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Homework – CIVL3420 Spring 20 20HW-1 (Chap 1&1) ( Due on 24 Feb, 2021)Q1. There are two raw water supply sources available for City T: river A and reservoir B. It is know that the former has a 7-day minimum flow rate of 2 m 3 ./sec and the latter’s max storage capacity and a static (dead) capaci...

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Homework(–(CIVL3420((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((Spring(2020( HW-1 (Chap 1.1&1.2)

(Due on 24 Feb, 2021)

Q1. There are two raw water supply sources available for City T: river A and reservoir B. It is know that the former has a 7-day minimum flow rate of 2 m3./sec and the latter’s max storage capacity and a static (dead) capacity are 9 million m3 and 1.5 million m3, respectively. The population of City T is anticipated to grow up to 120,000 in the next 20 years. The average water demand per capita per day (including industrial and other uses and losses) is currently 300 L/person/day and may increase to 120% in the next 20 years. The maximum daily water demand is 220% of the average daily demand while the max hourly demand is 400% of the average daily demand. You are assigned to design a water treatment plant for this city. Therefore, you need to solve the following problems: (1) Does the River A have enough water to supply? (5 pts) (2) Can enough water be taken from the reservoir in a continuous 3-month drought period without any rainfall? (10 pts) (3) What is the design capacity of water treatment plant (expressed in m3/day)? (5 pts) Suggested Solution Population = 120,000 Average Water Demand: 120% × 120,000 × 300 L / person!day Maximum Daily Demand: 220% × 43200 (1) Required River Flow Rate = 95,040 m3 / day = 1.1 m3/sec ∵ Smaller than the 7-day minimum flow rate of 2 m3/sec ∴ Enough (2) Total water demand for 3 months = 43200 × 3 × 30 Since the daily fluctuation in water demand is damped over long period, daily average water demand is used for calculation Amount of water available = (9 - 1.5) × 106 m3 ∵ Greater than the total water demand for 3 months ∴ Enough (3) Design capacity of the water treatment plant: Max. Daily Water Demand which is 95,040 m3 / day

Q2. Due to rapid growing economic, the government plans to upgrade the existing airport with more airport hotels and a large residential estate on an isolated island. No industry is allowed to be developed on the island. Please calculate the daily water consumption rate and sizing the capacity of an on-site wastewater treatment plant based on the following criteria and the tables of water usage in your class note. (30 pts) Criteria: • population of 55,000 • designed airport passenger of 160,000/day • total employee of 13,000 • parking space of 3,000 • 30 lavatories with a total using frequency of 120/min • three airport hotels with a designed capacity of 6,000 guests/day • 10 restaurants with a designed customer size of 50/hr in each • 80% of the consumed water becomes wastewater • and the unaccounted system losses should not be counted in wastewater.

Homework(–(CIVL3420((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((Spring(2020(

Use Domestic Industrial Public service Unaccounted system losses and leakage

Total

Flow, L/Capita · Day Percent based on Range Average average flow 150 ~ 500 230 36.5 40 ~ 380 260 41.3 20 ~ 80 40 6.3 40 ~ 150

250 ~ 1100

100

630

15.9

100

Flow, L/Unit · Day User Airport Hotel Office Public lavatory Restaurant (w/toilet) Shopping center

Unit

Range

Typical

10 ~ 20 150 ~ 230 30 ~ 50 30 ~ 75

15 190 40 60

User

10 ~ 25

20

Customer

10 ~ 40

25

4 ~ 12

8

30 ~ 50

40

Passenger Guest Employee Employee

Parking Space Employee

Figure 1. Planning Data for Water/Wastewater Treatment Facilities Suggested Solution Domestic: 230L / capita!day Public Service: 40L / capita!day System Losses and Leakage: 100L / capita!day Total Residential Water Demand: 370L / capita!day Calculation steps of water demand (1000L = 1 m3): Residential: 55,000 people × 370L / capita!day = 20,350 m3 / day Airport: Passages: 160,000 people / day × 15 L / person!day = 2,400 m3 / day Employee: 13,000 people × 40 L / person!day = 520 m3 / day Parking: 3,000 × 8 L / person!day = 24 m3 / day Lavatories: 120 × 60 × 24 people / day × 20 L / person!day = 3456 m3 / day Hotel: 6,000 guests / day × 190 L / person!day = 1,140 m3 / day Restaurant: 10 × 50 × 24 customer / day × 25 L / person!day = 300 m3 / day ∴ Total Daily Water Demand: 28,190 m3 / day System Losses and Leakages: 55,000 people × 100 L / capita!day Water Actually Consumed: (28,190 – 5,500) m3 / day = 22,690 m3 / day Design Wastewater Flow Rate = Loading capacity of the wastewater treatment plant: 0.8 × 22,690 m3 / day = 18,152 m3 / day

HW-1 (Chap 1.3&1.4) Q3 By comparing settled solid in primary settling tank of wastewater treatment plants (WWTP) elsewhere in the world and Hong Kong, it is found HK’s WWTP has the highest removal of solid in its 1o settler. Explain your answer by your observation during the lectures so far. (5 pts)

Homework(–(CIVL3420((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((Spring(2020(

Suggested Solution HK uses seawater toilet flushing for which there are many positivity ions that behaves like coagulant so the sedimentation process can be efficiently enhanced.

Q4. A municipal wastewater treatment plant receives an average flow of 35,000 m3/day. Estimate the loading rate (kilograms per day) of COD and TKN in terms of max, mean, and min. (10 pts) Suggested Solution From Table 1(Chap 1.3+1.4, p.11), the typical concentration (kg/m3) in sewage, we have: Strong Medium Weak COD 1 0.50 0.25 TKN 0.085 0.04 0.02 -3 3 (1mg/L = 10 kg/m ) Hence, loading rate (kg/d) = Flow rate (m3/d) × Conc. (kg/m3) Loading rate (kg/d) Maximum Mean COD 35,000 17,500 TKN 2,975 1,400

Minimum 8,750 700

Q5. Estimate the average sewage flow of Hong Kong (m3/day). Suggested Solution From Chap 1.3 &1.4 slide 3, the sewage is about 300 L/head/day. The total population in Hong Kong was recorded at 7.465 million people in 2019. (http://www.gov.hk/en/about/abouthk/factsheets/docs/population.pdf) Average sewage flow of Hong Kong (m3/day) in 2019 = Average sewage produced (m3/day/head) × Population in HK (head) = 2,239,500 m3/day (Using water consumption approach or using historical data by DSD to forecast are also accepted)

(10 pts)

Homework(–(CIVL3420((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((((Spring(2020( HW-1 (Chap 1.5) Q6. Calculate the TN of a solution containing 30 mg/L glycine (C2H5O2N), 100 mg/L NaNO3 and 50 mg/L NH4Cl. (5 pts) Suggested Solution

Q7. A water with a pH 10.3 contains HCO3- and CO32- at 6.1 and 6.0 mg/L, respectively. What is its total alkalinity in mg/L as CaCO3? (20 pts) Suggested Solution

Where

End of HW-1

Homework – CIVL3420

Spring 2021

HW-2 (Chap 2.1 Coagulation and Flocculation)

(Due on 19 Apr, 2021)

Q1. The design flow for a rapid mixing tank is 49,200 m3/d. There will have a mechanical mixer and the average alum dosage will be 30 mg/L. The theoretical mean hydraulic detention time of the tank will be 1 minute. Determine the following: (10 points) (a) The quantity of alum needed on a daily basis in kg/d, (b) The dimensions of the tank in meters for a tank with equal length, width, and depth, and (c) the power input required in kW for a G of 900 sec-1 for a water temperature of 10 ºC. (a)The quantity of alum needed on a daily basis in kg/d, The quantity of alum needed on a daily basis in 1000 𝐴𝑚𝑜𝑢𝑛𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 = 𝑄 × 𝐷𝑜𝑠𝑒 = 49,200 × 103 × 30 × 6 = 1476 𝑘𝑔/𝑑……2 points 10 The dimensions of the tank in meters for a tank with equal length, width, and depth, and (b)The dimensions of the tank in meters for a tank with equal length, with and depth 1 𝑉 = 𝑄𝑡𝑅 = 49.2 × 10 3 × = 34.17 𝑚3 ……3 points 1440 3

𝑙𝑒𝑛𝑔𝑡ℎ = 𝑤𝑖𝑑𝑡ℎ = 𝑑𝑒𝑝𝑡ℎ = √34.17 = 3.24𝑚……2 points (c)the power input required in kW for a G of 900 sec-1 for a water temperature of 10 ºC. 𝑃 = 𝜇𝑣𝐺 2 = 36.2 𝑘𝑊……3 points

Q2. Flocculation tanks are to be designed with the same total flow rate as Q1. The following conditions apply to the design (15 points):  water temperature of 10 ºC,  total mean detention time of 45 min,  basin depth of 3.5 m,  3 parallel trains of flocculators (each train receives one third of the total flow),  3 flocculation stages of the same dimensions for each train (so a total of 9 flocculators),  the first stage G is 50 sec-1,  the second stage G is 35 sec-1, and  the third stage G is 20 sec-1. Determine: (a) The dimensions of the tank for each stage as well as the overall dimensions, in meters, (b) the average G times detention time product (the “Gt” Camp parameter) for the overall process and compare to the Camp criterion, and (c) the power required for each stage, in kW. (a)The dimensions of the tank for each stage as well as the overall dimensions, in meters, 1 The whole volume:𝑉𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 𝑄 × 𝑡 = 49,200 × 45 × = 1,538 𝑚3 1440 𝑉𝑜𝑣𝑒𝑟𝑎𝑙𝑙 = 171 𝑚3 9 𝑉𝑜𝑣𝑒𝑟𝑎𝑙𝑙 1,538 = = = 439 𝑚2 𝑑𝑒𝑝𝑡ℎ 3.5

𝑉𝑡𝑎𝑛𝑘 = 𝐴𝑜𝑣𝑒𝑟𝑎𝑙𝑙 𝐴𝑡𝑎𝑛𝑘 =

𝑉𝑡𝑎𝑛𝑘 𝑑𝑒𝑝𝑡ℎ

=

171 3.5

= 48.8 𝑚2 ……5 points

Any combination of length and width that gives the above square meter areas is OK, as long as the L/W ration does not become too great (i.e. it should be below 10). For example, if you assume each floc tank is square, then the dimensions are just the square root of above

Homework – CIVL3420

Spring 2021

(b)the average G times detention time product (the “Gt” Camp parameter) for the overall process and compare to the Camp criterion, and overall G is just the average for all of the flocculator tank volume 50 + 35 + 20 = 35 𝑠 −1 𝐺= 3 𝐺𝑡 = 35 × 45 × 60 = 94 ,500 That is OK as it falls between 50,000 and 100,000. ……5 points (c)the power required for each stage, in kW. 𝑃 = 𝜇𝑉𝐺 2 = 1.307 × 10−3 × 1.71𝐺 2 Stage G P(kW) 1 50 0.56 2 35 0.27 3 20 0.089 ……5 points

HW-2 (Chap 2.2 Sedimentation) Q3. A laboratory settling analysis has been performed on a suspension exhibiting flocculent (type II) sedimentation with the results given in the following table. The column had a total depth of 240 cm, with sampling ports spaced every 0.60 m. The initial concentration of the suspension was 430 mg L−1. (15 points) (a) Prepare the following graphs (list data used in preparing the above graphs in tables.) i) Percentage SS removed versus time ii) Isoconcentration curves (b) If the sedimentation tank is designed for a SS removal of 60%, based on the above graphs, calculate the critical average settling velocity. Time (min) Concentration (mg/L) 60 cm 120 cm 180 cm 5 357 387 396 10 310 346 366 20 252 299 316 30 198 254 288 40 163 230 252 50 144 196 232 60 116 179 204 75 108 143 181 (a)Prepare the following graphs (list data used in preparing the above graphs in tables. iii) Percentage SS removed versus time iv) Isoconcentration curves Time SS removal (min) 60 cm 120 cm 180cm 5 16.98% 10.00% 7.91% 10 27.91% 19.53% 14.88% 20 41.40% 30.47% 26.51% 30 53.95% 40.93% 33.02% 40 62.09% 46.51% 41.40%

Homework – CIVL3420 50 60 70

Spring 2021

66.51% 73.02% 74.88%

54.42% 58.37% 66.74%

46.05% 52.56% 57.91%

……5 points

SS removal 5 10 20 30 40 50 60

Time (min) 60 cm 1.47 2.95 5.02 11.55 18.97 26.85 37.43

120 cm 2.50 5.00 10.43 19.57 29.11 46.25 62.92

180 cm 3.16 6.50 14.40 25.36 38.33 56.07 88.41

……5 points

(b)If the sedimentation tank is designed for a SS removal of 60%, based on the above graphs, calculate the critical average settling velocity. Average settling velocity = 150/75.665=1.98 cm/min ……5 points

Q4. The WSD of HKSAR is planning to install a new settling tank as an upgrade to their existing Pak Kong water treatment work in Sai Kung. Determine the overflow rate and detention time for a settling tank to remove 60% of the influent suspended solids from their design flow of 0.5 m3/s. (20 points) Note: Eckenfelder (1980) recommends that scale-up factors of 0.65 for overflow rate and 1.75 for detention time be used to design the tank.) For the 5% line, V0=180/3.16=56.96 cm/min The line x=3.16 min intersects with other isoconcentration lines at different points The corresponding removal percentage is RT5=11.08% (listed in the Excel table).

Homework – CIVL3420

Spring 2021

From the graph, when removal is 60%, the detention time is 69.42 min and overflow rate is 3.07 cm/min Applying the scale-up factors yields T0=69.42*1.75=121.49 min V0=3.07*0.65=2.0 cm/min=28.74 m/d Q5. Design the settling tank(s) for Pak Kong water treatment work expansion using the design overflow rate found in Q4. The maximum day design flow is 0.5m3/s. Assume a water temperature of 20C. Provide the following in your summary of the design: (20 points)  Q design  Number of tanks  Width of each tank  Length of each tank  Side water depth  Depth of sludge zone  L:D  vf  Reynolds number  Number of launders  Launder length  Weir loading  Type of sludge collector The surface area is As=43200/28.74=1503 m2 Two tanks are the minimum number. For this flow rate make trial calculations using six tanks. 1503/6=251m2

Homework – CIVL3420

Spring 2021

The maximum width for the chain-and flight sludge collector is 6 m increments. Assume a width of 4 m. L=251/4=62.75 m L/W=62.75/4=15.69>6, which is acceptable. Because the column depth used to calculate the overflow rate was 2.4 m, this is a starting point for setting the design depth. As allowance for the sludge depth of 1 m is added to this depth. In addition, the tank should be provided with 0.6 m of freeboard. The total depth of the tank is then 2.4 m+1m+0.6m=4m Side water depth (SWD)=3.0 m. If the sludge zone is not counted, the depth of the water is less than the design recommendation of 3 m. L/D=62.75/2.4=26.14 The ratio is acceptable. Vf=Q/A=0.5/(6 tanks*2.4 depth*4 width)=0.0087 m/s This is within the acceptable range of 0.005~0.018 m/s Rh=Ax/Pw=(2.4 m deep)(4m wide)/(2.4+2.4+4)m=1.09 m The viscosity is at the temperature of 20 ℃, the viscosity is 1.005*10-6m2/s and the Reynolds number is R=0.0087*1.09/1.005*10-6=9435 This is less than 20000 and acceptable. Provide launders for 1/3 of the tank length Llander=62.75/3=20.9 m Place them at 1 m intervals on center so that there are three in the tank Check the weir loading rate WL=42300 m3/d/(6 tanks)(62.75 m2)(2 sides)=57.37 m3/d·m This is well below the limit of 250 m3/d·m and acceptable. Sludge collector is chain and flight HW-2 (Chap 2.3 Filtration) Q6. A dual media is designated with the following specifications: (10 points) Item

Unit

Sand

Anthracite

Homework – CIVL3420

Spring 2021

Effective size

mm

Uniformity coefficient ~

0.5

1.6

1.3

1.3

Density

kg/m3

Depth

m

Sphericity

~

0.8

0.75

Porosity

~

0.4

0.5

2650

1700

0.3

0.7

Calculate the clean-bed head loss at a filtration rate of 10 m/h and temperature of 15 C.

𝐻 𝑙𝑜𝑠𝑠 = 0.069 + 0.083 = 0.152 𝑚 Q7. Explain why a low uniformity coefficient is important in rapid filtration. (10 points) (10 points)

End of HW-2c

(Due on 7 Apr, 2021)

HW-3 (Chap 2.4 disinfection)

Q1. (Problem 3 & 11 Chapter 16 in the text) a) What are the initial concentrations of CL2, HOCl, and OCl- in a water with a temperature of 20 ºC after dissolution and hydrolysis of 5000 mg/L of Cl2? The pH of the water is buffered at 6.00. b) A water contains the following concentration of chlorine related species: HOCl – 0.41mg/L; OCl- – 0.35mg/L; NH2Cl – 0.12 mg/L; and NHCl2 – 0.05 mg/L. What are the free, combined, and total residuals as Cl2? Solutions a) (20 points) + -6.00 For pH = 6.00, i.e.: [H ] = 10 mol/L 1mol = 0.0704 mol/L 71,000 mg

Initially, [Cl2] = 5000 mg/L× ‒

4'

+

Cl2 + H2O ↔ HOCl + Cl + H , ‒

+

HOCl ↔ OCl + H ‒ At pH = 6-8, Cl2 is nearly completely converted to HOCl and OCl , and [Cl2] = 0. ‒

Thus, [HOCl] + [OCl ] = 0.0704 mol/L 𝐾𝑎 =

[OCl−][H+ ] [HOCl]

= 10 −7.6

[OCl−] [HOCl]

=

𝐾𝑎

= [H+ ]

10−7.6 10−6.00

= 10 −1.6 = 0.025

(1)

6'

(2)

6'

‒

4'

From (1) and (2), [HOCl] = 0.0687 mol/L, [OCl ] = 0.0017 mol/L

b): (20 points) Free chlorine (HOCl and OCl-)

= 0.41 mg/L X

71mg Cl2

+ 0.35 mg/L X 52.5mg HOCl

71mg Cl2 51.5mg OCl −

= 1.04 mg/L as Cl2

8'

Combined chlorine (NH2Cl and NHCl2)

= 0.12 mg/L X

71mg Cl2 51.5mg NH2 Cl

+ 0.05 mg/L X

Total chlorine = 1.04 + 0.25 = 1.29 mg/L as Cl2

142mg Cl2 86mg NHCl2

= 0.25 mg/L as Cl2

8' 4'

Q2. What are the required detention times in a plug flow basin at a temperature of 5 ºC and pH of 7.0 to meet the SWTR for 99.9% reduction of G. lamblia for (a) free chlorine at 2.0mg/L and (b) chlorine dioxide at 2.0 mg/L? (20 points) From Table 2.4.9, at pH 7.0, 5ºC, to achieve 99.9% reduction of G. Lamlia, Ct1 = 165, when free chlorine is used; Ct2 = 26, when chlorine dioxide is used. (a) For free chlorine at 2.0 mg/L, t1= Ct1 / C = 165 / 2.0 = 82.5 min (b) For chlorine dioxide at 2.0 mg/L, t2= Ct2 / C = 26 / 2.0 = 13 min

6' 6' 4' 4'

Q3. A water contains the following concentration of chlorine related species: HOCl – 0.42mg/L; OCl- – 0.206mg/L; NH2Cl – 0.103 mg/L; and NHCl2 – 0.086 mg/L. What are the free, combined, and total residuals as Cl2? From the chlorine residual curve, the chlorine residual concentration was in the range of chlorine residual: chlorine dosage ratio= 1:2, then what contact time is required to achieve 99.9% kill of bacteria if the rate constant k = 0.5(L/mg)min-1? (40 points) (Chick’s-Watson’s law:

𝑑𝑁

= −k ∙ C ∙ N) Cl2+ H2O HOCl + H+ + Cl𝑑𝑡

(2’)

HOCl H+ + OClNH3+ HOCl NH2Cl NH3+ 2HOCl NHCl2

(2’)

Free chlorine (HOCl and OCl-): HOCl=0.42*71/52.5=0.568 mg/L as Cl2;

(4’)

OCl-=0.206*71/51.5=0.284mg/L as Cl2;

(4’)

Hence free chlorine= 0.568+ 0.284= 0.852 mg/L as Cl2.

(2’)

Combined chlorine (NH2Cl and NHCl2): NH2Cl= 0.103*71/51.5=0.142 mg/L as Cl2;

(4’)

NHCl2=0.086/86*2*71=0.142mg/L as Cl2;

(6’)

Hence total combined chlorine= 0.142+0.142=0.284mg/L as Cl2.

(2’)

Total chlorine residual = 0.852 + 0.284 = 1.136 mg/L as Cl2.

(4’).

So, chlorine dosage= 1.136 * 2=2.272 mg/L.

(4’)

𝑁

Solving the Chick’s-Watson’s laws, we get 𝑁 = 𝑒 −k𝐶𝑡 , 99.9%=3 log, 𝑜

10−3 = 𝑒 −0.5×2.272×𝑡

(4’)

Hence the contact time t=6.08 min.

(2’)

Q4. An experiment shows that a chlorine dose of 1.15mg/L yield a 99% kill of bacteria in 8 minutes. What contact time is required to achieve a 99.9% kill at a free available chlorine concentration of 0.05 g·m-3? (Assume that Chick’s- Watson’s Law hold with n = 1) (10 points)

Solution: 𝑑𝑁 Based on Chick’s-Watson’s laws𝑑𝑥 = −Λ ∙ C 𝑛 ∙ N: where n is given to be 1 𝑁

Solving above ODE, we get 𝑁 = 𝑒 −Λ𝐶𝑡 where Λ = k 𝑜

(1)

Given: 99% kill = 2 log ; C1= 1.15mg/L; time (t1) = ...