WASTEWATER SECTION August Instructor: J.B. Jones, P.E., PhD.

Transcription

1 WASTEWATER SECTION August 01 Instructor: J.B. Jones, P.E., PhD.

2 What's the format of the Civil PE exam? The civil PE exam is 8 hours long, divided into equal sessions, morning and afternoon. All questions are multiple-choice. This exam is structured in a "breadth and depth" format. In the morning session, all examinees work the same "breadth" exam, which consists of 40 questions drawn from all five areas of civil engineering listed below. Examinees must answer all 40 questions. In the afternoon, examinees choose to work one of five "depth" exam modules: Water Resources & Environmental, Geotechnical, Structural, Transportation, or Construction. Each depth module consists of 40 questions that test knowledge in the areas specified. Examinees must answer all 40 questions in the module they select. What type of the questions will there be? The questions are all multiple-choice, with four answer choices each. Nearly all questions are unique-that is, one problem statement followed by one question. There may be a few multi-part questions, where one problem statement is followed by or questions, but in these cases the answers to the questions will be independent from each other (i.e., the answers do not "cascade").

3 What does the Environmental & Water Resources Section cover? The exam topics are described by NCEES as follows: I. Hydraulics Closed Conduit (Session ) 15% II. Hydraulics Open Channel (Session ) 15% III. Hydraulics Hydrology (Session ) 15% IV. Groundwater & Well Fields (Session) 7.5% V. Wastewater Treatment (Session 1) 15% VI. Water Quality (Session 1) 15% VII. Water Treatment (Session ) 15%

4 TABLE OF CONTENTS WWTP Problems Typical Activated Sludge Plant #1 Wastewater Basin Sizing # Wastewater BOD Planning # Treatment Efficiency #4 Grit Chamber Design #5 Aeration Tank Detention Time #6 Trickling Filter Design #7 Trickling Filter Design # #8 Rotating Biological Contactor #9 Anaerobic Lagoon #10 Anaerobic Digester #11 Anaerobic Digester # #1 Sludge Stabilization #1- Sludge Solids Reduction #14 Wastewater Sludge Production Rate #15 Sludge Volume Ratios #16 Disinfection of Wastewater #17 - Disinfection of Wastewater # River Mixing and BOD Calculation #1 BOD Determination # Wastewater Effluent Mixing with Receiving Stream # Wastewater Effluent Mixing with Receiving Stream # #4 Dissolved Oxygen Sag in River System

5 Non Quantitative Short Answer 1 Viruses and Bacteria - Coliforms - Odor Constituents 4 Wastewater Loading Factor 5 Nitrogenous BOD 6 BOD Removal Rates 7 Sequential Batch Reactors 8 Sludge Thickening 9 Sewer Pipe Flow 10 Wastewater Peak Flow 11- Wastewater Generation 1- Potable water Usage and Wastewater Generation

6 CHAPTER TOPICS Chapter 7: Basic Microbiology Chapter 8: Wastewater Quantity and Quality Section 0: BOD Section : River Mixing Chapter 9: Wastewater Treatment: Equipment and Processes Stabilization Ponds Facultative Ponds Lagoons Sedimentation Trickling Filters RBC s Chapter 0: Activated Sludge & Sludge Processing Activated Sludge Process Recycle Rates Sludge Thickening Sludge Dewatering

7 WWTP Problems Loading, #/day Flow, MGD 8.4 Dose, mg/l lbs mg lb Chemical Feed Rate 4 day MG Flow, MGD Dose, 8. l mg l Dose Rate mg l Chemical lbs Feed Rate, day lb MG Flow, MGD 8.4 mg l

8 (1) Flow In (Q IN ) = Flow Out (Q OUT ) () Detention time (T) = Volume (V) / Flow (Q) () Flow (Q) = Volume (V) / Time (T) (4) Area Required = Flow/Loading Rate Write out your units in practice because Lecture 1 and are primarily an exercise in consistent unit formatting. The front insert on the PE manual is a powerful short cut for conversions.

9 Basics of Wastewater Strength Calculations Discussion (Refer to Section 8-0, page 8-8) The biochemical oxygen demand (BOD) is an empirical test in which laboratory procedures are used to determine the relative oxygen requirements of wastewaters and polluted waters. The BOD test is an empirical method, which involves the use of microorganisms. It is widely used for: - Measuring waste strength - Evaluating organic removal efficiencies - Assessing stream assimilative capacity The amount of oxygen consumed in the biological decomposition of waste is BOD (Biochemical Oxygen Demand). The BOD test measures: (1) Molecular oxygen consumed during a specific incubation period for the biochemical degradation of organic matter (carbonaceous BOD) () Oxygen used to oxidize inorganic matter such as sulfide and ferrous iron () Reduced forms of nitrogen (nitrogenous BOD) with an inhibitor BOD U BOD exerted t, days Look at Table 8. Page 8-6 Strong & Weak Domestic Sewages

10 INFLUENT BAR SCREEN GRIT TO LANDFILL PRIMARY SEDIMENTATION AIR AERATION TANK FINAL CLARIFIER Cl MIXED LIQUOR CHLORINATION (DECHLORINATION) PRIMARY SLUDGE & WAS GRIT CHAMBER WAS SLUDGE THICKENER PRIMARY SLUDGE SLUDGE DIGESTOR RAS RAS EFFLUENT TO RECEIVING STREAM SLUDGE DRYING BEDS DRY SLUDGE FOR DISPOSAL OR LAND APPLICATION DRAINAGE DRAINAGE TYPICAL ACTIVATED SLUDGE PLANT Figure 0.1 Typical Activated Sludge Plant (PE Manual)

11 1 Wastewater Basin Sizing (Slightly Modified) (a) What is the hydraulic residence hydraulic detention time for a rectangular tank with dimensions 4 m (W) by 15 m (L) by m (D) receiving a flow of 1000 m /day? (b) Tracer studies indicate the hydraulic efficiency of the tank is 85%. What is the corrected detention time? Stagnant Zone (a) (b) (c) (d) 1. h. h.8 h 4.6 h E fractional efficiency Q flow rate m /d t theoretical hydraulic detention time d t A actual hydraulic detention time d V volume m (a) 4 m15 m m 4 hr V t d.88 h, say hr Eqn 9.5 Q 1000 m d (b) t A t E.88 h h

12 - Wastewater BOD Planning Wastewater treatment guidelines for a planned community require that wastewater treatment capacity be provided based on four population equivalents (PE) per home. The community will eventually include 000 homes. What is the approximate biochemical oxygen demand (BOD) loading expected at the wastewater plant from the community? (a) (b) (c) (d) 1000 lbm BOD/day 1600 lbm BOD/day 00 lbm BOD/day 5,000 lbm BOD/day Assume that the typical person generates 0. lbm BOD/day-person (Refer to Page 8-, PE Manual) PE BOD loading rate per capita BOD generation rate The BOD loading rate is lbm BOD PE 0. 4 day person hom e The answer is (b). 000 hom es 1600 lbm BOD day

13 Treatment Efficiency The BOD of a wastewater entering the primary clarifier is 10 mg/l. If the BOD removal efficiency of the primary clarifier is 5% and the BOD removal efficiency of a single stage trickling filter is 80%, what is the effluent BOD? What is the overall efficiency? Compute the BOD of the wastewater leaving the primary clarifier (and entering the trickling filters). mg Pr imary effluent BOD l Compute the BOD leaving the trickling filter. mg Secondary effluent BOD l Compute overall efficiency. mg l mg l Overall efficiency % Think of it as a box. You do not need to know what is happening inside the box Just keep the treatment efficiencies in mind. 10 mg/l 7 mg/l Primary Clarifier 17 mg/l Trickling Filter

14 4 - Grit Chamber Design What is the required width of the rectangular horizontal flow grit chamber where the following conditions apply? Flow rate.0 x 10 6 gal/day Depth 4 ft Mean particle diameter 0.5 mm Grit specific gravity.65 Camp constant 0.05 Darcy friction factor 0.0 (a) (b) (c) (d) 0.6 feet 1.6 feet 8.8 feet 14 feet Refer to page 9-6, PE Manual d P mean particle diameter mm f Darcy friction factor g gravitational constant 9.81 m/s K Camp constant SG P grit specific gravity V horizontal velocity m/s 8 k g d P SGP 1 v Eqn 9- f m 1 m mm.65 1 sec 1000 mm v 0. m 0.0 s

15 A channel cross-section ft D depth ft Q flow rate gal/day W width ft Q = Cross Sectional Area x Velocity = width * depth * vel 6 gal ft.0 x 10 Q day 0.14 gal w d vel m ft sec 4 ft ,400 s m day ft The answer is (b).

16 5 - Aeration Tank Detention Time (Slightly Modified) (a) A flow of 6 MGD enters (6) aeration tanks 150 ft x 15 ft x 15 ft. Determine the detention time in hours. (b) A sewage treatment plant ran blowers for 4 hours at 5000 cfm. A BOD removal of 90% was obtained with a 5 MGD flow containing 10,000 lb BOD per day. How many cubic feet of air was required for each pound of BOD? (This is not a practical problem, but it reinforces units) Part (a) Tank Volume = 6 (150 ft) (15 ft) (15 ft) = 0,500 ft 6 gal 6 x 10 day Q IN 80,19 ft gal 7.48 ft day 0,500 ft hr Detention Time x hr ft d 80,19 d Part (b) lbs of BOD removed = 0.90 (10,000) = 9,000 lb/d Air Used ft min hr x 10 min hr d 6 ft d Therefore, ft lb BOD 6 7. x 10 ft 9,000 lb ft 800 lb BOD

17 6 - Trickling Filter Design Domestic sewage of 1.0 MGD has been passed through a primary clarifier. When it is applied to a trickling filter, the wastewater has a BOD concentration of 195 mg/l. Determine the acre-ft of filter required to effect 8% removal of BOD. Assume that the wastewater is not be recirculated. Using the NRC formula (Page 9-10 in PE Manual) 100 E Eqn 9-14 W V F where E W V F R = percentage of BOD removal at 0 0 C = BOD load applied, lbs/day = volume of filter, acre-ft = recirculation factor Acre-ft is one of those old 1 R sanitary engineering terms 1 0.1R = flow recirculate/raw wastewater flow For this problem: E = 8 lb mg W MGD 8.4 MG 166 L mg L V =? lb day F = 1.0 Then, V 1.0 Volume =.48 acre-ft

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19 7 Trickling Filter Design # A community of 15,000 has an average wastewater flow of 10 gpcd. The 5-day BOD is 195 mg/l at 0 0 C. The suspended solids content is 0 mg/l. The effluent BOD concentration is designed for 0 mg/l. BOD removal is 0% through the primary treatment stage. If dual trickling filters with recirculation are employed for secondary treatment, what is the required filter volume (acreft) Using the NRC formula (Page 9-10 in PE Manual) 100 E Eqn 9-14 W V F where E W V F R = percentage of BOD removal at 0 0 C = BOD load applied, lbs/day = volume of filter, acre-ft = recirculation factor 1 R 1 0.1R = flow recirculate/raw wastewater flow 6 15,000 people10 gpcd 1.8 x 10 gpd Q mg mg BOD applied to filter l l 17 0 E 1 78%, convert the filter load rate from mg/l to lbs/day 17 mg lb MG W MGD 057 l mg l lb day 1 R F 1.65 when R R 100 E where W V F V V 1. 1 ac ft

20 8 - Rotating Biological Contactor (Section 9-4) What is most nearly the required media total surface area for a rotating biological contactor (RBC) process selected to treat the following wastewater? flow rate influent BOD effluent BOD Hydraulic loading rate 50,000 gal/day 10 mg/l 0 mg/l gal/ft day (a) 10,000 ft (b) 150,000 ft (c) 15,000 ft (d) 60,000 ft See Page 9-11, Section 9-4. This will be a common design concept that Flow = Area x Loading Rate A h media surface area based on hydraulic loading ft HLR hydraulic loading rate gal/day-ft Q influent flow rate gal/day A h Q HLR gal 50,000 day gal.0 ft day 15,000 ft, The answer is (c).

21 9 - Anaerobic Lagoon (not in text) An anaerobic lagoon will pretreat a slaughterhouse wastewater that will be discharged to an existing facultative lagoon. Influent flow is.6 x 10 5 gal/day with a total BOD of 14,000 mg/l. The average waste temperature is 1 0 C, and the loading rate is 15 lbm BOD/10 ft -day. Site conditions limit the lagoon depth to 10 ft. What is the total required surface area of the anaerobic lagoon? (a).9 ac (b) 4.6 ac (c) 5.4 ac (d) 5.8 ac These two equations will be used repeatedly. Step 1: Determine the mass flow rate C BOD concentration mg/l Q volumetric flow rate gal/day m mass loading rate lbm/day lbs mg lb Chemical Feed Rate 4 day MG Flow, MGD Dose, 8. l mg l 5 gal 1 MG mg BOD lb MG m Q * C.6 x 10 14,000 x day 10 gal L mg L m 0, 65 lbm BOD day Step : Determine the required surface area. A surface area ac D liquid depth ft OLR organic loading rate lbm/10 ft -day Volume Loading Rate Mass Loading Area Depth Loading Rate Mass Loading

22 lbm BOD 1 ac 0,65 m day 4,560 ft A 4. 6 ac OLRD lbm BOD ft 10 ft day The answer is (b).

23 10 Anaerobic Digester (Section 0-19) A completely mixed anaerobic digester treats sludge containing primary and waste activated sludge. The sludge flow rate is 000 cubic feet/day and the sludge contains 5400 lb of volatile solids. Using a design loading rate of 0.16 lb- VSS/day/ft, determine the digestion period. SOLUTION Calculate the minimum volume based on volatile solids loading: Min Volume Storage Daily Plant Loading Allowed Unit Load Rate lb VSS 5400 day lb VSS 0.16 ft * day,750 ft Calculate the digestion period When a loading rate is supplied, it can usually be used to determine volume. V t D Eqn 9.5 Q,750 ft Digestion Period 17 days ft 000 day

24 11 Anaerobic Digester # (Not covered in Class) Estimate the sludge production rate for an anaerobic digester with the following characteristics: Total solids in digester influent = 700 mg/l Influent BOD U = 6600 lb/day Effluent BOD U = 00 lb/day Y = 0.5 lb cells/lb BOD k D = 0.0 day -1 mean cell residence time = 0 days Refer to Page 0-8 (PE Manual) P X Y 1 k MAX Q S0 D SRT S E Eqn 0.4 P X = sludge production (dry mass/time) Y MAX = max yield = 0.5 lb VSS/lb BOD 5 K D = endogenous decay coefficient = 0.0 d -1 SRT = sludge age = 0 d Q = wastewater flow rate S 0 = BOD 5 influent to activated sludge = BOD 5 in the plant effluent S E P X Y 1 k MAX Q S0 D SRT S E Remember that Loading (#/day) = Q * S lb VSS 0.5 lb BOD lb lb P X lb 1 d d d 0 d day

25 1 - Sludge Stabilization A waste biological sludge is dewatered to 5% solids and then stabilized with a lime dose rate of 0 g Ca (OH) /kg dry solids. The plant wastes 18,000 gallons of sludge daily at 10% solids. The locally available lime contains 4% inerts. What is most nearly the monthly mass of lime required to stabilize the sludge? (a) (b) (c) (d) 45,000 kg/month 58,000 kg/month 86,000 kg/month 190,000 kg/month f solids % m dry solids mass flow rate kg/day V wet sludge flow rate m /d sludge density kg/m Assume that for sludge at 10% solids, the sludge density is equal to that of water (1000 kg/m ). See Page 0-1, Eqn 0.46: gal kg m m V f 18, x kg d day m gal (This is the amount of sludge produced if all water were removed, i.e. only the dry portion remained) The lime requirement is Monthly Lime Re quirement days Daily Sludge Pr oduced mo unit weight dry solids Lime Dosage Rate 681 kg d d Ca OH 0 g month 0 kg dried solids g kg 86,058 kg month The answer is (c). Note that the 5% solids dewatering number was not used in the calculation.

26 1 - Sludge Solids Reduction A wastewater treatment plant process wastes sludge at 50,000 gal/day. The wasted sludge contains 1.% solids. What volume reduction can be realized by thickening & dewatering the sludge at 4% solids? (a) (b) (c) (d) 500 gal/day 600 gal/day 9,000 gal/day 48,000 gal/day Dry mass flow rate = (% solids) x (wet sludge density) x (volumetric flow rate) m f V f 1 1 V f V 1 1, drops out f V At 1.% solids, V 1 = 50,000 gal/day At 4 % solids, gal ,000 day V 500 gal 0.4 day Volume Reduction = V 1 V = 50, = 47,500 gal/day The answer is (d).

27 14 Wastewater Sludge Production Rate At a wastewater plant, the design flow is 6.0 MGD with an influent suspended solids concentration of 85 mg/l. At the primary clarifier, 60% of suspended solids are removed. Sludge production is 5% solids by weight and has a specific gravity of 1.0. Determine the sludge production rate in gallons/day. SOLUTION For future problems, it is reasonable to assume that SG = 1 mg lb MG Sludge removed 6.0 MGD l mg l lb day lb 8557 Weight day Volume 0, 100 Density lb 0.05 solids1.0 S G8.4 gal gal day

28 15 - Sludge Volume Ratios What is most nearly the sludge volume index for a mixed liquor suspended solids (MLSS) suspension at an initial concentration of 00 mg/l that settles to the 85 ml mark in a 1 L graduated cylinder after 0 minutes? (a) (b) (c) (d) 180 ml/g 70 ml/g 700 ml/g 6700 ml/g Refer to Page 0-5, PE Manual. The sludge volume index (SVI) is a measure of the sludge s settleability. SVI is a parameter for operation considerations to ensure that the sludge settles during clarification and does not carry over into a final treatment process. Mixed liquor suspended solids (MLSS) is the bacteria and other suspended material in the mixed liquor, which is measured as mg/l. Parameters MLSS mixed liquor suspended solids mg/l SV Sludge volume ml/l SVI Sludge volume ml/g ml mg SV L g SVI 175mL g Eqn 0.8 MLSS mg 00 L The answer is (a). Part (b): Calculate the TSS. 1000mg 1000 ml g L TSS mg L 5700 SVI ml g mg L

29 16 - Disinfection of Wastewater (Not covered in class) Disinfection of a wastewater using aqueous chlorine at a ph of 8.5 and a temperature of 1 0 C requires minutes to effect the desired percentage kill. How much time is required if the wastewater temperature is 17 0 C? (a) (b) (c) (d) 6.4 min 0 min 7 min min Refer to Metcalf & Eddy, page 0. The effect of temperature on the rate of kill can be represented by a form of the van t Hoff-Arrhenius relationship. Increasing the temperature results in a more rapid kill. In terms of the time, t required to effect a given percentage kill, the relationship is: ln t t 1 E T T R T T 1 where 1 t 1, t = time for given percentage kill at temperatures T 1 and T, 0 K respectively E = activation energy, J/mol (cal/mol) R = gas constant, 8.14 J/mol* 0 K (1.99 cal/ 0 K*mol) Source: Wastewater Engineering, rd edition, McGraw-Hill

30 Because decreasing temperature results in a longer reaction time, take 1 0 C as the reference temperature (T 1 ) and 17 0 C as the temperature of interest (T ) C 7 K C 7 K T 94 1 T 90 cal K 94 K t1 E T T1 ln mol t R T1 T cal K 90 K mol * K t1 t t ln 0.15 e t 1 7 min t t e 0.15 The answer is (c).

31 17 - Disinfection of a Wastestream # (Not covered in class) Disinfection is to be added to a sewage treatment plant with an average daily flow of 1 MGD, and an hourly peak factor of.5. The average coliform count, N 0 = 10,000 org/100 ml must be reduced to 00 org/100ml. Codes require a 15-minute contact time for peak hourly flow and 0-minute contact time for average daily flow, and a maximum chlorine dose rate of 15 mg/l. N N 0 & Eddy, page 8) T The equation 1 0. C T t describes the chlorination process. (Metcalf N T = number of coliforms organisms at time t N 0 = number of coliform organisms at time t 0 C t = total chlorine residual at time, t, mg/l T = residence time, minutes (a) Determine the total chlorine residual (mg/l) after 15 minutes contact time under peak conditions. (b) Determine the chlorine dosage rate in lb/day. Influent Effluent Cl feed

32 Part (a) N N T C t T N N 1 T, rearranges to 1 0. C t 0 T N N T C T t, rearranges to C T 1 N T 1 N 0 0. t ,000 C T mg L Part (b) If Q= 1 MGD and Cl dose is 15 mg/l (as a code requirement in the problem statement) lb lb MG Q * c day mg L mgd mg L lb day

33 River Mixing & BOD Calculations

34 Basics of Wastewater Strength Calculations Description of BOD Test Section 8-0) The method consists of placing a sample in a full, airtight bottle and incubating the bottle under specified conditions for a specific time. Dissolved oxygen (DO) is measured initially and after incubation. The BOD is computed from the difference between initial and final DO. Most wastewaters contain more oxygen demanding materials than the amount of DO available in air-saturated water. Therefore, it is necessary to dilute the sample before incubation to bring the oxygen demand and supply into appropriate balance. When dilution water is not seeded: BOD I F 5 Eqn 8.0 VSAMPLE V DO SAMPLE DO V DILUTION Industrial wastewater may lack sufficient microorganisms to oxidize the wastes. For such wastes, the dilution water is seeded with a population of microorganisms. BOD 5 DO I DOF DO VSAMPLE V V SAMPLE * I DILUTION DO * F Eqn 8.4 Where x = volume of seed in diluted sample/volume of seed in seed control

35 1 - BOD Determination Example For a BOD test, 75 ml of river water sample is used in the 00 ml of BOD bottles without seeding with three duplications. The initial DO in three BOD bottles read 8.86, 8.88, and 8.8 mg/l respectively. The DO levels after 5 days at 0 0 C incubation are 5.49, 5.65, and 5.5 mg/l, respectively. Find the 5-day BOD (BOD 5 ) for the river water. Determine average DO uptake and plug it into the BOD equation D 1 D x =[( ) + ( ) + ( )]/ =.0 mg/l BOD I F 5 Eqn 8.0 VSAMPLE V DO SAMPLE DO V DILUTION.0 mg L 1. mg L As a side reference, look at Page 8-6, Table 8.4 for various wastewater strength characteristics.

36 BOD Calculations as Related to Time and Temperature First Stage BOD is usually defined by: Where BOD T BOD U 1 exp K t Eqn 8.1 D BOD t = BOD in mg/l exerted after t days BOD u = Ultimate first stage BOD K D = coefficient of deoxygenation (reaction rate coefficient) t = time of days BOD U BOD exerted t, days When a waste is completely oxidized (i.e.treated or assimilated) so that it no longer creates an oxygen demand, the water has reached Ultimate BOD. Refer to this for Problem # in the next section.

37 River Mixing When wastewater is discharged to a receiving stream, the concentrations of various parameters of the mixture may be determined from the following: C C Q S S W W Eqn 8.5 Q S C Q W Q Q S = rate of flow in stream Q W = rate of flow of wastewater C S = concentration of parameter in stream above discharge point C W = concentration of parameter in waste flow before discharge This equation applies to oxygen content, solids, etc. NOT COVERED IN THIS CLASS The Streeter Phelps equation is generally used to define the oxygen deficit over time as follows: D T K D BODU e K K R K DT D e K R t Eqn 8.6 D T = dissolved oxygen deficit (mg/l) at time t (days) BOD U = ultimate first-stage BOD K D = deoxygenation coefficient (K 1 ) K R = reoxygenation coefficient (K ) D 0 = initial oxygen deficit (mg/l) Based on the above equation, the critical oxygen deficit is: D C K BOD K T D U exp D M Eqn 8.9 K R And T M is the time at which the minimum dissolved oxygen of the mixture occurs, and is found as follows: T M K R 1 K D K ln K R D D0 K R K 1 K D BODU D Eqn 8.8

38 - Wastewater Effluent Mixing with Receiving Stream A wastewater treatment plant (WWTP) discharges raw sewage during periods of high rainfall. Typical discharge flows are 15 x 10 6 gal/day with dissolved oxygen concentrations of 1.0 mg/l. During these periods, the river flows at 000 ft /sec with a dissolved oxygen concentration of 8.10 mg/l. What is most nearly the DO concentration in the river once complete mixing at the WWTP effluent has occurred? (a) (b) (c) (d) 4.65 mg/l 8.0 mg/l 8.05 mg/l 8.18 mg/l When wastewater is discharged to a receiving stream, the concentrations of various parameters of the mixture may be determined from the following: C C Q S S W W Eqn 8.5 Q S C Q W Q Q S = rate of flow in stream Q W = rate of flow of wastewater C S = concentration of parameter in stream above discharge point C W = concentration of parameter in waste flow before discharge Step 1: Convert WWTP flow to compatible units 6 gal ft day ft 15 x 10 day 7.48 gal 86,400 sec sec Step : Calculate the DO concentration C C S Q Q S S C W Q W Q W mg ft mg ft L sec L sec ft ft 000 sec sec 8.0 mg L The answer is (b).

39 - Wastewater Effluent Mixing with Receiving Stream # (Difficult) A secondary treated effluent from a 4.0 MGD wastewater treatment plant is discharged into a receiving stream. The wastewater has a BOD 5 = 0 mg/l. The receiving stream upstream from the point of wastewater discharge has a flow of 18 cfs and a BOD 5 = 4.0 mg/l. The BOD reaction rate constant is estimated at 0. d -1 (base e at 0 0 C). Determine the ultimate BOD U downstream of the receiving stream. Q STREAM = 18 cfs BOD 5 = 4 mg/l 1 cfs = 449 gpm = 646,560 gpd 18 cfs = 11.6 MGD Step 1: Determine BOD 5 after mixing Q = 4.0 MGD, BOD 5 = 0 mg/l From earlier notes, C C Q S S W W can be rewritten: Q S C Q W Q BOD 5 BOD WW Q WW Q WW BOD Q STREAM STREAM Q STREAM 0mg l 4 MGD 4mg l 11.6 MGD 15.6 MGD BOD mg L Step : BOD BOD T U 1 exp K t Eqn 8.1 D BOD5 8.1 BODU 1 K 1 exp 1 exp The answer is 1 mg/l mg L mg L

40 4 Dissolved Oxygen Sag in a River System (Not covered in class) Assume the following characteristics of the wastewater/river mixture at the point of discharge: Dissolved oxygen concentration 6 mg/l Ultimate BOD U 10 mg/l Temperature 0 0 C Reaeration rate constant (base e) 0.40 day -1 Deoxygenation rate constant (base e) 0. day -1 Assume that no other wastewater sources are discharged into the river. Determine the time (days) of the critical dissolved oxygen concentration from the point of wastewater discharge. WWTP X = 0 X* Distance for DO Sag Step 1: Determine the initial DO Deficit DO 0 = DO S DO = 9.08 (Table) 6 (given) =.08 mg/l Refer to Appendix.C, Page A-59 (slightly different value than mine) Step : Determine time, T MAX for DO sag using Streeter Phelps Equation T M K R K ln K 1 R 0 K D D D K R K 1 K D BODU D Eqn 8.8 K R K D reaeration cons tan t, 0.40 d 1 deoxygenation cons tan t, 0. d T M ln days

41 Non Quantitative Problems

42 1 Viruses and Bacteria What following statements are true regarding microbiology? (a) (b) (c) (d) Bacteria can be pathogenic Viruses are parasitic Rotifiers are chemoheterotrophs All are true Refer to Chapter 7 The answer is (d). - Coliforms Refer to Chapter 7-11 in PE Manual The coliform groups are considered good indicator organisms because: (a) (b) (c) (d) they apply to all types of water they are always present when pathogens are present, and normally absent otherwise they are easily detected by routine analytical methods all of the above Section 7-11, Page 7-4 The answer is (d).

43 Odor Constituents Wastewater Engineering: Treatment, Disposal, and Reuse, rd edition, by Tchobanoglous & Burton, McGraw Hill, 1991 What elements found in chemical compounds usually contribute to wastewater odor? (a) chlorine and iron (b) nitrogen and sulfur (c) nitrogen and iron (d) chlorine and sulfur Most compounds which contribute to odor contain nitrogen (amines) and sulfur (mercaptan and sulfides). It is referred to indirectly in Table 7. on page 7-5. The answer is (b). 4 Wastewater Loading Factor Page 8- What is the typical organic loading for domestic wastewater in the U.S.? (a) kg BOD/d per person (b) 0. kg BOD/d per person (c) 0.09 kg/bod/d per person (d) 0.4 kg/bod per person Refer to the top of page 8-. The answer is (c).

44 5 Nitrogenous BOD After 6-10 days of incubation during a BOD test, there is another phase of BOD exertion called nitrogenous demand. What is the cause? (a) oxidation of carbonaceous material (b) oxidation of ammonia to nitrates and nitrates (c) oxidation of phenolic compounds (d) oxidation of mutagenic substrates The answer is (b). See Page 8-9, right column, second paragraph. 6 BOD Removal Rates What is the typical biochemical oxygen demand (BOD) reduction realized by primary clarification? (a) 5-15% (b) 15-5% (c) 5-5% (d) 40-60% Refer to the Table 9.10, page 9-7. See also Page 9-, Section 5. The answer is (c).

45 7 Sequential Batch Reactors Page 0- What are the typical operation steps of sequential batch reactors? (a) fill, aerate, settle, decant (b) fill, aerate, settle, recirculate, decant (c) fill, settle, aerate, decant (d) fill, decant, aerate, recirculate Refer to the page 0-4. The answer is (a). 8 Sludge Thickening What are typical percent solids ranges for waste activated sludge and thickened sludge respectively? (a) WAS 0.5-1%, Thickened 4-5% (b) WAS 0.5-1%, Thickened 5-10% (c) WAS 1-%, Thickened 10-1% (d) WAS, 1-%, Thickened 0-5% Refer to the page 0-14, Section The answer is (a).

46 9 Sewer Pipe Flow Combined sewers are pipe systems that are designed to (a) convey domestic and industrial wastewater (b) convey wastewater and storm water (c) convey domestic wastewater, industrial wastewater, and storm water (d) convey only storm water Refer to the page 8-. The answer is (b). 10 Wastewater Peak Flow Over what period of time would you expect the wastewater to experience the highest average flow? (a) annually (b) seasonally (c) weekly (d) daily Refer to the page 8-, Table 8.1. The answer is (d).

47 11 Wastewater Generation What is the typical design flow for new sewer systems? (a) 80 gal/day/person (b) 100 gal/day/person (c) 150 gal/day/person (d) 180 gal/day/person Refer to the page 8-. The answer is (b). 1 Potable Water Usage & Wastewater Generation What is a typical percentage of domestic drinking water that contributes to wastewater? (a) 40-50% (b) 5-55% (c) 80-90%% (d) 70-80% Refer to the page 8-. The answer is (d).

48 POTABLE WATER SECTION August 01 Instructor: J.B. Jones, P.E., PhD.

49 TABLE OF CONTENTS Typical Water Treatment Plant #1 Water Hardness Calculation #1 # Water Hardness Calculation # # Chlorine Disinfection Dosage #1 #4 Chlorine Disinfection Dosage # #5 Motor Sizing for Flash Mixer #6 Jar Testing #7 Paddle Wheel Mixer for Flocculation Tank #8 Particle Settling Velocity #9 Clarifier Problem #10 Sedimentation Basin #1 #11 Clarifier Basin #1 Sand Filter Problem #1 #1 Sand Filter Problem # #14- Ion Exchange Calculation for Reactor Vessels #15 Zeolite Softening Problem #1 #16 Zeolite Softening Problem # #17 Reverse Osmosis Non Qualitative Problems #1 Algae # Hardness Calculations # Settling Basin Zones #4 Settling Basin Zones #5 Filter Bed Parameters

50 #6 Filter Bed Stratification #7 Granular Media Sequence #8 Treatment Plant Sequence #9 Jar Tests #10 Free Chlorine Disinfection #11 Mixing Physics #1 Mixing Physics ()

51 WATER SOURCE LOW SERVICE PUMP BACKWASH RECYCLE BACKWASH RECOVERY BASIN WASTE SLUDGE ph ADJUSTMENT, PRIMARY OXIDANT, COAGULANT RAPID MIX FLOCCULATION SEDIMENTATION FILTER AID SCREEN WASTE SLUDGE FILTER CLEARWELL HIGH SERVICE PUMP BACKWASH PUMP DISTRIBUTION SYSTEM TYPICAL WATER TREATMENT PLANT

52 1 - Water Hardness Calculation #1 PE Refresher Manual Page 5-1 and Appdx.B (Page A-57) The results of a well water analysis are given below: Ca + 51 mg/l Mg + 1 mg/l Na+ 5 mg/l SO mg/l Cl - 5 mg/l F mg/l NO - 14 mg/l ph 7.8 H S.4 mg/l as S Alkalinity 84 mg/l as CaCO Total Coliforms MPN. org/100 ml Turbidity 6. NTU Chlorine Demand 9.1 mg/l TDS mg/l Temperature 5 0 C Calculate the total hardness. We will do this the long way only to explain the shortcut method found in Appdx.B Hardness is the measure of soluble divalent metal cations (i.e. positive ions having a valence of ). The two main cations that cause hardness are calcium (Ca) and magnesium (Mg). Water hardness is generally expressed as mg/l of CaCO. (This will be discussed in work sample problems) Hardness can be categorized by either of two methods: o Calcium and magnesium hardness o Carbonate and noncarbonated hardness

53 Quick Refresher: Hardness is expressed as CaCO Equivalent Weight EW MW Charge EW CaCO EW CaCO Multiplier given in Appendix. B EW subs tan ce For this specific problem, Add calcium and magnesium concentrations in CaCO equivalents Calcium 51 mg mg as l 40 l CaCO or Use Appendix.C Calcium mg 51 mg CaCO l l.5 factor 17 as Magnesium mg 1 mg CaCO l l 4.1 factor 49 as Total Hardness 177 mg as CaCO l

54 Water Hardness Calculation # A town s water supply has the following ionic concentrations: Al mg/l Ca mg/l Cl mg/l CO 19 mg/l CO - 0 Fe mg/l Fl- 0 HCO mg/l Mg + 4. mg/l Na mg/l NO - 0 SO mg/l What is the total hardness? (a) 160 mg/l as CaCO (b) 00 mg/l as CaCO (c) 60 mg/l as CaCO (d) 00 mg/l as CaCO (Refer to Page 5-1, PE Manual) Part (1) Constituent mg/l as substance Factor from Appendix.c Hardness expressed as mg/l of CaCO Ca x.5 = 00.5 Mg + 4. x 4.1 = 99.6 Fe + 1 x 1.79 = 1.79 Hardness = 0 mg/l The answer is (d).

55 - Chlorine Disinfection Dosage #1 (not in PE Manual) A water is tested and found to have a chlorine demand of 6 mg/l. The desired chorine residual is 0. mg/l. How many pounds will be required daily to chlorinate a flow of 8 MGD? Cl dosage = Cl demand + Cl residual (See Section 8.6, Page 8-1) = 6 mg/l + 0. mg/l = 6. mg/l lbs day lb mg MG MG l d mg l 414 lb day 4 - Chlorine Disinfection Dosage for Potable Water # The flow rate in a water treatment process is 1. MGD. If the chlorine dosage is mg/l, how much calcium hypochlorite containing 70% available chlorine is required for disinfection? SOLUTION lb mg MG Available chlorine reqd MG 0 l d mg l 0 lb day Total chlorine reqd lb day lb day % Available strength should go in the denominator which makes the overall amount increase.

56 5 - Motor Sizing for Flash Mixer What standard motor size is required for a tank impeller flash mixer sized to treat 5 MGD at a temperature of 60 0 F. The design velocity gradient and residence time are 800 sec -1 and minutes respectively. The motor efficiency is 75%. (a) Determine the required volume Page 9-7, Eqn 9.5 Volume Q gal day ft gal day 1440 min 6 * t 5 x ft min 91 (b) Determine the power requirement, Page P G Eqn 6.5 V TANK P = mixing power, ft-lbf/sec = absolute viscosity,.59 x 10-5 lbf-sec/ft (Appendix 14 A) G = velocity gradient = 700 sec ft 14,056 lbf ft sec 5 lbf sec 1 P G VTANK.59 x ft sec ( c) Determine HP requirement Power Input to motor = Pump Power/Efficiency HP 14,056 lbf ft sec P 550 lbf ft sec Pa 4 HP 5HP E 0.75 See inside cover of book for conversion factor. I will cover this in Chapter 18 as well.

57 6 - Jar Testing (1.5 Problems and not very practical)) The results of jar tests and preliminary design of flash mixing for a new proposed water treatment plant are given in the following table. Parameter Criterion GT 4000 Temperature 10 0 C Total Design Flow 6 m /s Number of units Dimension of units 1.5 m square x m deep Configuration Unbaffled The power required for each flash mixer is most nearly: (a) (b) (c) (d) kw 4 kw 56 kw 61 kw In 194, T.R. Camp and P.C. Stein developed the relationship between the velocity gradient and power imparted to the water as P G Equation 6.5 V TANK P = mixing power, N*m/s, or W = absolute viscosity, x 10 - Pa * sec (Appendix 14 B) G = velocity gradient, sec -1 V TANK = basin volume, m Calculate the basin volume V TANK 1.5 m1.5 m m 4.5 m m 6 Flow per tan k Q s tan ks m s

58 Calculate the detention tank per tank s s m m Q V T Calculate G m s m s m m G kw m N s W m m s N x V G P TANK sec The answer is (b).

59 7 - Paddle Wheel Mixer for Flocculation Tank A flocculator tank with a volume of 100,000 ft uses a paddle wheel to mix the coagulant in 60 0 F water. The operating characteristics are as follows: Mean velocity gradient 60 sec -1 Paddle drag coefficient 1.75 Mixing Velocity 1. ft/sec (1) What is the theoretical power required to drive the paddle? (a) 1 hp (b) 16 hp (c) 0 hp (d) 40 hp () What is the drag force on the paddle? (a) 450 lbf (b) 500 lbf (c) 7000 lbf (d) 9000 lbf () What is the required paddle area? (a) 000 ft (b) 500 ft (c) 5000 ft (d) 8500 ft (a) Determine the power requirement Page 6-9, PE Manual P G V TANK P = mixing power, ft-lbf/sec = absolute viscosity,.59 x 10-5 lbf-sec/ft (Appendix 14 A) G = velocity gradient = 60 sec ,000 ft 849 lbf ft sec 5 lbf sec 1 P G VTANK.59 x ft sec

60 lbf ft 8,49 Water HP sec 15. 4hp lbf ft 550 hp sec The answer is (b). (b) Since work = force x distance, then power = force x velocity lbf ft 8,49 F P sec 7, lbf D vel ft sec The answer is (c). ( c) Using equation 6.0 (rearrange terms), page 6-10: A g F C D v. 7,077 lbf D, 478 lbf ft ft sec ft sec ft The answer is (b).

61 8 - Particle Settling Velocity A spherical sand particle has a specific gravity of.7 and a diameter of 0.1-mm. What is the settling velocity? (a) 0.05 ft/sec (b) 0.01 ft/sec (c) 1.0 ft/sec (d) ft/sec From P.E. Manual, page 6-5, the settling velocity is 0.05 ft/sec. The answer is (a).

62 9 - Clarifier Problem A circular clarifier is to be designed with the following characteristics. Flow rate Detention period Surface loading 1. MGD.5 hours 400 gal/ft -day (1) What is the approximate diameter? (a) 45 ft (b) 65 ft (c) 70 ft (d) 90 ft () What is the approximate depth? (a) 6 ft (b) 8 ft (c) 1 ft (d) 15 ft () If the initial flow rate is reduced to 0.8 MGD, what is the surface loading? (a) 100 gal/day-ft (b) 150 gal/day-ft (c) 00 gal/day-ft (d) 50 gal/day-ft Part (1): The surface area is Since A SURFACE A 4 D Flow Loading Rate 6 gal 1. x 10 day gal 400 ft day 50 ft 450 ft 64. ft Dia

63 The answer is (b). Part () 6 gal 1. x 10.5 hr day Volume Flow Time Q t hr 4 day The depth is 5 Volume 1.54 x 10 gal depth ASURFACE gal 50 ft 7.48 ft The answer is (a) x ft 5 gal Part () Loading Rate Flow Area 8 x 10 5,50 gal day ft 46 gal day ft The answer is (d).

64 10- Sedimentation Basin #1 A flow of.5 MGD from a coagulation/flocculation basin is settled in four rectangular sedimentation basins. The dimensions of each basin are 75 ft long x 15 ft wide x 1 ft deep. Calculate the detention time (hours), and horizontal flow velocity (ft/min) if the outlet is located at the other end of the basin. Calculate flow to each basin: Flow per Basin 6 gal.5 x 10 d 4 basin s 6.5 x 10 5 gpd Now, calculate the individual terms: Eqn 9.5 DetentionTime gal 75 ft15 ft1 ft Volume 7.48 ft t D Flow 5 gal 1 d 6.5 x 10 d 4 hr 4 hr Horizontal Vel Flow Cross Section Area 15 ft1 ft 5 gal d gal ft 6.5 x 10 min d 0. ft min

65 11 Clarifier Basin Two equally sized circular sedimentation basins handle a total design flow of.5 MGD. Each unit has an overflow rate of 700 gpd/ft and a detention time of four hours. (a) (b) Calculate the tank diameter. Calculate the weir loading if the influent is introduced through a vertical riser pipe in the center of the basin and flows radially outward. Part (a) Flow to one basin 6 gal.5 x 10 d units 1.5 x 10 6 gal d Surface Area Flow Overflow Rate 6 gal 1.5 x 10 d gal 700 d ft x 10 ft x 10 ft ft 4 A 4 Diameter 48 Part (b) 48 ft ft Weir length diameter gal 1.5 x 10 Weir loading d ft gpd ft

66 1 Sand Filter Problem #1 A water treatment plant has six square rapid sand filters. The flow rate is.5 gal/min-ft. Each filter has treatment capacity of 500,000 gal/day. Each filter is backwashed once a day for 1 min. The rate of rise during washing is 18 in/min. (1) What are the inside dimensions of each filter? (a) 6 ft x 6 ft (b) 8 ft x 8 ft (c) 10 ft x 10 ft (d) 1 ft x 1 ft () What percentage of the filtered water is used for backwashing? (a) % (b) 4% (c) 8% (d) 10% Part a: The area is A REQ Flow Loading Rate gal.5 min ft 500,000 4 gal day hr day 60 min hr 99. ft Use 10 ft x 10 ft area. The answer is ( c). Part b: The volume of water used during backwashing is FILTER t BACKWASH Volume A rate of rise Eqn 6.7 Volume ft min 100 ft ft day min day

67 ft gal day ft 1,464 gal day The percentage of backwash water then would be gal 1,464 day gal 500,000 day 0.06 (%) The answer is (a).

68 1 - Rapid Sand Filter Problem # A water treatment plant has four rapid sand filters. Each filter is designed for a capacity of 1 MGD. Backwashing is accomplished for 10 minutes at a rate of 15 gpm/ft once every 4 hours. The terminal head loss prior to backwashing is averages 4-10 feet. (a) If a loading rate of.5 gpm/ft is used, what are the filter dimensions if the tank is square? (b) How much water is required for backwashing and what percentage of the filtered water does this represent? Part (a) Filter Area Flow Loading Rate 6 gal 1 x 10 d gal min ft 198 min d ft Length 198 ft 14 ft Part (b) min gal Backwash Volume ft x 10 d min ft 4 gal d Percentage.97 x x % %

69 14- Simple Ion Exchange Blending A municipal plant processes water with a total initial hardness of 0 mg/l. The desired discharge hardness is 5 mg/l. If an ion exchange process is used, what is the bypass factor? See Section 6-, Page x, 0 mg/l 5 mg/l Resin Bed x, Bypass (0 mg/l) Page 6-18 x bypass fraction 1 x resin bed fraction Desired hardness 5 mg mg mg 0 l l l Solve for "x" 1 x0 x x %

70 15 - Zeolite Softening Problem # 1 Raw water analysis Ca(HCO ) 160 mg/l as Ca CO MgSO 4 70 mg/l as Ca CO CO 0 The water is to be softened using a zeolite process (ion exchange) with the following characteristics. The softener has an exchange capacity of 10,000 grains/ft and a salt requirement of 0.6 lbm per 1000 grains hardness removed. How much salt is required to soften the water to 80 mg/l. (a) 500 lbm/mg (b) 500 lbm/mg (c) 700 lbm/mg (d) 800 lbm/mg 7000 grains per pound is an arcane term but it can be found in the front jacket of the textbook. There are 7000 grains in a pound. The hardness removed is mg mg mg L L L 150 mg L If we are working with the units of million gallons (MG), lb mg Hardness removed 150 hardness 8.4 MG L mg L 150 lbm hardness / MG Convert lbm of hardness to lbm of salt required Salt required 0.6 lbm Salt lbm hardness gr Hardness MG water gr lbm 550 lbm Salt / MG The answer is (b).

71 16 - Zeolite Softening Problem # Raw water is to be reduced from 0 mg/l to 60 mg/l using a zeolite process. The volumetric flow rate is 1,000 gal/day. The zeolite process has a resin exchange capacity of 5,000 grains/ft and a zeolite volume of.5 ft. (1) What fraction of the water is bypassed around the process? (a) 0.0 (b) 0.5 (c) 0.0 (d) 0.40 Repeat of Problem #14 but with more questions added. Zeolite term used instead of Resin. () What is the removal rate of hardness for the resin filter? (a) 1.1 lb/hr (b).4 lb/hr (c).5 lb/hr (d) 4.8 lb/hr () What is the time between regenerations of the softener? (a) 8 hr (b) 16 hr (c) 4 hr (d) 0 hr Step (1): A bypass product is required. mg 60 fraction bypassed L mg 0 L fraction processed

72 0 mg/l 0 mg/l Zeolite (81%) 60 mg/l Bypass (19%) The answer is (a). Step (): The amount of hardness removed per vessel is gr.5 ft 5,000 ft Hardness reduction Resin Volume * Exchange Capacity 8. 9 lbm gr 7000 lbm lbs mg lb Chemical Feed Rate 4 day MG Flow, MGD Dose, 8. l mg l The hardness removal rate is Percentage of water processed lb MG mg MG day L mg L Q * conc * conversion hr 4 day lbm hr The answer is (a). Step (): Calculate Run Times Answer format is lbm/hr so additional conversion required. Hardness removed 8.9 lbm Run time 8. hr Hardness removal rate lbm 1.08 hr The answer is (a).

73 17- Reverse Osmosis (Look this over outside of class) A reverse osmosis (RO) system is required to treat a drinking water source that is subject to saltwater intrusion. The water and RO have the following characteristics: Desired Fresh Water Flow 0,000 m /d Permeate Recovery 75% Membrane Flux Rate 0.95 m /m -d Membrane Packing Density 800 m /m Determine the membrane volume required to treat this water. Step (1): Determine the required water flow with a permeate recovery fraction, f p, equal to 75%. Q Q f O P m 0,000 day 0.75 m 40,000 day Step (): Determine the required surface area for membrane: A M G A M required membrane area m membrane flux rate, 0.95 m day Q G m 40,000 day m 0.95 m day 4,105 m Just like all the previous design examples: Area x Loading rate = Flow Other treatment mechanisms solved this way are filters, weirs, clarifiers, digesters, ponds, and RBC s. Step (): Determine the membrane volume based on P M (membrane packing density): AM 4,105 m VM membrane volume 5.6 m PM m 800 m

74

75 Non Quantitative Problems

76 1 Algae What is the primary threat from algae to water supplies? (a) (b) (c) (d) color taste and odor parasites toxicity Refer to Page 6-, Section 6-6. The answer is (b). Hardness Calculations What are the common units of hardness? (a) meg L (b) mg L as Ca (c) mg L as CaCO (d) ppm The answer is (c). Settling Basin Zones (not in text) What are the four functional zones that define a sedimentation basin? (a) (b) (c) (d) flocculent, quiescent, overflow, sludge inlet, outlet, settling, sludge inlet, outlet, settling, overflow turbulent, quiescent, overflow, underdrain The answer is (b).

77 4 Settling Basin Zones What is a typical overflow rate for sedimentation basins? (a) (b) (c) (d) 10 m 10 5 m 5 50 m m m m m m d d d d Refer to the first sentence on the top of Page 6-6. The answer is (c).

78 5 Filter Bed Parameters The loading rate for granular media filters is determined by the flow rate divided by what other parameter? (a) (b) (c) (d) filter bed volume filter bed surface area filter bed volume times media porosity filter bed surface area times media porosity See Page 6-1, Section 4. The answer is (b). 6 Filter Bed Stratification In a multimedia filter, how does filter media remain stratified following backwash? (a) (b) (c) (d) The different media are separated by baffles to prevent mixing Mixing occurs and the media is eventually replaced after many backwash cycles The media have different specific gravities and settle at different rates The media remain stratified because they mix only at the interface layers Refer to the AWWA Operator Manuals The answer is (c).

79 7 Granular Media Sequence What most completely describes the operating modes, in sequence, for conventional granular media filtration? (a) (b) (c) (d) filtration and backwash filtration, backwash, and conditioning filtration, draining, backwash, and wasting filtration, draining, backwash, wasting, and conditioning Refer to the AWWA Operator Manuals The filtration process occurs in () phases that are repeated continuously: Filtration water flows downward through the filter media and particles are retained Backwash flow is reversed to expand the bed Conditioning term used as the first flush of water is cycled through the bed. This water is wasted or returned to the front of the plant. The answer is (b). 8 Treatment Plant Sequence What is the usual sequence of water treatment unit processes? (a) flash mix, flocculation, sedimentation, filtration, chlorination (b) filtration, flash mix, flocculation, sedimentation, chlorination (c) chlorination, filtration, flash mix, flocculation, sedimentation (d) sedimentation, flocculation, filtration, chlorination, flash mix Refer to Class Handouts The answer is (a).

80 9 Jar Tests What is the basic design parameter for flocculation basins? (a) (b) (c) (d) velocity-time gradient mixing power solids flux hydraulic loading rate The velocity time gradient is Gt. The answer is (a). 10 Free Chlorine Disinfection Which compound does not provide a free chlorine residual? (a) CaOCl (b) NH Cl (c) NaOCl (d) HOCl See Page 6-0. Free chlorine exists in the form of HOCl and OCl. The answer is (b).

81 11 Mixing Physics The velocity gradient, G, is expressed usually in units of sec -1. It relates power, tank volume, and velocity. This equation is typically applied to what treatment sequence? (a) (b) (c) (d) filtration flash mixer disinfection sedimentation See Page 6-9. The answer is (b). 1 Mixing Physics () Typical values for the velocity gradient, G, for rapid mixers are usually in the range of (a) sec -1 (b) sec -1 (c) sec -1 (d) sec -1 See Page The answer is (a).

82 CLOSED PIPE HYDRAULICS August 01 Instructor: J.B. Jones, P.E., PhD.

83 TABLE OF CONTENTS #1 Continuity Equation # Tank Discharge via Orifice # Darcy Weisbach Head Loss #4 - Hazen Williams Head Loss #5 Equivalent Length Method by Darcy Weisbach #6 Equivalent Length Method with Elevation Change (Darcy Weisbach) #7 Equivalent Length Method with Elevation Change (Hazen Williams) #8 Pump Affinity Laws #9 Pump Affinity Laws # #10 - Multiple Stage Pumps #11 Pump and System Curves Non-Quantitative Problems #1 Bernoulli s Equation # Velocity Head # Pipe Schedule #4 Determining Friction Loss #5 Moody Diagram #6 Relative Roughness #7 Pump Curves #8 Water Network #9 Water Network # #10 Hazen Williams Question #11 Water Hammer #1 Basic Flow #1 Pump Speed Change #14 Cavitation

84 CHAPTER TOPICS Chapter 14: Fluid Properties Chapter 15: Fluid Statics Section 4: Fluid Height equivalent to Pressure Section 1: Hydrostatic Forces on a Dam Chapter 16: Fluid Flow Parameters Fluid Energy Units Bernoulli Equation Pitot Tube Laminar vs. Turbulent Flow Energy Grade Line Chapter 17: Fluid Dynamics Conservation of Mass Darcy Friction Minor Losses Energy & Hydraulic Grade line with Friction Discharge from Tanks Series Pipe Systems Venturi Meter Orifice Meter Water Hammer Chapter 18: Hydraulic Machines Centrifugal Pumps Terminology Pump Power Cavitation System Curves Performance Curves Affinity Laws

85 1 Continuity Equation Water flows at 4 ft/sec in a 4-in diameter pipe that is connected through a reducer to a -inch diameter pipe. What is the flow velocity in the -inch diameter pipe? (a) (b) (c) (d) 5 ft/sec 6 ft/sec 7 ft/sec 9 ft/sec Because the specific information regarding the type of pipe is not given, assume pipe diameters given are actual inside diameters. A 1, A d 1,d cross-sectional area of upstream and downstream pipe in inside diameter of upstream and downstream pipe in 4in d1 A in 4 4 d in A in 4 4 V 1 velocity in upstream pipe ft/sec V velocity in downstream pipe ft/sec ft 1.56 in 4 A1 V1 sec V 7.1 ft / sec Eqn. 17. A 7.07 in The answer is (c).

86 Tank Discharge via Orifice (Not very typical) A full cylindrical tank 60 feet high has a constant diameter of 0 feet. The tank has a 4 in (100 mm) diameter hole in the bottom. The coefficient of discharge for the hole is How long will it take for the water level to drop from 60 feet to 0 feet? (a) (b) (c) (d) 1000 sec 700 sec 1400 sec 100 sec 4in d 1in ft 0 A ft A t d 4 t 0 ft ft 60 ft 0 ft The time to drop from 60 feet to 0 feet is given by Equation Insert terms and original equation At z z t C A g d 1 O 15 ft 60 ft 0 ft t 10 sec The answer is (d) ft. ft sec

87 Darcy-Weisbach Head Loss (Long but practical) Water at 60 0 F flows in a - inch diameter iron pipe (surface roughness, ε, is ) at a flow rate of 15 gallons per minute. Using the Darcy-Weisbach relation, the head loss in a 00 feet section of pipe will most nearly be: (a) (b) (c) (d) 5 ft 11 ft 9 ft 4 ft The D-W equation gives a frictional head loss for fluid flow based on a friction factor. Friction factor is a function of the Reynolds number and the relative roughness of the pipe. Step 1: Compute velocity, Re, and relative roughness inches1 ft 1 inches 0. ft D 5 It might make sense to split this equation into two parts because unit conversion can be confusing. Gallons to cubic feet and minutes to seconds. v Q A 15 gpm0.14 ft gal1 min 60 sec 0.5 ft ft sec Referring to Appendix 14.A on page A-15, use a kinematic viscosity of water of 1.17 x 10-5 ft /sec. Re D vel 0.5 ft5.7 ft sec x 10 sec x ft Eqn. 16. D ft 0.5 ft 0.004

88 Step : Calculate friction factor and head loss DO NOT USE THIS FORMULA f friction factor Eqn f log 0.5 D Re log x INSTEAD, If you are given the pipe material but not the roughness, typical roughness values can be found in Appendix 17.A and Table 17.. It is easy to lose a zero along the way. Go to Appendix A-17B on Page A-. Using the D values listed along the top row, find the right Reynolds number range. For this example, look on A-5, select the D value of Interpolate between the Reynolds number ranges to obtain a friction factor of 0.08 which matches the value found with Eqn Eqn You will see that we can estimate the friction factor quickly because the third decimal place will not change the answer. Learn the rudiments of this table and it will you save a lot of time. ft ft 5.7 f LV sec h f 11. D g ft 0.5 ft. sec ft The answer is (b).

89 p 1 V1 p V Z1 Z g g h L Head loss will be calculated either through Hazen Williams or Darcy Weisbach method.

90 1 ft of water weighs 6.4 lbs. Over a surface area of 144 inches, that equates to 0.4 psi/ft for every foot of water column. Conversely, it takes.1 feet of water column to create 1 psi.

91

92 4 Hazen Williams Head Loss Water at 60 0 F flows in a in diameter pipe (Hazen-Williams coefficient, C, is 100) at a flow rate of 15 gallons per minute. Using the Darcy-Weisbach relation, the head loss in a 00 feet section of pipe will most nearly be: (a) (b) (c) (d) 4 ft 8 ft 15 ft 46 ft The H-W equation gives a frictional head loss for turbulent flow independent of the Reynolds number. It gives good results for liquids with a viscosity similar to that of water at 60 0 F. Step 1: Compute velocity inches1 ft 1 inches 0. ft D 5 v Q A 15 gpm0.168 ft gal1 min 60 sec 0.5 ft ft sec Step : Compute head loss Pay attention to units v L h f 15. ft Eqn C D The answer is (c).

93 5 - Equivalent Length Method by Darcy Weisbach Water travels through 00 feet of -inch Schedule-80 PVC pipe at a velocity of 5 ft/sec. The pipe includes 15 couplings, regular elbows, regular elbows, 6 tees (straight flow), and 1 globe valves. Assume the equivalent length for the fittings will be the same as screwed steel pipe (Appdx A-17.D). The water temperature is 60 0 F. Using Darcy Weisbach, what is the total head loss in the system from all sources? (a) 7 ft (b) ft (c) 95 ft (d) 7 ft For an equivalent length of threaded -inch schedule 80 PVC fittings, the following characteristics apply: Refer to Page A-1, Appendix 17.D, Fitting Quantity Unit Equivalent Length (ft) Total Equivalent Length (ft) coupling ell ell Straight tee Globe valve For friction head loss in the pipe (including equivalent length of fittings), D inside diameter for -in Sch 80 Pipe roughness coefficient for PVC pipe

94 D = 1.99 in Appendix 16.D = ft Table 17., page 17-4 D in ft1 ft 1.99 in It is not necessary to use the true inside diameter. A nominal value of would suffice. Re Reynolds number V flow velocity ft/sec kinematic viscosity ft /sec = 1.17 x 10-5 ft/sec at 60 0 F Re D V 1.99 in 1.17 x 10 5 ft 5 sec ft 1in sec ft 6.6 x 10 4 Eqn 16. f friction factor Eqn f log D Re log x (Alternative is to look at tables in A-17.)

95 Note that the friction factor could also be determined using a Moody diagram with the same input values for using Reynolds number and relative roughness. Refer to the Moody diagram on Page Eqn Estimating from Appdx 17.B gives f = 0.0 which is close enough. ft ft5 f L Le V sec h f D g 1 ft ft in. 1 in sec ft The answer is (b).

96 6 - Equivalent Length Method with Elevation Change (Darcy Weisbach) Water is pumped from a lower reservoir (Elevation = 5 feet) to an upland reservoir (Elevation = 4 feet). The water travels through 00 feet of -inch Schedule-80 PVC pipe at a velocity of 5 ft/sec. The pipe includes 15 couplings, regular elbows, regular elbows, 6 tees (straight flow), and globe valves. Assume the equivalent length for the fittings will be the same as screwed steel pipe (Appdx A-17.D). The water temperature is 60 0 F. Using Darcy Weisbach, what is the total head loss in the system from all sources? (a) (b) (c) (d) 7 ft 85 ft 95 ft 41 ft For an equivalent length of threaded -inch schedule 80 PVC fittings, the following characteristics apply: Refer to Page A-1, Appendix 17.D, Fitting Quantity Unit Equivalent Length (ft) Total Equivalent Length (ft) coupling ell ell Straight tee Globe valve I will skip the intermediate steps which are a repeat of the previous problem. Eqn Estimating from Appdx 17.B gives f = 0.0 which is close enough. ft ft5 f L Le V sec h f ft D g 1 ft ft in. 1 in sec

97 For total head loss, h 4 5 ft ft ft hz h f 41 Eqn 18.7 The answer is (d). Elev = 4 feet Elev = 5 feet 00 feet of pipe

98 7 - Equivalent Length Method with Elevation Change (Hazen Williams) Water is pumped from a lower reservoir (Elevation = 5 feet) to an upland reservoir (Elevation = 4 feet). The water travels through 00 feet of -inch Schedule-80 PVC pipe at a velocity of 5 ft/sec. The pipe includes 15 couplings, regular elbows, regular elbows, 6 tees (straight flow), and globe valves. Assume the equivalent length for the fittings will be the same as screwed steel pipe (Appdx A-17.D). The water temperature is 60 0 F. Using Hazen Williams, what is the total head loss in the system from all sources? (a) 4 ft (b) 85 ft (c) ft (d) 14 ft v L h f 5 ft Eqn C D For total head loss, Where did C = 140 come from? Look in Appendix 17.A. h 4 5 ft 5 ft ft hz h f 4 Eqn 18.7 The answer is (a). Elev = 4 feet Elev = 5 feet 00 feet of pipe

99 8 Pump Affinity Laws A pump intended to run at 1750 rpm when driven by a 0.5 hp motor. What is the required power rating of a motor that will turn the pump at 000 rpm? (a) (b) (c) (d) 0.5 hp 0.45 hp 0.65 hp 0.75 hp Similarity (Section 18-7) and Affinity (Section 18-6) equations are quick and easy ways to solve pump problems even if you are unfamiliar with pumps. 5 D 000 rpm P P1 P D rpm 0.5 hp 0. hp Eqn 18.5 The answer is (d).

100 9 Pump Affinity Laws # A centrifugal pump operates at a speed of 1750 rpm and is rated at 850 gal/min for 78% efficiency and 180 ft of head. For constant efficiency and head, what is most nearly the flow rate if the pump is operated at 00 rpm? (a) (b) (c) (d) 50 gal/min 680 gal/min 80 gal/min 1100 gal/min Refer to the affinity laws stated in Section 18.6, page Q n Eqn Q 1 n 1 Q Q1 n n 1 gal 850 min 00 rpm 1750 rpm min 1069 gal The answer is (d).

101 10 - Multiple Stage Pumps (Not very typical) A submersible pump operated at 1750 rpm is needed to deliver 800 gal/min from a 50 ft deep well. At 9% efficiency, the average specific speed of the pump is 00. How many stages are required for the pump? (a) (b) (c) (d) 1 stage stages 6 stages 4 stages This will be a variation of Eqn This is not covered in the PE manual. h elevation head ft n number of stages N S specific speed Q pump discharge gpm w rotating speed rpm w Q N S Modification of Eqn h n h n 0.75 w N S Q N S n h w Q 4 50 ft 00 gal 1750 rpm 800 min stages, say 6 The answer is (c).

102 11 Pump & System Curves PART A: A pumping station is lifting water from a smaller reservoir to another tank. Water travels through 000 feet of 4-inch pipe. If the pipe is changed to -inch pipe, the friction loss will (a) (b) (c) (d) not change, and the flow rate will remain the same increase, and the flow rate will decrease increase, and the flow rate will increase decrease, and the flow rate will decrease Refer to page 18-18, Section 18.: For those unfamiliar with pumps, this can be a little daunting to interpret pump and system curves. Pump Curve TDH, ft 4 Discharge, gpm Regardless of which method we use to calculate the friction losses in pipe (either the DW or HQ method), smaller pipe will increase the energy losses. Therefore, the friction loss will be greater in the -inch pipe. Although the same pump curve applies, the system curve is different. The zero flow head is the same, but the curve is shifted upwards. This will shift the operating point upward and to the left. The head added by the pump will increase; the flow rate will decrease. The answer is (b).

103 PART B: Referring to the previous problem, the pump is designed to pump 50 gal/min at 0 feet of head. What approximate hydraulic power does the pump develop? (a) (b) (c) (d) 1 hp hp hp 5 hp Refer to Table 18.5, page 18-8: Take a moment to understand how Table 18.5 is set up. gal 0 ft ha Q SG min WHP 1. 9 hp gal ft gal ft hp min hp min The answer is (b). PART C: What is the brake horsepower if the pump efficiency is 70%? BHP WHP n P 1.9 hp hp

104 Non Quantitative Problems

105 #1 Bernoulli s Equation What three parameters are included in the Bernoulli equation for an ideal fluid flowing in a closed conduit? (a) (b) (c) (d) Elevation, pressure, viscosity Elevation, pressure, velocity Elevation, velocity, viscosity Friction, pressure, velocity Refer to Eqn on Page 16-. The answer is (b). # Velocity Head For laminar flow in pipes, how does the kinetic energy change when the flow is halved? (a) It decreases by a factor of (b) It decreases by a factor of 4 (c) There is no change since kinetic energy is a function of pressure, not velocity (d) There is no change since kinetic energy is a function of elevation, not velocity Refer to Eqn on Page 16-. Remember velocity = Q/A The answer is (b).

106 # Pipe Schedule As the schedule increases for common pipe, how do pipe diameters change? (a) (b) (c) (d) The wall thickness does not change as the outside and inside diameters get larger The inside diameter does not change as the outside diameter gets larger The inside and outside diameters change depending on the pipe material The outside diameter remains constant requiring the inside diameter to decrease Refer to Appendix 16 a, b, and c. The answer is (d). #4 Determining Friction Loss Which equation is not useful for determining friction head loss in a pressurized pipe? (a) (b) (c) (d) Hazen Williams Darcy Weisbach Van t Hoff-Arrhenius Chezy Manning The van t Arrhenius is used to make temperature corrections to kinetic reaction rate coefficients. The answer is (c).

107 #5 Moody Diagram If the Reynolds number and relative roughness are known, what information is available from a Moody diagram? (a) (b) (c) (d) Pipe friction head loss pressure Hazen Williams coefficient Friction factor Refer to Figure 17.4 on page The answer is (d). #6 Relative Roughness What parameters define relative roughness for a pipe? (a) (b) (c) (d) Absolute roughness and Hazen Williams coefficient Absolute roughness and pipe diameter Friction factor and Hazen Williams coefficient Friction factor and pipe diameter Look at Moody Diagram on page Re lative roughness D The answer is (b).

108 #7 Pump Curves What parameters are determined at the intersection of a pump curve and a system curve? (a) (b) (c) (d) Head and horsepower Head and efficiency Efficiency and flow rate Head and flow rate The intersection of the pump curve and system curve defines the head and flow rate operating conditions. The answer is (d). #8 Water Network Two pipes of equal length branch from a common pipe and terminate at a common pipe. The pipe diameters are 6-inches and 1 inches respectively. Which statement is most correct? (a) (b) (c) (d) The head loss in the smaller pipe is twice that of the larger pipe The head loss in the smaller pipe is four times that of the larger pipe The head loss in the smaller pipe is half that of the large pipe The head loss in the smaller pipe is the same as that of the larger pipe Refer to Section 17-1 on page 17-. The answer is (d).

109 #9 Water Network # In a pipe network, water can flow though multiple routes between nodes. Which statement is most correct? (a) (b) (c) (d) The inflow and outflow at a node does not have to be equal The pressure drop between any two nodes, regardless of the flow path, are equal The pipe velocities between any two nodes, regardless of the flow path, are equal The pipe flow rates between any two nodes, regardless of the flow path, are equal Refer to Section 17-9 on page The answer is (b). #10 Hazen Williams Question A best estimate for the Hazen-Williams resistance coefficient for a new 10-inch diameter PVC pipe is most nearly: (a) 110 (b) 140 (c) 100 (d) 80 Refer to fine print in Appendix A-17 on Page A-5. The answer is (b).

110 #11 Water Hammer What is not an effective strategy for reducing water hammer? (a) unanchored pipe joints (b) surge tanks (c) slowly closing valves (d) air chambers Refer to Page 17-8, first paragraph above Figure The answer is (a). #1 Basic Flow Water flows in a horizontal pipe between points A and B. The pipe is a constant diameter. Which statement is not true? (a) ρ A = ρ B (b) p A = p B (c) v A = v B (d) z A = z B Density, velocity and elevation do not change per the problem statement. The answer is (b). There will be energy lost due to the surface roughness of the pipe.

111 #1 Pump Speed Change When a speed of a given pump is increased, what is the effect on the pump curve? (a) upward shift but no change in shape (b) downward shift but no change in shape (c) upward shift and change in shape (d) downward shift and change in shape Refer to Page 18-17, Figure The answer is (c). #14 Cavitation During a pump cavitation event, what condition has likely occurred with respect to the inlet and outlet pressures in the pump? (a) absolute pressure at inlet is below the vapor pressure (b) absolute pressure at the outlet is below the vapor pressure (c) absolute pressure is equal on both sides (d) absolute pressure at the inlet is less than the outlet Refer to Page 18-14, Section 16. The answer is (a).

112 OPEN CHANNEL FLOW August 01 Instructor: J.B. Jones, P.E., PhD.

113 TABLE OF CONTENTS #1 Rectangular Channel using Manning s Equation # Rectangular Channel Depth of Flow by Trial & Error # Head Loss in Trapezoidal Channel #4 Circular Pipe General Calculations #5 Trapezoidal Channel with Earthen Lining #6 Cipoletti Weir #7 Parshall Flume #8 Hydraulic Jump Head Loss Calculation #9 Hydraulic Jump Flow Calculation #10 Hydraulic Jump over Spillway #11 Culvert Flow Type Identification #1- Sluice Gate Non Quantitative Problems #1 Weirs # Wetted Perimeter # Sewer Pipe Slopes #4 Froude Number #5 Hydraulic Radius #6 Most Efficient Cross Section for a Rectangular Channel #7 Most Efficient Cross Section for a Trapezoidal Channel #8 Maximum Discharge in a Circular Pipe #9 Hydraulic Radius Change #10 Hydrostatic Forces

114 CHAPTER 19 TOPICS Note: Chapter 17, Fluid Dynamics will not be covered explicitly. Types of Flow Refer to Table 19.1 Flow Parameters Manning equation - Normal Depths Energy Relationships (Bernoulli s Equation) Weirs & Spillways Parshall Flumes Specific Energy Relationships Alternate Depths & Critical Flow Hydraulic Jumps Culvert Flow Types (1 6)

115 1 Rectangular Channel using Manning s Equation Water flows in an open rectangular channel that is 5 feet wide with a normal water depth of 1 inches. The channel is concrete lined along its entire length and has a constant slope of 0.%. What is the flow rate of the water in the channel? (a) 10 ft /sec (b) 0 ft /sec (c) 15 ft /sec (d) 5 ft /sec Refer to Appendix 19.A, page A-46: The Manning roughness coefficient for concrete is Using Table 19., page 19-, R b d b d, where b = 5 ft and d = 1 ft 1.49 b d 1.49 Q A R S b d S Eqn. 19.1(b) n n b d Q ft sec The answer is (b).

116 Rectangular Channel, Depth of Flow by Trial & Error A natural channel in good condition flume with a rectangular cross section is ft wide. The flume carries 4 ft /sec of water down a 1% slope. What is the depth of flow? (a) (b) (c) (d) 0. ft 0.6 ft 1.0 ft 1.6 ft Refer to Appendix 19.A, page A-46: The Manning roughness coefficient for is Using Table 19., page 19-, R d d d 1 d 1.49 A R S Eqn 19.1 n Q ft 4 sec d 1 d d d d 1 5 d, d 1 By trial and error d 0.6 ft The answer is (b).

117 Head Loss in Trapezoidal Channel What is the head loss per unit length in a smooth earthen trapezoidal channel with a base width of m, a water depth of 0.5 m, a flow velocity of. m/s and 1-to-1 side slopes? (a) (b) (c) (d) m/m m/m 0.01 m/m 0.4 m/m Page 19- For a unit length of smooth earthen channel, it is reasonable to assume that flow is uniform and that all head loss is due to friction. With these assumptions, the head loss is the product of the channel length and slope, with the slope determined using the Manning equation. b channel base width m d water depth m R H hydraulic radius m angle from the horizontal to the side wall degree For to 1-to-1 side slopes, the angle from the horizontal to the sidewall is m0.5 msin m cos 45 0 msin m bd sin d cos R H m bsin d h f friction head loss m/unit length L channel length m n Manning coefficient - V water velocity m/s

118 Use 1 m for the unit length of the channel. The Manning roughness coefficient for a smooth earthen lining is Refer to Appendix 19.A, page A-46. h m 1 m L n v s 0.01 m m length of channel Eqn. 19.0a 4 4 R 0.66m H f / The answer is (c).

119 4 Circular Pipe A 0 in diameter concrete pipe (n = 0.01) was installed on a slope. (a) What is the original full flow capacity? (A) (B) (C) (D) 6.5 ft /sec 1.0 ft /sec 8.8 ft /sec 4. ft /sec Using Eqn 19.1, 1.49 A R S Eqn. 19.1(b) n Q Q in in 1 ft 0 in in 1 ft ft 1.0 sec The answer is (b). (b) If PVC pipe were used, what would the increase in full flow capacity be? (A) % (B) % (C) 44% (D) 55% Q is inversely proportional to n, n The answer is (c) CONCRETE QPVC QCONCRETE QCONCRETE n PVC Q CONCRETE

120 5 Trapezoidal Channel with Earthen Lining Two reservoirs differ in elevation by 5 m and are connected by a 10 km long earth lined channel with a constant slope. The channel bottom is m wide with a normal water depth of 1 m and the channel has a -to-1 horizontal to vertical sides. What is the flow rate in the channel? (a) 5 m /s (b) 10 m /s (c) 1 m /s (d) 18 m /s side slope angle measured from the horizontal degrees For -to-1 horizontal-to vertical slope, tan A cross-sectional area of channel m b base width m d water depth m Refer to Table 19., page 19- A b d d tan m 1 m tan m 5. m 0 0 P wetted perimeter m d 1 m P b m 8. m 0 sin tan 18.4 R A 5.0 m 0. m H P 8. m 60

121 S channel slope m/m L change in distance m z change in elevation m z 5 m S L m 10 km1000 km m m n Manning roughness coefficient - Q flow rate m /s For an earth-lined channel, the Manning roughness coefficient is (Appendix 19.A, page A-46) 5.0 m 0.60 m 1 1 Q A R S n m m Q 9.9 m s The answer is (b).

122 6 Cipoletti Weir What is the flow rate through a 4-inch Cipoletti weir when the water height above the notch is 10 inches? (a) (b) (c) (d).8 ft /sec 4.1 ft /sec 5.1 ft /sec 1 ft /sec Refer to Page b weir length ft h head above notch ft Q flow rate ft /sec The discharge equation giving flow rate in ft /sec for a Cipoletti weir when b and h are given in feet is: 1 ft 1 ft Q.67b H.674in 10in 5.1 ft 1in 1in Eqn (b) sec The answer is (c).

123 7 Parshall Flume A Parshall flume has a throat width of 4 ft. The upstream head measured from the throat floor is 4 in. (a) What is the flow rate? (A) (B) (C) (D) 40 ft /sec 50 ft /sec 60 ft /sec 65 ft /sec Refer to Section 19-19, page 19-14: ft n 1.5 b Eqn n 4 in ft Q K b H a 44 ft 47.8 Eqn in sec 1 ft The answer is (b)

124 8 Hydraulic Jump Head Loss Calculation A hydraulic jump with a stilling pool is selected to dissipate energy over a spillway prior to the water entering a natural river channel. The spillway is.5 meters wide and the hydraulic jump occurs when the water depth at the toe of the spillway is 0.15 m for a flow of.0 m /sec. What is most nearly the head dissipated? (a) (b) (c) (d) 1.4 m 1.8 m. m 4. m Refer to Page 19-, Section d 1 upstream depth m Q water flow m /sec v 1 upstream water velocity m/s w channel width m m.0 Q v sec 1 8 m d w m.5 m s d downstream water depth m g gravitational constant 9.81 m/s v downstream water velocity m/s d v1 d1 0.5d 0. d1 Eqn g 1 5

125 m m s d m 4 m 9.81 s m 1. m The specific energy lost in the jump can be solved using Eqn d d1 1.4 m 0.15 m 4 d d m1.4 m E. m 1 The answer is (c).

126 9 - Hydraulic Jump Flow Calculation Water flowing in a rectangular channel 6 ft wide experiences a hydraulic jump. The depth of flow upstream from the jump is 1 ft. The depth of flow downstream of the jump is ft. What quantity is flowing? (a) 45 ft /sec (b) 54 ft /sec (c) 7 ft /sec (d) 8 ft /sec Refer to Page 19- for an illustration of conjugate depths. v 1 g d d1 d d Eqn (rectangular channels) 1 ft. ft v sec 1 ft 1 ft 1 ft ft 1.9 sec Q v A ft 1.9 sec 6 ft1 ft ft 8 sec The answer is (d).

127 10 Hydraulic Jump over Spillway A hydraulic jump forms at the toe of the spillway. The water surface levels are 0. ft and 5 ft before and after the jump, respectively. The velocity before the jump is 45 ft/sec. What is the energy loss in the jump? (a) (b) (c) (d) 17 ft-lbf/lbm 5 ft-lbf/lbm 5 ft-lbf/lbm 45 ft-lbf/lbm Refer to Page 19- d d1 5 ft 0. ft 4 d d 4 0. ft5 ft E 17 ft Eqn The answer is (a).

128 11 Culvert Flow Type A 0 in diameter concrete culvert is 00 ft long and laid at a slope of The culvert entrance is flush and square edged. The tailwater level at the outlet is just at the crown of the barrel, and the headwater is.0 ft above the crown of the culvert s inlet. What type of flow is this? (a) Type 1 (b) Type (c) Type 4 (d) Type 6 Refer to Figure in Pipe Diameter D. 5 in 1 ft ft 00 ft slope ft h ft submergence.5 ft pipe diameter 7 1 Determine the culvert flow type: h1 z D 7 ft 00 ft ft. The tailwater submerges the culvert outlet, so this is a type-6 flow. The answer is (d).

129 1 Sluice Gate (not likely) A radial gate is used to control flow into a waterway turnout and prevent erosion of a section downstream channel. The gate is operated partially open. The normal flow in the waterway channel is 40 ft /sec with feet of available head at the gate. The gate discharge coefficient is What is the required area of the gate opening? (a) 1.8 ft (b).0 ft (c).8 ft (d) 5.4 ft A 0 area of gate opening ft C d gate discharge coefficient -- g gravitational acceleration. ft /sec h available head at the gate ft Q discharge through the gate ft /sec Because a partially open gate has the characteristics of a submerged orifice, the following equation applies. v Cd g h Eqn (Page 17-17) rearranges to, 0 Q 0 A0 Cd g h, then solve for A 0 A ft 40 sec ft sec 0. 8 C d Q gh ft ft The answer is (c).

130 Non Quantitative Problems

131 #1 Weirs What weir types are best suited for measuring very small flows? (a) (b) (c) (d) Cipoletti proportional rectangular V-notch Note: I got this one wrong. I picked (d). I have looked on the web for a better explanation. Anyway, I toss this one in with some hesitation. The answer is (b). # Wetted Perimeter Water is flowing in a rectangular channel with a normal depth of 5 feet and a base of 5 feet. What is most nearly the wetted perimeter of the channel? (a) (b) (c) (d) 1 feet 5 feet 9 feet 4 feet The wetted perimeter is the length of the channel cross section that has water contact. For a rectangular channel, 5 ft ft P W b d 5 ft 5 The answer is (b).

132 # Sewer Pipe Slopes (wording is confusing) How is the recommended minimum slope influenced by sewer pipe diameter? (a) (b) (c) (d) Pipe slope is determined by minimum scour velocities and is not influenced by pipe diameter Shallower slopes are used for larger diameter pipes because they are more accessible for cleaning Shallower slopes will maintain minimum scour velocities as pipe diameters increase Slopes must increase to maintain minimum scour velocities as sewer diameters increase For a specified scour velocity: As diameter increases, the required slope decreases. Using Eqn 19.1, 1.49 R S Eqn. 19.1(b) n V Remember R = D/4 for circular pipe at full and half flow The answer is (c).

133 #4 Froude Number What is the relationship between Froude number and critical flow in a rectangular channel? (a) (b) (c) (d) Fr = 0 at critical flow Fr is positive at critical flow Fr is negative for critical flow Fr = 1 at critical flow See Section 19-7, page Also refer to Figure 19.1 on page The answer is (d). #5 Hydraulic Radius For a circular pipe, if the hydraulic radius flowing half full is 6 inches, what is the hydraulic radius when the pipe flows full? (a) (b) (c) (d) 1 inches 9 inches inches 6 inches The answer is (d).

134 #6 Most Efficient Cross Section for a Rectangular Channel What proportion of depth to width gives the most hydraulically efficient section for uniform flow in an open rectangular channel? (a) (b) (c) (d) d = 0.67 w d = 0. w d = 0.5 w d = 0.50 w Refer to Section 19-1, page The answer is (d). #7 Most Efficient Cross Section for a Trapezoidal Channel What proportion of depth to hydraulic radius gives the most hydraulically efficient section for uniform flow in an open trapezoidal channel? (a) (b) (c) (d) d = R d = R d =.5 R d =.5 R Refer to Section 0-1, page The answer is (a).

135 #8 Maximum Discharge in a Circular Pipe A closed circular pipe is flowing partially full. At what depth, in terms of the height of the pipe d, will maximum discharge occur? (a) (b) (c) (d) 0.95 d 1.0 d 1.0 d 0.90 d Refer to Section 19-7, page The answer is (a). #9 Hydraulic Radius Change If the hydraulic radius increases, what is the effect on velocity? (a) (b) (c) (d) velocity increases velocity decreases velocity could increase or decrease velocity remains the same Refer to Section 6, page The answer is (a).

136 #10 Hydrostatic Forces On a rectangular vertical plane surface, how is hydrostatic force distributed? (a) increases linearly with depth (b) decreases linearly with depth (c) even distribution with depth (d) non-uniform distribution with depth Refer to Section 7, page The answer is (a).

137 HYDROLOGY August 01 Instructor: J.B. Jones, P.E., PhD.

138 TABLE OF CONTENTS #1 Rainfall Data Averaging # Normal Ratio Method # Hydrology Short Answer #4 Rainfall Intensity based on Return Period #5 Calculation of Risk #6 SCS Curve Number #7 - Detention Basin #8 Peak Runoff with Different Land Uses #9 Vegetative Cover SCS Determination #10 Time of Concentration with Steel Formula #11 Peak Runoff using Steel Formula #1 Short Calculations for Hydrographs Non-Qualitative Problems #1 IDF Curves # Detention Basin #1 # Detention Basin # #4 Detention Basin #

139 CHAPTER 0 TOPICS Hydrologic Cycle Precipitation Data & Estimation Methods Time of Concentration Rainfall Intensity Flood Prediction Hydrographs Unit Hydrographs Synthetic Unit Hydrographs NRCS Espey Rational Method for Peak Flow NRCS Curve Numbers Reservoir Sizing

140 1 Rainfall Data Averaging When would it be appropriate to estimate missing rainfall data using a simple average surrounding stations? (a) (b) (c) (d) A simple average is never appropriate. The normal-ratio or other similar weighted methods should be used. It is always appropriate to estimate missing data using the simple average. It is only appropriate when precipitation measurements at the surrounding stations are within 10% of each other. It is only appropriate when the stations included in the average are within a 5-mile radius of one another. Refer to Section 0.4, Paragraph 1 & : Normally, missing rainfall data can be estimated using a simple average when the precipitation measurements at surrounding stations are within 10% of each other. When the precipitation shows differences greater than 10%, the normalratio or a similar weighted method should be used. The answer is (c).

141 Normal Ratio Method (Read on your own) Rainfall records for four precipitation stations are summarized in the table. Stations B, C, and D are those located in closest proximity to Station A. What is the estimated precipitation at Station C for 1991? Annual Precipitation Annual precipitation for year indicated (in) Station (in) A B C D (a) (b) (c) (d) 1 inches 7 inches 40 inches 48 inches Because the normal annual precipitation between Station C and the other three stations varies by more than 10%, the normal ratio method should be used for computing an estimate of the missing record. N normal annual precipitation inches P annual precipitation for year of interest inches Subscripts A, B, C, and D refer to the stations. Refer to 1991 data points. N C PA PB PD PC 7 inches N A N B N Eqn 0. D The answer is (b).

142 Hydrology Short Answer (Read on your own) In general, converting an area from natural grassland to ¼ acre lot single family housing will have what effect on time of concentration and amount of runoff? (a) (b) (c) (d) Decrease time of concentration and increase runoff Decrease time of concentration and runoff Increase time of concentration and decrease runoff Increase time of concentration and runoff Natural grassland in an undeveloped condition is less impervious than ¼ acre lots. Therefore, runoff will increase. Due to the increase in impervious area, the time of concentration will decrease. The correct answer is (a).

143 4 Return Period Rainfall Intensity The 5-year return period rainfall frequency- depth-duration curves for a coastal region is shown in the illustration. For a mean annual precipitation (P MA ) of 7 inches, what is the rainfall intensity for a.5-hour storm? (a) (b) (c) (d) 0.40 in/hr 0.60 in/hr 1.5 in/hr 1. in/hr

144 d rainfall depth in t storm duration hr From the illustration, for mean annual precipitation of 7 inches and a storm duration of.5 hours, the rainfall depth is about 1.5 inches. i rainfall intensity in/hr i d 1.5 in 0. in t.5 hr 60 hr The answer is (b).

145 5 Acceptable Risk A manufacturing facility is willing to accept only 1% risk of flooding during its 50 yr design life. What is the annual probability that flooding will occur during the facility design life? (a) 0.01% (b) 0.0% (c) 0.0 % (d) 1% Refer to Page 0-6: Example 0- solves a similar problem for the probability of a flood occurring within the useful life of a plant. This example calculates the probability of a flood occurring during a year s time (i.e. annual probability). n period of interest yr PF annual probability of a flood event R acceptable risk of a flood event occurring R 1% P F Rn % The answer is (b).

146 6- SCS Curve Number Assume the SCS curve number for an area is 50 and that precipitation during a 4-hour storm was 5 inches. The average depth of water (inches) running off the area is most nearly: (a) (b) (c) (d) 0.7 inches 1. inches.5 inches.1 inches Refer to Section 0.16 Q = runoff (inches) CN = curve number S = potential maximum retention of water by soil (inches) P = accumulated rainfall (inches) For CN = S 10 Eqn 0.4, page 0-17 CN Solving for S, 1000 S inches 50 For S = 10 inches and P= 5.0 inches, the runoff depth is P 0. S x 10 Q inches Eqn 0.44 P 0.8 S x 10 The correct answer is (a).

147 7 Detention Basin A community requires developers to store the entire runoff of a 4 hour 10 yr storm on site. The design rainfall at this location is 4 inches. A developer is proposing a 100 acres subdivision on ½ acre lots. What is most nearly the required detention pond volume if the subdivision will be located on a parcel with hydrologic soil group C? (a) (b) (c) (d) 11 ac-ft 17 ac-ft ac-ft 6 ac-ft Using the NRCS method, a curve number, CN, is identified for each combination of land use and soil type. The maximum water storage can be calculated using the NRCS method 1000 S 10 Eqn 0.4, page 0-19 CN Referring to Table 0.4, the curve number is S in 80 Using the NRCS P 0.S P 0.8S Q. 0 inches Eqn 0.44 Over the entire area of the development, the total volume of water to be stored is: 1 ft Q A inches inches100 acres ac ft The answer is (b).

148 8 Peak Runoff with Different Land Uses (a little long) A 11-acre drainage area has the following characteristics and 10-year storm frequency-intensity-duration curve. What is the peak runoff from the drainage area for the 0-minute duration, 10- year storm? Land Use Area (%) Apartments 0 Landscaped open space (park) 5 Light industrial 45 (a) (b) (c) (d) 4 ac-ft/hr 6 ac-ft/hr 7 ac-ft/hr 10 ac-ft/hr

149 Refer to Appendix 0.A on Page A-54. I am assuming these values for the illustration of this type of problem. You would be supplied more definitive values for C. Land Use Area (%) C Land area (ac) Apartments open space (park) Light industrial A C land area for each use (subscript j designates each land use) runoff coefficient for each land use ac C AVE C j A A j j ac ac ac 11 ac 0.40 i storm intensity in/hr From the illustration, the storm intensity is 6 in/hr Q runoff ac-ft/hr Equation 0.6 on Page 0-14: in 1 ft Q C AVE i A hr in ac ac ft hr The answer is (b).

150 9 Vegetative Cover SCS Determination Assume all soils in a drainage basin are in the Soil Conservation Service (SCS) hydrologic soil Group B. Also assume that the vegetative covers are in good condition. The land use is parks and open space. The SCS Runoff Curve Number (CN) for the entire area is most closely approximated by: (a) 4 (b) 54 (c) 61 (d) 81 Determine the SCS Runoff Curve Number for the entire area. For Soil Group B, with good vegetative cover in urban, fully developed open space (parks, lawns), the appropriate SCS Curve Number can be found on Page CN = 61 The answer is (c).

151 10 Time of Concentration with Steel Formula A rainfall-depth-duration curve for a watershed is shown in the following illustration. The values for the Steel formula constants K and b are 180 inmin/hr and 5 minutes. What is the time of concentration for the -hr duration 50-year storm? (a) (b) (c) (d) 15 min 6 min 4 min 75 min d depth of rainfall cm i intensity cm/hg t storm duration hr From the illustration, for a 50-year recurrence interval and h duration storm, the rainfall depth is 9 cm. i d 9 cm 4. cm t h 5 hr

152 For the Steel formula, refer to Equation 0.1 on page 0-5. However, units must be in units of in/hr. cm 1 in i in hr.54 cm hr b constant min K constant in-min/hr T C time of concentration min The Steel formula is i K Eqn t b C t C K i b in min 180 hr in 1.8 hr 5 min 75 min The answer is (d).

153 11 Peak Runoff using Steel Formula A 00 acre drainage area has a suggested runoff coefficient of 0.75 and a time of concentration of 45 minutes. The drainage area is in Steel region #, and a yr storm is to be used for design purposes. What is the peak runoff? (a) (b) (c) (d) 180 ft /sec 80 ft /sec 460 ft /sec 50 ft /sec Use Table 0., page 0-6 K = 106 b = 17 i K Eqn t b C 106 i in hr, pay attention to units Using the Rational Equation Q C i A Eqn 0.6 Q P in hr acres ft 85 sec The answer is (b).

154 1 Short Calculations for Hydrographs (read on your own) A -hour storm over a 00-km area produces a total runoff volume of 6 x 10 6 m with a peak discharge of 400 m /sec. 1. What is the total excess precipitation? (a) (b) (c) (d) 1.4 cm.5 cm.0 cm 4.0 cm V A D P AVE Eqn V 6 x 10 m PAVE 0.00 m. 0 cm AD m 00 km 1000 km The answer is (c).. What is the unit hydrograph peak discharge? (a) (b) (c) (d) 100 m /s-cm 10 m /s-cm 10 m /s-cm 00 m /s-cm Refer to Page 0-8: A unit hydrograph discharge is the peak discharge divided by the excess precipitation.

155 Q QP P m s cm m 1 s cm (Refer to Ex. 0-) The answer is (b).. If a -hour storm producing 5 cm of runoff is to be used to design a culvert, what is the design flood hydrograph volume? (a) 1 x 10 6 m (b) x 10 6 m (c) 1 x 10 7 m (d) x 10 7 m The design flood hydrograph volume for a 5 cm storm is determined by multiplying the unit hydrograph volume by 5. For the unit hydrograph, 6 V 6 x 10 m 6 m V HYDROGRAPH x 10 P.0 cm cm For the 5 cm storm, V m cm 6 x 10 m 7 5 cm 1 x 10 The answer is (c).

156 Non Quantitative Problems

157 1 IDF Curves (Read on your own) How are rainfall intensity and duration and storm frequency related? (a) Intensity decreases as frequency decreases for a constant duration (b) Intensity increases as duration increases for constant frequency (c) Less frequent storms have higher intensity for a constant duration (d) More frequent storms have longer duration for a constant intensity Refer to Figure 0.6, page 0-6 Rainfall and intensity are inversely related. Less frequent storms have a longer return period and are characterized by higher intensity and longer durations, but not usually both. Therefore, less frequent storms have higher intensities for a constant duration. The answer is (c). Detention Basin #1 What method of sizing a storm water detention basin would be appropriate for a small water shed of 0 acres and limited hydrologic data? (a) Modified rational method (b) Non-sequential drought method (c) Routing method (d) Stochastic simulation method See Page 0-0, Section 0-18, 4 th paragraph. The answer is (a).

158 Detention Basin # Which of the following must be considered when designing a stormwater detention basin? (a) (b) (c) (d) Design storm recurrence interval Peak outflow from the basin Storage volume of the basin All of the above See Page 0-0 The design storm is chosen by assumption of risk, economics, and regulations. Peak outflow depends on current and future downstream uses and channel capacity. Storage volume is the largest factor in terms of cost and liability. The answer is (d). 4 Detention Basin # Which of the following, besides storage volume and outlet structure, should be considered in the design of a stormwater detention pond? (a) (b) (c) (d) Downstream impacts Impact of extreme flood events on the structure An effective sediment and erosion control plan All of the above The answer is (d).

159 Groundwater August 01 Instructor: J.B. Jones, P.E., PhD.