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Unformatted text preview: L"? Determine the dimensions of the coef
ﬁcients A and B which appear in the dimen
sionally homogeneous equation dlx dx
W+Adt+Bx—0 where .x’ is a length and I is time. 7 dt1 4*
{1.7"} m 7'3 + mm = o
Since Cac/p ‘l'erm musf have ﬁe same dimemfms:
[A] [LT'] :3 [1.7"] 5I1m‘ __ 0V mmmmm  I 1.10 i'l‘he pressure difference, Ap, across a I IcdsitthL'ZT), p the blood density (ML‘3), D“ ' partial blockage in an artery (called asteriosis) is the artery diameter, A0 the area of the unolb
approximated by the equation structed artery. and Al the area of the stenosis.
#V AU 2 Determine the dimensions of the constants KL, 3P = Kr “5 + K“ — 1) PW and K“. Would this equation be valid in any sys— tern of units? where V is the blood velocity, ,u the blood Vis—
2.
1/ [£0 .. t] V
A P = k V A95— + k“ A. f m M [till—Hm] «italif ‘53 l Ellil" Since each term mast have ﬁve same dime/45mg, My and K“ are d/Jrnensi'm/ess, 774145, We 63mm”
1': a 7enem/ hamzoyewoas quail[on 474% Moa/c/Le
va/ic/ m my ﬂows/skml sysk’m 075 emit5. es. 1.51 The viscosity of a ﬂuid plays a very important role in
determining how a ﬂuid ﬂows. (See Video V1.1.) The value of
the viscosity depends not only on the speciﬁc fluid but also on
the ﬂuid temperature. Some experiments show that when a
liquid, under the action of a constant driving pressure, is forced
with a low velocity, V, through a small horizontal tube, the
velocity is given by the equation V = K/p. In this equation K
is a constant for a given tube and pressure, and y is the dynamic
viscosity. For a particular liquid of interest, the viscosity is given
by Andrade’s equation (Eq. 1.11) with D = :5 X 10'7 lb  s/ft2
and B = 4000 °R. By what percentage will the velocity increase
as the liquid temperature is increased from 40 "F to 100 °F?
Assume all other factors remain constant. V ° "' Vq. ? ' =
’1: Increase in V = I ’,°° ° actor: and £10m Eismdcz) ‘
%‘—i}:wo = .... xléo
K/Iu‘io' "/0 Increase in V = R‘ﬁm Andmde’: eguaein'w a”
z ‘7 : (‘la’Fi‘féo) .
/U'fo° ' 5x10 8 if .
000
and /(/1 .7" 'Eﬂl'Dqe (loo‘Ff‘fw)
l 0‘
Thus I, From 5%, (31' « _., boo
, 5X“) e 5'00 “'tncras‘e m V "=“ _ i 0,,
[5 e 5xlb7e, sea = 13421. As shown in Video V1.2, the “no slip" condition
means that a ﬂuid “sticks” to a solid surface. This is true for
both ﬁxed and moving Surfaces. Let two layers of ﬂuid be
dragged along by the motion of an upper plate as Shown in Fig.
P154. The bottom plate is stationary. The top ﬂuid puts a shear stint bit the upset slate, and ma lower ﬂuid puts a than was on the bottom plate. Determine the ratio of these two shear
stresses. ' ‘ ' Fluid 1 Fiuid 2 H2 mls+ I FIGURE P1.54 a a (may; to,» Surface:
Far «clMidL : M5 2. ZOE
7; ‘‘ A ('33 )aam and” W m M) = M”
Thus ’ ﬂ _ , _
(who? sump“: = 20 %z : 1
TL?on swim. L0 £1 1; /— w, 15‘! A layer of water ﬂows down an inclined
ﬁxed surface with the velocity proﬁle shown in Fig. H.517. Determine the magnitude and dhrec~
tion of the shearing stress that the water exerts
on the ﬁxed surface for U = 2 m/s and h =
0.1 m. T—‘ﬂ '7" FIGURE P153 77mg.J ai '/’he #Ikec/ surﬁace (31:0) 43 7:0 «5 50 771472‘ 3 {2 lm)
TA'ﬂ = é/JX/ow (‘21:) _—_ 9:, 943? x “32 J! «cf/:77 fl; d/recfm” all f/aw 1 155/ logy 1.84 As shown in Video V1.5, surface tension forces
can be strong enough to allow a doubleedge steel razor blade to “ﬂoat” on water, but a single—edge blade will sink.
Assume that the surface tension forces act at an angle 6 rel ative to the water surface as shown in Fig. P134. (3) The
mass of the doubleedge blade is 0.64 X 10’3kg, and the
total length of its sides is 206 mm. Determine the value of
9 required to maintain equilibrium between the blade weight
and the resultant surface tension force. (b) The mass of'the ‘ F i G U R E 91 ' 3t}
singleedge blade is 2.61 X 10"3kg, and the total length of its sides is 154 mm. Explain why this blade sinks. Support your answer with the necessary calculations. Surface tension
force ' (a) Z Fuevltal :0
{LU : TSl'n 9
where v = mnHadex' g Cmd T: (T‘x Jemfl‘h 0% Sides. (OW KID3&3) (ml Mtge): (73L; “7%){0 20L m) sine
StnB‘ 3‘ oLl’5
Q}: 2,45" (E) For SIny/e~ed3e blade ‘3
2d :Mbladex 2 (2.4:! KID I‘M/52’)
:: 0.025!» N Ctl’ld
T sine = [Tx leng'lla mil/ml me = (7,3W,0«2 N/fm){0./5H‘/m) 5m 9
: 0.01:3 5M9
In order acor blade “xio "qt/oat” %< Time. Since Maximum: Value acoY‘ 5M9 15 l} H” ‘icollows
"but %>Tsm9 4nd 5in5/cedge blade will sink, ...
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This note was uploaded on 01/30/2012 for the course ME 3340 taught by Professor Smith during the Spring '09 term at Georgia Tech.
 Spring '09
 Smith

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