Exam #1. A. Answer any 1 of the following 2 questions. CEE 371 March 10, Please grade the following questions: 1 or 2

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1 CEE 371 March 10, 2009 Exam #1 Clsed Bk, ne sheet f ntes allwed Please answer ne questin frm the first tw, ne frm the secnd tw and ne frm the last three. The ttal ptential number f pints is 100. Shw all wrk. Be neat, and bx-in yur answer. A. Answer any 1 f the fllwing 2 questins Please grade the fllwing questins: 1 r 2 3 r 4 5, 6 r 7 Circle ne frm each rw 1. Basic Hydraulics (40 pints) As shwn in the schematic belw, water can flw frm the strage tank thrugh the pipes t satisfy the demands at ndes C & D. Relevant data fr the system are shwn in the table belw. Calculate results fr all missing values in the table. Explain and shw all wrk, and assume a Hazen Williams C value f 100 fr all pipes. B 2 C 4 A 1 3 D PIPES NODES N. Length (ft) Diam (in) Flw (gpm) h f (ft) N. Elev. (ft) HGL (ft) Pressure (psi) Demand (gpm) A B C D 350 Step # use Hazen - Williams equatin 1 Q = h f CD L

2 1 calculate flw fr pipe #3 2 nte that headlss fr pipe #2 must be the same as fr pipe #3 (they are parallel pipes) 3 calculate flw fr pipe #2 4 flw in pipe #1 is sum f flws fr #2 and #3 5 flw in pipe #4 is equal t #1 minus demand at nde C 6 demand at nde D must be equal t flw in pipe #4 7 calculate headlss in pipe #1 8 calculate headlss in pipe #4 use HGL and Bernulli Equatin Step # 9 calculate pressure at nde A 10 determine HGL value fr nde B 11 determine HGL value fr nde C 12 determine HGL value fr nde D 13 calculate pressure at nde B 14 calculate pressure at nde C 15 calculate pressure at nde D Q L h L = 10.5 C D HGL = Z HGL = HGL hl P γ Answers in Red belw PIPES NODES Length Diam Flw h f Elev. HGL Pressure Demand N. (ft) (in) (gpm) (ft) N. (ft) (ft) (psi) (gpm) A B C D Water Distributin Pipe Systems (40 pints) a. Fr the pipe system shwn belw (Figure 1; including pipes #1-#4), determine the length f a single equivalent pipe that has a diameter f 10 inches. Use the Hazen Williams equatin and assume that C HW = 120 fr all pipes. Start by assuming a flw f 1000 gpm in pipe #2. Please shw all steps. 2

3 A 1 B 2 3 C 4 Figure 1. Pipe System fr equivalent pipe prblem Table 1. Pipe Data fr Figure 1 Pipe System Pipe 1 Pipe 2 Pipe 3 Pipe 4 Length ft Diameter in Steps 1. Determine headlss in pipe #2; which is the nde B-C headlss Q L h L = C D (1000) 1100 = (120) (10) = 7.87 ft 2. Establish flw fr pipe 3 and 4 based n the nde B-C headlss. Fr pipe # hl CD Q = L = gpm Fr pipe # hl CD Q = L = gpm 3. Determine the ttal flw frm nde B t C Q = = 2713 gpm While nt necessary fr this prblem, yu may calculate the length f 10 inch pipe that is equivalent t pipes 2, 3 and 4. The crrect length wuld be 174 ft. 4. Calculate headlss in pipe 1 based n the recgnitin it must carry the ttal flw fr pipes 2-4 3

4 h L Q L =.5 4. C D = ft Determine the verall headlss by adding the headlss fr nde A-B and B-C h L = = ft 6. Back calculate a 10 pipe length that has that headlss at the ttal system flw hlc D L = 10.5Q 24.67(120) (10) = 10.5(2713) = 544 ft 4.87 B. Answer any 1 f the fllwing 2 questins 3. Ppulatin and Water Use (40 pints) Table 1 cntains ppulatin data fr Durham NC. In 2005, Durham s Divisin f Water Supply and Treatment prvided an average f MGD t its custmers. Table 1. Ppulatin fr Durham NC, Year Ppulatin , , , , , , , , , , , , ,816 a. Using this histrical ppulatin, make ppulatin prjectins fr 2015 and 2030 fr Durham. Use tw different mathematical mdels f yur chice. Discuss the 4

5 prs and cns f the tw mdels yu selected. Clearly explain yur apprach and state all assumptins. b. Using yur ppulatin prjectins frm part a, estimate the average daily and maximum daily demands (in MGD) fr 2015 and c. Calculate the fire demand needed fr Durham fr 2030 using yur ppulatin prjectins frm part a. Express answers in units f gpm and MGD. Slutin: First yu need a strategy t calculate average demand based n predicted ppulatin. One valid apprach is t assume the per capita average demand will remain cnstant (i.e., apply the value yu have fr 2005). This is preferable t use f a natinal average value such as 180 gpcd. Then yu need a strategy t calculate maximum day demand. A valid apprach here wuld be t use the 1.8 rati as dne in class. Finally yu need the fire flw equatin. While nt perfect fr this example, it is a reasnable way f estimating fire flws in the absence f ther data Per capita usage specific t Durham based n 2005 data: 27,650, ,514 Max day demand requires use f a natinal average (e.g., 1.8 x avg demand), since we dn t have any Durham-specific data Fire flw can be determined frm the standard equatin: I then Set t 0 = 1890, and prceeded as fllws fr each mdel: 1. calibrate mdels (evaluating cefficients) 2. apply mdels t predict ppulatin in 2015 and calculate average demand, maximum demand and fire flw fr each predictin applying and calibrating the mdels requires that yu decide which data t use. There are many valid ways f ding this: select al data and d a least squares linear regressin (this is what I ve dne belw) inspect the data and select tw r mre data pints that seem t represent current trends; them may be the last tw data pints, r pssible sme frm the mid 1900s. I wuld nt, hwever just select the first and last data pint. Belw I shw calculatins and graphs fr 4 different mdels. Yu nly needed t d the calculatins fr tw f these. There are many ways f calibrating yur mdel. The simplest is t select 2 dates and use nly thse ppulatin data. While easy, this methd des nt take fullest advantage f 5 ( P )( 1 0. P ) Q =

6 the cmplete data set. A mre pwerful apprach is t use a least squares linear regressin f the linearized data. Linear Mdel t-t0 Y = Y + K Demand (MGD) Fire Flw Ppulati Year n Avg Max (gpm) (MGD) , , , , , , , , , , , , , a ( t - t ) (1c) , , , , Y = Ka= Using nly the last tw datapints wuld get yu a Ka f y = x Series1 6

7 Expnential Mdel Demand (MGD) Fire Flw t-t0 Year Ppulatin Avg Max (gpm) (MGD) ln(pp) , , , , , , , , , , , , , , , , , ln Y = ln Y + K e (t - t ) (2c) LnY = Ke= y = x R² = Series1 7

8 Declining Grwth Mdel z = 2,000,000 Demand (MGD) Fire Flw t-t0 Year Pp = Y Avg Max (gpm) (MGD) ln(z-y) , , , , , , , , , , , , , , , , , Y = Y + ( Z Y )(1 e K d ( t t ) ) (3d) Ln(Z-Y) = Kd= ln(z - Y) = ln( Z Y 0 ) - K d t (3b) y = x R² = Series1 8

9 Lgistics Mdel Demand (MGD) Fire Flw t-t0 Year Pp = Y Avg Max (gpm) (MGD) lgistics % errr ,485 5,485 0% ,679 7,754 16% ,241 10,944-40% ,719 15,412-29% ,037 21,638-58% ,195 30,249-50% ,368 42,038-43% ,642 57,953-32% ,768 79,039-22% , ,294 5% , ,414 3% , ,461-3% , ,346-2% , , , , , ,018 N t K = K N 1+ N0 0 e rt = N + KN rt ( K N ) e K = 600,000 pp r = yr-1 N = 5,485 pp Mdel Actual Water Distributin System Strage (40 pints) a. Given the hurly average demand rates shwn belw (in gpm), calculate the unifrm 24 hur supply (r pumping) rate and the required equalizing strage vlume (in millin gallns). Prepare a cumulative demand graph fr the prblem. Find the equalizing strage using the cumulative demand graph. b. Make an estimate f the ttal required distributin strage vlume fr this cmmunity. Assume that the infrmatin fr part a is fr the average day flw fr a cmmunity f 20,000 peple and that the rati f Q max day t Q average day is 1.8 fr this cmmunity. Als assume that this cmmunity has a backup 9

10 supply it can use in emergencies. Fr design purpses assume a fire duratin f 10 hurs. Clearly state any additinal assumptins yu make. 12 midnight nn AM PM midnight 1000 Answer: Preliminary calculatin fr demand graphs: TIME DEMAND (GPM) hr supply 24 hr supply Cumulative Demand (gal) Cumulative Supply (gal) midnight a.m nn p.m

11 Avg Q 2452 Cumulative flw graph fr calculating Equalizing Strage 4.0e+6 3.5e+6 24 hur pumping 4.0e+6 3.5e+6 Cumulative Demand (gal) 3.0e+6 2.5e+6 2.0e+6 1.5e+6 1.0e MG required strage 3.0e+6 2.5e+6 2.0e+6 1.5e+6 1.0e+6 5.0e+5 5.0e Time f Day (0-24 hrs) Figure 2. Cumulative Demand Figure 2. Cumulative Demand with 24 Hur Pumping Assumptin: use 24 hr pumping; multiple surces s that emergency strage isn t needed. Fire Flw Q fire = 1020 = 1020 ( P )( P ) ( 20 )( ) = 4358gpm = 6.27MGD V fire = Duratin * Q fire = (10/24) days * 6.27 MGD = 2.61 MG 11

12 Equalizing strage based n average daily flw (determined in part 1), must be adjusted fr max daily flw Q equal =1.8 x 0.75 MG = 1.34 MG Ttal Required Strage Q tt = = 3.95 MG C. Answer ne f the fllwing 3 questins 5. Cst Estimatin (20 pints) Bids fr cnstructin f the new 7500 ft lng transmissin main are taken and a yung CEE 371 engineer infrms Oakdale s experienced Directr f Public Wrks that the lw bidder s cst is $2,080,000. The Directr is clearly unhappy and says, They must be nuts! Just a shrt while back in 1970 (ENRCCI = 1381) in Milltwn (a neighbring cmmunity) we installed 2.5 miles f that same size pipe fr nly $220,000 The engineer replies, I think they have given a fair price. Wh d yu agree with? Supprt yur answer quantitatively and state assumptins (Feb 2009 ENRCCI = 8533). Slutin: The directr has a valid pint. Even with the increase in the CCI, the cst per ft in current (2009) dllars was much lwer fr the 1970 Milltwn prject. length cst per ft Year CCI Bid ft mi as bid 2009 dllars $ 220, $ $ $ 2,080, $ $ Multiple Chice. Circle the answer that is mst crrect. (20 pints ttal; 2.5 pints each) a. A typical design perid fr a water transmissin main is 1. 1 year 2. 5 years years years years 12

13 b. A twn has a present ppulatin f 45,000 peple and has grwn by 5000 peple ver the last 10 years. If yu use a linear (arithmetic) mdel fr grwth, what ppulatin wuld yu predict fr 20 years int the future? 1. 50, , , , nne f the abve c. Fr questin b, what wuld the ppulatin be if yu assume expnential grwth? 1. 50, , , , nne f the abve d. A cmmunity has a ppulatin f 17,500. What fire demand wuld yu design fr in MGD? MGD MGD MGD 4. 43,600 MGD 5. nne f the abve e. The average daily demand fr the cmmunity per questin d is 2.1 MGD. Their average daily per capita demand is gpcd gpcd gpcd gpcd 5. nne f the abve f. An estimate f the maximum daily demand fr the cmmunity per questins d and e is MGD MGD MGD MGD 5. nne f the abve g. Drinking water distributin systems 1. Are best designed as grids with many lps 2. Need nt prvide adequate water strage within the pipes themselves 3. Shuld be cnstructed with the smallest pipes that can adequately deliver fire flw and maximum daily demand 4. Shuld nt prvide pressures abve 150 psi 5. All f the abve 6. Nne f the abve h. Cnnecting identical pumps in parallel 1. Results in a higher shutff head dwnstream f the pumps than if they were in series 2. Results in a higher flw rate at nrmal perating heads, than if they were in series 13 Nte that K e = yr-1 Which gives y=56,953 ~ 57,000 Fr 20 years in the future

14 3. Is always the mst ecnmical slutin 4. Is never dne because f high maintenance csts 5. All f the abve 6. Nne f the abve 7. Pwer (20 pints) Water is pumped 8 miles frm a reservir at an elevatin f 120 ft t a secnd reservir at an elevatin f 180 ft. The pipeline cnnecting the reservirs is 48 inches in diameter. It is cncrete with a C f 100, the flw is 25 MGD, and the pump efficiency is 82%. What is the mnthly pwer bill if electricity csts 10 cents per kilwatt-hur? (ignre minr lsses) Slutin: Use the Hazen-Williams and pwer equatins: Q L h L = C D Where Q is in MGD, and D is in feet Then use the Bernulli equatin: 2 2 Recgnizing that velcity drps ut, and that the pressure in an pen reservir is zer, this becmes pwer equatin: MDG( lb ) ft 3 Qγh ft ft P = = η 0.82 MGD watts = 4.0x10 ft lb / s T 5 And finally the results: 1 calculate headlss hl = ft = m 2 Calculate the ttal head ht = ft = m 3 calculate pwer P = watts = KW watts 4 cst Energy = KW-H/mnth Cst = $ 29,230 per mnth 14

15 Gd stuff t knw Cnversins 7.48 galln = 1.0 ft 3 1 gal = x10-3 m 3 1 MGD = 694 gal/min = ft 3 /s = 43.8 L/s 1 ft 3 /s = 449 gal/min g = 32 ft/s 2 W=γ = 62.4 lb/ft 3 = 9.8 N/L 1 hp = 550 ft-lbs/s = 0.75 kw 1 mile = 5280 feet 1 ft = m 1 watt = 1 N-m/s 1 psi pressure = 2.3 vertical feet f water (head) At 60 ºF, ν = x 10-5 ft 2 /s HGL = Z + P/γ EGL = V 2 /2g + Z + P/γ [V 2 /2g + Z + P/γ] 1 + H pump = [V 2 /2g + Z + P/γ] 2 + H L 1-2 Hazen-Williams equatin (circular pipe) Q in cfs, V in ft/s, D in ft: Q = C D 2.63 S 0.54 S = h f /L = 4.73 (Q )/(C D 4.87 ) Q in gpm, D in inches: Q = C D 2.63 S 0.54 S = h f /L = 10.5 (Q )/(C D 4.87 ) Darcy-Weisbach equatin: h f = f (L/D) (V 2 /2g) Q = (g π 2 /8) 0.5 f -0.5 D 2.5 S 0.5 Re = V D/ν Pump Pwer: P = (γ Q H)/η Ppulatin Prjectin Mdels: Y = Y + K Linear: dy/dt = K a ( t - t ) a Expnential: dy/dt = K e Y lny - lny K 2 1 e t 2 - t1 ln Y = ln Y ( ) 0 + K e t t0 Decreasing Rate f Increase: Z - Y2 - ln Z - Y1 K ( K (Z - Y) K Y Y ( )(1 D t t = 0 d d = = 0 + Z Y0 e ) dt t t Fireflw (Q) based n ppulatin (P) Q = 1020 P 1/2 ( P 1/2 ) (Q in gpm, P in 1000s)

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