Assignment 6 Solutions for Environmental Engineering | CE 321 | Assignments Civil Engineering

DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE

Homework Assignment 6 SOLUTIONS

CE 321, Fall 2008

1. [Design Problem] A rectangular clarifier in a wastewater plant is to be designed to settle 2000

m3/day with a surface loading rate of 32 m3/day per m2. The tank is to be 2.4 m deep and 4 m

wide. How long should it be and what detention time would it have?

Surface loading rate = Q/As

So As = Q/SLR = 2000 m3/d / 32 m/d = 62.5 m2

As = L*W, so L = As/W = 62.5 m2/4 m = 15.6 m long

HRT = Vol/Q = 62.5 m2*2.4 m/2000 m3/d = 0.075 d = 1.8 hrs

2. [Design Problem] A settling tank for a 2 MGD wastewater plant is to be designed to have a

surface loading rate of 800 gal/day per ft2. The tank must have a minimum retention time of 2 hrs

and must be at least 11 ft deep to allow proper settling of waste sludge. If the tank is circular,

what should its diameter and depth be?

As = Q/SLR = 2,000,000 gal/d / 800 gal/d-ft2 = 2500 ft2

Area = 1/4πd2 = 2500 ft2, so d = 56.4 ft

HRT = Vol/Q = (2500 ft2*7.48 gal/ft3*depth) / 2,000,000 gal/d must be ≥ 0.0833 days

Solve for depth, and you find depth ≥ 8.9 ft.

But for settling, we must have a depth of at least 11 ft.

So we set the design depth at 11 ft, which will increase the HRT to:

HRT = Vol/Q = (2500 ft2*7.48 gal/ft3*11 ft) / 2,000,000 gal/d = 0.103 d = 2.5 hr

3. Water containing 125 mg/L of suspended solids flows into a clarifier (settling tank) at 18,500

m3/day. The effluent from the top of the tank has no suspended solids. Solids are continuously

pumped from the bottom of the tank at a concentration of 8000 mg/L. What are the flow rates out

the top and bottom of the tank, assuming no accumulation of solids in the tank? What is the mass

of solids being removed from the tank per hour?

Apply the mass balance equation to the solids:

QinCin = QeffCeff + QbotCbot

Since effluent has no suspended solids, Ceff=0, and thus

QinCin = QbotCbot

Qbot=QinCin/Cbot = 18,500*125/8000 = 289.1 m3/day

Now apply a mass balance to the flow:

Qin = Qeff + Qbot

Qeff = Qin - Qbot = 18,500 - 289 = 18,211 m3/day

Solids flux = QinCin = QbotCbot =

289 m3/day*8000 mg/L*1000 L/m3*1 kg/1,000,000 mg*1day/24 hrs = 96.3 kg/day

4. A wastewater contains a pollutant with an initial concentration of 1200 mg/L, which will be

treated in a batch reactor. The reaction is assumed to be first order, with a rate constant of -2.5

/day. Plot the concentration in the reactor for the first 2 days. How long will it take for

reductions of 75%, 90%, and 95% of the pollutant?

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DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE

Homework Assignment 6 SOLUTIONS

CE 321, Fall 2008

1. [Design Problem] A rectangular clarifier in a wastewater plant is to be designed to settle 2000

m^3 /day with a surface loading rate of 32 m^3 /day per m^2. The tank is to be 2.4 m deep and 4 m

wide. How long should it be and what detention time would it have?

Surface loading rate = Q/ASo A s A (^) s = L*W, so L = As^ = Q/SLR = 2000 ms/W = 62.5 m^3 /d / 32 m/d = 62.5 m^2 /4 m = 15.6 m long^2 HRT = Vol/Q = 62.5 m 2 *2.4 m/2000 m 3 /d = 0.075 d = 1.8 hrs

2. [Design Problem] A settling tank for a 2 MGD wastewater plant is to be designed to have a

surface loading rate of 800 gal/day per ft 2. The tank must have a minimum retention time of 2 hrs

and must be at least 11 ft deep to allow proper settling of waste sludge. If the tank is circular,

what should its diameter and depth be?

A (^) s = Q/SLR = 2,000,000 gal/d / 800 gal/d-ft 2 = 2500 ft 2 Area = 1/4πd 2 = 2500 ft 2 , so d = 56.4 ft HRT = Vol/Q = (2500 ftSolve for depth, and you find depth 2 *7.48 gal/ft 3 ≥ *depth) / 2,000,000 gal/d must be 8.9 ft. ≥ 0.0833 days

But for settling, we must have a depth of at least 11 ft. So we set the design depth at 11 ft, which will increase the HRT to:HRT = Vol/Q = (2500 ft 2 *7.48 gal/ft 3 *11 ft) / 2,000,000 gal/d = 0.103 d = 2.5 hr

3. Water containing 125 mg/L of suspended solids flows into a clarifier (settling tank) at 18,

m^3 /day. The effluent from the top of the tank has no suspended solids. Solids are continuously

pumped from the bottom of the tank at a concentration of 8000 mg/L. What are the flow rates out

the top and bottom of the tank, assuming no accumulation of solids in the tank? What is the mass

of solids being removed from the tank per hour?

Apply the mass balance equation to the solids: QSince effluent has no suspended solids, C (^) inC (^) in = QeffCeff + Qbot C (^) bot Q (^) inC (^) in = QbotCbot eff=0, and thus Q (^) bot =Q (^) inC (^) in/Cbot = 18,500125/8000 = 289.1 m^3 /day Now apply a mass balance to the flow: QQ (^) in = Qeff + Q (^) bot eff = Qin - Q^ bot = 18,500 - 289 = 18,211 m^3 /day Solids flux = QinC (^) in = QbotCbot = 289 m^3 /day8000 mg/L*1000 L/m^3 1 kg/1,000,000 mg1day/24 hrs = 96.3 kg/day

4. A wastewater contains a pollutant with an initial concentration of 1200 mg/L, which will be

treated in a batch reactor. The reaction is assumed to be first order, with a rate constant of -2.

/day. Plot the concentration in the reactor for the first 2 days. How long will it take for

reductions of 75%, 90%, and 95% of the pollutant?

Batch reaction, non-steady state solution: C = C Plot C = 1200 mg/L e 0 e-kt -2.5t (^) gives a standard exponential decay curve (plot not shown)

Reduction of 75% means the pollutant is reduced to 25% of its original concentration, so C/CSolving the equation for t gives: 0 = 0.

t = -1/kln(C/Ct = -1/kln(C/C (^) 0) = -1/2.5*ln(0.25) = 0.55 days = 13.2 hr

= -1/2.5ln(0.10) = 0.92 days = 22.1 hr t = -1/kln(C/C (^) 0) = -1/2.5*ln(0.05) = 1.2 days = 28.8 hr

5. [Design Problem] For the wastewater of the previous problem, determine the HRT and

required volume for a CSTR for 90% removal and a flow rate of 0.05 m^3 /sec.

Use design equation for a CSTR, based on steady-state conditions:HRT = 1/k(C in/C^ out - 1) = 1/kfr/(1 - fr) = 1/2.5 day^ -1^ (0.9/0.1) = 3.6 days Vol = HRTQ = 3.6 day0.05 m 3 /sec86,400 sec/day = 15,552 m 3

6. For a reaction order of 0, that is, rxn rate = -kC^0 = -k , what is the difference in required

volumes for a CSTR and PFR? (hint: assume steady-state, and find the expressions for HRT)

Need to find design equations for CSTR and PFR, based on steady-state conditions and 0-order rxn. For a PFR, which is equivalent to a batch reactor with t=HRT, we have: VdC/dt = -kV, so dC = -kdt Integrating the left side from CC in to C (^) out , and the right side from 0 to HRT, we get: Solving for HRT:out^ - C^ in^ = -kHRT HRT = (C (^) in - C (^) out )/k For a CSTR, we have: 0 = QC0 = (C in - QCout - kV = Q(C (^) in - C (^) out ) - kV Solving for HRT:in^ - C^ out^ ) - kV/Q = (C^ in^ - C^ out^ ) - kHRT HRT = (C (^) in - C (^) out )/k Since the HRTs are the same for PFR and CSTR, the required volumes will also be the same. This result is because the reaction rate is a constant, independent of concentration. For any reaction orderthe PFR volume will be less than CSTR volume. greater than 0 ,

Assignment 6 Solutions for Environmental Engineering | CE 321, Assignments of Civil Engineering

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Download Assignment 6 Solutions for Environmental Engineering | CE 321 and more Assignments Civil Engineering in PDF only on Docsity!

DEPARTMENT OF CIVIL & ENVIRONMENTAL ENGINEERING; LAFAYETTE COLLEGE

Homework Assignment 6 SOLUTIONS

CE 321, Fall 2008

1. [Design Problem] A rectangular clarifier in a wastewater plant is to be designed to settle 2000

m^3 /day with a surface loading rate of 32 m^3 /day per m^2. The tank is to be 2.4 m deep and 4 m

wide. How long should it be and what detention time would it have?

2. [Design Problem] A settling tank for a 2 MGD wastewater plant is to be designed to have a

surface loading rate of 800 gal/day per ft 2. The tank must have a minimum retention time of 2 hrs

and must be at least 11 ft deep to allow proper settling of waste sludge. If the tank is circular,

what should its diameter and depth be?

3. Water containing 125 mg/L of suspended solids flows into a clarifier (settling tank) at 18,

m^3 /day. The effluent from the top of the tank has no suspended solids. Solids are continuously

pumped from the bottom of the tank at a concentration of 8000 mg/L. What are the flow rates out

the top and bottom of the tank, assuming no accumulation of solids in the tank? What is the mass

of solids being removed from the tank per hour?

4. A wastewater contains a pollutant with an initial concentration of 1200 mg/L, which will be

treated in a batch reactor. The reaction is assumed to be first order, with a rate constant of -2.

/day. Plot the concentration in the reactor for the first 2 days. How long will it take for

reductions of 75%, 90%, and 95% of the pollutant?

5. [Design Problem] For the wastewater of the previous problem, determine the HRT and

required volume for a CSTR for 90% removal and a flow rate of 0.05 m^3 /sec.

6. For a reaction order of 0, that is, rxn rate = -kC^0 = -k , what is the difference in required

volumes for a CSTR and PFR? (hint: assume steady-state, and find the expressions for HRT)

7. [Design Problem] A wastewater flow of 1 MGD will be treated with 150,000-gal PFR

reactors. The reaction is first order, with a rate constant of 2.5 /day. How many PFRs connected

in series (one after the next) are necessary for reductions of 75%, 90%, and 95% of the pollutant?

How many must be connected in parallel (the total flow is split evenly between them), for the

same % reductions?