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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 ReynaIda numbers, for example: the ﬂew ia turbulent since the
inertia fl'Ji'CE'} are large relative to the viscous, forces: and thus the aislenus fern: ea cannot prevent the random
and rapid ﬂuctuaticna 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 ereas—aeetien a K .5: Re 4.4: where D__ _ 4:21: _ 20b
” p 2(a+b} (ax—b} is, the hydraulic diameter. 31“: 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 ilJﬂrm‘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 pipe 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 erossseeticma] are; and diLiding the uesult by l aiﬂee = 1" A = [1" ' 2:].4c . «DEE r IL‘IﬂK ' SIQE 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‘ﬁs. respectively. Analysis The pressure drop across the pipe and the cros ssectional 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‘ [Hoistsf
IESIILL iisqsosts lbin‘ﬁ a); 120 a} (b) For uphill ﬂow with an inclination of 33'". we have El = +3 D“. and se'  32.21hinﬁ."s3
ltbf .' '. i}. nocriz 5:11.0109 ftats ‘1 psi IN .2 \_ 1 11st
i. E —. =15: psi
14¢1bﬁ‘ft‘ ...3:.3 lbmﬁ"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'asjlgisoa)I. lpsi Vupsu; = 33.1meftss] “‘
11st arouses ftals (N .‘l [cj For downhill flow with an inclination of 2D". we hate B = 2 El", and  20o {aP— ng sin {assol “amam = 113’“: 1 K _[1os —(—16.2] pawns1: a): “noisea3 32.2 ionsas.l x”
liﬂljﬂtllE 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 = _ = —‘:= 9.24 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_. 332 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
lcg"ms. 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.3fidm.'s A: _ n92 4 _ ﬁTIiDﬂltﬂjz 4
,oE’D _ {999.1kg‘m 5:1{6366 m"sj{D.Dr1 m)
,u Luann—3 lcg"ms
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 mm , are Lt’2 30 sass 3 hi =—I=f— =s_c+15?3 “1 ﬂ=2d4m pg o 2g 0.04m 2(9.Eln.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—rR'} The velocity proﬁle in thi5 case 13 given 1.1}; 'L uﬂrj=4(1—rj £le C'ompanug the two relations above give3 the maximum velocitjr 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)mommemo: 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"nts.
ﬁnalists The pressure drop across the pipe and the crosssectional 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+ ‘lkginfsl Eamonm3
“'3” 123a: 123I:C.2=llcg:'lnsj{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 roams 3 H15 in) sin 8”]:[Illtlli not llzg m—‘sl reasonsmoms 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..=ms]l;13mj I. leam3 =2.24>:1ﬂ_5 mats The ﬂow rate is the highest for downhill flow case, as expected. The arerage ﬂuid s'elocitjf' 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'IIEICIllTni"ﬂﬂlﬂlﬁ m} _ y {1'24 kg"rns 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. grasitjf' opposes uphill flow. but helps downhill ﬂow. Granty 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 wellrounded and sharpe 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 sharpedged 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=T3'[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 = 1.34x1o3 mtrs Wailrounded 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 liftschill"1 m3."s. Note that the frictional
losses cause the ﬂow rate to decrease by 1.5% for wellrounded entrance. and 18.5% for the sharpedged
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}. Properﬁes 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 rf P. _
—'+Q;T+Z]+fluumIJ =  —ﬂ']  +IE+h1umj+fli —> —'+ﬂ'_1—=—_+Gr1  +111
PS —§ ' PS 3 PE —3 PE 
Solving for P2 and substituting,
[WEIursz l
P2 =a +p=——gs..>
l 2 .I
 1.051s Rinses 3 1 ’  = (3 no lcPa) + (was kg"mj jelw —{9.s1m.'=. 3(187 m}l=l is 2 JKIDGG kgm."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—TG]
—'+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~lag311’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.
 Fall '08
 SMITH

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