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Unformatted text preview: 6.62 Water at 20°C is to he pumped
through 2134]!) ft of pipe from reservoir 1 to
E "u a rate of 3' ft3'i‘s, as shown in Fig. 136.62.
[f the pipe is east iron of diameter 6 in and
the pump is 7’5 pereent efﬁcient, what
horsepower pump is needed? Solution: For water at 213°C, take ,0 =
1.94 slugfftg' and ,u = liJQE—J' slugi‘fts. For
cast iron, take E : 111311035 ft, or t‘i‘d =
ﬂﬂjﬂﬁﬁi’toi‘iﬂi = ilﬂﬂi’i‘. Compute V, Re,
and f: 3 ft
Vzg —= 15.3 —; a _ marsh 212 s we _ 1.94t153n6ﬂE]
,u ZINE—5 Re: =Tﬂ9ﬂﬂD e‘i’d=ﬂ.ﬂ]i'i', medy =ﬂﬂ22? The energy equation, with pl = p; and V, 2 V2 = 1], yields an expression for punip head: .2 q 2
hJrl1 =az+r£L=ison+uossT Emu] “"31 =iso+sso=4son
P P d 2g 149113912}
1 h as“: s
Power: 1?:ng Pzwziismmssozsmhp Ans. r; {IE5 15.6? A straight lﬂcni commercial—steel pipe is 1 km long and is laid on a constant slope
of 5°. Water at EDDIE? flows downward, due to gravity only. Estintate the flow rate iit ntg'i'h.
What happens if the pipe length is 2 km”? Solution: For water at 211°C, take ,0 = 9'98 ltghn3 and ,H = 11.1341] kgl'illS. If the flow is
due to gravity only, then the head loss exactly balances the elevation change: .2
hf: dz=Lsint=l= fLL or f‘vr2 = ..gdsint=i'=2[9.Sl][ﬂ.l]sin 5°=ﬂ.l?l dEg’ Thus the flow rate is independent of the pipe length L if laid on a constant slope. Ans.
For commercial steel, take E = 0.0415 nun, or EM 2 ilﬂlIHJJo. Begin by guessing fully— rough ﬂow for the friction factor, and iterate ‘v' and Re and f: +1121 “2 scams or
f=ﬂ.ﬂ]64, v=' J 4,332; RE:#
ladies 5 Gm, fb,ether = ﬂﬂlT‘J, HIM,” = 3.09 nu‘s, Re : FDEDIJIJ {converged}
Then o = {FIMJEDJ Frans}. = 12.12243 m3fs = 32mm]. Ans. = SEEDDD H.651 For Prob. 6.152 suppose the only punip available can deliver only SI} hp to the ﬂuid.
What is the proper pipe size in inches to maintain the 3 ft"i’s flow rate? Solution: For water at 211°C, take ,p = 1.94 slugi‘ft3 and ,H = lﬂr'QIE—ii slugi'fts. For cast
iron, take E: ﬂﬂﬂﬂﬁﬁ ft. We can‘t specify H'n’ because we don‘t know d. The energy analysis
above is correct and should he modified to replace ‘v’ by Q: 22 2, 3 22 hp=120+f£w=12ﬁ+f“mﬂm=1gﬂ+453i
a 2g :1 2:32.21. :15
P  sosso 4521r  Butalso up: Wfr#235=]2ﬂ+ or: ci”=.i.94r ,ng _ 62mm] _
Guess f zilﬂﬁ, calculate d, EM and Re and get a better f and iterate: r=o.o2o, a=[3.94io.o2i]”5=o.eo2n, Re=ﬂ= Amman] and 2(2.an—5ito.eo2i’
: access if [LED1 or Re = 5891104], =D.ﬂﬂl4l, lvloody chart: fblm,r = 1111213 [repeat] We are nearly converged. The final solution is f=ﬂ.ﬂ2]?, d = +1612 ft = 13 in Am. I535 You wish to water yottr garden with wmjct
lﬂlﬂ ft of g—in—diameter hose whose rough—
nessﬁis llﬂll iit. What will he tlte delivery,
in ft'lt’s, if the gage pressure at the fattcet is
61] lhfflttg'? If there is no nozzle {just an
open hose exit], what is the maximum
horizontal distance the exitjet will carry? Solution: For water, talte ,p: 1.94 slugifta and it: EDGE—5 slugrfts. We are given Ei'n' =
{Hill lt‘liiE] = ﬂﬂllﬁ. For constant area hose, V, = V; and energy yields so 144 r L V2 too V2
“mesh or: H=issrt=r——=r ,
pg l.94l31.2 d 2g {starts spas =ﬂ.ﬂ=163, V=lllﬂ E, Re=4E4ﬂIﬂ
s or W2=4.6=1. Guess r=r,,”,,,,,, then fbemr = 11.13479, y'ﬁnal : 9.91 this [converged] Tlte hose deliyery then is Q: {#415fo 1212?}.9] ]= [LIE] 1 Hails. Arts. {at Front elementary particle—trajectory theory, the maximum horizontal distance X travelled
by the jet occurs at I9: 45° {see figure} and is X = Vgig = {9.91}2E[32.2}I= M15 ft Ans. [h],
wltich is pitiful. You need anneale on the hose to increase the exit yelocity. 15.33 For the system of Fig. P655, let it: =
SI} nt and L = 135 m of cast—iron pipe.
What is the pipe diameter for which the
flow rate will he 7" m'U'h'? Soltttiott: For water, talte ,p = 993 ltgt‘ma
andrr= EDD] ltgt‘ms. For cast iron, talte E:
11.26 mm, but :1 is unknown. The energy
equation is simply “9 P555 srLo2 _ srrissrrosmﬁ _ sass—st sz=som=h = . _ _ , or t:t=o.osstir”5
f It‘gds 12 {9.31 lds a5
1:5 4,00 e
ottess r = cos, a = [toss irons] = aosss m, Re = E = eases, E = noose? lterate: fbem = llﬂi’i‘l, dbemr = 11133136 nt, Rem,r = EﬂTﬂﬂ, Htﬂbﬂm = 11110851], etc. The
process converges to f = 1111361 (I 2: 13.0305 Itt. Airs. ...
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 Spring '08
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