CE2385 Final Exam Review – Full solutions

Chapter 1 Review
1. Basic Concepts
Filling a basin with volume V that is simultaneously filling at a rate Qin and draining at a rate Qout (could be a
bathtub filling and draining, a storm retention pond with rain and a stream that leaves, etc.).
a) How would you set up an equation to calculate when the
basin would be full (assume that the basin starts empty)?
𝑉𝑉
𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 =
𝑄𝑄𝑖𝑖𝑖𝑖 − 𝑄𝑄𝑜𝑜𝑜𝑜𝑜𝑜
b) How would you calculate the outflow volume for the time the
basin was filling?
𝑉𝑉𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑄𝑄𝑜𝑜𝑜𝑜𝑜𝑜 𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
2. Ideal Gas Conversions
The air quality standard of NO2 is 100 ppb (100x1000) ppm.
(a) Express as mg/m3 assuming ideal gas at standard temperature & pressure (1 atm and 25 oC.)
1000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑔𝑔
𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 𝑀𝑀𝑀𝑀 100 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
∙ 46
𝑚𝑚𝑚𝑚𝑚𝑚
= =
𝑚𝑚3 24.465 𝐿𝐿
24.465
𝑚𝑚𝑚𝑚𝑚𝑚
(b) Express as mg/m3 if given the P = 1.014 atm and T=20 oC
1000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑔𝑔
𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 𝑀𝑀𝑀𝑀 273.15𝐾𝐾 𝑃𝑃 (𝑎𝑎𝑎𝑎𝑎𝑎) 100 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
∙ 46
𝑚𝑚𝑚𝑚𝑚𝑚 273.15𝐾𝐾 1.014 𝑎𝑎𝑎𝑎𝑎𝑎
= ∙ ∙ = ∙ ∙
𝑚𝑚3 22.414 𝑇𝑇(𝐾𝐾) 1 𝑎𝑎𝑎𝑎𝑎𝑎 22.414 293 𝐾𝐾 1 𝑎𝑎𝑎𝑎𝑎𝑎
(c) Express as mg/m3 if given T=20 oC and elevation is 4000 ft.
𝑚𝑚
𝑃𝑃 = 𝑃𝑃0 − 1.15 × 10−4 𝐻𝐻 = 1 − (1.15 × 10−4 ) ∙ 4000 𝑓𝑓𝑓𝑓 ∙ = 0.88 𝑎𝑎𝑎𝑎𝑎𝑎
3.2808 𝑓𝑓𝑓𝑓
1000 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑔𝑔
𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 𝑀𝑀𝑀𝑀 273.15𝐾𝐾 𝑃𝑃 (𝑎𝑎𝑎𝑎𝑎𝑎) 100 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
∙ 46
𝑚𝑚𝑚𝑚𝑚𝑚 273.15𝐾𝐾 0.88 𝑎𝑎𝑎𝑎𝑎𝑎
= ∙ ∙ = ∙ ∙
𝑚𝑚3 22.414 𝑇𝑇(𝐾𝐾) 1 𝑎𝑎𝑎𝑎𝑎𝑎 22.414 293 𝐾𝐾 1 𝑎𝑎𝑎𝑎𝑎𝑎
(d) If given a concentration as mg/m3, can you convert to ppm?

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CE2385 Final Exam Review – Full solutions

3. Conservative Pollutant & Steady State

a) What is Q3?
𝑄𝑄3 = 𝑄𝑄1 + 𝑄𝑄2 = 15 𝑀𝑀𝑀𝑀𝑀𝑀
b) What is C3?
C1Q1 + C2Q2 = C3Q3; solve for C3
𝑄𝑄1 𝐶𝐶1 + 𝑄𝑄2 𝐶𝐶2 𝑚𝑚𝑚𝑚
𝐶𝐶3 = = 5.33
𝑄𝑄3 𝐿𝐿

c) How much of the contaminant in lb/day would be

measured downstream of the mixture?
𝑙𝑙𝑙𝑙
𝑀𝑀3 = 𝑄𝑄3 𝐶𝐶3 = 667.63 𝑑𝑑𝑑𝑑𝑑𝑑 (there are unit conversions needed to get this result)
4. More Complicated Steady State & with Conservative Pollutants:
Given a problem with a diagram like this: can you figure out Qs & Cs?

The key parts are to use all of your mass balance knowledge!

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CE2385 Final Exam Review – Full solutions
What is the flow in the river right before the farm stream mixes with it (Qbefore)?
Qbefore = Q1 – 70 acre-ft/day (of course some unit conversions are needed)
What is the concentration in the river right before it mixes with the farm stream (Qbefore)?
**Key thing: the problem would state that there are no new sources or sinks (places where the pollutant is removed)
between Q1 and this mixing point. So, C1 = Cbefore = 1 mg/L
What is Qs? Add the two mixing streams (Qfarm + Qbefore)
𝑄𝑄𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝐶𝐶𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 +𝑄𝑄𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝐶𝐶𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
What is Cs? Use the standard mass balance equation with two streams. 𝐶𝐶𝑠𝑠 =
𝑄𝑄3

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CE2385 Final Exam Review – Full solutions
5. Non-Conservative Pollutants & Steady State

Consider a 10x106 m3 lake fed by a polluted stream having a flow rate of 5.0 m3/s and pollutant concentration
equal to 10.0 mg/L. There is also a sewage outfall that discharges 0.5 m3/s of wastewater having a pollutant
concentration of 100 mg/L and a coefficient of 0.20/day. Assuming the pollutant is completely mixed in the
lake and assuming no evaporation or other water losses or gains, find the steady-state pollutant concentration in
the lake. Report in mg/L and round to two decimal places.

With perfect mixing & steady state, we know that Qs = Qin1 + Qin2
Mass Balance: 0 = Inputs - Output - Decay
Or: Inputs = Output + Decay
Cin1Qin1 + Cin2Qin2 = CsQs + kCV
10mg/L*5 m3/s + 100 mg/L*0.5m3/s = Cs*5.5 m3/s + kCV
Solve for Cs & make sure to keep track of units!
𝑄𝑄1 𝐶𝐶1 + 𝑄𝑄2 𝐶𝐶2 𝑚𝑚𝑚𝑚
𝐶𝐶3 = = 3.49
𝑄𝑄3 + 𝑘𝑘𝑘𝑘 𝐿𝐿
6. Step-Change Problems (non-steady state)
Working with the same lake as in problem 4, suppose the pollutant stream was sent to a treatment plant that
removes 95% of the contaminant and has 100% efficiency (all of the flow entering the treatment plant goes to the
lake). Calculate a) the new steady state concentration and b) the concentration of the lake after 10 days.
Solution:

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CE2385 Final Exam Review – Full solutions

Part a) You’ll need to know Qwtp and Cwtp – can you do it?
Qwtp = 0.5 m3/s, Cwtp = (1-0.95)*100 mg/L = 5 mg/L
𝑄𝑄𝐶𝐶𝑖𝑖 +𝑘𝑘𝑔𝑔 𝑉𝑉
𝐶𝐶∞ = Start with this equation & modify for what is given (two inputs)
𝑄𝑄+𝑘𝑘𝑑𝑑 𝑉𝑉

𝑄𝑄1 𝐶𝐶1 + 𝑄𝑄2 𝐶𝐶2 𝑄𝑄1 𝐶𝐶1 + 𝑄𝑄𝑤𝑤𝑤𝑤𝑤𝑤 𝐶𝐶𝑤𝑤𝑤𝑤𝑤𝑤

𝐶𝐶∞ = = = 1.83 𝑚𝑚𝑚𝑚/𝐿𝐿
𝑄𝑄3 + 𝑘𝑘𝑘𝑘 𝑄𝑄3 + 𝑘𝑘𝑘𝑘
Part b)
𝑄𝑄
𝐶𝐶(𝑡𝑡) = 𝐶𝐶∞ + (𝐶𝐶0 − 𝐶𝐶∞ )exp �− �𝑘𝑘𝑑𝑑 + � 𝑡𝑡 �
𝑉𝑉
 What is C0?

1.83𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚 0.2 5.5 𝑚𝑚3 86,400𝑠𝑠

𝐶𝐶(10 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑) = + �3.49 − 1.83 � ∙ 𝑒𝑒𝑒𝑒𝑒𝑒 ��− � �+ ∙ ∙ � 10𝑑𝑑𝑑𝑑𝑑𝑑�
𝐿𝐿 𝐿𝐿 𝐿𝐿 𝑑𝑑𝑑𝑑𝑑𝑑 𝑠𝑠 10 × 106 𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑
= 1.97 𝑚𝑚𝑚𝑚/𝐿𝐿

Chapter 9 Review
1. Energy and carbon savings from recycling
Problem:
Imagine there is a recycling program with the following information that recycles 76 tons of aluminum, 12,773 tons
of cardboard, 5821 tons of mixed plastics, and 3331 tons of office paper. Calculate a) the total amount recycled (in
tons), b) the total greenhouse gas savings (in MTCE) and c) the total energy savings (in MMBTU). Table 9.8 and 9.9
data are in the equation sheet. Round all answers to the nearest digit.
Solution:
I suggest that you make a table on your calculation sheet like this:
Amt Recycled MTCE/ton MMBTU/ton a) MTCE b) MMBTU
(ton)
Aluminum 76 3.71 206.95 281 15695
Cardboard 12773 0.96 15.65 12262 199894
Mixed Plastics 5821 0.42 52.94 2445 308150
Office Paper 3331 1.31 10.09 4363 33606
25000 19351 557345
a) 25000
b) 19351
c) 557345
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CE2385 Final Exam Review – Full solutions

2. Energy and CO2 emissions as vehicle equivalents

Find the carbon and energy savings associated with recycling 2 million tons of aluminum. Compare this with the
energy needs and carbon emissions from a vehicle. The vehicle information is: average mileage 25 mpg, driven
12,000 miles per year, burns 0.125 MMBTU per gallon, and emits 4.2 metric tons of CO2 each year. Calculate a) the
number of vehicle equivalents based on energy savings and b) the number of vehicle equivalents based on CO2
emissions. Enter answers as millions rounded the nearest digit.
Solution:
Savings from aluminum recycling:
Energy Savings = 2,000,000 ton * 206.95 MMBTU/ton = 413.9*10^6 MMBTU
CO2 Emissions Savings = 2,000,000 ton * 3.71 MTCE ton * (44/12) = 27.2*10^6 metric ton CO2
Vehicle energy & CO2 emissions:
Energy usage = 12,000 miles/25 mpg*0.125 MMBTU/gal = 60 MMBTU
CO2 emissions = 4.25 metric tons CO2
Vehicle equivalents:
Energy usage: vehicle equiv. = 413.9*10^6 MMBTU/60 MMBTU = 6,898,333
CO2 emissions: vehicle equiv. = 27.2*10^6 metric ton CO2/4.25 metric tons CO2 = 6,401,569
3. Landfill Calculations
Suppose a city of 50,000 people generates 40,000 tons of MSW per year. At current recovery and recycling rates,
22% of that is recovered or recycled and the rest goes to a landfill. Suppose also that the landfill density is
1,000 lb/yd3, cell depth is 10 ft, and 80% of the cell is MSW. Find a) Lift area required per year and b) If the current
landfill covers 50 acres, including 10 acres needed for access roads and auxiliary facilities, and two more lifts are
envisioned, how long would it take to complete this landfill?
Solution:
Amount of waste coming to landfill: (1-0.22)40,000 ton/yr = 31,200 ton/yr
Part a)
31200 𝑡𝑡𝑡𝑡𝑡𝑡 𝑦𝑦𝑦𝑦3 2000 𝑙𝑙𝑙𝑙
𝑉𝑉𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = ∙ 1000 𝑙𝑙𝑙𝑙 ∙ = 62,400 yd3
𝑦𝑦𝑦𝑦 𝑡𝑡𝑡𝑡𝑡𝑡
62400 𝑦𝑦𝑦𝑦3
𝑉𝑉𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = = 78,000 yd3 (or 2.11 x 106 ft3)
0.8
A = V/h = V/lift
78,000 𝑦𝑦𝑦𝑦3 3 𝑓𝑓𝑓𝑓 3
𝐴𝐴 = ∙ � 𝑦𝑦𝑦𝑦 � = 210,600 ft2/yr = 4.8 acre/year
10 𝑓𝑓𝑓𝑓

Part b)
Area available for landfill is 50 acre minus the 10 acre for roads & auxiliary needs (40 acre)
40 𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 𝑦𝑦𝑦𝑦
𝑇𝑇𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙
∙ 2 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 ∙ 4.8𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 16.5 years

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CE2385 Final Exam Review – Full solutions
Chapter 2 Problems
Problem similar to examples 2.4 & 2.5, Problem 2.10
Calculate the gross heat and net heat of combustion released when 5 kg of methane (CH4) is burned.

[NOTE: on the test, you would be given Table 2.1 on a tables sheet]]

Steps:

1. Write a balanced equation for combustion of CH4

2. Calculate ∆Hgross, then use 5 kg
3. Calculate ∆Hnet, then use 5 kg

Step 1. Balanced reaction will have CH4 and O2 reacting to produce CO2 and H2O. You can balance the equation any
way you’d like, but I’m going to use the method provided at the top of page 3 on the equation sheet.

CH4 + (1+4/4) O2  CO2 + (4/2) H2O

CH4 + 2 O2  CO2 + 2 H2O

Step 2. Remember that ∆Hgross implies that water will be in the liquid form. Look up the standard enthalpies in
Table 2.1.
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘
∆𝐻𝐻𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = �2 �−285.8 � + �−393.5 �� − ��−74.9 � + 2(0)� = −890.2
𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚
1000 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐻𝐻4 890.2 𝑘𝑘𝑘𝑘
5 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 ∙ ∙ ∙ = 278,188 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 16 𝑔𝑔 𝐶𝐶𝐻𝐻4 𝑚𝑚𝑚𝑚𝑚𝑚

Step 3. Remember that ∆Hnet implies that water will be in the gas form. Look up the standard enthalpies in Table
2.1.
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘
∆𝐻𝐻𝑛𝑛𝑛𝑛𝑛𝑛 = �2 �−241.8 � + �−393.5 �� − ��−74.9 � + 2(0)� = −802.2
𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚
1000 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐻𝐻4 802.2 𝑘𝑘𝑘𝑘
5 𝑘𝑘𝑘𝑘 𝐶𝐶𝐻𝐻4 ∙ ∙ ∙ = 250,688 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 16 𝑔𝑔 𝐶𝐶𝐻𝐻4 𝑚𝑚𝑚𝑚𝑚𝑚
Problem 2.11 (a)
For the following possible alternative automobile fuels, express their higher heating value (HHV) in BTU/gal:
0
Methanol (CH3OH), density 6.7 lb/gal, 𝐻𝐻298 = -238.6 kJ/mol
Steps:

1. Write a balanced equation for combustion of CH3OH

2. Calculate ∆Hgross
3. Convert to BTU/gal

Step 1. Balanced reaction will have CH3OH and O2 reacting to produce CO2 and H2O. You can balance the equation
any way you’d like, but I’m going to use the method provided at the top of page 3 on the equation sheet. You will
need to put CH3OH in the same form to be able to do this.

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CE2385 Final Exam Review – Full solutions

CH4O + (1+4/4-1/2) O2  CO2 + (4/2) H2O

CH4O + 1.5 O2  CO2 + 2 H2O

Step 2. Remember that HHV is ∆Hgross, which implies that water will be in the liquid form. Look up the standard
enthalpies in Table 2.1.
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘
∆𝐻𝐻𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 = �2 �−285.8 � + �−393.5 �� − ��−238.6 � + 2(0)� = −726.5
𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚
Step 3. Convert units.
𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐻𝐻3 𝑂𝑂𝑂𝑂 1000 𝑔𝑔 𝑙𝑙𝑙𝑙 𝐵𝐵𝐵𝐵𝐵𝐵
𝐻𝐻𝐻𝐻𝐻𝐻 = 726.5 ∙ ∙ ∙ 6.7 ∙ = 65,388 𝐵𝐵𝐵𝐵𝐵𝐵/𝑔𝑔𝑔𝑔𝑔𝑔
𝑚𝑚𝑚𝑚𝑚𝑚 32 𝑔𝑔 𝐶𝐶𝐻𝐻3 𝑂𝑂𝑂𝑂 2.205 𝑙𝑙𝑙𝑙 𝑔𝑔𝑔𝑔𝑔𝑔 1.055 𝑘𝑘𝑘𝑘
Problem similar to Example 2.10
Calculate the concentration of oxygen (in mg/L) in water at Ascarate Lake in El Paso at 20oC and using an elevation of
3,740 ft. (hint: oxygen accounts for about 21% of air by volume)

[NOTE: on the test, you would be give Table 2.4 on a tables sheet]

Steps:

1. Know this is a Henry’s Law problem. Calculate Pg.

2. Calculate [O2]

Step 1. [𝑂𝑂2 ] = 𝐾𝐾𝐻𝐻 𝑃𝑃𝑔𝑔

𝑃𝑃𝑔𝑔 ≡ 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶

𝑚𝑚
𝑃𝑃 = 1 𝑎𝑎𝑎𝑎𝑎𝑎 − 1.15 × 10−4 ∙ 3740𝑓𝑓𝑓𝑓 ∙ = 0.87 𝑎𝑎𝑎𝑎𝑎𝑎
3.2808 𝑓𝑓𝑓𝑓

21
𝐶𝐶 = = 0.21
100

𝑃𝑃𝑔𝑔 ≡ 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎 𝐶𝐶 = 0.21 ∙ 0.87 𝑎𝑎𝑎𝑎𝑎𝑎 = 0.18 atm

Step 2. [𝑂𝑂2 ] = 𝐾𝐾𝐻𝐻 𝑃𝑃𝑔𝑔

𝑚𝑚𝑚𝑚𝑚𝑚 32 𝑔𝑔 𝑂𝑂2 1000 𝑚𝑚𝑚𝑚
[𝑂𝑂2 ] = 0.0013840 ∙ 0.18 𝑎𝑎𝑎𝑎𝑎𝑎 ∙ ∙ = 8.09 𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑚𝑚𝑚𝑚𝑚𝑚 𝑂𝑂2 𝑔𝑔
Problem similar to 2.29
Scientists estimate the concentration of CO2 in the atmosphere in 2019 was about 409.8 ppm. If the accumulation of CO2
in the atmosphere in the atmosphere continues at the same rate, it will probably exceed 900 ppm the end of this
century. Calculate the pH of rainwater (neglecting the effect of any other gases) at 25 oC both now and by the end of this
century. Hint: Use this charge balance & simplify knowing that rain water is acidic: [H+] = [HCO3-] + 2[CO32-] + [OH-]

[NOTE: on the test, you would be give Table 2.4 on a tables sheet]

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CE2385 Final Exam Review – Full solutions
Steps:

1. Calculate [CO2] using Henry’s Law at 409.8 ppm and 900 ppm
2. Calculate [H+] using simplified charge balance & Kw
3. Calculate pH using pH = -log [H+]
𝑚𝑚𝑚𝑚𝑚𝑚
Step 1. [𝐶𝐶𝑂𝑂2 ] = 𝐾𝐾𝐻𝐻 𝑃𝑃𝑔𝑔 = 0.033363 ∙ 𝑃𝑃
𝐿𝐿∙𝑎𝑎𝑎𝑎𝑎𝑎 𝑔𝑔

At 409.8 ppm:
409.8
𝑃𝑃𝑔𝑔 = 𝑎𝑎𝑎𝑎𝑎𝑎
106
𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚
[𝐶𝐶𝑂𝑂2 ] = 𝐾𝐾𝐻𝐻 𝑃𝑃𝑔𝑔 = 0.033363 ∙ �409.8 × 10−6 � ∙ 1𝑎𝑎𝑎𝑎𝑎𝑎 = 1.37 × 10−5
𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿 𝐿𝐿
At 900 ppm:
900
𝑃𝑃𝑔𝑔 = 𝑎𝑎𝑎𝑎𝑎𝑎
106
𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚
[𝐶𝐶𝑂𝑂2 ] = 𝐾𝐾𝐻𝐻 𝑃𝑃𝑔𝑔 = 0.033363 ∙ �900 × 10−6 � ∙ 1𝑎𝑎𝑎𝑎𝑎𝑎 = 3.00 × 10−5
𝐿𝐿 ∙ 𝑎𝑎𝑎𝑎𝑎𝑎 𝐿𝐿 𝐿𝐿
Steps 2-3. Knowing that rain water is acidic & looking at pKa values, we can assume that the [CO32-] is negligible.

[H+] = [HCO3-] + [OH-]

10−14
Recall that: Kw = [H+] [OH-] = 10-14, or [OH-] = [H+]

Solve K1 equation for HCO3-:

𝐾𝐾1 [𝐶𝐶𝐶𝐶2 ] 10−6.35 [CO2]

[𝐻𝐻𝐻𝐻𝐻𝐻3 − ] = =
[𝐻𝐻 + ] [𝐻𝐻 + ]
10−6.35 [CO2] 10−14
[H+] = + [H+]
[H+]

10−6.35 [CO2]+10−14
= [H+]

[H+]2 = 10−6.35 [CO2] + 10−14

At 409.8 ppm:
𝑚𝑚𝑚𝑚𝑚𝑚
[H+]2 = 10−6.35 1.37 × 10−5 + 10−14
𝐿𝐿

[H+]=2.48 x 10-6 M, so pH = -log(2.48 x 10-6)= 5.61

At 900 ppm:
𝑚𝑚𝑚𝑚𝑚𝑚
[H+]2 = 10−6.35 3.00 × 10−5 𝐿𝐿
+ 10−14

[H+]=3.66 x 10-6 M, so pH = -log(3.66 x 10-6)= 5.44

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CE2385 Final Exam Review – Full solutions
Variations on this type of a problem could be to give you a ratio of two of the components. If you were given a
ratio of CO32-/H2CO3 (which is the same as CO32-/CO2), how would you approach this? Remember, square
brackets imply molar concentrations (mol/L).
[H+][HCO3-] [H+][HCO3-]
Look at K1: 𝐾𝐾1 = = [H
[CO2(aq)] 2 CO3(aq)]

You could rearrange for the HCO3 as a function of K1.

[CO2(aq)]𝐾𝐾1
[HCO3-] =
[H+]
And insert it into K2.

[H+][CO32-] [H+][CO32-][H+]
𝐾𝐾2 = =
[HCO3-] [CO2(aq)]𝐾𝐾1
Do you see the ratio? Everything at this point would be known, assuming you remember how to find [H+] from the
pH, CO2 from the Henry’s Law equation, and K1 and K2 using the pK relationship (K = 10-pKa).
On another type of problem, if you’re given a ratio of NH3/NH4, how might you approach a problem? In this case,
you’d do a similar approach, but use Ka for ammonia & calculate NH3 using Henry’s Law.
The key parts to remember on these types of problems is to see whether you can figure out how to use Ka equations
to get an equation where you’ll have the ratio. And recall the pH relationships for hydrogen & hydroxide.
Problem with ammonia stripping
A wastewater treatment plant uses sodium hydroxide to raise the pH to 9.5 and the atmospheric concentration of
ammonia is 5x10-10 atm. What will be the equilibrium concentration of NH4+ in wastewater after air stripping.
KH ammonia = 57 M/atm at 25oC

[NOTE: on the test, you would be give Table 2.2 on a tables sheet]

Steps:

1. Recall the reaction is NH4+  NH3(g) + H+

2. Recall that this is a Henry’s gas problem: [𝑁𝑁𝐻𝐻3 ] = 𝐾𝐾𝐻𝐻 𝑃𝑃𝑔𝑔
�𝑁𝑁𝐻𝐻3 + �[𝐻𝐻 + ]
3. Use Ka for NH4+: 𝐾𝐾𝑎𝑎 = = 10−9.26
�𝑁𝑁𝐻𝐻4 + �
4. Solve for NH4

Try it! (The answer is 1.64 x 10-8 mol/L)

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CE2385 Final Exam Review – Full solutions
Chapter 4 Problems
Relative Risk, Attributable Risk, Odds Ratio
Suppose 40 out of 400 rats exposed to a potential carcinogen develop tumors. A control group of 300 rats not exposed
to the carcinogen develops only 10 tumors. Based on these data, compute the relative risk, attributable risk, and odds
ratio.

Steps:

1. Fill out risk matrix with data from problem

2. Calculate the risks and odd ratio using the equation sheet.

Risks With Disease Without

Disease

Exposed 40 360

Not Exposed 10 290

Rest of this is in Lecture 7b – any questions?

Variations on this: 10% of 400 people exposed to the carcinogen got cancer. 3.33% of 300 in a control group got
cancer. Know how to fill out the matrix & you’ll do well on this type of problem.

Problem 4.18
One way to estimate maximum acceptable concentrations of toxicants in drinking water or air is to pick an acceptable
lifetime risk and calculate the concentration that would give that risk assuming agreed-on exposures such as the
residential factors for an average adult. Find the acceptable concentrations for the following substances:

a) Benzene in drinking water (mg/L), at a lifetime acceptable risk of 1x10-5

b) Trichloroethylene in air (mg/m3), at a lifetime acceptable risk of 1x10-6
c) Benzene in air (mg/m3), at a lifetime acceptable risk of 1x10-5
d) Vinyl chloride in drinking water (mg/L), at a lifetime acceptable risk of 1x10-4

[NOTE: on the test, you would be given Tables 4.9 and 4.10 on a tables sheet]

For each of these problems with water, the following general equation can be used.
𝑚𝑚𝑚𝑚 𝐿𝐿 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝐶𝐶 𝐿𝐿 .∙ 2 ∙ (30 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ∙ 350 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) � �∙ ∙
𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 70 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = ∙ 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
70 𝑘𝑘𝑘𝑘
So, solving for concentration (C), the equation becomes:
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 ∙ 70 𝑘𝑘𝑘𝑘
𝐶𝐶 =
𝐿𝐿 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
2 ∙ (30 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ∙ 350 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) � �∙ ∙ ∙ 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 70 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

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CE2385 Final Exam Review – Full solutions
For each of the problems with air, the following general equation can be used.

𝑚𝑚𝑚𝑚 𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

𝐶𝐶 � 3 � ∙ 20 ∙ (30 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ∙ 350 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) � �∙ ∙
𝑚𝑚 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 70 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = ∙ 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
70 𝑘𝑘𝑘𝑘
So, solving for concentration (C), the equation becomes:
𝑅𝑅𝑖𝑖𝑖𝑖𝑖𝑖 ∙ 70 𝑘𝑘𝑘𝑘
𝐶𝐶 =
𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
20 ∙ (30 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ∙ 350 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) � �∙ ∙ ∙ 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹
𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 70 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
What about if you have a consumption problem with fish and a contaminant concentration in water?
This is the equation you’ll want to use for CDI:
𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
𝑚𝑚𝑚𝑚 𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � � ∙ 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 � �∙ ∙
𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑ℎ (𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝐶𝐶𝐶𝐶𝐶𝐶 � �=
𝑘𝑘𝑘𝑘 ∙ 𝑑𝑑𝑑𝑑𝑑𝑑 𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊ℎ𝑡𝑡 (𝑘𝑘𝑘𝑘)
How do you calculate the Intake Rate?
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵
Start with the BCF equation: 𝐵𝐵𝐵𝐵𝐵𝐵 =
𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊𝑊

You’re given the Concentration in water & BCF is a table lookup. If you solve for concentration in fish & do the
appropriate unit conversions, you’ll wind up with something in mg/kg.
Next, you use the table lookup to see how much (weight) of fish is eaten each day (54 g/day).
Example: DDT at 20 ppb (0.020 mg/L) and exposure is the full (70-year) lifetime of a person.
DDTfish = BCF x Concentrationwater
𝐿𝐿 𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚
𝐷𝐷𝐷𝐷𝐷𝐷(𝑓𝑓𝑓𝑓𝑓𝑓ℎ) = 54,000
× 0.020 = 1,080
𝑘𝑘𝑘𝑘 𝐿𝐿 𝑘𝑘𝑘𝑘
NOTE: the highlighted section isn’t needed, but sometimes it’s good to use the full equation on the equation sheet to
make sure you get to the correct answer. Potency Factor for DDT is 0.34 (mg/kg-day)-1
𝑔𝑔 𝑓𝑓𝑓𝑓𝑓𝑓ℎ 𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑𝑑𝑑 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
54 . ∙ 1,080 ∙ (70 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 ∙ 365𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑/𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦) � �∙ ∙ 0.34
𝑑𝑑𝑑𝑑𝑑𝑑 1000 𝑔𝑔 𝑘𝑘𝑘𝑘 𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 70 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = ∙ 𝑚𝑚𝑚𝑚 = 0.28
70 𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘 ∙ 𝑑𝑑𝑑𝑑𝑑𝑑
BE READY FOR: Variations on these types of problems – different weight, different type of exposure (industrial,
residential), different age at death, being told to use all of the values in the table (exposure frequency & days per
year will be different). Pay attention to what is given & you will do well.

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CE2385 Final Exam Review – Full solutions
Chapter 5 Problems
Problem 5.5 – calculating BOD & removal
6.0 − 2.0
𝐵𝐵𝐵𝐵𝐵𝐵𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 = = 240 𝑚𝑚𝑚𝑚/𝐿𝐿
5/300
9.0 − 4.0
𝐵𝐵𝐵𝐵𝐵𝐵𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = = 100 𝑚𝑚𝑚𝑚/𝐿𝐿
15/300
240−100
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 = = 58.3% This is much less than 85%, so it’s not working properly
240

Problem 5.28 – BOD problem when DOinit = DOsat

For part a) goal is to find distance when the DO is at a minimum. This is at the critical time (and critical distance), so
know that this is the equation to start with:

1 𝑘𝑘𝑟𝑟 𝐷𝐷0 (𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑 )

𝑡𝑡𝑐𝑐 = ln � �1 − ��
𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑 𝑘𝑘𝑑𝑑 𝑘𝑘𝑑𝑑 𝐿𝐿0
𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚
𝐷𝐷0 = 𝐷𝐷𝐷𝐷𝑠𝑠𝑠𝑠𝑠𝑠 − 𝐷𝐷𝐷𝐷 = 10 − 10 =0
𝐿𝐿 𝐿𝐿
So, the simplified equation is:
1 𝑘𝑘𝑟𝑟
𝑡𝑡𝑐𝑐 = ln � �
𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑 𝑘𝑘𝑑𝑑
𝑥𝑥 = 𝑣𝑣 ∙ 𝑡𝑡
For part b), know that you will want to start with this equation, which is for any deficit at time t:
𝑘𝑘𝑑𝑑 𝐿𝐿0
𝐷𝐷 = (𝑒𝑒 −𝑘𝑘𝑑𝑑 𝑡𝑡 − 𝑒𝑒 −𝑘𝑘𝑟𝑟 𝑡𝑡 ) + 𝐷𝐷0 (𝑒𝑒 −𝑘𝑘𝑟𝑟 𝑡𝑡 )
𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑
So, since D0 = 0, the simplified equation is below (insert the tc for the time since the Dmax is when the DO is at a
minimum)
𝑘𝑘𝑑𝑑 𝐿𝐿0
𝐷𝐷𝑚𝑚𝑚𝑚𝑚𝑚 = �𝑒𝑒 −𝑘𝑘𝑑𝑑 𝑡𝑡 − 𝑒𝑒 −𝑘𝑘𝑟𝑟 𝑡𝑡 �
𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑
Remember – this is D, which is a deficit! IT IS NOT THE DO (dissolved oxygen)
Dmax = DOsat – DOmin
So, you have Dmax and were given DOmax,, so now you solve for DOmin.

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CE2385 Final Exam Review – Full solutions
BOD, DO, critical time (similar to problem 5.31)
A city of 50000 people deposits 0.30 m3/s of sewage having BOD of 6.4 mg/L and 1.0 mg/L of DO into a river that has a
flow rate of 0.9 m3/s and a flow speed of 0.65 m/s. Just upstream of the release point the river has a BOD of 7.0 mg/L
and a DO of 6.0 mg/L. The saturation value of DO is 8.0 mg/L. The de-oxygenation coefficient kd is 0.20/day and the
reaeration coefficient kr is 0.37/day. Assume complete mixing instantaneously. Find the time to minimum DO and the
minimum DO concentration.

First step. Look at initial conditions (do a mass balance w/o decay)
𝐷𝐷𝐷𝐷𝑟𝑟 ∙𝑄𝑄𝑟𝑟 +𝐷𝐷𝐷𝐷𝑤𝑤 ∙𝑄𝑄𝑤𝑤
𝐷𝐷𝐷𝐷∙(𝑄𝑄𝑟𝑟 +𝑄𝑄𝑤𝑤 )
Solve for the unknown (DO):
𝑚𝑚3 𝑚𝑚𝑚𝑚 𝑚𝑚3 𝑚𝑚𝑚𝑚
𝐷𝐷𝐷𝐷𝑟𝑟 ∙ 𝑄𝑄𝑟𝑟 + 𝐷𝐷𝐷𝐷𝑤𝑤 ∙ 𝑄𝑄𝑤𝑤 0.3 𝑠𝑠 ∙ 1.0 𝐿𝐿 + 0.9 𝑠𝑠 ∙ 7.0 𝐿𝐿 𝑚𝑚𝑚𝑚
𝐷𝐷𝐷𝐷 = = 3 3 = 4.75
(𝑄𝑄𝑟𝑟 + 𝑄𝑄𝑤𝑤 ) 𝑚𝑚 𝑚𝑚 𝐿𝐿
0.3 𝑠𝑠 + 0.9 𝑠𝑠

The initial deficit (D0) is then: D0 = 8.0 mg/L – 4.75 mg/L = 3.25 mg/L
Initial BOD can be calculated by remembering that BOD is the difference in DO values and are represented by L.
Initial BOD = L0.
𝑚𝑚3 𝑚𝑚𝑚𝑚 𝑚𝑚3 𝑚𝑚𝑚𝑚
0.3 ∙ 6.4 𝐿𝐿 + 0.9 ∙ 7.0 𝐿𝐿 𝒎𝒎𝒎𝒎
𝑳𝑳𝟎𝟎 = 𝑠𝑠 𝑠𝑠 = 𝟔𝟔. 𝟖𝟖𝟖𝟖
𝑚𝑚 3 𝑚𝑚 3 𝑳𝑳
0.3 + 0.9
𝑠𝑠 𝑠𝑠

The minimum DO is located at the critical time (tc)

1 𝑘𝑘𝑟𝑟 𝐷𝐷0 (𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑 )
𝑡𝑡𝑐𝑐 = ln � �1 − ��
𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑 𝑘𝑘𝑑𝑑 𝑘𝑘𝑑𝑑 𝐿𝐿0
0.37 𝑚𝑚𝑚𝑚 0.37 0.20
1 3.25 𝐿𝐿 ∙ � − �
𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑
𝑡𝑡𝑐𝑐 = ln � �1 − ��
0.37 0.20 0.2 0.20 𝑚𝑚𝑚𝑚
− ∙ 6.85 𝐿𝐿
𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑
𝑡𝑡𝑐𝑐 = 0.582 days
The maximum deficit can be calculated using equation 5.27:
0.20 𝑚𝑚𝑚𝑚
𝑘𝑘𝑑𝑑 𝐿𝐿0 ∙ 6.85 𝐿𝐿 0.20 0.37
𝑑𝑑𝑑𝑑𝑑𝑑 − ∙0.582 𝑑𝑑𝑑𝑑𝑑𝑑 − ∙0.582 𝑑𝑑𝑑𝑑𝑑𝑑
𝐷𝐷𝑚𝑚𝑚𝑚𝑚𝑚 = �𝑒𝑒 −𝑘𝑘𝑑𝑑 𝑡𝑡 − 𝑒𝑒 −𝑘𝑘𝑟𝑟 𝑡𝑡 � + 𝐷𝐷0 �𝑒𝑒 −𝑘𝑘𝑟𝑟 𝑡𝑡 � = �𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑 �
𝑘𝑘𝑟𝑟 − 𝑘𝑘𝑑𝑑 0.37 0.20
−
𝑑𝑑𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑𝑑𝑑
𝑚𝑚𝑚𝑚 −0.37∙0.582 𝑑𝑑𝑑𝑑𝑑𝑑
+ 3.25 �𝑒𝑒 𝑑𝑑𝑑𝑑𝑑𝑑 �
𝐿𝐿
𝐷𝐷𝑚𝑚𝑚𝑚𝑚𝑚 =3.3 mg/L. So, the DOmin = 8.0 mg/L – 3.25 mg/L = 4.75 mg/L

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CE2385 Final Exam Review – Full solutions
Problem 5.47 – pump drawdown example & goal is to find the hydraulic conductivity
=5000 m3/day

B=30 m
=0.3 m
=3 m

=15 m
=150 m

Use equation for confined aquifer, and solve for K:

𝑟𝑟
𝑄𝑄 ln �𝑟𝑟1 �
2
𝐾𝐾 =
2𝜋𝜋𝜋𝜋 (ℎ𝑐𝑐,1 − ℎ𝑐𝑐,2 )
ℎ𝑐𝑐,1 = 30𝑚𝑚 − 0.3𝑚𝑚 = 29.7𝑚𝑚
ℎ𝑐𝑐,2 = 30𝑚𝑚 − 3𝑚𝑚 = 27𝑚𝑚
𝑚𝑚3 150 𝑚𝑚
5000 ln � �
𝑑𝑑𝑑𝑑𝑑𝑑 15 𝑚𝑚
𝐾𝐾 = = 22.62 𝑚𝑚/𝑑𝑑𝑑𝑑𝑑𝑑
2𝜋𝜋 ∙ 30𝑚𝑚(29.7𝑚𝑚 − 27𝑚𝑚)
Problem 5.53 – single well capture curve:

Start with equation 5.58 & solve for Q (recall that 45o= π/4):
𝜋𝜋
𝑄𝑄 𝜙𝜙 𝑄𝑄 3𝑄𝑄 100𝑚𝑚
𝑦𝑦 = �1 − � = �1 − 4 � = =
2𝐵𝐵𝐵𝐵 𝜋𝜋 2𝐵𝐵𝐵𝐵 𝜋𝜋 8𝐵𝐵𝐵𝐵 2
𝑑𝑑ℎ
Solve for Q & use Darcy’s law for v (𝑣𝑣 = 𝐾𝐾 )
𝑑𝑑𝑑𝑑

𝑑𝑑ℎ 10−4 𝑚𝑚
8𝐵𝐵𝐵𝐵 8𝐵𝐵𝐵𝐵 8 ∙ 20𝑚𝑚 ∙ �1.0 × 𝑠𝑠 � ∙ 0.0015
𝑄𝑄 = 50𝑚𝑚 ∙ = 50𝑚𝑚 ∙ 𝑑𝑑𝑑𝑑 = 50𝑚𝑚 ∙ = 0.004
3 3 3
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CE2385 Final Exam Review – Full solutions
Problem 5.54 – DNAPL leak
Part a) Use Darcy velocity equation & solve for K:
𝑚𝑚
𝑣𝑣′𝜂𝜂 0.09 𝑑𝑑𝑑𝑑𝑑𝑑 ∙ 0.34
𝐾𝐾 = = = 61.2 𝑚𝑚/𝑑𝑑𝑑𝑑𝑑𝑑
𝑑𝑑ℎ 0.0005
𝑑𝑑𝑑𝑑
Part b) Use seepage velocity equation to solve for the velocity of the contaminant:
𝑅𝑅 0.09 𝑚𝑚/𝑑𝑑
𝑣𝑣𝑠𝑠 = = = 0.015 𝑚𝑚/𝑑𝑑𝑑𝑑𝑑𝑑
𝑣𝑣 ′ 6
𝑚𝑚
Calculate distance after one year (365 days): 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 0.015 ∙ 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 5.48 𝑚𝑚
𝑑𝑑𝑑𝑑𝑑𝑑

Part c) Calculate the amount of water in affected aquifer:

𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴 𝑜𝑜𝑜𝑜 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 = 𝑉𝑉 × 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = (2.0 𝑚𝑚 × 4.0 𝑚𝑚 × 5.48 𝑚𝑚) ∙ 0.34 = 14.91𝑚𝑚3
Calculate the maximum amount of PCE that would dissolve based on the solubility in Table 5.14 (150 mg/L = 150 g/m3):
𝑔𝑔
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 150 × 14.91 𝑚𝑚3 = 2236 𝑔𝑔
𝑚𝑚3
Calculate the mass leaked using specific gravity of PCE in Table 5.14 (1.63):
𝑚𝑚𝑚𝑚 𝐿𝐿 𝑘𝑘𝑘𝑘 1000 𝑔𝑔
𝑚𝑚𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙𝑙 = 100 ∙ ∙ 365 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 ∙ 1.63 ∙ 1 ∙ = 59,495 𝑔𝑔
𝑑𝑑𝑑𝑑𝑑𝑑 1000 𝑚𝑚𝑚𝑚 𝐿𝐿 𝑘𝑘𝑘𝑘
THIS IS AN IMPORTANT CONCEPT: Since the mass leaked is greater than what can dissolve into the water, the
concentration is based on the solubility. As described in the text, about 10% of the solubility is typical. So, the
concentration is 0.10*150 mg/L = 15 mg/L

Part d) The problem gives the width of the capture curve is 2m, which is equal to Q/2Bv, and remembering that the
velocity can be represented by 𝑣𝑣 = 𝑣𝑣′𝜂𝜂.

5.48m

2m

𝑄𝑄 𝑄𝑄
Look at how the thickness of the plume is equal to . Recall that you are given that this equals 2m. So, solve = 2𝑚𝑚
2𝐵𝐵𝐵𝐵 2𝐵𝐵𝐵𝐵
for Q: 𝑄𝑄 = 2𝑚𝑚 ∙ 2𝐵𝐵𝐵𝐵. Given B = 4 m, so:
𝑚𝑚
𝑄𝑄 = 2𝑚𝑚 ∙ 2 ∙ 4 𝑚𝑚 ∙ 0.09 ∙ 0.34 = 0.490 𝑚𝑚3 /𝑑𝑑𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑𝑑𝑑
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CE2385 Final Exam Review – Full solutions
0.490 𝑚𝑚3 𝑔𝑔 𝑔𝑔
Part e) Calculate the TCE flux (concentration is 15 mg/L = 15 g/m3): 𝑇𝑇𝑇𝑇𝑇𝑇 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = × 15 = 7.35
𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑

(I did a detailed derivation of contaminant flux in Lecture 9d (bonus lecture))

𝑚𝑚 59,495 𝑔𝑔
𝑡𝑡 = 𝑇𝑇𝑇𝑇𝑇𝑇𝑃𝑃𝑃𝑃𝑃𝑃
𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓
= 𝑔𝑔 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 22.2 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦
7.35 ∙365
𝑑𝑑𝑑𝑑𝑑𝑑 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦

Chapter 6 Problems
Problem 6.4 a-b:

a) must assume that specific gravity given is at the same temperature as for water (20 oC).

So, 𝜌𝜌𝑝𝑝 = 𝑆𝑆. 𝐺𝐺.× 𝜌𝜌𝐻𝐻2𝑂𝑂 (𝑎𝑎𝑎𝑎 20) = 1.4 × 998.2 𝑘𝑘𝑘𝑘/𝑚𝑚3 . If they use 1000 kg/m3, it’s OK. Just make a note saying the correct
process.
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 2
2
𝑔𝑔(𝜌𝜌𝑝𝑝 −𝜌𝜌)𝑑𝑑𝑝𝑝 9.807 𝑚𝑚/𝑠𝑠2 �1.4∙998.2 3 −998.2 3 ��1.0×10−5 𝑚𝑚�
𝑣𝑣𝑠𝑠 = 18𝜇𝜇
= 𝑚𝑚 𝑚𝑚
18(0.00100 𝑘𝑘𝑘𝑘/𝑚𝑚∙𝑠𝑠)
= 2.18 × 10−5 𝑚𝑚/𝑠𝑠

𝑚𝑚3 𝑑𝑑𝑑𝑑𝑑𝑑
𝑄𝑄 7500 ∙
c) 𝑣𝑣0 = = 𝑑𝑑𝑑𝑑𝑑𝑑 86400 𝑠𝑠
= 2.89 × 10−4 𝑚𝑚/𝑠𝑠
𝐴𝐴𝑏𝑏 10 𝑚𝑚×30 𝑚𝑚

Since vo > vs, less than 100% of the particles will be removed.

6.5 Solve for vs using ρp=1.2 g/mL = 1200 kg/m3:

𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 2
9.807 𝑚𝑚/𝑠𝑠2 �1200 3 −998.2 3 ��1.0×10−5 𝑚𝑚�
𝑣𝑣𝑠𝑠 = 𝑚𝑚 𝑚𝑚
18(0.00100 𝑘𝑘𝑘𝑘/𝑚𝑚∙𝑠𝑠)
= 1.1 × 10−5 𝑚𝑚/𝑠𝑠

Now, set vo = vs and solve for width (length = 5w):

𝑚𝑚 𝑄𝑄 0.100 𝑚𝑚3 /𝑠𝑠

𝑣𝑣0 = 1.1 × 10−5 = =
𝑠𝑠 𝐴𝐴𝑏𝑏 5𝑤𝑤 ∙ 𝑤𝑤
1/2
0.100 𝑚𝑚3 /𝑠𝑠
𝑤𝑤 = � 𝑚𝑚 � = 42.7 𝑚𝑚
5 ∙ �1.1 × 10−5 �
𝑠𝑠
6.11
Accum. = Inlet – Outlet + Reaction

0 = QN0 - QN - kNVb

𝑄𝑄𝑁𝑁0 = 𝑄𝑄𝑄𝑄 + 𝑘𝑘 ∗ 𝑁𝑁𝑉𝑉𝑏𝑏

Solve for Q (showing full process I used):
𝑁𝑁
𝑄𝑄 ∙ = 𝑄𝑄 + 𝑘𝑘 ∗ 𝑉𝑉𝑏𝑏
𝑁𝑁0
𝑁𝑁
𝑄𝑄 ∙ � − 1� = 𝑘𝑘 ∗ 𝑉𝑉𝑏𝑏
𝑁𝑁0

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CE2385 Final Exam Review – Full solutions
𝑘𝑘 ∗ 𝑉𝑉𝑏𝑏 7.8/𝑠𝑠𝑠𝑠𝑠𝑠 ∙ 2.00 𝐿𝐿
𝑄𝑄 = =
𝑁𝑁 𝑁𝑁
�𝑁𝑁 − 1� �𝑁𝑁 − 1�
0 0

Use relationship between % removal and log removal to calculate N/N0:

𝑃𝑃 = (1 − 10−𝐿𝐿 ) × 100
99.9 = (1 − 10−𝐿𝐿 ) × 100
0.999 = (1 − 10−𝐿𝐿 )
0.001 = 10−𝐿𝐿
− log(0.001) = 𝐿𝐿 = 3
𝑁𝑁 𝑁𝑁
3 = log10 �𝑁𝑁0 �  103 = 𝑁𝑁0
𝑡𝑡 𝑡𝑡
Calculate flow rate:
7.8/𝑠𝑠𝑠𝑠𝑠𝑠 ∙ 2.00 𝐿𝐿 7.8/𝑠𝑠𝑠𝑠𝑠𝑠 ∙ 2.00 𝐿𝐿
𝑄𝑄 = = = 0.0156 𝐿𝐿/𝑠𝑠
𝑁𝑁 (1000 − 1)
�𝑁𝑁 − 1�
0
The volume produced in 10 hours would then be:
𝐿𝐿 3600 𝑠𝑠
𝑉𝑉(10 ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) = 10 ℎ𝑟𝑟 ∙ 0.0156 ∙ = 562 𝐿𝐿
𝑠𝑠 ℎ𝑟𝑟

Review making a bar graph for softening problems

CH NCH
Concentrations
Ca-CH Mg-NCH
must be in
Mg-CH meq/L or mg/L
Ca 2+
Mg2+
HCO3-
as CaCO3

6.18
Calculate the equivalent weight for each ion, then the equivalent concentration for each. It is helpful to make the bar
diagram for problems like these. Problem provides 3.1 mM HCO3, which is 189.1 mg/L (3.1 mmol/L * 61 mg/mmol).
Conc
Ion (mg/L) EW Conc (meq/L)
CO2 6 22 0.273
Ca 2+
50 20 2.500
Mg2+ 20 12.15 1.646
4.1461

Na +
5 23 0.217
HCO3- 189.1 61 3.100
SO4 2-
85 48 1.771
pH 7.6

Total Hardness (TH) = 2.5 + 1.646 = 4.146 meq/L

Carbonate hardness (CH) = HCO3- = 3.10 meq/L
Ca-CH = 0
Mg-CH = CH – Ca-CH = 3.10 meq/L – 2.50 meq/L = 0.60 meq/L
Mg-NCH = TH - Ca-CH - Mg-CH = 4.146 meq/L – 2.5 meq/L – 0.6 meq/L = 1.046 meq/L

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Use the reactions given in 6.37 to 6.41 to calculate the amount of lime or soda ash is needed for softening and for how
much CaCO3 or Mg(OH)2 will precipitate in the process. Assume excess lime is 0.4 meq/L. Look at Example 6-6 and 6-6b
(extra slides in the Helpful files folder where this is discussed in more detail) and also this section of the video for more
details

New table summarizing the data is as follows:

Component meq/L Lime Soda ash CaCO3(s) Mg(OH)2(s)

(meq/L) (meq/L) (meq/L) meq/L
CO2 0.273 0.273 0 0.273
Ca-CH 2.50 2.50 0 5
Mg-CH 0.60 1.20 0 1.2 0.6
Ca-NCH 0 0 0 0
Mg-NCH 1.046 1.046 1.046 1.046 1.046
Excess 0.40
Totals (meq/L) 5.419 7.519 1.646
Equivalent weight of lime (Ca(OH)2) is 74/2 = 37 mg/meq; Equivalent weight of soda ash (Na2CO3) is 106/2=53 mg/meq
Part a)
𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � � = 5.419 ∙ 37 = 200.5 𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴𝐴𝐴ℎ 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � � = 1.046 ∙ 53 = 55.4 𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚
Part b)

Equivalent weight of CaCO3 is 50 mg/meq; Equivalent weight of Mg(OH)2 is 29.15 mg/meq

𝑘𝑘𝑘𝑘 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑔𝑔 𝑔𝑔𝑔𝑔𝑔𝑔 3.785 𝐿𝐿 𝑘𝑘𝑘𝑘
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 � � = ��7.519 ∙ 50 � + �1.646 ∙ 29.15 �� ∙ 15 × 106 ∙ ∙ 6
𝑑𝑑 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚 𝑑𝑑𝑑𝑑𝑑𝑑 𝑔𝑔𝑔𝑔𝑔𝑔 10 𝑚𝑚𝑚𝑚
𝑘𝑘𝑘𝑘
= 24,068
𝑑𝑑𝑑𝑑𝑑𝑑

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CE2385 Final Exam Review – Full solutions
6.19 Calculate the CO2 concentration using pK1 = 6.35:

[H+][HCO3-]
𝐾𝐾1 =
[CO2(aq)]

Solving for CO2(aq) using the HCO3- given in the problem (165 mg/L = 2.705 mol/L) and pH 7.5:
[H+][HCO3-] 10−7.5 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿 ∙ 2.705 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿 −4
𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚
[CO2(aq)] = = = 1.91 × 10 = 0.191
𝐾𝐾1 10−6.35 𝐿𝐿 𝐿𝐿
Need this in meq/L, so:
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 2 𝑚𝑚𝑚𝑚𝑚𝑚
(𝐶𝐶𝐶𝐶2 ) = 0.191 ∙ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 = 0.383 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿

Alternate method:
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 44 𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚
(𝐶𝐶𝐶𝐶2 ) = 0.191 ∙ ∙ = 0.383 𝑚𝑚𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 22 𝑚𝑚𝑚𝑚
CO2 equivalent weight & others are in a table on the equation sheet (page 5)
Calculate the equivalent weight for each ion, then the equivalent concentration for each. It is helpful to make the bar
diagram for problems like these. The book solution uses everything as mg/L as CaCO3, so I’m including the calculations in
both meq/L and mg/L as CaCO3. I’ll provide the book solution afterwards as well.

Ion Conc (mg/L) EW Conc (meq/L) Conc (mg/L as CaCO3)

CO2 8.4 22 0.38 19.1
Ca2+ 90 20 4.50 225.0
Mg2+ 30 12.15 2.47 123.5
Na+ 72 23 3.13 156.5
K+ 6 39 0.15 7.7
Cl- 120 35.45 3.39 169.3
SO42- 225 48 4.69 234.4
HCO3- 165 61 2.70 135.2
pH 7.5

Total Hardness (TH) = 4.5+2.47 = 6.97 meq/L

Carbonate hardness (CH) = HCO3- = 2.7 meq/L
Noncarbonate hardness (NCH) = TH – CH = 6.97-2.7 = 4.26 meq/L
Ca-CH = 2.7 meq/L

Ca-NCH = Ca-H – CH = 4.5-2.7 = 1.80 meq/L

Mg-CH = 0 meq/L

Mg-NCH = TH – CH – Ca-NCH = 6.97 meq/L – 2.7 meq/L – 1.8 meq/L = 2.47 meq/L

Use the reactions given in 6.37 to 6.41 to calculate the amount of lime or soda ash is needed for softening and for how
much CaCO3 or Mg(OH)2 will precipitate in the process. Assume excess lime is 0.4 meq/L (20 mg/L as CaCO3).

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CE2385 Final Exam Review – Full solutions
New table summarizing the data is as follows:

Component meq/L Lime Soda ash

(meq/L) (meq/L)
CO2 0.38 0.38
Ca-CH 2.70 2.70 0
Mg-CH 0.00 0.00 0
Ca-NCH 1.80 0.000 1.795
Mg-NCH 2.47 2.469 2.469
Excess 0.40
Totals (meq/L) 5.957 4.264

Equivalent weight of lime (Ca(OH)2) is 74/2 = 37 mg/meq; Equivalent weight of soda ash (Na2CO3) is 106/2=53 mg/meq
Part a)
𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚
𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � � = 5.957 ∙ 37 = 220.4 𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚
Part b)
𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚
𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝐴𝐴𝐴𝐴ℎ 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 � � = 4.264 ∙ 53 = 226.0 𝑚𝑚𝑚𝑚/𝐿𝐿
𝐿𝐿 𝐿𝐿 𝑚𝑚𝑚𝑚𝑚𝑚

Chapter 7 Problems
Problem 7.2

Consider a new 38% efficient 600-MW power plant burning 9,000 BTU/lb coal containing 1% sulfur. If a 70% efficient
scrubber is used, what would be the emission rate of sulfur (lb/hr)

Step 1: Set up the problem by drawing a diagram & labeling the streams with what you know from the problem

Step 2: Calculate Input Sulfur in lb/hr

𝑙𝑙𝑙𝑙 600 𝑀𝑀𝑀𝑀 1000 𝑘𝑘𝑘𝑘 3412 𝐵𝐵𝐵𝐵𝐵𝐵 𝑙𝑙𝑙𝑙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 0.01 𝑙𝑙𝑙𝑙 𝑆𝑆
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 � � = ∙ ∙ ∙ ∙ = 5986 𝑙𝑙𝑙𝑙 𝑆𝑆/ℎ𝑟𝑟
ℎ𝑟𝑟 0.38 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ𝑟𝑟 9000 𝐵𝐵𝐵𝐵𝐵𝐵 𝑙𝑙𝑙𝑙 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐
𝑙𝑙𝑙𝑙 𝑆𝑆 𝑆𝑆
𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 � � = (1 − 0.70) ∙ 5986 𝑙𝑙𝑙𝑙 = 1796 𝑙𝑙𝑙𝑙
ℎ𝑟𝑟 ℎ𝑟𝑟 ℎ𝑟𝑟

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CE2385 Final Exam Review – Full solutions
Chapter 7.6 Problem (Day 2) – What AQI descriptor (Good, moderate, etc) should be reported for air quality on the
following days? Answer: Very unhealthy, triggered by both O3 and NO2.

Pollutant Day 1 Concentration

O3, 1-hr (ppm) 0.22

CO, 8-hr (ppm) 15

PM2.5 (mg/m3) 150

PM10 (mg/m3) 300

SO2 (ppm) 0.20

NO2 (ppm) 0.7

Example 7.4 – Settling Velocity of a Spherical Particle.

Find the settling velocity of a spherical droplet of water with diameter 2 mm, and estimate the residence time of such
particles if they are uniformly distributed in the lower 1,000 m of atmosphere and their removal rate is determined by
how fast they settle in still air.

Step 1. Calculate the settling velocity using equation 7.24 (and assume 20 oC):
𝑘𝑘𝑘𝑘 𝑘𝑘𝑘𝑘 𝑔𝑔
Diameter of the particle is 2 mm = 2 × 10−6 𝑚𝑚, 𝜌𝜌 = 997 ≅ 1000 = 106 𝑚𝑚3,
𝑚𝑚3 𝑚𝑚3
η=0.0172 g/m s (page 391), g=9.807m/s2
−6 2 6 𝑔𝑔 2
𝑑𝑑2 𝜌𝜌𝜌𝜌 (2 × 10 𝑚𝑚) ∙ 10 𝑚𝑚3 ∙ 9.807 𝑚𝑚/𝑠𝑠
𝑣𝑣 = = 𝑔𝑔� = 1.27 × 10−4 𝑚𝑚/𝑠𝑠
18𝜂𝜂 18 ∙ 0.0172 𝑚𝑚 ∙ 𝑠𝑠

Step 2. Use Simple box model to estimate residence time, t, of N particles uniformly distributed:
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑁𝑁 (𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝)
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ( )= ∙ 𝑣𝑣
𝑠𝑠𝑠𝑠𝑠𝑠 ℎ
Step 3. Residence time definition:
𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 𝑖𝑖𝑖𝑖 𝑏𝑏𝑏𝑏𝑏𝑏 𝑁𝑁 ℎ 1000 𝑚𝑚 6
𝑑𝑑𝑑𝑑𝑑𝑑
𝜏𝜏 = = = = = 7.9 × 10 𝑠𝑠𝑠𝑠𝑠𝑠 ∙ = 91 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑
𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝑜𝑜𝑜𝑜 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑜𝑜𝑜𝑜 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑁𝑁𝑁𝑁/ℎ 𝑣𝑣 1.27 × 10−4 𝑚𝑚/𝑠𝑠 86400 𝑠𝑠𝑠𝑠𝑠𝑠

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CE2385 Final Exam Review – Full solutions
Example 7.12 – A Power Plant Plume

A 40% efficient, 1,000 MW (106 kW) coal-fired power plant emits SO2 at the legally
allowable rate of 0.6 lb SO2 per million Btu of heat into the plant. The stack has an
effective height of 300 m. An anemometer on a 10-m pole measures 2.5 m/s of wind,
and it is a cloudy summer day. Predict the ground-level concentration of SO2 4 km
directly downwind.

Step 1: Set up the problem by drawing a diagram & labeling

the streams with what you know from the problem

Step 2: Calculate Input Power in Btu/hr:

𝐵𝐵𝐵𝐵𝐵𝐵 103 𝑘𝑘𝑘𝑘 3412 𝐵𝐵𝐵𝐵𝐵𝐵

𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼𝐼 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃 � � = 2500 𝑀𝑀𝑀𝑀 ∙ ∙ = 8.53 × 109 𝐵𝐵𝐵𝐵𝐵𝐵/ℎ𝑟𝑟
ℎ𝑟𝑟 𝑀𝑀𝑀𝑀 𝑘𝑘𝑘𝑘ℎ𝑟𝑟
Step 3: Calculate SO2 emission rate (mg/s), Q:

𝐵𝐵𝐵𝐵𝐵𝐵 0.6 𝑙𝑙𝑙𝑙 𝑆𝑆𝑆𝑆2 𝑘𝑘𝑘𝑘 109 𝜇𝜇𝜇𝜇

𝑄𝑄 = 8.53 × 109 ∙ ∙ ∙
ℎ𝑟𝑟 106 𝐵𝐵𝐵𝐵𝐵𝐵 2.205 𝑙𝑙𝑙𝑙 𝑘𝑘𝑘𝑘
ℎ𝑟𝑟
∙ = 6.54 × 108 𝜇𝜇𝜇𝜇 𝑆𝑆𝑆𝑆2 /𝑠𝑠
3600 𝑠𝑠
Step 4: Calculate wind speed at plume height (H):

𝐻𝐻 𝑝𝑝 𝑚𝑚 300 𝑚𝑚 0.20
𝑢𝑢𝐻𝐻 = 𝑢𝑢𝑎𝑎 � � = 2.5 ∙ � � = 4.9 𝑚𝑚/𝑠𝑠
𝑧𝑧𝑎𝑎 𝑠𝑠 10 𝑚𝑚
Step 5: Solve equation 7.49 at 4 km. Use Table 7.9 to
get σy= 359 m and σz= 216 m (cloudy summer day = slight solar insolation, use 2.5 m/s wind speed in Table 7.7 to know
classification is C).
−𝐻𝐻 2
𝑄𝑄 2
𝐶𝐶(𝑥𝑥, 𝑦𝑦) = × 𝑒𝑒 2𝜎𝜎𝑧𝑧
𝜋𝜋𝑢𝑢𝐻𝐻 𝜎𝜎𝑦𝑦 𝜎𝜎𝑧𝑧
2
6.54 × 108 𝜇𝜇𝜇𝜇 𝑆𝑆𝑆𝑆2 /𝑠𝑠 −300 𝑚𝑚
𝐶𝐶(4,0) = × 𝑒𝑒 2(216𝑚𝑚)2 = 206 𝜇𝜇𝜇𝜇/𝑚𝑚3
𝑚𝑚
𝜋𝜋 ∙ 4.9 𝑠𝑠 ∙ 359𝑚𝑚 ∙ 216 𝑚𝑚
𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝∙𝑀𝑀𝑀𝑀
Step 5: Convert to ppmv using this equation =
𝑚𝑚3 24.465
𝑚𝑚𝑚𝑚
206 𝜇𝜇𝜇𝜇/𝑚𝑚3 ∙ ∙ 24.465
103 𝜇𝜇𝜇𝜇
𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 = = 0.1095 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝
46 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚

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CE2385 Final Exam Review – Full solutions
Yuja Quiz Question with log-log graph.

Suppose you have the same SO2 emitted as

in Examples 7-12 and 7-13 (6.54 ×
108 𝜇𝜇𝜇𝜇 𝑆𝑆𝑆𝑆2 /𝑠𝑠), the wind speed is 5.2 m/s,
and you use figure 07-52 for a different case
where the stability is neutral and you still
want to check the max concentration 4 km
downstream.

a) What stability class is appropriate

for this problem (A-F). You will have
this table (Table 7.7) available to
you, so you need to interpret what is
in the question & in the table. Class
D is for neutral (could be called
stable also) atmospheric conditions.

b) What is the concentration at 4 km?

Use this equation & the log-log plot (you will have it on the exam with the other tables) to calculate maximum
concentration:
𝑄𝑄 𝐶𝐶𝑢𝑢𝐻𝐻
𝐶𝐶𝑚𝑚𝑚𝑚𝑚𝑚 = � �
𝑢𝑢𝐻𝐻 𝑄𝑄 𝑚𝑚𝑚𝑚𝑚𝑚
𝜇𝜇𝜇𝜇 𝑆𝑆𝑆𝑆2
6.54 × 108 5.2 × 10−7
𝐶𝐶𝑚𝑚𝑚𝑚𝑚𝑚 = 𝑠𝑠 ∙
5.2 𝑚𝑚 𝑚𝑚2
𝑠𝑠
𝐶𝐶𝑚𝑚𝑚𝑚𝑚𝑚 = 65.4 𝜇𝜇𝜇𝜇/𝑚𝑚3

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CE2385 Final Exam Review – Full solutions
Example 7.14 – Peak Downwind Concentration.

A large power plant has a 250-m stack with inside radius 2m. The exit velocity of the stack gases is estimated at 15 m/s,
at a temperature of 140 oC (413 K). Ambient temperature is 25 oC (298 K), and winds at stack height are estimated to be
5 m/s. Estimate the effective height of the stack if (a) the atmosphere is stable with temperature increasing at the rate
of 2 oC/km, and (b) the atmosphere is slightly unstable,
class C.

Part a)

Key part is to know that you want to use the ∆h

1
𝐹𝐹 3
equation �∆ℎ = 2.6 � 𝑆𝑆� � so that you can add it to
𝑢𝑢ℎ

the stack height

Step 1: Calculate the buoyancy parameter F from:

𝑇𝑇𝑎𝑎 𝑚𝑚 𝑚𝑚 298𝐾𝐾
𝐹𝐹 = 𝑔𝑔𝑟𝑟 2 𝑣𝑣𝑎𝑎 �1 − � = 9.807 2 ∙ (2𝑚𝑚)2 ∙ 15 ∙ �1 − � = 164 𝑚𝑚4 /𝑠𝑠 3
𝑇𝑇𝑠𝑠 𝑠𝑠 𝑠𝑠 413 𝐾𝐾

Step 2: Calculate S and ∆h:

𝑔𝑔 ∆𝑇𝑇𝑎𝑎 9.807 𝑚𝑚/𝑠𝑠 2 𝑜𝑜𝐶𝐶 𝑘𝑘𝑘𝑘 𝑜𝑜𝑜𝑜

𝑆𝑆 = � + 0.01 𝑜𝑜𝐶𝐶/𝑚𝑚� = �2 ∙ + 0.01 � = 0.0004/𝑠𝑠 2
𝑇𝑇𝑎𝑎 ∆𝑧𝑧 298 𝐾𝐾 𝑘𝑘𝑘𝑘 1000 𝑚𝑚 𝑘𝑘𝑘𝑘
1/3
𝐹𝐹 1/3 164 𝑚𝑚4 /𝑠𝑠 3
∆ℎ = 2.6 � � = 2.6 � � = 113 𝑚𝑚
𝑢𝑢ℎ 𝑆𝑆 5 𝑚𝑚/𝑠𝑠 ∙ 0.0004/𝑠𝑠 2

Effective stack height = 250 m + 113 m = 363 m

Problem 7.45 –Yuja Quiz

A stack with effective height of 45 m emits SO2 at the rate of 150 g/s. Winds are estimated at 5 m/s at the stack height,
the stability class is C, and there is an inversion at 100 m. Estimate the ground-level concentration in mg/m3 at the point
where reflections begin to occur from the inversion.

Hints for equations you will need.

• Use 7.56 to determine σz.
• Use Figure 07-53 for interpretation.
• In this problem, L = 100 m, H = 45 m
• Use equation 7.49 to solve for C (XL, 0)

a) What is the table lookup value for σy?

𝜎𝜎𝑧𝑧 = 0.47(𝐿𝐿 − 𝐻𝐻) = 0.47(100 − 45) = 26𝑚𝑚

Look at Table 7-9 (will be provided for the exam)

At 𝜎𝜎𝑧𝑧 = 26𝑚𝑚 (for Class C), x=0.4 km, 𝜎𝜎𝑦𝑦 = 46 𝑚𝑚

b) What is the ground-level concentration in mg/m3 at XL?

−𝐻𝐻2
𝑄𝑄 2𝜎𝜎2
𝐶𝐶(𝑥𝑥, 𝑦𝑦) = × 𝑒𝑒 𝑧𝑧  use this equation because x = XL. (pay attention to the problem statement!)
𝜋𝜋𝑢𝑢𝐻𝐻 𝜎𝜎𝑦𝑦 𝜎𝜎𝑧𝑧

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CE2385 Final Exam Review – Full solutions
𝑔𝑔 1000 𝑚𝑚𝑚𝑚
150 ∙ −452
𝑠𝑠 𝑔𝑔
𝐶𝐶(𝑋𝑋𝐿𝐿 , 0) = × 𝑒𝑒 2 = 1.8 𝑚𝑚𝑚𝑚/𝑚𝑚3
2∙26
𝜋𝜋 ∙ 5 𝑚𝑚/𝑠𝑠 ∙ 46𝑚𝑚 ∙ 26𝑚𝑚
If you want to do the second part (distance is twice the XL distance), you approach it in a similar manner.

x = 2*XL = 0.8 km, so using Table 7.9, 𝜎𝜎𝑦𝑦 = 85

Now, to solve for the concentration, you will use a different equation:
𝑄𝑄
𝐶𝐶(𝑥𝑥, 0) = (2𝜋𝜋)1/2  use this equation because x = 2XL. (pay attention to the problem statement!)
𝑢𝑢𝐻𝐻 𝜎𝜎𝑦𝑦 𝐿𝐿)

Example 7.18 – Indoor Air Quality

An unvented, portable, radiant heater, fueled with kerosene is tested under controlled laboratory conditions. After
running the heater for two hours in a test chamber with a 46.0 m3 volume and an infiltration rate of 0.25 ach (air
changes per hour), the concentration of CO reaches 20 ppm. Initial CO in the lab is 0, and the ambient CO level is
negligible throughout the run. Initial CO in the lab is 0, and the ambient CO level is negligible throughout the run.
Treating CO as a conservative pollutant, find the rate at which the heater emits CO. If the heater were to be used in a
small home to heat 120 m3 of space having 0.4 ach, predict the steady-state concentration.
𝑚𝑚𝑚𝑚 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝∙𝑀𝑀𝑀𝑀
Calculate concentration using this equation =
𝑚𝑚3 24.465

𝑚𝑚𝑚𝑚 20 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 ∙ 38 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚 𝑚𝑚𝑚𝑚

𝐶𝐶 � 3 � = = 22.9 3
𝑚𝑚 24.465 𝑚𝑚
Simplify (Ca=0) & rearrange C(t) for conservative pollutant (K=0) to solve for the heater’s emission rate S at t=2hours:
𝑚𝑚𝑚𝑚
𝑛𝑛𝑛𝑛𝑛𝑛(𝑡𝑡) 0.25 𝑎𝑎𝑎𝑎ℎ ∙ 46 𝑚𝑚3 ∙ 22.9 3
𝑆𝑆 = = 𝑚𝑚 = 669 𝑚𝑚𝑚𝑚/ℎ𝑟𝑟
1 − 𝑒𝑒 −𝑛𝑛𝑛𝑛 0.25
− ∙2ℎ𝑟𝑟
1 − 𝑒𝑒 ℎ𝑟𝑟
Solve for 𝐶𝐶∞ knowing that Ca=0 and K=0:

𝑆𝑆/𝑉𝑉 669 𝑚𝑚𝑚𝑚/ℎ𝑟𝑟/120𝑚𝑚3 𝑚𝑚𝑚𝑚

𝐶𝐶∞ = = = 13.9 3 (12.1 𝑝𝑝𝑝𝑝𝑝𝑝)
𝑛𝑛 0.25 𝑎𝑎𝑎𝑎ℎ 𝑚𝑚

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