CEE 345 - Benjamin - Winter 2012 - Homework 1 Solutions

CEE 345 - Benjamin - Winter 2012 - Homework 1 Solutions -...

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Unformatted text preview: CEE 345, Part 2, HWl Solutions Note: The following solutions are a combination of those from the solutions manual for the textbook and others that I have prepared myself (hence the different formats and fonts). Please ask Shara or me for help if you have questions about any of them. . This problem requires that we pick some reasonable values for the roughness of the “very rough” pipe that is currently in place and the diameter of the plastic pipe that is to be inserted into it. Values for the roughness of new pipes are given in Table 8.1. The largest values in the table are 0.03 ft for riveted steel and 0.01 ft for concrete. The existing, corroded pipe is likely to be even rougher than these new pipes, but to be conservative, we can choose the value of 0.3 ft. The diameter of a plastic pipe that will fit inside the corroded pipe will depend on how tuberculated the rusted pipe is, but presumably the plastic pipe could be a least 5 inches in diameter, thereby allowing a 0.5 inch gap between it and the wall of the existing pipe. ‘ Next, we can choose a representative velocity of 5 ft/s in the original pipe, and assume a water temperature of 60°F, at which the kinematic Viscosity is 1.21x10"5 ft2/s. Because of the smaller diameter of the plastic pipe, the corresponding velocity that would provide the same flow rate is 7.2 ft/s. These values yield Reynolds numbers of 2.07x105 and 2.48x105 in the original and plastic pipes, respectively. The corresponding friction factors can then be found from the Moody diagram and are approximately 0.034 and 0.015, respectively. Finally, using the Darcy-Weisbach equation, we can determine the headloss per foot of pipe length (hL/Z) as (flD)( VZ/Zg). The resulting values are 0.026 for the corroded pipe and 0.029 for the smooth, plastic pipe. Thus, for these guesses of the roughness and the plastic pipe diameter, insertion of the plastic pipe would not be beneficial. However, if the equivalent roughness of the corroded pipe is substantially greater than 0.03 ft, or if a larger diameter plastic pipe could be inserted, the switch would be beneficial. 3-4-6' To conserve water and energy. a “flow ‘ ' Fm reducer “Sher reducer” is installed in the shower head as shown in Fig. P8..46. If the pressure at point (1) remains _ constant and all 19$§._cxcep_t forthatin the “flow reducer“ are neglected, determine the value of ' $313555", the loss coefl‘icient (based on the velocity in the pipe) of the “flow reducer” if its presence is to reduce the flowrate by a factor of 2. Neglect grav- ity- ‘ I __ _ I FIGURE P8.46 \ ‘\ “if, Q g 462 2 2. WHbe {be reducer éé‘l'g-{é’ +2, 3% +%+Zz ’ where f2= 0, Z,=Zz 4G I x = -——-—-— = 7330 737 70,1 7r(%§g)‘ fl): #46”? and V2"? and QM?) K" V,‘) :e 9(1467’622— 7330“)=8.ovx10‘(002 w Mere p’rfffgé ’1 #3 Will) Me How reducer Me f/owl‘afe is reduced bya fedor of five. 77703, V, =%(733Q) and K‘at’U‘WWQJ MY}; (2) 1543006, by 5mm £273. 0, (1), and (3) we 462‘th 8.07xlo‘pQ2= 2Lp[(flg_—JZQ)2+~(IQ-I)(7-§3Q)z] or ; KL= 9.00 6’}? 13?? 4B9 'I‘he‘pump shown in Pig. P8375 «1% of 250 ft to as: water. Determine the power that the pump tom: wa- ter. The difference in elevation of the two ponds is 2% ft. Pipe length a 500 It Pipe diameter 3 0.75 n Pipe roughness : 0 3:. " {£941- leg-11 M,» = 22. so 7%)! MW; Z/xf .11 40.: +44“) +5,o+/)_K # f ‘ 1‘} 0 ‘23- vz ‘ ' : 5,00 + a. 1; ['{075 ~/‘2"9]2(32.:L) 250 200 or v 1 “lg (667f+/2..9)V =3zzo / E = 3V0 , 0,9443%) V(o.7sff) [9/39 Re 1“ . 2.3mm" 0r (2);, ‘ Re = 6.22 X/oyl/ and from Fig, 8.20: : Re _ 5 r . _ (I) (:0 7773M and ermr .ro/ml/M. Assume f: 0.02 —-—o V=//./jf—’l “fie : 5, 7””! ——Lif= 0.0/1 990.02 ‘ ' ‘ (z) I _ (0 ¥ 4) Avme f— 0.0/2“ V= Izfi‘ £75 ——~ 1% * 7.7X/a‘——v f = 0,0/z/ 2' 0.0/2. T 7W1», V= M4 404/ 3 i =3 3’pr =(e’7-c44 5%) -,¥/o.7sfi)‘(/2,efi)(zsofi) = Mme“ 1? s “,0 {fl = M540 Mow/y 6/)4/7’ ) 6y 6’, 35’ 560/) ’1‘} Hhe Cb/ebrook eqUafion) obfa/n Va: fol/om. Ham Eq. ( 1)) i V:[3zzo/(u7.£ +1150] y‘J W/I/c/I when comb/heal «115% Kg. (2172/95 (4) Re 3 6.flfl04[3220/(567ff/2.%’)]% : 3'53x/06/(A’a’7f +/2,9)Jz’ Am; Me 50/»er (27114790,; ‘th 5/0 :01; , _ 2.51 (5) fi fl 20109 Rel/f" ) ‘ ; By Gambia/[)9 Eqs (Hand (5) We obfain a why/o eymh'm lbw/why only a“: J. _,_ - ,_2.s:(662f+/2.é)’i] w l0q[—-3'53 Xl‘oé' I Using a wmpufe raaf— Mali/y [Ingram ‘la solve qué) 7/1/03 f ‘= 0.0/13, cam/kfenf wi/b m above {rib/and error Inc-Mad. mm w The flowmte between tank A and itank B shown in Fig. P8. 100 is to be increased by (i.e.. from Q to 1.30Q) by the addition of a; sec- ond pipe (indicated by the dotted lines) running from node C to tank B. If the elevation of the free surface in tank A is 25 ft above that in tank 3. determine the diameter, D, of this new pipe. Neglect minor losses and assume that the frietion factor for each pipe is 0.02. WI) {be single fl/fl‘” 4f+g+za Diamete: l). 500 ft long FIGURE P8.100 _ v“ z a =4‘*i‘é*ze*fléz‘é+é£% a; where {530847) lge’légi'rOJzfi‘zsfll E's-=0, and {fr-V, (szhceDfiflg) 2 777”, 2’9: (5+I1)El{1.?l or 25f/g(0.02) (600f500)ff M «m «mungezmusnawwaamv‘v ‘* “‘73: y i i (7%!) 262.2%) I. ’ 3 ° v, = 6.05 gt Hence, Mv, =£6%H)’(6-05§) = we £1 Wi/l) -f/7e sechp/pe /. 300.198 #3) = [.54 2:1? This) 0’=/,54_§_3=02+Q‘4 0,. M:% =#§%}; ‘ I Br {ll/in, flowing from 14 fo§B Ming/i Moe: I and 2’ - = I V2 ' I W 259‘IZLI + ‘2, Isis—L f}; .03; a; (see 57.0)) or 5&1: , 600}? (zafi§)1+ . so?“ V V: 2 (002) (132.1"; 2(32.2;§£) (0 02) (E, H3 2(321g) Hence; V2: 2.60? ' ; 0/14 ' ‘ .3 01... ,91 14 = 5563. {972.60% =0. 5// 2? W: «23 4—0145410.“ *’ we? For fluid flowing from )9 {03 I‘ll/my}: pipfls / 404% 3 zfl=l7q+ ‘3 =£T€fg2 fifi-gfi’wé”, g=%=’:~03£= _I_;~1_ I 77795; ; [.3] )2 ‘g D”- Dy. 3 600 Fl (2 85%): 500g i 25!! (0.02) T61”) 2 (32.251) No.02.) 03 2 (32.1%) or D3 = 0. 66sz No/e -' k/i/é Me pervade/'5 give/J, fie 30/0790” Ii: «70/79 semi/[Va {a ‘ I‘Wfldfl‘f 9mm lie #70 éFo/ctv/cflficwj 5. Designate the pipes from reservoirs A, B, and C, as pipes l, 2, and 3, respectively. We know from the system geometry that the headloss from the top of reservoir A to the junction point must be 12 ft less than that from the top of reservoir B to the junction point, and 33 ft more than the headloss from the top of reservoir C to the junction point. Assuming that all this headloss occurs in the pipes, we can write: 1qu2 2 11Ll +12 ft hi“3 = h“ —33 ft Thus, we can guess a value for hm, compute the corresponding hL values in pipes 2 and 3, and use the Darcy Weisbach equation to determine the velocities and flow rates in the three pipes. Then, we can test to see if continuity is satisfied at the junction, i.e., if the flow rate into the junction equals the flow rate out. If it doesn't, we can adjust the guess of hm until this criterion is met. By programming the critical calculations into a spreadsheet, it is fairly easy to find the correct result by a brute force, trial and error approach. Alternatively, if we want to be a bit more sophisticated about the search, we can use the equation from the handout [Ahj 2 *n2 Q” /2 to home in on the correct result more quickly. This result is: l 1' Lu VA 2 1.89 ft/s from the reservoir to the junction VB 2 3.3l ft/s from the reservoir to the junction IQ = 4.09 ft/s from the junction to the reservoir 6. We can address this problem in two steps, first finding a pipe that is equivalent to the three pipes in parallel (#2, 3, and 4), and then finding one that is equivalent to the series of pipes that includes #1, #5, and the one that is equivalent to #2, 3, and 4. All the calculations are summarized in the following table, which is copied from a spreadsheet, with K being the coefficient in the expression h,‘ = KQ". In this case, the Hazen-Williams equation in BG units is used to relate headloss to flow rate, so n is 1.85 and K is 4.73 ( 1/0487 ) /C;?,? . When identifying the properties of a pipe equivalent to the three pipes in parallel, we can specify any two of the three parameters: 1, D, and CW. I chose to specify 1 = 3000 ft and C Hw: 100, leading to the result that the diameter was 20.75 in. Then a pipe with l = 5000 ft and CW»: 100 that would be equivalent to equivalent to this pipe plus the other two in series was identified, with the outcome that such a pipe equivalent to the whole network would have a diameter of 20.4 in. n 1.85 Pipe Length Diameter CHW K [Cl/,1 A 1ft)»+ (m) 1 L 2750 21 130 0.105 3.387 2 3250 12 130 1.888 0.709 E 3 2250 * 10 130 g 3.176 0.535 4 2750 14 130 0.754 1.165 5 B 4750 27 130 r 0.053 4.885 LEq 2+3+4 3000 20.75 100 g 0.197 2.410 Eq_1-4 5000 20.42 100 0.354 7. The distribution of the flow between the two branches in the middle of the system can be ascertained by making guesses for Q1 and Q2 that satisfy the continuity equation at both junctions, and then using the Hardy-Cross approach to improve the guesses until the headlosses through the two branches equal one another. Starting with a guess of 40 m3 per second flowing through each branch, after four iterations, the headlosses through the two branches differed by less than 1%. At this point, the flow rates through the two branches in the headloss in the entire system worse follows: Q 218.69 m3/s Q2 261.31m3/s h L,Iol 2 56.6 m ...
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This note was uploaded on 03/31/2012 for the course CEE 345 taught by Professor Batho,m during the Spring '08 term at University of Washington.

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CEE 345 - Benjamin - Winter 2012 - Homework 1 Solutions -...

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