Homework Assignment #7 with Solutions - Environmental Engineering | CE 321 | Assignments Civil Engineering

DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE

Homework Assignment 7, SOLUTIONS

CE 321, Fall 2008

1. At the Easton Water Treatment Plant on the day of our visit, four parallel rectangular

clarifiers, each 140 ft long, 17 ft wide, and 18 ft deep were being used to settle solids from an

approximately 8 MGD flow. Calculate the HRT and surface loading rate per tank.

HRT = Vol/Q = (140 ft*17 ft*18 ft)*7.48 gal/ft3/2,000,000 gpd = 0.16 d = 3.85 hrs

Surface loading rate = Q/As = 2,000,000 gpd*1 day/(60*24 min) / (17 ft*140 ft) = 0.58 gpm/ft2

2. A 1.5 million gallon clearwell operating as a plug-flow reactor (PFR) is used to disinfect a

flow of 8 MGD, resulting in a 99.9% bacteria removal. What would the % removal be if short-

circuiting of flow in the tank reduced the HRT by 2 hrs?

Plug Flow Reactor: C = C0 e-kHRT

Ln(C/C0) = -k HRT k = -Ln(C/C0)/HRT

HRT = V/Q = 1.5 MG/8 MGD = 0.1875 days = 4.5 hrs

So k = -Ln(C/C0)/HRT = -Ln(0.001)/4.5 hrs = 1.535 hr-1

With short-circuiting, the new HRT is 2.5 hrs, so C/C0 = e-1.535*2.5 = 0.0215. Thus % removal is now = 97.85%

3. Hypochlorous acid (HOCl) dissociates into hypochlorite ions (OCl-) and hydrogen ions (H+)

during chlorination. As discussed in class, the equilibrium constant for this reaction is 10-7.5.

Plot the ratio of [OCl-] to [HOCl] as a function of pH ranging from 6 to 9. Given that HOCl is a

much more effective disinfectant than OCl-, which is more advantageous for disinfection:

slightly acidic or slightly basic water?

This is an aqueous acid-base equilibrium reaction:

HOCl ↔ H+ + OCL- KA=10-7.5

The equilibrium expression is:

KA = [H+][OCL-] / [HOCl] So [OCL-] / [HOCl] = 10-7.5 / [H+]

Since pH = -log[H+], you can plug in values for pH, and calculate the right hand side, which is the ratio you

are looking for. When you plot these numbers, you'll find that [OCL-] / [HOCl] = 1 at pH=pKA=7.5.

If pH < 7.5, the ratio is less than 1, and if pH > 7.5, the ratio is greater than 1. So lower pH (acidic) favors

HOCl, which is the better disinfectant.

4. A major natural disaster has occurred and you need clean drinking water. There is no

sunshine and your stove isn’t working, but you find a container of liquid bleach. How much

bleach (in teaspoons) should you add to a 5-gal bucket of water for a OCl- concentration of 2

mg/L in the water? (Notes: the bleach is 6% by weight NaOCl, the density of the bleach is ~1.1

g/mL, 1 teaspoon = 5 mL)

First find mg of OCl- needed: 2 mg/L * 5 gal * 3.785 L/gal = 37.85 mg OCl-

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Download Homework Assignment #7 with Solutions - Environmental Engineering | CE 321 and more Assignments Civil Engineering in PDF only on Docsity!

DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE

Homework Assignment 7, SOLUTIONS

CE 321, Fall 2008

1. At the Easton Water Treatment Plant on the day of our visit, four parallel rectangular

clarifiers, each 140 ft long, 17 ft wide, and 18 ft deep were being used to settle solids from an

approximately 8 MGD flow. Calculate the HRT and surface loading rate per tank.

HRT = Vol/Q = (140 ft17 ft18 ft)7.48 gal/ft 3 /2,000,000 gpd = 0.16 d = 3.85 hrs Surface loading rate = Q/As = 2,000,000 gpd1 day/(6024 min) / (17 ft140 ft) = 0.58 gpm/ft 2

2. A 1.5 million gallon clearwell operating as a plug-flow reactor (PFR) is used to disinfect a

flow of 8 MGD, resulting in a 99.9% bacteria removal. What would the % removal be if short-

circuiting of flow in the tank reduced the HRT by 2 hrs?

Plug Flow Reactor: C = C 0 e-kHRT

Ln(C/C0) = - k HRT k = - Ln(C/C0)/HRT HRT = V/Q = 1.5 MG/8 MGD = 0.1875 days = 4.5 hrs So k = - Ln(C/C0)/HRT = -Ln(0.001)/4.5 hrs = 1.535 hr -

With short-circuiting, the new HRT is 2.5 hrs, so C/C 0 = e-1.535*2.5^ = 0.0215. Thus % removal is now = 97.85%

3. Hypochlorous acid (HOCl) dissociates into hypochlorite ions (OCl

) and hydrogen ions (H

during chlorination. As discussed in class, the equilibrium constant for this reaction is 10

-7.

Plot the ratio of [OCl-] to [HOCl] as a function of pH ranging from 6 to 9. Given that HOCl is a

much more effective disinfectant than OCl-, which is more advantageous for disinfection:

slightly acidic or slightly basic water?

This is an aqueous acid-base equilibrium reaction: HOCl ↔ H+^ + OCL-^ K (^) A =10 -7.

The equilibrium expression is: K (^) A = [H+^ ][OCL-^ ] / [HOCl] So [OCL-^ ] / [HOCl] = 10 -7.5^ / [H+^ ]

Since pH = -log[H +^ ], you can plug in values for pH, and calculate the right hand side, which is the ratio you are looking for. When you plot these numbers, you'll find that [OCL-^ ] / [HOCl] = 1 at pH=pKA =7.5.

If pH < 7.5, the ratio is less than 1, and if pH > 7.5, the ratio is greater than 1. So lower pH (acidic) favors HOCl, which is the better disinfectant.

4. A major natural disaster has occurred and you need clean drinking water. There is no

sunshine and your stove isn’t working, but you find a container of liquid bleach. How much

bleach (in teaspoons) should you add to a 5-gal bucket of water for a OCl-^ concentration of 2

mg/L in the water? (Notes: the bleach is 6% by weight NaOCl, the density of the bleach is ~1.

g/mL, 1 teaspoon = 5 mL)

First find mg of OCl-^ needed: 2 mg/L * 5 gal * 3.785 L/gal = 37.85 mg OCl-

Now calculate how much NaOCl bleach you need to get 37.85 mg OCl-

NaOCl dissociates completely in water: NaOCl → Na+^ + OCl-

X teaspoons * 5 mL/teaspoon * 1.1 g bleach/mL * 6 g NaOCl/100 g bleach * (51.5 g OCl-/74.5 g NaOCl) *

1000mg/g = 37.85 mg OCl-

X = 0.165 teaspoons (less than a quarter teaspoon). Note: this is less than the EPA estimate on the handout I gave you which is based on a higher dose. Remember that effectiveness depends on both dose and contact time (CT)

5. A groundwater supply of 3 MGD has 120 mg/L of Ca

and 50 mg/L of Mg

. Using the

following two softening reactions, how much lime (Ca(OH) 2 ) in kg must be added per day to

precipitate the Ca+2^ and Mg +2? (you may assume that HCO 3 -^ is plentiful and will not limit the

precipitation reaction)

Ca^2 + 2 HCO 3 + Ca ( OH ) 2 → 2 CaCO 3 ( s )+ 2 H 2 O

Mg^2 + 2 HCO 3 + 2 Ca ( OH ) 2 → 2 CaCO 3 ( s ) + Mg ( OH ) 2 ( s )+ 2 H 2 O

Q*C gives the mass flow rate (e.g., mass/time)

Ca+2: 120 mg/L x (1 mol Ca/40,100 mg Ca) x (1 mol lime/1 mol Ca) x (74,100 mg Lime/mol lime) x ( kg/1,000,000 mg) x (3,000,000 gal/day) x (3.785 L/gal) = 2,518 kg/day (a lot)

Mg+2: 50 mg/L x (1 mol Mg/24,300 mg Mg) x (2 mol lime/1 mol Mg) x (74,100 mg Lime/mol lime) x ( kg/1,000,000 mg) x (3,000,000 gal/day) x (3.785 L/gal) = 3,463 kg/day (even more!)

Total = 5981 kg/day

6. Problem 10-10 [Design problem]

Just for context, Q = 0.8 m^3 /s = 18.25 MGD = more than twice the size of Easton WTP

SLR = 110 m^3 /d/sq m = Q/As = 0.8 m^3 /s86,400 s/day/(N10mX10m) N = 6.28 so use 7. But you could likely get by with 6 because 110 m^3 /d/sq m is a conservative SLR for filtration (see p436 of text and class notes giving typical SLRs of 2-4 gpm/sq ft which works out to 120-240 m 3 /d/sq m ) For N = 6 we would have an SLR of 115 m 3 /d/sq m, which is still at the low end of the range.

7. Problem 10-11 [Design problem]

SLR =300 m^3 /d/sq m = 0.8 m 3 /s86,400 s/day/(N10mX10m) N = 2.3 This is not many filters for a pretty large treatment plant. Text (see p436) suggests a minimum of 4 filters for large (Q > 0.5 m^3 /s) treatment plants, so use N = 4

8. Problem 10-16 (use the data in Table 10.4) [Design problem]

Using the data in Table 10-4, we need at CT value of 144 mg/L*min for the conditions in the problem (3-log removal = 99.9% removal). Since C = 1.20 mg/L, then T = HRT = 144/1.2 = 120 min or 2 hrs

Vol = HRT*Q = 120 min * 2.409 m 3 /s * 60 s/min = 17,345 m 3 (= 4.5 million gals – that’s a big tank but Q is very large – about 55 MGD)

Homework Assignment #7 with Solutions - Environmental Engineering | CE 321, Assignments of Civil Engineering

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Partial preview of the text

Download Homework Assignment #7 with Solutions - Environmental Engineering | CE 321 and more Assignments Civil Engineering in PDF only on Docsity!

DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE

Homework Assignment 7, SOLUTIONS

CE 321, Fall 2008

1. At the Easton Water Treatment Plant on the day of our visit, four parallel rectangular

clarifiers, each 140 ft long, 17 ft wide, and 18 ft deep were being used to settle solids from an

approximately 8 MGD flow. Calculate the HRT and surface loading rate per tank.

2. A 1.5 million gallon clearwell operating as a plug-flow reactor (PFR) is used to disinfect a

flow of 8 MGD, resulting in a 99.9% bacteria removal. What would the % removal be if short-

circuiting of flow in the tank reduced the HRT by 2 hrs?

3. Hypochlorous acid (HOCl) dissociates into hypochlorite ions (OCl

) and hydrogen ions (H

during chlorination. As discussed in class, the equilibrium constant for this reaction is 10

Plot the ratio of [OCl-] to [HOCl] as a function of pH ranging from 6 to 9. Given that HOCl is a

much more effective disinfectant than OCl-, which is more advantageous for disinfection:

slightly acidic or slightly basic water?

4. A major natural disaster has occurred and you need clean drinking water. There is no

sunshine and your stove isn’t working, but you find a container of liquid bleach. How much

bleach (in teaspoons) should you add to a 5-gal bucket of water for a OCl-^ concentration of 2

mg/L in the water? (Notes: the bleach is 6% by weight NaOCl, the density of the bleach is ~1.

g/mL, 1 teaspoon = 5 mL)

X teaspoons * 5 mL/teaspoon * 1.1 g bleach/mL * 6 g NaOCl/100 g bleach * (51.5 g OCl-/74.5 g NaOCl) *

5. A groundwater supply of 3 MGD has 120 mg/L of Ca

and 50 mg/L of Mg

. Using the

following two softening reactions, how much lime (Ca(OH) 2 ) in kg must be added per day to

precipitate the Ca+2^ and Mg +2? (you may assume that HCO 3 -^ is plentiful and will not limit the

precipitation reaction)

Ca^2 + 2 HCO 3 + Ca ( OH ) 2 → 2 CaCO 3 ( s )+ 2 H 2 O

Mg^2 + 2 HCO 3 + 2 Ca ( OH ) 2 → 2 CaCO 3 ( s ) + Mg ( OH ) 2 ( s )+ 2 H 2 O

6. Problem 10-10 [Design problem]

7. Problem 10-11 [Design problem]

8. Problem 10-16 (use the data in Table 10.4) [Design problem]