homes. Assume to population of the U.S. is 300,000,000 people. For one person: risk = (1 cancer/ 8000 person – rems) * (1 rem/ 10^3 mrem)* 400 mrem/yr = 5 x10-5^ /yr For U.S. population # of cancers = 3 x 10^8 * 5 x10-5^ /yr = 1.50 x10^4 /yr
- An underground storage tank has been leaking for years, contaminating groundwater and causing contaminant concentration beneath site of 0.30 mg/L. The contamination is moving with velocity = 0.50 ft/day toward public drinking water well 1 mile away. The half-life of contaminant = 10 years. The PF for the contaminant is 0.02 kg-day/mg. What is the life time cancer risk for one person who drank the water for 10 years? BREAK INTO 3 PROBLEMS:
- VELOCITY - time of arrival
- CONCENTRATION - Estimate steady state pollutant concentration expected at water well. C(t) = C(0)*e-Kt
- RISK - Estimate lifetime cancer risk - assume 70 kg person drank 2 L/day for 350 days out of each year
- estimate steady state pollutant concentration at well time required to travel 1 mile is Time to well = 5280 ft/ (0.5ft/day) = 10,560 days
- reaction rate coefficient = K= 0.693/ = 0.693/(10 yr * 365 days/yr) = 1.9x10-4/day In 10,560 days, concentration is C(t) = C(0) * e-Kt^ = 0.30 mg / L * e -(1.9x10-4*10,560 d)^ =0.040 mg / L
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Chapter 5 - Surface Water
- A liter water sample collected for analysis accidentally has an insect weighing 0.1 g trapped in the bottle. The initial DO is 10 mg/L. Assume that 10% of the insect's fresh weight is biodegradable and has the formula CH 2 O. Assume microbes are present to decompose. If the laboratory analyzes the sample after decomposition, what DO will be measured?
CH 2 O + O 2 CO 2 + H 2 O
Molecular weight of CH 2 O = 12+2+16 = 30 g
CH 2 O in insect = 0.01 g/ (30g/mole) = 3.3 x 10-4^ moles
O 2 : 3.3 x 10-4^ moles (^) insect *(1mole (^) O2/ 1 mole (^) insect) * 32 g/ 1 mole (^) O2 = 10.67 mg/L
Therefore biodegradation will use up all the DO and there will be part of the insect undecomposed.
- 5 day BOD test: A 15.0-ml sample of sewage is mixed with water filling 300-ml bottle. The mixed sample has DOi = 8.0 mg/L. When measured at 5 days, DO 5 = 5.0 mg/L. What is BOD 5?
BOD 5 = (DOi – DOf) / P
= (8.0 mg/L – 5.0 mg/L) / (15 mL / 300 mL)
= 60 mg/L
- A standard five-day BOD test is run using a mix consisting of four parts distilled water and one part wastewater. The initial DO of the mix is 9.0 mg/L and the DO after five days is 1.0 mg/L. What is BOD 5?
BOD 5 = 9 mg/L - 1 mg/L = 40 mg/L 1/
- The following data have been obtained in a BOD test that is made to determine how well a wastewater treatment plant is operating. What percentage of the BOD is being removed by this treatment plant? If this is a secondary treatment plant that is supposed to remove 90 % of the BOD, is the plant operating properly?
DOi (mg/L) DOf (mg/L) Volume of wastewater (mL)
Volume of dilution water (mL)
Untreated sewage 7.0 1.5 4 296
Treated sewage 8.0 3.2 12 288
Untreated sewage: BOD 5 = ( 7 mg/L - 1.5 mg/L) / (4/300) = 412.5 mg/L
Treated sewage: BOD 5 = ( 8 mg/L – 3.2 mg/L) / (12/300) = 120.0 mg/L
Removing 412.5 – 120.0 = 292.5 mg/L or 71% - not operating properly
- The dilution factor, P, for unseeded waste and water = 0.030 and the DOi = 9.0 mg/L. At 5 days, DO 5 = 3.0 mg/L. The reaction constant is kd = 0.22/day What is BOD 5? What is the ultimate CBOD? What is the remaining oxygen demand after 5 days?
P = 0. DOi = 9.0 mg/L DO 5 = 3.0 mg/L kd = 0.22/day BOD 5 = (DOi – DOf) / P
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=(9.0 – 3.0)/ 0.030 = 200 mg/L BODt =Lo (1-e-kdt) ultimate CBOD = Lo = 200 mg/L / (1- e-0.22*5) = 300 mg/L Remaining oxygen demand at 5 days L 5 = 300 – 200 = 100 mg/L
- If the BOD 5 for a wastewater is 200 mg/L and the ultimate BOD is 300 mg/L, what is the reaction rate constant k?
BOD 5 = Lo (1-e-k*t)
200 mg/L = 300 mg/L * (1 – e – k*5d)
0.6667 = 1-e-k*5 d
0.33333 = e-k*5 d
ln (0.33333) = -k*5d
k= 0.220/d
- A ssume a sample of domestic wastewater with 30 mg/L of nitrogen as TKN. Assume very few new cells of bacteria are formed during nitrification of waste and NBOD can be found by stochiometric analysis. Find nitrogenous oxygen demand.
nitrogenous oxygen demand. 2NH 3 + 3O 2 2NO 2 -^ + 2H+^ + 2H 2 O 2NO 2 -^ + O 2 2NO 3 - Combine the two equations NH 3 + 2O 2 NO 3 -^ + H+^ + H 2 O m.w. 17g 2*32g -N = 14 g NBOD =[ 30 mg/L (^) N * 17gNH3/14g (^) N] * 1moleNH3/17gNH3 * 2mole (^) O/1moleNH3 * 32gO/1 mole (^) O = 137 mg/L of O 2
- A wastewater treatment plant discharges 1.1 m^3 /s of treated effluent with a BOD = 50.0 mg/L into stream. The stream has a Q = 8.70 m^3 /s and BOD = 6.0 mg/L. The deoxygenation constant, kd = 0.20/day. Estimate ultimate carbonaceous BOD of river just downstream from outfall. Assume CSTR.
Q 1 L 1 + Q 2 L 2 = (Q 1 + Q 2 )* Lo 8.7 * 6.0 + 1.1 * 50.0 = (1.1 + 8.7)*BOD Ultimate carbonaceous BOD = Lo = 10.9 mg/L velocity = 0.3 m/s - BOD 30,000 m downstream? t = d/v = 30000m / 0.3 m/s *1 hr/3600s * 1 day/24 hr = 1.16 days Lt = Lo e-kdt^ = 10.9 mg/L * e – (0.2/day * 1.16 day)^ = 8.7 mg/L
- A stream has a DO of 9 mg/L, an ultimate biological oxygen demand of 12 mg/L and an average flow of 0.2 m^3 /s. Industrial waste with 0.0 DO and an ultimate biological oxygen demand of 20,000 mg/L and a flow rate of 0.007 m^3 /s is discharged into the stream. What is the ultimate biological oxygen demand immediately below the point of discharge? What is the dissolved oxygen in the stream immediately below the point of discharge?
BOD (or Lo): 0.2 m^3 /s * 12 mg/L + 0.007 m^3 /s * 20,000 mg/L = 0.207 m^3 /s * BOD BOD = 688 mg/L DO: 0.2 m^3 /s * 9 mg/L + 0.007 m^3 /s * 0.0 mg/L = 0.207 m^3 /s * DO DO = 8.70 mg/L
K. Bower
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10 m^3 /s * 0.005 mg/L + 0.4 m^3 /s * C 2 = 10.4 m^3 /s * 0.020 mg/L + 1 x 10^8 m^2 * 15m/yr * 0.020 mg/L * (1 yr / 365 d) * (1 d / 24 hr) * (1 hr / 3600 s)
C 2 = 2.90 mg/L
K. Bower
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Chapter 5 - Groundwater
- Suppose there is a saturated sand aquifer. What is the volume of water contained in a volume of sand that is 3.0 m^3? What volume would be released from a volume of sand that is 3.0 m^3? The porosity is .34 and the effective porosity is .25. The volume of water is V= n * Vt = .34 * 3.0 m^3 = 1.02 m^3 The volume of water released is V = ne * Vt = .25 * 3.0 m^3 = .75 m^3
- A confined aquifer, 20.0 m thick has two wells spaced 500 m apart along the direction of flow. One well has a piezometric level at 50 m above a datum; the second well has a level of 48 m above datum. What is the flow rate per meter of width into the aquifer? The hydraulic conductivity is 50 m/day.
q = kA dh/dx = 50 m/day (20 m1 m ) * (50m-48m)/500m = 4 m^3 /day per width of aquifer
- Suppose the groundwater head changes by 32 m within a horizontal distance of 300 m. The hydraulic gradient is h / x = 32 m / 300m = 0.
The hydraulic conductivity in the aquifer is 75 m/day and the effective porosity is 0.20. What is the Darcy velocity of the groundwater?
v’ = k h / x = 75 m/day * 0.107 = 8.02 m/day
What is the average actual velocity of the groundwater?
v = 75 m/d x 0.107 = 40.1 m/day
If the cross-sectional area of the aquifer is 100,000 m^2 , what is the discharge of the aquifer?
q = vA = vA*n = 40.1 m/day * 100,000 m^2 * 0.20 = 802,000 m^3 /day
- If the limestone aquifer under the farmhouse is 200 m thick and has an effective porosity of 0.15, how much water (km^3 ) can be stored in an area of 0.1 km by 0.1 km of the aquifer?
Vrock = l x w x h = 200m x 100 m x 100 m = 2.00x 10^6 m^3
V (^) pores = Vwater = Vrock x n = 2.00x 10^6 m^3 x 0.15 = 3.00 x 10^5 m^3 = 0.000300 km^3
- The hydraulic gradient of an unconfined aquifer is 0.00295. The hydraulic conductivity of the aquifer is 150 m/day and the effective porosity is 0.32. a. What is the average velocity of the groundwater?
v = k h / x = 150 m/day * 0.00295 = 1.38 m/day n 0.
b. If the nitrogen travels with the same velocity as the groundwater, how many years will it take to travel 1 km?
