Sol 7 - 6.62 Water at 20°C is to he pumped through 2134]!)...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
This is the end of the preview. Sign up to access the rest of the document.

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‘ft-s. For cast iron, take E : 111311-035 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: =Tfl9flfl-D 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 h|Jrl1 =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 lfl-cni 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 l-tghn3 and ,H = 11.1341] kgl'ill-S. 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 ilfllIHJJ-o. 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'ft-s. 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 [LED-1 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 slugrft-s. 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‘m-s. 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 = 1111-0851], etc. The process converges to f = 1111361 (I 2: 13.0305 Itt. Airs. ...
View Full Document

Page1 / 4

Sol 7 - 6.62 Water at 20°C is to he pumped through 2134]!)...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online