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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 DarcyWeisbach 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. 346' 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+~(IQI)(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 leg11 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’7c44 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 IncMad. 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 {frV, (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)’(605§) = 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 ﬁﬁgﬁ’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 HazenWilliams
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_14 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 HardyCross 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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 Spring '08
 Batho,M
 pipe, Darcy–Weisbach equation, plastic pipe

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