# CE411 final - Name Civil Engineering 41-— Final Exam 22...

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Unformatted text preview: Name: Civil Engineering 41] -— Final Exam 22 December 2005, "£230 ~ 9:30 a.m. This is a closed book, closed notes exam. Problem 1 (20 points) A pipeline extends between two reservoirs as shown below with h] = 80 m and h; = 40 m. The pipeline is horizontal and consists of three pipes in series each 500 m long. Pipe 1 and 3 are 300 mm in diameter and pipe 2 is 200 mm in diameter. All pipes have the same frictional parameter f=0.02. The entrance and exit loss coefﬁcients are 0.5 and 1.0 respectively. Compute the discharge rate and the minimum pressure in the pipeline‘ neglecting minor losses at the junctions. l v‘ [L02 93W @0966? 5’30 1“- 3®+ "glam"? ’ﬁL-Qx'f' £th +’£L'f;.+ "ﬂag a. .L ’- 3 L 2. Li'igﬁghe/Tq/Q 1%sz ?G%DE_ Q . 00 : IgKOu-D‘QXS (\$22- &2— 714x93 x (0.3); 2 1.5 E _>E._"“’__c_,—_*2>< ‘50" 51’ 3‘— ;253977 52 I : ﬁ"x?.’8?< (0-)); ; Etc/w a: _ .-Z __ TC" K‘?3><(0\3)S 622‘ 0 9k 3 62'; # (2L tires : ’9ll—6ut r" KM ‘ D‘t '" ‘ ﬁx is we)“ ' l “I t .5; a. 3’5": : ’aLCx 3 [(454 11304532“: box 5" ragga); rt’vcﬁaxmgﬂ“ “at :5? t”??? 5‘35: 80 t: #0 + (Ell—[0'1 "l— gcfo‘Cf-f‘ Zzghicf-Y-fr 3%?)CQL g i 62:0./{0Cf H075 {(Dho Pressure. to Fm & ,.. - .-l-'-"'mr-’-"‘ '-"’ r _ . _ _. ‘1 ' "‘”--'"--:?m f"- :O @ "837+ "if if}??? " 9‘7 +§9=§u3*”,fﬂ I 3) “i’ﬁLExtl/L mblem 2 (20 points) As a water resources engineer, you are designated to design a pump station in a water distribution system to withdraw water from elevation 100 ft to 200 ft. The length and diameter of /" the pipeline are L #500 ft and D = 1 ft, réspeetively. The roughness of the pipe is k; = 0.0048 inch. (Neglect minor losses). You've decided to choose a product from Peerless Pump Company. type 6AE14. running at speed 11 = 1750 rpm. There are 5 different models in this series. with diameters: 10.38, 1 1.0. 12.0 13.0 and 14.0 inches. respectively. Their performance curves and the manufacture NPSHR are given below. The designed discharge is approximated Q = 1,625 gpm. Which model will you choose, and what is the required motor power (the brake horse power)? (Note: 1 gpm —' 2.228X l 0'3 cfs, viscosity ofwater: v = 1.059x10'5 ftzz’s. speciﬁc gravity ot'water: r: 62.4 tram-i. g = 32.2 ftfs2) HORIZONTAL SPLIT CASE PUIIPB ECTION 1240 j == 1 750 A -Cun;m regret it - - iiﬁi m. J, m "4'43: VELOCITY HEAD mewceo - ._ RPM :3: ' E‘ﬁ Im- """'".;.:: Tom. eve AREA 30.63 50. m. I II _ ______._-I::: - fun-.5333 . _:. smears smarsm. Elm; . PERFORMANCE AT to SP. GR. lmnmmumluuuulmlummu . semaaraaram . m...- . III" III I... 3; Man: ﬁm'" :ﬁ?‘ 260 mi“..m..'“!!!'"" " “""" u-...___ ‘MIQWIIIII IIIII I. u "m '92::“easier-easmurnﬁﬁm - wﬁﬁieaiﬁimﬁﬁmmmm - f l..'.. a ' '-..'l..-.I-D_ ‘1. IIIHHI'JQ ’C'. Imn'un'Iw i..." w n gnu-run. 2 - case-on-nmusmnur "Ir nu :l-iiiutumn-..—niuquuuunar I uni a:' I "Hum. “C‘Iﬂur “a” " 'l 'QI'J'II-l --.'h' ‘-'J-“; .‘3 .l I. l'zln". Eggs; .3 I . I II 'ilIIII =LFII'I5II-:\TI:::I..!.I ‘ _ ‘ _ l - . . _ . "enigma: in: . I I an. l...‘."-‘7 l“ a .U. I I III "In :-.I .jnma .«m- at new I )I‘ ._I "IIII9II l—T'H-FI'“. ﬁ“"..hl .Il- ' . . ill.u..f"!l 1 . ! IIIIII I."l-:'.“u‘m “ “rt-lam; ;" Inn n...:-- I. ..!-;“ . . amazes. III-"I'll. In. anti-n.“ «gm-:3Ri- I “Hallucination-In laamuIIqu-nl-ma .II‘ 'HIIm-m I -. I... “I... I“. I I n I ‘I lining-1m“..- m lllllllllllmlll mi-mmﬁﬁﬁm‘iﬂugg-g ""- m — U-ll—“III-Oi ' out.nan-3:via-tiniii-iiiiiﬁiiii======= I===:===:=Enncnl -- ii; i' .- - _ a III 'II -‘t— . thug --‘- I‘ ‘ I I —-‘. 5”“.“.-”-“-m- ‘- us. moms PEF‘I MINUTE I ﬁgs-m .._. .! LLEH ' .1 ,. [\J Pol/Dora)? Pump: HP = zs+v9wf :(oO'fOﬂ/Zx 4:00 X l ZXBJQ :- 20? 6‘4 % (1“) 77-42 wan ‘ ' I a afar j Format (3 655‘ f 62: /6J§(§}Pm)[ar3~735(0f5)J H7 /o?6"f (ft) Zn W 0515‘ méak , 0&30 ﬂaum CW lkmpéae/W 16W we ==) Cm_ «mpwww awe, e’ﬁt‘ccmcj a. 0‘36 [/1 .gjkt'=;-A-t/«Q. z: 60 r. we {arm M36 : -_—; 530g 52% {fé/ﬁa' X .979; [Cf5)>< _ ggox DB 6 ﬂag-\$3, (w ééal. Problem 3 (10 points} .- Consider again the system in Problem 2. The atmospheric pressure is 30 ft, the vapor pressure is 0.82 ft. What is the maximum elevation (with respect to the lower reservoir) that the pump can be installed? At WMﬁMtgﬁamﬁD/I, mam col/(M ta. me, coma ﬂ/PSHAZ/VPSHR Mpg/{Q @ [H “62]: [(4660, “5°19 69%)] :2 [5 (ft) m P —- O WPS%A:Zstﬂ%‘“”%/ﬂ)fhf+% :Zs+ (§o~031)*-26 .'/I f _, , i‘roblem 4 ( 20 points) An old water distribution system has one pipe with length 2L and diameter D. Due to the increase of system load. a new pipe has to be installed in parallel with the old one to increase the total discharge. The new pipe is made of the same material (assume the same frictional factorf ) as the old one, the length is L and the diameter is D. If the elevation difference of the system will be kept same. in what percentage will the discharge be increased? (Neglect all the minor losses). _ Problem 5 (15 points) A wide rectangular channel with a Manning’s coefﬁcient n of 0.015 carries a discharge (per unit r width) 0154 ft3/(s fl). The channel slope changes abruptly from 0.001 to 0.02. Classify and sketch the flow proﬁle near the change in slope. Drama PM M26 Wm _- g: 5% («Del/SJ The mod 3’0 =3 g7g =ﬂ: “1%? CW " Problem 6 (15 points) Estimate the peak discharge for a watershed that has a 0.1 probability of being equaled or exceeded in any year. The watershed is characterized by a drainage area of 1.0 mil, 3 time concentration of 25 minutes, and a runoff coefﬁcient C of 0.75. An IDF relationship for this watershed is shown below. Mean hm: nsily‘ im‘hu diumr 5 [1} 2(1 3|) 61') 1m 2011 6t!" H40 MIX] Rainfall Duration. minutes 0% probabtit‘éﬁéj‘ b30443 W @W mead/‘5 5L ...
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• A Closed Book, diameter, minor losses, exit loss coefﬁcients, .-Z __ TC, l— gcfo‘Cf-f‘ Zzghicf-Y-fr

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