Fluid Mechanics PS5_sol

Fluid Mechanics PS5_sol - B—EC Reynolds number ia the...

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Unformatted text preview: B—EC Reynolds number ia the ratio of the inertial force; te- viscnua forces, and it aen’ea a3 a criteria fer determining the flow regime. At large Reyna-Ida numbers, for example: the flew ia turbulent since the inertia fl'Ji'CE'} are large relative to the viscous, forces: and thus the aisle-nus fern: ea cannot prevent the random and rapid fluctuatic-na nf'the fluid. It is defined as follows: _I’D la" [0} For flaw in a circular tube of inner diameter D: Re [59} Fur flew in a rectangular duct of ere-as—aeetie-n a K .5: Re 4.4: where D__ _ 4:21: _ 20b ” p 2(a+b} (ax—b} is, the hydraulic diameter. 3-1“: The region flonl the tube inlet to the point at which the boundary layer merges at the eenterline is called the hydmalwamto ailments region: and the length of this region is called ilJ-flrm‘yndmk entry length. The entry length is much longer in laminar flow than it is in turbulent flow. But at very low Reynolds numbers: L. is TE:le small [Lil = 1.2L} at Re = ID}. 340‘: The wall shear mess t“- is. highest at the tube inlet where the thickness. of the houndalj: layer it. zen; and decreases graduth ten the fillly developed value. The same is true fer turbulent flow. B—I9C Yes: the volume flow rate in a circular pip-e 1with laminar flow can be determined by measuring the velocity at the eentei'line ill the full}r dEL'EleEd region multiplying it by the eross-seeticma] are; and diLiding the uesult by l aiflee = 1" A = [1" -' 2:].4c . «DEE r IL‘IflK ' S-IQE The pressure readings across a pipe are given. The flow rates are to be determined for 3 different orientations of horizontal. uphill. and downhill flow. si _-lssimipn'ons l The flow is steady and incompressible. 1 The entrance effects are negligible, and thus the flow is full}r developed. 3 The flow is laininar (to be verified}. 4 The pipe int'ots'es no components such as bends. T.‘als'es. and connectors. 5 The piping section inrolt'es no worl: des‘ices such as pumps and nirhines. Properties The density: and dynamic viscosity of oil are giren to be p = 36.8 lbin-‘fi! and ii = DUI-'8 tbm-‘fi-s. respectively. Analysis The pressure drop across the pipe and the cros s-sectional area of' the pipe are dP=P1 —P; = 120—14: 1'36 psi .4. = .le :‘4 = miss .512 it}2 4 = [1.001354 a2 {a} The flow rate for all three cases can be determined from Vt [oP—pgl sin sumo4 i=125fi 123.:11: 1.s'here '9 is the angle the pipe makes with the horizontal. For the horizontal case. B = Cl and thus sin 6 = D. Therefore. _ seas” _ tisspsompss'izm‘ [Hoist-sf IESIILL iisqsosts lbin-‘fi- a); 120 a} (b) For uphill flow with an inclination of 33'". we have El = +3 D“. and se' | 32.21hin-fi."s3 ltbf .-' '-. i}. nocriz 5:11.0109 ftats ‘1 psi IN .2 \_ 1 11st i. E —. =15: psi 14¢1bfi-‘ft‘ .--..3:-.3 lbm-fi-"s /. H51ll'=| . n1: "fin: is. .-'s' :- sinlL ‘3 pg” ' 5 :SISSlb 'ftJ 3"”‘Pt j 1th D (dP—pglsin Hist)" _ issue _ (1Uti—‘ld..‘- psi].rl:|C|.5.-"12 ail [iiiissai mastersism-'a-sjlgisoa)I. lpsi Vupsu; = 33.1me-ftss] “‘ 11st arouses ftals (N .‘l [cj For downhill flow with an inclination of 2-D". we hate B = -2 El", and | 20o {aP— ng sin {as-sol “am-am = 113’“: 1 K _[1os —(—16.2] pawns-1: a): “noise-a3 32.2 ions-as.l x” liflljfltll-E lbin-‘fi- gurus} _ 1psi ltbf The flow rate is the highest for downhill flow case. as expected. The rerage fluid s'elocitjs' and the Reynolds number in this case are v 0.0126 a-‘s y = _ = —‘:= 9.2-4 as .4. 0.001354ft' _ ,oI’D _ {56.s1um.-'s33{9.34 fi-"sj{[l.i.-'12 ft) _ Jet CLUE—ES Ibiza-"ft -s 1.s'liich is less than 1300. Therefore. the flow is In minor for all three cases. and the analysis ah ore is valid. =s.s1zs flats Re T3? Discussion Kote that the flow is driven by the combined effect of' pressure difference and grant—f. As can be seen flood the calculated rates above. gravity opposes uphill flow. but helps downhill flow. Gravity has no effect on the flow rate in the horizontal case. Downhill flow can occur es'en in the absence of' an applied pressure difference. _-_..r... _ _._.. ..._.r_. 3-32 The flow rate through a spec1fied water pipe is given. The pressure drop, the head loss. and the pumping power requirements are to be dete1mined. s' Assimrpn'sns l The flow is steady and incompressible. 1 The entrance effects are negligible, and thus the flow is full}; developed. 3 The pipe involves no components such as bends. valves, and connectors. 4 The piping section involves no worlc devices such as pumps and turbines. Propern'cs The density and dynamic visco site of water are given to he on = 999.1 lcg-"m3 and In = 1.13331 llII'J l-cg-"m-s. respectively. The roughness of stainless steel is [1.0132 mm. Analyst's First we calculate the average velocity and the Reynolds number to determine the flow regime: I’=i V ——flm3m "5 =d.3|fidm.-'s A: _ n92 4 _ fiTIiDfl-l-tfljz 4 ,oE’D _ {999.1kg-‘m 5:1{6366 m-"sj{D.Dr1 m) ,u Luann—3 lcg-"m-s which is greater than 40m. Therefore. the ‘flow is turbulent. The relative roughness of the pipe is 2 m'5 s a :‘D=¥=ixlfl" [1.04 m The friction factor can be determined from the Mood}; cha1t. but to avoid the reading error. we determine it from the lC‘olebroolc equation using an equation solver [or an iterative scheme}, Re= _ \_ . . x .' .' j ‘3' 1 5 L =—lfl]og- 5' D+ 25L —> L =—I.Ulog- “E — " 1. “ff II. 3.? Rats NH" I.__ 3.; 2.235s1o3: z It givesf= {infill 5T3. Then the pressure drop, head loss, and the required power input become L ’2 some! 3 sass 3“ - '~ 3P=3LPI =f_PI =.;}_.;}1573 3am wli flhgag kpa n 2 aw. in 2 ._1o:+o lcg - m-"s _.t 1 m-m- ,- are Lt’2 30 sass 3 hi =—I=f— =s_c+15?3 “1 fl=2d4m pg o 2g 0.04m 2(9.El|n.-s') _ ' _ 5 -. a "I “"9" H—19'IkW Rpm—VdP—(Uflflfim .-sj{a39k_Pa} _ _ . '._l L'Pa -m:'.-'s , Therefore. usequ power input in the amount of 1.91 kl. " is needed to overcome the flictional losses in the pipe. Discussion The fliction factor could also be determined easile fiom the explicit Haaland relation. It would givef= 0.9155, which is sufficiently close to {2213157. Also. the friction factor corresponding to e = t} in this case is 13.0153. which indicates that stainless steel pipes in this case can he assumed to be smooth with an error of about 2%. Also, the power input detelmined is the mechanical power that needs to he imparted to the fluid. The shaft power will he more than this due to pump lJJEfi'lClEflC'ff'; the electrical power input will be even more due to motor inefficiency. 3—35 The velocit}r profile in fully developed laminar flotv in a circular pipe is given. The average and ruminum velocitiee as well as the flotv rate are to be determined. Asstuflpn'om The flotv is stead}: laminar, and fully developed. Almiyiii The velocrt}r profile in fitll}r developed laminar flotv in a circular pipe is giver. by I" J" 2 .- 'I . "- UV? = “male. 1‘ my} = 4(l—r-R'} The velocity profile in thi5 case 13 given 1.1}; 'L uflrj=4(1—rj £le C'ompanug the two relations above give-3 the maximum velocitj-r to be um}; = -l Ill-'1. Then the average velocity and volume flow rate become u . aim-"5 =&=_=2mf5 2 -.| Tr ‘ .2'. . ti = Fm .4r =r-’a.lg (3R1)mom-memo: m311=fl.0[r251 m3}; _-_..r... _ _._.. _.r_. 343 The pressure of oil in a pipe which discharges into the atmosphere is measured at a certain location. The flow rates are to be determined for 3 different orientations. Assumptions 1 The flow is steady and incompressible. 1 The entrance effects are negligible, and thus the flow is full}r developed. 3 The flow is laminar (to be verified}. 4 The pipe ins‘ols‘es no components such as hends, takes. and connectors. 51 The piping section involves no work devices such as pumps and turbine s. Propem'cs The densit}r and dynamic 's'iSCDSil'f of oil are gis'en to he ,3 = ETIS lcg-"rn3 and p. = 0.24 lcg-"nt-s. finalists The pressure drop across the pipe and the cross-sectional area are dP= P1 — P2 =13 —EE=4'.-' kPa A: = .le 4 = moms 111)]:1 =1.?s? you—4 all LJI {a} The flow rate for all three cases can be determined front. ICl'il p; _ [dP—ngsinEjra‘J'l lESgrL where E! is the angle the pipe makes with the horizontal. For the horizontal case, B = U and thus sin El = 0. Therefore, r' ‘-. Q _oPaD'l _ I:=l.7l:Pfl},TI:C'.C'l}-1uj+ ‘lkg-infsl Eamon-m3 “'3” 123a: 123I:C|.2=llcg:'ln-sj{liinj 11s 1km '-. z i=1.52x1[l_5 mats [5} For uphill flow with an inclination of 3". we hate 3 = +3”. and . [as — pg: sin toad4 “Jpllill = .' s._ [HT-'DCfl Pa —(s?s kg-‘ml roam-s 3 H15 in) sin 8”]:[Illtlli not llzg- m—‘sl reasons-moms m] _ 1mm! =1.[ll]:-<1l]_5 malls [cj For downhill flow with an inclination of 8”, we hate El = -E°. and g _ (a? — ,iEL sin Qjafl 4 doanhfll _ 128%: _ [{4 END Pa — {ass agar! ){931 111’s: 1:1 3 n1) s1nfl—3”}],rl:fl'.fl 13 in): 1 kg -n:L-'s2 N” 12s{u.:4kg..=m-s]l;13mj I. lea-m3 =2.24>:1fl_5 mats The flow rate is the highest for downhill flow case, as expected. The arerage fluid s'elocitj-f' and the Reynolds number in this case are c 2.24x1o'5m3s w _ |. =—=__'—_4.=U.1.¢.'1IL'S A: 1.36:le in‘ _ ,oIIt _ {3.76 lag-"ni'E'IIEICIll-Tni-"flfllfllfi m} _ y {1'24- kg-"rn-s 4-. I Re which is less than 2300. Therefore. the flow is laminar for all three cases. and the analysis above is valid. Discussion Kote that the flow is dris'en by the combined effect of pressure difference and grant—3'. .fiis can he seen flow the calculated rates above. grasitj-f' opposes uphill flow. but helps downhill flow. Grant-y has no effect on the flow rate in the holizontal case. ___..r... _ _ ._.. 8—53 Water is to be withdrawn from a water reservoir by drilling a hole at the bottom surface. The flow rate of water through the hole is to be determined for the well-rounded and sharp-e dged entrance cases. Assimlpn'ons l The flow is steady and incompressible. 1 The rescivoiris open to the atmosphere so that the pressure is atmospheric pressure at the free sulface. 3 The effect of the lcinetic energ}r correction factor is disregarded, and thus or, = l. Analyst's The loss coefficient is K; = 0.5 for the sharp-edged entrance, and K; = GU33 for the 1well—rounded entrance. We take point 1 at the free stuface ofthe reservoir and point 2 at the exit ofthe hole. We also take the reference level at the exit of the hole {:3 = '3'). Noting that the fluid at both points is open to the atmosphere (and thus P1 = P2 = PM) and that the fluid velocityr at the free surface is zero [F] = U}, the energy equation for a control volume between these tu-‘o points [in terms ofheads) simplifies to P‘ + F11+ +P P] + F32 + +iw +‘ —> I32 h —a—: ...= a1 31:1':J‘| :=nr-. +. FE ] 2g ] puip..| Pg _ 2E _ 1111032.- I l _ 25 L 1a-‘here the head loss is expressed as bi = K; . Substituting and solving for I’; gives —3 F3- Ff ,1 _ :l=fl'] _ —KL —? 2g31=T-3'[or2+fiti} —} Es Es since on = 1. Note that in the special case offi'j = '3', it reduces to the Toricelli equation F] = .ullgz; __ as expected. Then the volume flow rate becomes L}: I I, =flfliiofa I 135] d r .n Substituting the numerical values, the flow rate for both cases are determined to be '— yflfim {23:l _.~.r[n:+_r:+1jm}2 I'Eflgfllnn-‘sljfimj 4 'Illl—KI 4 t; 1+o.c+3 = 1.34x1o-3 mtrs Wail-rounded entrance: = -'I. :03... Irsgzl =3(G_Glim}1 ||2[9.Elm."s3}(3m} =‘1.‘|1:<1E|'3m3ls 4 tII1+gL s ‘It 1—n.5 Sharp—edged snout: cs: = Discussion The flow rate in the case of frictionless flow (K; = U} is lifts-chill"1 m3."s. Note that the frictional losses cause the flow rate to decrease by 1.5% for well-rounded entrance. and 18.5% for the sharp-edged entrance. 3—61 r—‘L horizontal water pipe has an abrupt expansion. The water velocity and pressure in the smaller diameter pipe are given. The pressure after the expansion and the error that would have occurred if the Bernoulli Equation had been used are to be determined. Assuurpfi'urrs lThe flow is steady. horizontal, and incompressible. I The flow at both the inlet and the outlet is fully developed and turbulent with kinetic energy corrections factors of n] = c1; = 1.06 {given}. Proper-fies TWe take the density ofwater to be p = 1901] kg-‘m3. Analyst's Noting that p = const. [incompressible flow}: the downstream velocity of'water is 2 , , a no: :‘4 or? ans _ ml=mI —> pr-'].41=prgal —> r = 1r-'-_= 1. r-'-_= glfl=flflflufisfil=15m=s A; aflj :‘4 Di (Dlfimj' The loss coefficient for sudden expansion and the head loss can be calculated floor I a] z 1 .2 - a x: ' :l - ' D' l' 0.03" KI=i1_ seal =|1_ 1: =.1— 2 =D.:'~62§ s Alarm,- s D}; 1- D'l'fi r in] [I _. 2 hi =3; +=flfljfilfijfl= 2.3Tm —S spams?) Noting that :1 = :2 and there are no pumps or turbines involved the energy equation for the expansion section can he expressed in terms ofheads as P. if P. 1’3 P- r-f P. _ —'+Q;T+Z]+fluumI-J = - —fl'] - +IE+h1umj+fli —> —'+fl'_1—=—_+Gr1 - +111 PS —§ ' PS 3 PE —3 PE - Solving for P2 and substituting, [WEI-ursz l P2 =a +p=——gs..> l 2 .I - 1.051s Rinses 3 1 -’ - = (3 no l-cPa) + (was kg-"mj jelw —{9.s1m.-'=.- 3(187 m}l=l is 2 JKIDGG kg-m."sl..\1 kjfim' = 322 kPa Therefore. despite the head (and pressm'ej loss: the pressure increases from EDIE? kFa to 321 kPa after the expansion. This is due to the conversion of dynamic pressure to static pressure when the velocity is decreased. When the head loss is disregarded: the downstream pressure is determined from the Bernoulli equation to be P- F3 P1 r2 P- 1": P. I’1 I’j—T-G] —'+l——'l='—2+—"2 —’ —'+l='—2 —} 5:34‘1‘311- as Es as Es as 13 as is z Substituting. . Io 3— 15 3“ 1m lkP '- P2 =[3m1.5E*.a}+|:1I:J+:u:1~lag-311’j;M — 4l=34TkPa 2 ._1mm kg - m-‘s st'm‘ Error =lefl— P3 = 34? — 322 = 25 IrPa Note that the use ofthe Bernoulli equation results in an error of [3 47 — 3232312 = 0.073 or 18%. Therefore, the error in the Bernoulli equation is Discussion It is common knowledge that higher pressure upstream is necessary to cause flow, and it may come as a surprise that the downstream pressure has Increased after the abrupt expansion: despite the loss. This is because the sum ofthe three Bernoulli terms which comprise the total head: consisting of pressure head, velocity head. and elevation head. namely [Ping + *5 f"2."g+ :]. drives the flow. With a geometric flow expansion: initially higher velocity head is converted to downstream pressure head, and this increase outweighs the non—convertible and non—recoverable head loss term. ‘. / ...
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This note was uploaded on 10/15/2008 for the course ME 307 taught by Professor Smith during the Fall '08 term at Howard.

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Fluid Mechanics PS5_sol - B—EC Reynolds number ia the...

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