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Unformatted text preview: 1. MAE 101B — Summer Session II 2009
Homework Assignment #2 Due Friday August 18 Ignoring frictional losses and assuming steady ﬂow (tank is very large) and parallel
streamlines at the exit, what happens to the exit velocity and volume ﬂow rate when
you add a diffuser? (a) Express exit velocity and volume discharge in terms of
pressure head, H, and exit area A2. (13) Express exit velocity and volume discharge in
terms of pressure head, H, and exit area A3. (c) Neglecting friction which is the
correct statement: discharge ﬂow rate at T(top) is greater, equal or less than B
(bottom). Figure (a) Figure (b) Figure _(_c) afﬁrm. Problem I Given: A tank of water with different exit conditions, but each with penile} sireMines
at the exit. Figure (21) Figure (b) Find: (a) What is the exit velocity and volume flow mite of water of figure (:1), expressed
in terms 02" {1113333113 head, H+ and diffuser area, A2. (1')) What happens to the exit valwiry and volume flow raie (if water when a
diffuser is addecig expressed in terms; Elf pressure head, 1}, and diffuser area, A; (ﬁgure (13)). {e} Neglecting friction which is the correct statemem for ﬁgure (e): discharge flow
rate at T(mp] is greater, :2un er Iess than B (bettom). Assumglions: {'1} No frictional bases (2} Steady ﬂow, 32 {i
6%: Equations: (1] Energy equation {‘Bemeeiii since frictionless ﬂow aieng a streamline) {3 V:
— “l”? 'f" g: = mm!
p ._
Sciation:
(a)
1. Swing aieng a streamline from point 1 to 2 in figure (a). use equetien (I).
i + 8H : & + p p 2 2. Note 1’. = $33 = atm. Then plug into new equation and ﬁnd: V: = [23.8 )m 3‘ Since Q = IVA: Q = V2132 = A2 2(2ng {Evangeiista ToﬁicolEi OMS164‘?) — Student of Galiiieo, his early work on vacuums. led
to the invention of manometer, derived frictionless expression for discharge of water
from tank. Know if frictionless that 11 = M g 12 and v = g1 so H = % vzlg or v = (2gH}'ﬂ}
(b) 4. Aéong the. same limes as for part (a), but with a new output area. m :>. Going along ti. streamline from point 1 to 3 in figure (13}, use equation ( l). p l9 2 {3. Note Pl = P3 = aim. Theo plug into new equation and ﬁnd: V3 =(2gl‘1']”2
7. Since Q = VA. Q = v3.5.3 3 A; {ngm 30. with a diffuser the velocity does not change bul the voiume flow is increased by
A3ng. Note that the velocity at 2 must be greater than 5113 by continuity
(incompressible). Consequently the pressure at 2 111113le below ambient pressure. (0)
8. Same!
This was a favorite questioo of a noted hyéro’log’st. Geoxge J. Pissing, and still is often
asked of graduate students during oral exams. The flow rate only éepencls on d, the
pressure head: v = ago)”. Another way of thinking about this is if the discharge ﬂows
were different. then if the spout was moved around to attach to the hole at B — if the flows
were not exactly equal then one would have made a perpetual motion machine. 2. Prove that squeezing the end of your garden hose will only increase the exit ﬂow
speed (V; f V1 > 1) if which of the following conditions is satisﬁed: f L I D > 1;
f L I D = I ; or f L f D < 1. Assume: turbulent ﬂow; all losses are due to pipe frictiOn;
V12 << V22 ; and “f” for piping is the same with and without the nozzle. IIIIIIIII I;
/ D f
E ............. .. J . _ ’ . ........... .. _ ._ _._‘——'I._ _ _.__ 3h —“" 15— Problem 2 Given: Pinched garden hosed! when friction factor, “f”. is 00213221111 Find: Shaw fer which ﬂow Ctmdititm the exit ﬂow wif} incmaxe: fL I D > l; {L 1’ D = l;
f E. I D < i: Assumgiinns: (i) No body fumes, g :0
(:2) Incumpressibie flow
(3) Turbuiem' ﬂaw, cc; =32 =1 Eiguations:
(I) Bernoulli with 31836 i038: head I053=hfl ={ﬁnka £4» 321]“[p3 +er if; +gzz]~Zh, +21% (2) Maéorhead loss (friction factor) :11, = f £5 Solulism: 3*
I. First find ; m” {or the upper came (no nozzle) by plugging in equation (2} and
p assumptitm (3) to equation (I) and mating Va = V; in this case because Ag = A3
and by {musewade!) of mass. poi/“AB 2 93192242. filly; 2: __ pram +IV2§ =paWXK
I} 2 ’7 p go Do the same 31; {I} for the pipe and nozzle, but note Va :t V! a: V1. ’ w D 2 p 2 p 2 2 2 p . Combine the two equations just calculated and find: y ., V} V3 L L
Divide by VZ'ND' on lhe Rafthand side: 6;:1"V—'2£1“ f —]:l = f E
0 " a» . Rearrange abcve and find: I. f 5
v3 v3 ' '
a 1" +[amfi’1]
V; 1‘3
. z, 7 W L .
So 1f 2' <:<: V; , than V”: 2: . So far V; > V; (and mmemhenng V; :> V0 by
{5 the physics of the probiem), {L i I} 3v 1. 3. Air at 15°C is ﬂowing from a large clean room through pipe (a), shown below, with
diameter 150mm. What is the improvement in flow rate, Q, when adding a diffuser,
as shown by the modified pipe, (13)? The diameter of pipe (b) from 12 is the same as for pipe(a), and pipe (b) has a squareedged entrance as does pipe (a).
(a) (b) Problem 3 Given: Flaw from dean mam out a pipe as shown halow“ $35 ....... WWW .... u‘ ....... .... .w = I . _ = f glﬂL ' I /
x : _. KL)"
M 1.3“
3 .
,I 1.5—
3“ I
3' 1:31—
? 1.3w
i i
1
13m WM ~ “3'39
J4
V2 . . .
A?“ L3? L1” J". E ?__.i_ l .  a __L_ _
0 w Q 623 QB 13} .15 2.43 SD 6.53 iﬁ 5.13 113.33
Find: improvement: Lxlﬂﬂgéz ﬂewamass sang»..st
Assumgtians: (1) Vl = Figure 8.15 E O
(E) (13 = a} (3.} 32 = 0
('4) Uniform flow across each cross xeclion
(,5) Neglect frictional losses Eguatinns:
. . V3" 1 W
(l) Tolai head loss: hi: =[ﬂa§«aI+é+gzl][p‘ +532 ' +g81]=h.z +53 p p 2 i»:
. . 173
(Z) MIHOI‘ head loss: hm = Km — + hf $.er
.2 . ._
1'73 A 2
(3} Head loss in diffuser: famfﬁm 2 .3... I.“ __E MC”
’ " 2 A2 ’ Solution: 1. First solve for the flow rate in the origina} pipe (a). Cambine equatiens (i) and
(2) with assumptions (E), (2), and (3} to ﬁmf: 72 "9 _3
ﬂ+a.L+gsa H p3+32V3 +33: =19,“ V2
)1) 2 P 2 ' 2 h.) Use the given information ta ﬁnd: ' kg _ m. N
p. — p2 = gymsgm: = 999—. 9.81—2 (omzsm): 2&5 3 1
m _ s m 3. Leak up the entrance lures coefﬁcient for this pipe in Table 8.2: szﬂéﬁ. 4. Lack up the density ()fair for 15°C in Tnbie AID: pm.r : 1.21:3 In? Elk ., Hz
. . — 2 . U ' 2
Ns'aw calculate the avezage velocny: V, M 25.211113”
* lﬁkgfm‘ 0.5+: 6. Next solve the ﬂaw rate for the modifietﬁ case. $323 33E of the equations and assumptions to simplify the tow} head 10331:}: I’ _2 —J —3 .—; 3
V 'l'jl 1 ‘l
p_i+gl1€l_+gzi w ﬁ+awi+gza =Kmﬂt_ +1“: 1— j}; "C?
p 2 p * 2  2 0 A3 t I By consen'atitm of mass: VgAg = V3133 330 V3 : V3 (mfﬁu) 8. Again look up Km. it is the same as before. Km = 0.5 9. Since the pressure dmp is fixed we want it; ﬁnd the maximum (213 to get the
maximum velocity. Calculata NR = {3.45 f (0.152?) = 64 Using lhix look at
Figure 8.15 and for NR = 6 ﬁnd the maximum C3,. This is at abom: AR 2 2"?
and CF, = 0.62. If). Plug the51: values into the latest equation and find: 2 *2 1
29:“191 =.Vl_ ﬂ_5»§«]—~[A2} ~062
p L A: 3 A} __ 24.5me2 2
IZkgfm] 1.543552 133
J = 6.8m I s
I I. New cnmpare 17}th rates; knowing Q = VgAl and Qumdmcd = V21 mdAz
EMA: "1723»: 6.831%“w .Emm Q’" — Q xl{}(}% : —:—_—‘x100% = —.‘t‘I [10% = 35.3%
Q VBAZ 5.2»: I 3 So the impl'twement by mudifying the pipe output is 30.8% If! 4. Water ﬂows down a 1.2 inch diameter pipe at 78°F with a ﬂow rate of 0.04 ftsr'S. The
pressure at the top and bottom are the same, as shown below. What is the slope of the
pipe? Assume this is turbulent ﬂow in a smooth pipe. . I." MW“
' r
I 'f’ “a.
H.._ H...
M“MV.\M\¥T : a“
my“. H .. ..‘ Jr ’ am ~ ~
tttt my/  , i’mhlem 4 Given: Smomh, turbuleni pipe flow: D = 1.2. inches? '2‘ = 78W, Q = 0.604 113)», p1 = pg Find: Slope of the pipe _ _! .1_
'*. f Assumgtions: (3) Fully developed, a, :53: ' . "'“w. _ ﬁcw _ \\\\ «:92 95:,
EquaLiuns:
. 1?, V5 1?, VB
(1) Handle53:11}. = —+aJ—+g21 w —+a, " +ng
‘ p 2 p ' 2
L 3'73 (2) Major head 105$ (friction factor) = h! = f (3} Fur smth turbuient pipe ﬂow: f m :33:
e . Seiut'ionz 1. Combine equations (3) and (2) and assumption {'1} to find {:zgZEVL‘ L172 V3 , "f '
f——=[ﬂ+al#+g:i}[p—‘+ag ‘ +gzgj=ﬂa ~31) D 2 p 2 p 2
zlng __f 173
L W 2gb
3x
— . f.
2. NowrearrangeQ:VAlu find: V zganw=5ﬂQﬁh A {D " [mm 1f: T
21' — :r
2 '2 121'}: 3‘ Interpohte Tabie AK? [0 ﬁnd the kinematic: viscosity at 73°F. VWF = 'E .86xk0'5_ﬁ3 1’s and Vim”, 2 9.32.1r1lf}""‘(‘f72 {'3 80°F «70%? _ 78°F  ":00?
9.3le 0*“ ft: is — 2 mm" fr 1 3.: VW —1.06xm"5ﬁ9 Ix rm = 9.58x10"" ftI 2's UI Caicuiate the Reynolds number: Re. : Q «t 1 _ (5.09fm)[12£n ﬂ}
_ 12in
v 9.58xm'ﬁﬁ2m = 53:30" 03:6 Find the friction Factm‘ usinv a nation (3). f = _; = {1021
'3 q (5.3x10‘1“—‘
Plug these into the first equalinn:
3 ‘ 3'? —3 53 _ I
‘ ‘va—20Jf12! ( wﬁm‘ =ﬁ.169
L 25:}; Elm if? 282.213353} 2 l2??? Tm Em} {he slope (it) a iitlle geumetry: slope = sin'lﬂalzﬂfL) = shit“). 16*?) = 9A?" 5. 10 From White, assume Newtonian ﬂuid and constant viscosity and incompressible:
(1) please note other assumptions; (2) please list equations needed to solve problem (and
only those equations); (3) please describe in words how you will solve the problem; if
you like provide a numerical answer. 5 = 6.12 Given that flow is laminar and you know Ap, L Q, D so what is it?
Further Assumptions: steady and fully developed. Equations: (1) hf = 32tt[V]Lf(szg) [from cons. of momentum and energy] (2) ﬁsh = AP (3) Re = pUDKu) Substitute (2) into (1) and solve for )1
Substitute it into (3) to check if ﬂow is laminar (Re = 16, u = 0.292 kgf(mvs) 6 = 6.15 You know Ap, p, u, L, D and want to know if V is possible for vertical
flow? Further Assumptions: steady and that straw is barely inserted into milkshake so pressure
is atmosriheric (p1 = pm) and location (1) while just beneath the surface is far enough
from the straw inlet so V1: 0. Equations: (1) pif(pg) + oerzf(2g) + 21 = p2f(pg) + OLVfKZg) + z; + hf [cons of energy] V, = 0 (note V1 is near surface but far enough from straw entrance that V] = 0), z] = 0, AZ
= 0.3, Ap = 3000 Pa, V; = Vtube, OL = 2 since laminar hf = 0.255m — 0.3m — thfi’g < 0 which is impossible since hf can not be negative. If straw was reduced from 30 cm to 15 cm then possible for hf to be positive. If straw was horizontal there will always exist some V so that hf is positive regardless of
the length of the straw. 7 = 6.25 For ethanol take 9 = 789 kgi’m3 and u = 0.0012 kg/(ms); Assume steady and
laminar then check. (I) pﬂpg) + oanKZg) + 21 = p2!(pg) + ochKZg) + 22 + hf [cons of energy] 13, = p2, V. = V; = 0; solve for h (= 0.9m) and substitute into (2) (2) hr = 32u[VlLJ(D2pg) = 128u[QJLf(nD4pg) [from cons. of momentum and energy] Solve for Q (= 0.00684 m3fh) Check Re number
(3) Re = pUDKu) = 4pr(1tuD) = 795 LAMINAR 8 = 6.56 For galvanized pipe take a = 0.15 mm, covert D and Q into metric. Find in, use ﬂ, = 0.0157 for your calculation of pipe length between pumping stations.
Assume can use Moody Diagram (steady and fully developed); assume no minor losses
(1) Re = pUDfu = p(QfA)UDr’p = 292,500
(2) 81D = 0.000123
fD ~ 0015? from Moody Diagram (3) Ap = f (L!D)(pV2)f2 know everything but L, solving for L get 188,000 m = 117 miles
(4) Power = QApr’ Efficiency = 26.5 Mega Wats (35,500 hp) 9 = 6.63 For water at 20°C take p = 998 kg/m3 and p = 0.001 kg/(ms) and 6 = 0.0015 mm for drawn tubing. Assume steady and no minor losses (1) p1f(pg) + (1V12K2g) + 21 = p2l(pg) + OLVfKZg) + 22 + h [cons of energy] p1: pg: atm.;V1= 0 h. = f(LfD)V2f(2g) = AZ — wag) [v2;(2g)][1 + (0.810.04)f] = 1.8m; v2 = 35.32;’(1 + 20f) Guess f= 0.015 then V = 5.21 mls and Re = 208000 But for this Re and ED Moody chart gives f = 0.0158 so use this f to calculate new V
from V2 = 35.32;”(1 + 20f) and get V = 5.18 ms and new Re = 207000 which converges
on Moody Diagram. Thus V = 5.18 ms and Q = V x Area = 0.00651 msfs 10 = 6.64 If liquid in problem 6.63 was viscous enough to produce laminar flow what should be
used for hf? hr = 32uLW(pgD2) instead of hf = f(LlD)V2!(2g) Note if substitute f = 64I’Re then both equations are the same. #5  6.12 A 5mmdiqmeter capillary tube is‘ used as a viscometer for oils. When the ﬂow rate is 0.0?l mo'h. the measured pressure drop per unit length is 375 kPas’m. Estimate the viscosity of the ﬂuid. 15 the flow laminar? Can you also estimate the density
of the fluid? Solution: Assume laminar flow and use the pressure drop formula {6.13}:
$pr = 8Qpﬂ n18}; 3?5000Pa.='m 2 310.0713 anometntonozsr}; u = 0.292 kgfrns
Guess p = 900 kgt‘m3 [consistent with Table 1.4]
Reg = pVDEu e 4pQ."[':I:uD] == 4(900}!.0?l.53600};'[{3.I415)[D.292'}(0.005i : if:
0K Laminar [t is not possible to ﬁnd density from this data, laminar pipe ﬂow is independent of
density. Suppose did not guess laminar ﬂow:
pu'ipgi i a.'v'."'.='{2ga ' Z: = pﬂpgi + ozvglﬁtig]  23 + [‘1p {no minor iosses] atm. pressure at both ends oftube and assume fuliy developed so .52 = ’rhiLr'D) st'ﬂg}
pgAz : on I pf},(i_."D} \5:."3: opr'L = (MED) V3122 V = DEA
Solve for f}, and get about 4 Looking at IVloodj.r Chart {extrapolated for low Re} get Re about [6 11,. = mum Vl."{2g} {assume smooth pipe} { f: 4 is offchart, but ifextrapolated Laminarﬂow line would ﬁnd Re  16!. #6 — 6.!5 (a: L = 30 cm py‘fpg) I HIVJJHZg'] t z; py‘lﬁpg} .' aﬂ"ffﬂg;  23  hr (no minor losses)
{just beneath the shake surface with [3 ~ pm, and far enough from straw that V ~ 0]
0 I 0  o = (3000 Pa}:"[{1200 kgr'msﬂgjll "1.95331 : vimlg + 0.3 +11. hr: 0.255 m 0.3 m 49mm“? : 0 which is impossible (b) L I [S cm pif'lpg) l aIV’;:."I[2g'] i z; = py'mg) i u3V;:."l2g)  z:  by {no minor losses} (inst beneath the shake surface with p ~ pat... and far enough from straw that V  0] 0 + o  0: (3000 Pa]!‘[(l20{) kgr'm‘MMI misgi] l vﬂmgg : 0.15 mt h,—= 0.255 In 0l S m —\*':,m.r'g : 0 which is possible Assume laminar so hr = JZHLV‘lpgﬁll = 32(6,0'}t’0.ISWI[(120{)}{9.81110.008)? = 3823‘!
3323va = 0.355 m — 0.15 m  Viubgg Solve for Vim. and get me = 0.002%? [not sensitive to whoihor a is l or 2 because Vmbcl is
so small) #7 — 6.25
pE."{pg) I aIVJ‘ﬁiZg] + 2; 2 py'mg} l c13¥"3"."t3g} ' 2; ‘ hr (no minor losses}
{1 5 is at top rank surface [2) is at bottom tank ofﬁce so V. ~ V3 ~0 and p;  p; ~ 0
hf: 32uLVJ‘ngD3) : [ammo*{npgoﬂ : IESIDBUIE)“.2)Q‘[n(?89)(9_81){ﬂ.002)"] : [2' 4.31: on
(J 1.0.x 10'“ m worm m‘ h Rel, = pVDs'y = 4an’{ 11ij = 1'95 LAMINAR #8 — 6.56 (a: 70 x If)“ gallooSSday 1 3.07 m“:‘s: V QEA = 3.01% 311221354} 3 2.63 unis;
Rm, 2 pVDflJ I (910212.63): l.22]."{0.0[J]) = 292.5000 53D = 0.!5mmI‘1220mm = 0.000123 From Moody Diagram f1, ~ 0.015? {note instead ofusing the Moody Chan could use Eq 6.49 or 6.48; o‘gf 6.48 provides I], = ODISSS} f]; : {op"URI E": {3 V1] solving for L gel 188.000 to   [7' miles {by Powet = gap"Emceency = 3.0? [113.55 (85: 10"‘Pa — 4x10‘Pa1."o.38 = 2.65 x 10T Watts #9  6.63 it'did not have Moody Plot could use Eq 6.39 to solve for 11, Eq. 6.39 11, = (1.3 logIReua'691)‘:
me energy equation have V = {35.32.50 '9' 2O?”1 5 50 in principal can calculate Re in
terms of f0, substitute into Eq. 6.38 or Eq‘ 6.39 and solve for 1]}. #ll] — 6.64 Another way oflooking at if  can write hf: fr, {'LJ'D] Virugi regardless it'laminar or
turbulent ﬂow. If laminar know i}; = I34."Re so hr = [ﬂip'{pVDm L;"D]IV1.'(2g) = 33pLW{'pDIgJ. so don‘t
have To use Moody plot In ﬁnd i}; vs RED relationship. 11 . The typical shape of small cumulous clouds is as indicated“ in the ﬁgure below (top
of cloud further left than bottom of cloud). Based on boundary layer ideas, explain why it
is clear that the wind is blowing from right to left as indicated. Given: Cioud formations Shawn to right
Find: Why cloud is “leaning”
Solution: Just like a bonndary layer on a plate, the veiocity in the annosphettic boundary layer
increases with ai’titude (at least generaliy]. Thus the top portion of the cloud travels faster
than its base  the clouds tend to tip towards the direction of the wind. That is, the wind is
from ﬁght to left. I —————_—__.._—______________ 12. An atmospheric boundary layer is formed with the wind blows over the earth’s
surface. Typically, such velocity proﬁles can be written as a power law: 11 = ay“,
where the constants “a” and “n” depend on the roughness of the terrain. As is
indicated in the ﬁgure below, typical values are n = 0.40 for urban areas, n =' 0.28
for woodland or suburban areas, and n = 0.16 for ﬂat open country. (a) If the
velocity is 20 ftfs at the bottom of the sail on your boat (y = 4 ft), what is the
velocity at the top of the mast (y = 30 ft)? (b) If the average velocity is 10 mph on
the tenth ﬂoor of an urban building, what is the average velocity on the sixtieth
floor? Pmblem‘lz Given: Aimaspheric boundary layer wind u=ay Fimi: a) 130mm of 83% u =. 20.1% at y = 4ft, ﬁné u at mp of mas; y = 39 ft
b) 10"” Elm: of urban huildéng t3 = 10mph, what is average velncity on 6%?” flow”? Assumgzions: l) Ska:15? flaw
ﬂuaticns: Given in gmﬁlem: 'u : 3y“ Solution: 8) 1. Create the ratin of veiocity at the top of the sail to the veiocity a: the bottom of the 5311
2. Plug in given equaii‘mas for u, and “value of .n in. open country.
3. Calculate u at £1}: tsp of the$31}. b) 1. Again create: the ratia) 9f velocity at the 10"“ ﬂﬁor'to lhai on the 60‘”.
2. Plug in given equation and value of n in urban areas.
3. Caiculaz'e u on me. :50“ floor. ' Numberg:
(a) a = C ya'm J Mam {Tia a mmfam‘
This} 5;  0‘15 . 9"”
. ..  . H WH __ 
i=6?) 0" “3‘29? “225%: (In) a a E yaw!1 W/JBN 5 is a emskw} new, a  0’” M
734%) w 13.A plate is oriented parallel to the free stream as indicated below. If the
boundary layer is laminar, determine the ratio of drag for case (a) to case (b)
using the boundary layer equations. “"1 .. I. is
PLATE
#2
.U
r
H
m {a} (b) Given: Hates oriented parallel to the free stream as indicated below, each with a laminar
boundary layer rimg Find: Ratio of drag force. for case {a} to case (b) ..._ _..__.—( Assumgtions: (I) No pressure gradient Egnations: w w}
(l) 5‘ka friction mqﬁgigm = Cf = I 12.; 2 9.730 tier
(2) ﬁ‘fazfmr drangﬂ _ LEW mﬁm deA
Solution:
.7 f} 1 0.730
1. Rearrange equation (1) toﬁnd rm. z” = G  3" ,3, {W :_pU2
I R3: 2 Re:
2 2. Plug this into equation (2} and integrate. 2 0.730 r.
l
Lami mrfm'l' fwd” 2 l p—Ux
V ﬂ 3. Compare case {a} to case (b), where for case (a); b 2 4E, and L 2 .E' ; for case (b): a, = bdx £0,732: new} E1226, and Lad1?. FD (case a) 3.7%4f)1ll,tt(f}pU3 _ 2 Fatwa b) _ 9..73{§}‘llﬂ{4f)pUJ “ 14.1f the drag on one side of a flat plate parallel to the upstream ﬂow is FD(U)
when the upstream velocity is U, what will be the drag when the upstream
velocity is 2U; or UIZ? Assume laminar ﬂow. Given: Laminar ﬂow ever plate parallel to ﬂew Find: New drag force if: (a) U m 21}; or (b) D = U12 Assumgtﬁms: (1) 1% pressure gmdient Emions:
(a) Skin fffcrian coquﬁefem = C! = I f” = 0330
_ Re;
{2) ﬁicn’on dragAPE, : J TFM
palm acmer
solution:
I Rearrange equatienﬂﬂeﬁad run I r... = 0330 m; f :1 pg: 0'30 _pU2 “Rex * 2 2 2. Plug this into equation (2) and integrate. {173% II ,on
1!
.3. (a) Cmnpare drag force fer U = 2131:; original. F990):ng— m} FDCZU}=2«83F3(U)
my) 033a: MW 4. {13) Compare drag force fer U = U1? :0 eﬁginal. 3
a
04735 pr[—} 0’
F . 2
D(UI2)2____:‘/E a F5(2a')=e.354Fp{U)
PU 3 FECU} I snake 3“ _ _ 1 2 3
F” " Imam waft“ " 39” Mr —0.?3b .uLpU aﬂ—rqh ...
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 Summer '09
 Rohr
 Fluid Dynamics, ........., Pressure head, Express exit velocity

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