wk-11-hmwk

# wk-11-hmwk - WWW 1,9,1 ‘ = 50 95L Q = 150 ﬂab = 3 ft...

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Unformatted text preview: WWW. 1,9,1 ‘. = 50 95L Q = 150 ﬂab = 3 ft . Section (1) urn Water is supplied at 1501:7th and 60 psi to a hydraulic turbine through a 3—ft inside diameter inlet pipe as indicated in Fig. P5.110. The turbine discharge pipe has a 4-ft inside diam— eter. The static pressure at section (2), 10 ft below the turbine inlet, is 10-iu. Hg vacuum. If the turbine develops 2500 hp, de- term ine the power lost between sections (1) and (2). 10‘ ft p2v=W10 in. H3 vacuum ; 02.74.“. FIGURE PS-IIO ’ Section (2) For [/ow boa/ween secﬁ'an: (U and/2), 535-82 Iced; 1% = ' " I i" — V); I — ‘ II power /05: ,oQ[(/; +ﬂi‘; 2,) + 6715] maﬁ ( ) I. Hem given a’miz “d M g :(v/a H (ii-éJW-W s/ngﬂzzz {IX / ’5 )z .. 7031!: (L—lz w s= 5w e» 19 r 5‘ 5/50 3 W=Q=£:{M)=2/.zzﬂ A, 7117,‘ ‘ 77(379)‘ 5 If Ham wﬁierl/aﬁ‘on of mass {/57- 5’3) L 1 V1 = *4/51” = V 92 = ﬂarzz ﬁ)§ii) = M"? ’i" A» o; i 5 Mr 5 From 5,. / r ‘ 3 703 .lé - . flu: W H3 (éoléllﬂw’lq +( ‘; power [05: .. (’94 _: )ﬂ50 3. j in} 16/. fr ___fi_______ M (5-50 ff, [Q /.74’ 51:41:) 5 - 5/0 . +73 ‘ ’ z 1 ' 1» 32.2 fi)ﬂa H) /L + [(2mz 15+) {WV 7.9% ’13. 5‘ i slay. ff -‘ 5 WI-f" ‘ :1 1v 5’ _. 2500/90 W pan/0v (as: = 30 5P i 52/26 § 20 5.12 Oil (80 = 0.88) ﬂows in an inclined pipe at a rate of5 ft3/s as shown in Fig. P5122. If the differential reading in the mercury manometer is 3 ft, calculate the power that the pump supplies to the oil if head losses are negligible. ‘ Link; “Me Cdnﬁrol Volume ShaWn ‘ and W 531997 93144524,,[é523‘0‘ we 7m": The power Supp/fed 59 #1: pump V5 {7% oil [5 14M 53.5-25.- iera/f '7 :IQ 4: = \$6 3’ Q 4; 128/“ do V: Q = —% we 767‘ A 7:4! f3 3 "-I \{2 53st tug-37ft and KR -‘ :25-53’? ‘5 .5 2. 127/202 \$6M If ‘ Fwy-m Hie Manama/er efmﬁgh (52¢ rec/199v, 2.6) we 797‘: 3+3”. f/ +3 I - (Eff/+ij s [07‘ TAM: hi7 “’7 ;1¢//7‘33*.'2~ﬁ+ﬁ+z.)== fa; ail 33;, 2;” Wf‘ +3Zf—IJ’3‘I’ 212’ {9'2 23’” 2;? )3”: anUé/rn‘vy 53;. 0) and [3) We 9:311. V2 343;, +(324‘2551 -éﬂ)«4 fill/3; 1:121 _— +23; #Eiféytgv WI”: 22-2,+‘3ff%1 ./ - A+ Eggaggﬂjc—Zg -/)+[2§.55ﬂ) = 52 7 H '5’ 2(32.2 5.11] Gasoline (50 z 0.) ﬂows through a pump at 0.12 m3/s as indicated in Fig. P5111. The loss between motions (l) and (2)15 loss w th m 0.3 What will Elm differean in pres— “—*‘ " ‘ sures batwecn sections (1) and (2) be if 20 kW is deliwered by the pump to the ﬂuid? ‘ Q = 0.12 «#15 Section (2) 02 = 0.2 m 3 m Pump 3 I Section (1) FIGURE PSJH D1 = 0.1 m From Eg. 5M2 we 997‘ 75% {he ﬂow 74”»: Jeeﬁono) 7‘2 ﬁts/woka ) nef 1;) 2 z ‘ 8—47. :/9 f-Z('?&~2,)— wxéal‘f 7" /05i/ 579m #15 volume ﬂan/rah: w: aé/a/h m3 2‘ v1: 7%: +09; 3 (4/233); a 3.3425,»: 7;” 77“(o.z».) 5 and 19pm cameruaﬁ'an of r3255 (€3.53) if ﬁ/lm; ﬁra/ 2. ‘ 2. V = K- g; = lg f." =‘ ‘rg'sz ’3!) = [5.23 2 I e?— ‘7 (oi/m)» .5 Also . WM]? 20 Mr» z "cflh a 1W0 5- Iv new. a " “'= 24 '3 J F mmﬁﬁ ﬁz)(a./z a?) 4'! z ‘ lass "’ 03:1 =(0'3) ng‘zgsaff/ N .-.- £5702 M"! 7’ 2 17-23) 7k} 52- ﬁdié Water ﬂows by gravity from ear; t9 dumber a5 .; eketched in Fig. P5116 at the steady rate of 80 3pm. What ie the loss in available energy associated with this ﬂow? If this same amount of loss is assooiated with pumping the ﬂuid from the lower lake to the higher one at the same flowrate, estimate the amount of pumping power required. : om ft? I z 72. _ Q 3" m ._...————-——— 5 FIGURE PS.11.6" _ a, (603,-: )(7'1‘3 £7?) . For %he [low M semi/mo (at) 7‘0 "5ecz‘z'dn(é) 55. 5732 leads 7‘19 loss = (ii—2 V3121?“ saﬂ /__’é,_ =/6/0 7‘7"!) 9 b ( “X Xmﬂ) __ m] 5‘ Ear pumped f/ow 74m sedan 5 7‘0 sccﬁ'on a 53.532 yield; mm; (ca 2“ - 25) + loss] = ﬂy? {If}: [9, Hz? 3 062‘ in 79' +1510 1414 2'. Shy 5.11 7 A Hip motor is required by an air ven- tilating fan to produce a 24-in.-idiameter stream of air having a uniform speed of 40 ft/s. Deter- mine the aerodynamic eﬁiciency of the fan. 0 The aerodynamic efﬁeflmc} 0/ ﬁne 79ml 7 I ,5 ,Z a idea/ power rein/“yea” tic/MM pan/er Niall;¢q/ 721.1. 10944.! :hm‘f power reﬁnerith “an” I5 0.75' A; . 7726 idea/ ﬁle/f power refwﬂd) Mala/l 1'5 oé/a/Bed 15am 53.5.32 76r7f/aw Wif/muf /o:5 across 7718 ﬁn; 7.2m: 1 3 2 - . 2. ‘ z. - := m L/auf slalq V .— [07"ng ‘5”: (2.38x/035/13)77‘(sz) a Z ouf owl jit’uf - . If f. H3 4 or MW: 6241355,» Xena?“ I 73654 25 5 /}’ . 7 _ 0.4/35" 5/; — 0. g ' 0.75“ ’7/1 — —-" I 5 [2 Section (2) 5.12! Water is to be moved from one large reservoir to another at a higher elevation as in- 3-;n._;nside dicated in Fig. P5.121. The loss in available en- diameter pipe ergy associated with 2.5 _f_t3/ 5 being pimped from sections (1) to (2) is 61V2/2 where V is- the av— erage velocity of water in the 8-in.-inside diam- eter piping involved, Determine the amount of shaft power required. ‘ 59V Hae [/ow 14pm Secrqonfl) 7‘v Sec/fon(2) 53. 5192 lead/1 7‘0 «1 WM? w nef in From ﬂee Valume 7%me we oév‘m'h 2 V :r. a '5 a z = I .... _ ﬁ ﬂ 7717 .__.__.__.I, z , 2/627. 4/ Z?“ 5’”? 1,: /2/_‘n_. 771a!) 757»: £25. / i {1- ‘W’ = (#94 5/“/5)(/2.5 §7ﬂ32.2;§f)/5‘0 ff) 2 r6479" ﬁt 5’ : ncf in + (60/1/62 f! ’ / /é ./ ._____.___:_2 ﬂ ,4 2 my. 23* £50 __ ‘W 5’— 5.15:}: = 28 l: “Chaﬁ- :8 her I!» 5436 5.123 Water is to be pumped from the large tank shown in Fig. Nozzle area = 0.01 m2 P1123 with an exit velocity of 6 m/s, It was determinedithat the original pump (pump 1) that supplies 1 kW of power tolthe wa- ter did not produce the desired velocity. Hence, it is proposed that an additional pump (pump 2) be installed as indicated to in- crease the ﬂowrate t0 the desired value. How much power must pump 2 add to the water? The head loss for this flow is It], “ 250922, where 11,. is in to when Q is in m3/s. IFIGUFIE £35123 I V: __ ._ 1 ‘V7' %“LZI +hP h1- "' hpghﬁzz *3; , Were use”: so mm=ﬂza=aolm7ww =0.06”’3/.9 :bpynul I’MFZ T/IW, wif/7 b; = 25<9622=250mvé>1=a¢0m ﬁfe/ﬂaws f/Mi‘ p 2 (int/s) I 4473”, I 3' W3 =d‘0/1F :(mox/fﬁ/Ms)(0.,04%)(%73m) : 2.7gx/03LVgL’ =2.7e kw Therefore ) ’ M = Mum,” 4. %Umpz :2.76’ W) MM Mémﬂ = [kW Hence, I WpUMpz : 2'78 kW‘ l/cM/ 5 [’78/(W ———___._. 3 .5438 ...
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wk-11-hmwk - WWW 1,9,1 ‘ = 50 95L Q = 150 ﬂab = 3 ft...

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