CEE 371 Octber 8, 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) 1 4000 18 A 450 560 0 2 1000 12 B 320 0 3 1200 10 14 C 380 750 4 2500 14 D 350 Step # use Hazen - Williams equatin 1 Q = 0.54 2.63 h f 0.281 CD L
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 + 4.87 HGL = HGL hl 2 1 1 2 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) 1 4000 18 3014.93 17.65 A 450 560 47.67 0 2 1000 12 1931.36 14 B 320 542.35 96.35 0 3 1200 10 1083.57 14 C 380 528.35 64.29 750 4 2500 14 2264.93 22.10 D 350 506.25 67.71 2264.93 2. 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 12 inches. Use the Hazen Williams equatin and assume that C HW = 120 fr all pipes. Start by assuming a flw f 1200 gpm in pipe #2. Please shw all steps. 2
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 900 1100 800 1000 ft Diameter 16 10 8 10 in Steps 1. Determine headlss in pipe #2; which is the nde B-C headlss Q L h L = 10.5 4.87 C D (1200) 1100 = 10.5 4.87 (120) (10) = 11.03 ft 2. Establish flw fr pipe 3 and 4 based n the nde B-C headlss. Fr pipe #3 0.54 2.63 hl CD Q = 0.281 0. 54 L = 792.2gpm Fr pipe #4 0.54 2.63 hl CD Q = 0.281 0. 54 L = 1263.4gpm 3. Determine the ttal flw frm nde B t C Q = 1200 + 792.2 + 1263.4 = 3256 gpm While nt necessary fr this prblem, yu may calculate the length f 12 inch pipe that is equivalent t pipes 2, 3 and 4. The crrect length wuld be 421 ft. 4. Calculate headlss in pipe 1 based n the recgnitin it must carry the ttal flw fr pipes 2-4 3
h L Q L =.5 4. C D = 5.80 ft 10 87 5. Determine the verall headlss by adding the headlss fr nde A-B and B-C h L = 5.80 + 11.03 = 16.83 ft 6. Back calculate a 12 pipe length that has that headlss at the ttal system flw. 4.87 hlc D L = 10.5Q 16.83(120) (12) = 10.5(3256) = 644 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 Hadley MA. In 2000, Hadley prvided an average f 0.28 MGD t its custmers. At that time 55% f the Hadley ppulatin was n city water. Table 1. Ppulatin fr Hadley, MA, 1930-2000 Year Ppulatin 1930 2682 1940 2576 1950 2639 1960 3099 1970 3760 1980 4125 1990 4231 2000 4793 a. Using this histrical ppulatin, make ppulatin prjectins fr 2015 and 2030 fr Hadley. Use tw different mathematical mdels f yur chice. Discuss the 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 2030. 4
c. Calculate the fire demand needed fr Hadley 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 2000; taking int accunt that 55% f the ppulatin is using the city water). This is preferable t use f a natinal average value such as 180 gpcd. Next yu need t decide if the percent n city water will remain at 55% r if it will change (if it des change, it s likely t g up). Finally, 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 Hadley based n 2000 data: 280,000 106.2 4,793 0.55 Max day demand requires use f a natinal average (e.g., 1.8 x avg demand), since we dn t have any Hadley-specific data Fire flw can be determined frm the standard equatin: I then Set t 0 = 1930, and prceeded as fllws fr each mdel: 1. calibrate mdels (evaluating cefficients) 2. apply mdels t predict ppulatin in 2015 and 2030 3. 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 late 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 the cmplete data set. A mre pwerful apprach is t use a least squares linear regressin f the linearized data. 5 ( P )( 1 0. P ) Q =1020 01 Y = Y + K a ( t - t ) (1c)
Linear Mdel Demand (MGD) Fire Flw t-t0 Year Ppulatin Avg Max (gpm) (MGD) 0 1930 2,682 10 1940 2576 20 1950 2639 30 1960 3099 40 1970 3760 Y = Y + Ka ( t - t ) (1c) 50 1980 4125 60 1990 4231 70 2000 4793 0.28 85 2015 5,165 0.30 0.54 2,265 3.26 100 2030 5,668 0.33 0.60 2,371 3.42 Y = 2314.3 Ka= 33.537 ( P )( 1 0. P ) Q = 1020 01 Using all f the data gets yu a Ka f 33.54 6,000 5,000 y = 33.537x + 2314.3 4,000 3,000 2,000 Series2 Linear (Series2) 1,000 0 0 10 20 30 40 50 60 70 80 Expnential Mdel Demand (MGD) Fire Flw t-t0 Year Ppulatin Avg Max (gpm) (MGD) ln(pp) 0 1930 2682 7.8943181 10 1940 2576 7.8539931 20 1950 2639 7.8781553 30 1960 3099 8.0388348 40 1970 3760 8.2321742 50 1980 4125 8.3248213 60 1990 4231 8.3501937 70 2000 4793 0.28 8.4749118 85 2015 5,488 0.32 0.58 2,334 3.36 8.6103 100 2030 6,338 0.37 0.67 2,503 3.61 8.7543 ln Y = ln Y + K e (t - t ) (2c) LnY = 7.7943 Ke= 0.0096 6
8.6 8.5 8.4 8.3 8.2 8.1 8 7.9 7.8 7.7 Hadley y = 0.0096x + 7.7943 0 20 40 60 80 Hadley Linear (Hadley) Declining Grwth Mdel z = 7,000 Demand (MGD) Fire Flw t-t0 Year Pp = Y Avg Max (gpm) (MGD) ln(z-y) 0 1930 2682 8.3705476 10 1940 2576 8.3947995 20 1950 2639 8.3804567 30 1960 3099 8.2689882 40 1970 3760 8.0833286 50 1980 4125 7.963808 60 1990 4231 7.9262415 70 2000 4793 0.28 7.6993894 85 2015 4,938 0.29 0.52 2,216 3.19 7.6314 100 2030 5,228 0.31 0.55 2,279 3.28 7.4800 Y = Y + ( Z Y )(1 e ln(z - Y) = ln( Z Y 0 ) - K K ( t t) d d t ) (3b) (3d) Ln(Z-Y) = 8.4891326 Kd= -0.0100911 7
8.6 8.5 8.4 8.3 8.2 8.1 8 7.9 7.8 7.7 7.6 Hadley y = 0.0101x + 8.4891 0 20 40 60 80 Hadley Linear (Hadley) Lgistics Mdel Demand (MGD) Fire Flw t-t0 Year Pp = Y Avg Max (gpm) (MGD) lgistics % errr 0 1930 2682 2,682 0% 10 1940 2576 2,934 14% 20 1950 2639 3,192 21% 30 1960 3099 3,454 11% 40 1970 3760 3,716-1% 50 1980 4125 3,976-4% 60 1990 4231 4,231 0% 70 2000 4793 0.28 4,477-7% 85 2015 4,828 0.28 0.51 2,192 3.16 4,828 100 2030 5,150 0.30 0.54 2,262 3.26 5,150 K N t = K N 0 1 + N0 e rt = N + KN rt ( K N ) e K = 7,000 pp r = 0.015 yr-1 N = 2,682 pp 8
6,000 5,000 4,000 3,000 2,000 Mdel Actual 1,000 0 0 20 40 60 80 4. 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 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 1000 12 nn 3150 1 AM 950 1 PM 3250 2 900 2 2830 3 875 3 2732 4 850 4 3050 5 900 5 3350 6 1840 6 3515 7 4100 7 4500 8 3850 8 4345 9 2850 9 2610 10 2200 10 1100 11 3050 11 1050 12 midnight 1000 9
Answer: Preliminary calculatin fr demand graphs: TIME DEMAND (GPM) 24 hr supply 24 hr supply Cumulative Demand (gal) Cumulative Supply (gal) midnight 12 1000 0 0 a.m. 1 950 58500 147118 2 900 114000 294235 3 875 167250 441353 4 850 219000 588470 5 900 271500 735588 6 1840 353700 882705 7 4100 531900 1029823 8 3850 770400 1176940 9 2850 971400 1324058 10 2200 1122900 1471175 11 3050 1280400 1618293 nn 12 3150 1466400 1765410 1 3250 1658400 1912528 p.m. 2 2830 1840800 2059645 3 2732 2007660 2206763 4 3050 2181120 2353880 5 3350 2373120 2500998 6 3515 2579070 2648115 7 4500 2819520 2795233 8 4345 3084870 2942350 9 2610 3293520 3089468 10 1100 3404820 3236585 11 1050 3469320 3383703 12 1000 3530820 3530820 Avg Q 2452 Cumulative flw graph fr calculating Equalizing Strage 10
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+6 0.747 MG required strage 3.0e+6 2.5e+6 2.0e+6 1.5e+6 1.0e+6 5.0e+5 5.0e+5 0.0 0.0 0 2 4 6 8 10 12 14 16 18 20 22 24 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 )( 1 0.01 P ) ( 20 )( 1 0.01 20 ) = 4358gpm = 6.27MGD V fire = Duratin * Q fire = (10/24) days * 6.27 MGD = 2.61 MG 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 = 2.61 + 1.34 = 3.95 MG 11
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 $720,000. 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 (Sept 2009 ENRCCI = 8586). Slutin: The engineer is crrect. The new bid price is a bit better than the 1970 price, accunting fr the differences in length and increases in the CCI. length cst per ft Year CCI Bid ft mi as bid 2009 dllars 1970 1381 $ 220,000 13200 2.5 $ 16.67 $ 103.62 2009 8586 $ 720,000 7500 $ 96.00 $ 96.00 6. 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 3. 25 years 4. 50 years 5. 300 years b. A twn has a present ppulatin f 45,000 peple and has grwn by 10,000 peple ver the last 20 years. If yu use an expnential mdel fr grwth, what ppulatin wuld yu predict fr 20 years int the future? 1. 47,982 2. 55,000 3. 57,857 4. 65,441 5. nne f the abve Nte that K e = 0.012566 yr-1 Which gives y=57,857.14 Fr 20 years in the future c. Fr questin b, what wuld the ppulatin be if yu assume linear grwth? 1. 47,982 2. 55,000 3. 57,857 12
4. 65,441 5. nne f the abve d. A cmmunity has a ppulatin f 37,500. What fire demand wuld yu design fr in MGD? 1. 5.9 MGD 2. 62.8 MGD 3. 0.6 MGD 4. 43,600 MGD 5. nne f the abve e. The average daily demand fr the cmmunity per questin d is 7.3 MGD. Their average daily per capita demand is abut 1. 195 gpcd 2. 175 gpcd 3. 155 gpcd 4. 135 gpcd 5. nne f the abve f. A gd estimate f the maximum daily demand fr the cmmunity in questins d and e is. 1. 5 MGD 2. 18 MGD 3. 21 MGD 4. 29 MGD 5. nne f the abve g. Drinking water distributin systems 1. Are best designed as grids with many lps 2. Must prvide adequate water strage within the pipes themselves 3. Shuld be cnstructed based n the average daily demand 4. Shuld prvide pressures f at least 150 psi 5. All f the abve 6. Nne f the abve h. Cnnecting identical pumps in series 1. Results in a higher shutff head dwnstream f the pumps than if they were in parallel 2. Results in a higher flw rate at nrmal perating heads, than if they were in parallel 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 7 miles frm a reservir at an elevatin f 290 ft t a secnd reservir at an elevatin f 880 ft. The pipeline cnnecting the reservirs is 36 inches in diameter. It is cncrete with a C f 100, the flw is 20 MGD, and the pump 13
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 = 10.6 4.87 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 20 62.4 3 684.7 T ft ft watts 6 P = = 1.54 1.3558 = 2.186x10 watts η 0.82 MGD ft lb / s And finally the results: 1 calculate headlss hl = 94.703 ft = 28.86541 m h 2 Calculate the ttal head ht = 684.703 ft = 208.6974 m 3 calculate pwer P = 2185507.81 watts = 2185.508 KW 4cst Energy = 1596513.46 KW-H/mnth Cst = $ 159,651 per mnth 14
Gd stuff t knw Cnversins 7.48 galln = 1.0 ft 3 1 gal = 3.7854x10-3 m 3 1 MGD = 694 gal/min = 1.547 ft 3 /s = 43.8 L/s 1 ft3/s = 449 gal/min g = 32 ft/s2 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 = 0.3048 m 1 watt = 1 N-m/s 1 psi pressure = 2.3 vertical feet f water (head) At 60 ºF, ν = 1.217 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 = 0.432 C D2.63 S0.54 Q in gpm, D in inches: Q = 0.281 C D2.63 S0.54 S = hf/l = 4.73 (Q)/(C D4.87) S = hf/l = 10.5 (Q)/(C D4.87) Darcy-Weisbach equatin: h f = f (L/D) (V 2 /2g) Q = (g π2/8)0.5 f-0.5 D2.5 S0.5 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 lny2 - lny1 K e = t 2 - t1 lny = lny0 + 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 (1-0.01 P 1/2 ) (Q in gpm, P in 1000s) 2 1 15
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