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) 1 5000 16 A 450 560 0 2 800 10 B 320 0 3 1000 8 14 C 380 500 4 2500 12 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 5000 16 2013.68 18.56 A 450 560 47.67 0 2 800 10 1348.80 14 B 320 541.44 95.96 0 3 1000 8 664.88 14 C 380 527.44 63.89 500 4 2500 12 1513.68 22.21 D 350 505.23 67.27 1513.68 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 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
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 12 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 (1000) 1100 = 10.5 4.87 (120) (10) = 7.87 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 = 660.16gpm Fr pipe #4 0.54 2.63 hl CD Q = 0.281 0. 54 L = 1052.87gpm 3. Determine the ttal flw frm nde B t C Q = 1000 + 660 + 1053 = 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
h L Q L =.5 4. C D = 16.80 ft 10 87 5. Determine the verall headlss by adding the headlss fr nde A-B and B-C h L = 7.87 + 16.80 = 24.67 ft 6. Back calculate a 10 pipe length that has that headlss at the ttal system flw. 4.87 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 27.65 MGD t its custmers. Table 1. Ppulatin fr Durham NC, 1890-2005 Year Ppulatin 1890 5,485 1900 6,679 1910 18,241 1920 21,719 1930 52,037 1940 60,195 1950 73,368 1960 84,642 1970 100,768 1980 100,831 1990 136,611 2000 187,035 2005 208,816 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
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. 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: 27,650,000 132.4 208,514 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 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 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 =1020 01
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) 0 1890 5,485 10 1900 6,679 20 1910 18,241 30 1920 21,719 40 1930 52,037 50 1940 60,195 60 1950 73,368 70 1960 84,642 80 1970 100,768 90 1980 100,831 100 1990 136,611 110 2000 187,035 115 2005 208,816 27.65 a ( t - t ) (1c) 125 2015 188,514 24.96 44.93 12,082 17.41 140 2030 213,118 28.22 50.80 12,717 18.32 Y = -16524 Ka= 1640.3 Using nly the last tw datapints wuld get yu a Ka f 4356 250000 200000 150000 100000 50000 0 50000 y = 1640.3x 16524 0 20 40 60 80 100 120 140 Series1 6
Expnential Mdel Demand (MGD) Fire Flw t-t0 Year Ppulatin Avg Max (gpm) (MGD) ln(pp) 0 1890 5,485 8.6097724 10 1900 6,679 8.8067236 20 1910 18,241 9.8114271 30 1920 21,719 9.9859427 40 1930 52,037 10.85971 50 1940 60,195 11.005345 60 1950 73,368 11.203243 70 1960 84,642 11.346186 80 1970 100,768 11.520576 90 1980 100,831 11.521201 100 1990 136,611 11.824893 110 2000 187,035 12.139051 115 2005 208,816 27.65 12.249209 125 2015 352,992 46.74 84.13 15,563 22.43 12.7742 140 2030 550,290 72.87 131.16 18,314 26.39 13.2182 ln Y = ln Y + K e (t - t ) (2c) LnY = 9.0742 Ke= 0.0296 14.54 14.52 14.5 14.48 14.46 14.44 14.42 14.4 14.38 y = 0.0009x + 14.518 R² = 0.9129 0 20 40 60 80 100 120 140 Series1 7
Declining Grwth Mdel z = 2,000,000 Demand (MGD) Fire Flw t-t0 Year Pp = Y Avg Max (gpm) (MGD) ln(z-y) 0 1890 5,485 14.505911 10 1900 6,679 14.505313 20 1910 18,241 14.499495 30 1920 21,719 14.497739 40 1930 52,037 14.482295 50 1940 60,195 14.478098 60 1950 73,368 14.471284 70 1960 84,642 14.465415 80 1970 100,768 14.45696 90 1980 100,831 14.456927 100 1990 136,611 14.437907 110 2000 187,035 14.410474 115 2005 208,816 27.65 14.398387 125 2015 187,288 24.80 44.64 12,049 17.36 14.4103 140 2030 210,548 27.88 50.18 12,653 18.23 14.3974 Y = Y + ( Z Y )(1 e K d ( t t ) ) (3d) Ln(Z-Y) = 14.51796 Kd= -0.000861 ln(z - Y) = ln( Z Y 0 ) - K d t (3b) 14.54 14.52 14.5 14.48 14.46 14.44 14.42 14.4 14.38 y = 0.0009x + 14.518 R² = 0.9129 0 20 40 60 80 100 120 140 Series1 8
Lgistics Mdel Demand (MGD) Fire Flw t-t0 Year Pp = Y Avg Max (gpm) (MGD) lgistics % errr 0 1890 5,485 5,485 0% 10 1900 6,679 7,754 16% 20 1910 18,241 10,944-40% 30 1920 21,719 15,412-29% 40 1930 52,037 21,638-58% 50 1940 60,195 30,249-50% 60 1950 73,368 42,038-43% 70 1960 84,642 57,953-32% 80 1970 100,768 79,039-22% 90 1980 100,831 106,294 5% 100 1990 136,611 140,414 3% 110 2000 187,035 181,461-3% 115 2005 208,816 27.65 204,346-2% 125 2015 253,762 33.60 60.48 13,660 19.68 253,762 140 2030 332,018 43.96 79.13 15,199 21.90 332,018 N t K = K N 1+ N0 0 e rt = N + KN rt ( K N ) e K = 600,000 pp r = 0.035 yr-1 N = 5,485 pp 250000 200000 150000 100000 Mdel Actual 50000 0 0 20 40 60 80 100 120 140 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 9
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 Answer: Preliminary calculatin fr demand graphs: TIME DEMAND (GPM) 10 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 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 11
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 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 1970 1381 $ 220,000 13200 2.5 $ 16.67 $ 102.98 2009 8533 $ 2,080,000 7500 $ 277.33 $ 277.33 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 12
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. 50,000 2. 55,000 3. 57,000 4. 65,000 5. nne f the abve c. Fr questin b, what wuld the ppulatin be if yu assume expnential grwth? 1. 50,000 2. 55,000 3. 57,000 4. 65,000 5. nne f the abve d. A cmmunity has a ppulatin f 17,500. 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 2.1 MGD. Their average daily per capita demand is 1. 1750 gpcd 2. 175 gpcd 3. 120 gpcd 4. 80 gpcd 5. nne f the abve f. An estimate f the maximum daily demand fr the cmmunity per questins d and e is. 1. 2.1 MGD 2. 3.8 MGD 3. 5.9 MGD 4. 21 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 = 0.0118 yr-1 Which gives y=56,953 ~ 57,000 Fr 20 years in the future
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 = 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 25 62.4 3 100.289 ft ft P = = 1.54 1.3558 η 0.82 MGD watts = 4.0x10 ft lb / s T 5 And finally the results: 1 calculate headlss hl = 40.289 ft = 12.28009 m 2 Calculate the ttal head ht = 100.289 ft = 30.56809 m 3 calculate pwer P = 400141.534 watts = 400.1415 KW watts 4 cst Energy = 292303.391 KW-H/mnth Cst = $ 29,230 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 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 = 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 D 2.63 S 0.54 S = h f /L = 4.73 (Q )/(C D 4.87 ) Q in gpm, D in inches: Q = 0.281 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 (1-0.01 P 1/2 ) (Q in gpm, P in 1000s) 2 1 15
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