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Material Type: Assignment; Professor: Bower; Class: Environmental Geology; Subject: Geology; University: Eastern Illinois University; Term: Spring 2010;
Typology: Assignments
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Gel 4335 K. Bower
Page 1
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:
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
K. Bower
Page 2
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.
K. Bower
Page 4
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
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
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
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
Input rate = output rate
Solving for Cm
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
K. Bower
Page 5
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
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
1500 L/min
3000 L/min (^) 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
K. Bower
Page 7
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
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
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
K. Bower
Page 8
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
V = 8 x 10 5 m^3 C 3
K. Bower
Page 10
pH = -log (3.5x10-6) = 5.
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
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.
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
K. Bower
Page 11
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
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))
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.
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
K. Bower
Page 13
Chapter 4 - Risk Analysis
(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%
((1 / (10^6 * 1 hr)) * (40 hr/1 wk * 50 wk/1 yr * 25 yr) *100 = 5%
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
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
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-
K. Bower
Page 14
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
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-
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
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
K. Bower
Page 16
Chapter 5 - Surface Water
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.
BOD 5 = (DOi – DOf) / P
= (8.0 mg/L – 5.0 mg/L) / (15 mL / 300 mL)
= 60 mg/L
BOD 5 = 9 mg/L - 1 mg/L = 40 mg/L 1/
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
P = 0. DOi = 9.0 mg/L DO 5 = 3.0 mg/L kd = 0.22/day BOD 5 = (DOi – DOf) / P
K. Bower
Page 17
=(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
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
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
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
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
Page 19
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
Page 20
Chapter 5 - Groundwater
q = kA dh/dx = 50 m/day (20 m1 m ) * (50m-48m)/500m = 4 m^3 /day per width of aquifer
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
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
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?