Introduction to Fluid Mechanic
Introduction to Fluid Mechanic CE 321
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This 25 page Class Notes was uploaded by Jolie Shields on Saturday September 19, 2015. The Class Notes belongs to CE 321 at Michigan State University taught by Roger Wallace in Fall. Since its upload, it has received 54 views. For similar materials see /class/207369/ce-321-michigan-state-university in Civil Engineering at Michigan State University.
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Date Created: 09/19/15
Lecture 17 Energy Equation 2 First Law for a CV 83 a I epdv ZeaulpoutAautVout ZeinpinAinViiz Qnet in Weelz n Cv First Law for a CV I 60 Z eout 0 out Aout Vout Z em pm Am CV Qnet in Wmajl netm Z pmAm Kn Z pouleul Vout First Law for a CV 59 p V2 u gZ poutAoutVout at 0 2 out CV V2 Z u7gzj Om1me Qnelm 39Wmajmelm Simplifications qu 7 v If the time rate of change of the total stored energy 6 of the contents of the CV is zero steady flow loinAinVin pouleulVoul V2 V2 m Z710111 z 7in Mgzouz Zin 390 out 390 in 239 Qnelin Wsha nelin 1D Energy Equation for Steady Flows pout10m Vow 2 2 V V V V out m m hour hin 2 gZ0ut Zin Qnet in WSW net in Steady Flow with Zero Shaft Power V V V2 V2 out uin Mgzoul Zin 390 out 390 in 2 Qnel in netin Heat transfer per unit mass I m 2 2 pout V out pin in v v gzout 2 gZin out um gnelin p p x J V loss Steady Incompressible Frictionless Flow out irt qnetin 0 loss Steady Incompressible Flow with rncuon V2 V2 p0 0 gzom 2 pm gzm loss 0 2 Work per unit mass W shaft net in W shaft net in m Steady Incompressible Flow with Irlcuon and enart vvorK 2 2 pout V out pin V in 0 gzoul 2 gZin wshaft net in loss Dividing both sides by g 2 2 pout V out pin V in 2 zom 2 zm 115 hL 7 g 7 g h wshaft net in VVshaft net in VVshaft net in S g mg 7Q 11L 2 Head Loss lossg Energy per unit weight NmN m pm 2 pout out Vin 2 zmhS hL out Problem 550 550 A IOOft wide river with a owrale of 2400 ft3s ows over a rock pile as shown in Fig P550 Determine the direction of flow and the head loss associated with the ow across the rock pile 550 Solution 72 abbrmn q le dlnzlc nn aquot 751m m will 612714 a 076 It ML emf77 ewe749s 3 757 and C cuqk e Aead or If in head lax is pox24w aw a umea dyad af flaw if Wtd If 2 head o nym 39vc witcl mi 04yI cay foJQ39 c our nxxume Jr c 39rn offM wmj 50 4551me 26 79m r m ray76 e W My p0a 7Q 7057 2 n M kek awncl we 9ch SIhf Ei 529 DJ no Irtl WrL mmuum z I zy32139V ZI1A t 25 I0 19 I L 550 solution 39 3 V Q 2400 t 6 if I 4 L J M v Q g M1 12 25 z A z Q oa ff 7 7 z ff f A 1 m L 29 27 2332 z zzig AL 032 f and rhae AL I law743m MAUIWWI rgn 7 e 19 FIN fr cmd 7 11 Problem 564 564 Water ows by gmvily from one Iaku to unul 39 as sketched in Fig 1564 at the sleady rate nl ll0gpmWha1 is the loss in available energy associaied with this ow If this sauna amount of loss is associated widi pumping the uid from the lower lake to ihc higher one at me same owratc estimate lhe amount of pumping power required 421 unila 3 S F 6027M or 23 fow 19m yeam a 7 0 sec mb 155556 eads 7 72 055 921 26gt222 f509 16 15 5 5M For pumped fow 79 quot 5867919 1 sec 39on at 556 yield ml pr g 39a 2b 55 Lq gj zu g zigfoy g nef I 0r H m 7 l3 M 2 153 5 quot Sheff 3 L my nef 1h 563 A pump transfers water from the upper reservuir lo the lower one as shown in Fig P563ai The difference in eleva tion between the two reservgirs is 100 ft he friction head loss in the piping is given by KLV22g where V is the average uid velocity in the pipe and KL is the loss coef cient which is con sidered constant The relation between the head added to the water by the pump and the owrate Q Lhrougli the pump is given in Fig 1356311 IfKL 20 and the pipe diameter is 4 in what is the owrate through the pump Pump head ft of water EV le aw n sec 39m l 7 17 569639 2 53 5575 quot7 hr 112 AL oa397 7L m Fm F49 550 1 we conclude fmf 200 mo 42 AP From file pm cm swirlamen 1 KV 5 L2 Pump head It 01 watev A ML z 2 4071121 IN L gag f 2621 1r 41quot j g 7m Cpm m nj 75 a 4143 we 0 4073 at 00 Q 300 0 n 75 reef of 4 mf mes phyp u Scnfc Q 75 1 13 5 568 Water is to be moved from one large reservoir to another at a higher elevation as indicated in Fig P568 The loss in avail able energy associated with 25 ft s being pumped from sections 1 to 2 is 61 V22 ftzsl where V is the average velocity of wa ter in the Sin inside diameter piping involved Determine the amount of shaft power required Section 2 8 in inside diameter pipe Section 1 E 145 aw 14w sec 39mm 13927 scc m 2 1956 lead 7 1 5 V WWH a 271 21 MM ya 20 62 l nefin From fhe volume flanmic w ughh a 25 flquot V1 p392 539zzszj 9 Z M lz 77114 rm 639 39 I m 3 Wm I 94 lt s s ij iz zgffo new in l 6162 19 6 7 39 39 xu m i w 5 S f Z4 A Shq g 566 Gasoline 6 068 ows through a pump M 012 m3s as indicated in Fig P566 The loss between sections 1 and 2 is equal to 0 3V 2 What will the difference in pres sures between sections 1 and 2 be if 20 kW is delivered by the pump to the fluid Q 012 m3s Section 2 D2 02 rn Section 1 FVIM a 556 w 92 74y IA Flow m Itch vn 7 seama v V It li ff r M 2 3 A m rm 1 Wumc guree we webL 3 Z V 2 i W2 g 322 2 7 A 7712 2 7 72112quot m4 5 mum72m of 4 53511 M Mm m 2 m 1523 1 V V127 K i h 02m 393 01mgt