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Unformatted text preview: M45 101 4, Y/omawohé l
So/HZ'IOAS Problem 1 A number of common substances are Tar Sand “Silly Putty” Jello
Modeling clay Toothpaste
Wax Shaving cream Some of these materials exhibit characteristics of both solid and ﬂuid behavior under different conditions. Explain and give examples. Given: Common Substances Tar Sand “Silly Putty” Jello
Modeling clay Toothpaste
Wax Shaving cream Some of these substances exhibit characteristics of solids and ﬂuids under different conditions. Find: Explain and give examples. Solution: Tar, Wax, and Jello behave as solids at room temperature or below at ordinary pressures. At high pressures or over long periods, they exhibit fluid characteristics. At higher temperatures, all three
liquefy and become viscous fluids.
Modeling clay and silly putty show fluid behavior when sheared slowly. However, they fracture under suddenly
applied stress, which is a characteristic of solids.
Toothpaste behaves as a solid when at rest in the tube. When the tube is squeezed hard, toothpaste “flows” out the
spout, showing fluid behavior. Shaving cream behaves similarly. Sand acts solid when in repose (a sand “pile”). However, it “flows” from a spout or down a steep incline. Problem .2 TJ ery small particles movingin fluids are known to experience a drag force proportional to speed Consider a
particle of net weight W droppedin a ﬂuid. The particle experiences a drag force, FD: RF, Where V is the particle speed Determine the time required for the particle to accelerate from rest to 95 percent ofits terminal speed, F1, in terms of'Fc, W, andg. Given: Small particle accelerating from rest in aﬂuid. Netweightis W, resisting force FD = 1:1? , Where E?
is speed
Find: Time requiredto reach 95 percent ofterminal speed Vt. Solution: Consider the particle to he a system. Apply N ewton's second law. Basic equation: EFy= may «’35 w Hirahhic, Assumptions:
1. Wis net weight 2. Resisting force acts opposite to V ThEﬂ ZFy=W—kv =may=m£=ﬂﬂ
dt g dt
d'if k
or —= 1——V
dt a W :1
d?
Separatingvariahles, 1 1: = gdt
‘ﬁ V
Integ'ating, noting that velocity is zero initially, L d? =  —1ﬂ1:1 — —V::ﬂ MW ' Problem 3 In Chapter 9 we will study aerodynamics and learn that the drag force F9 on a body is given by
1 F}; = ipVZACD Hence Ute drag depends on speed V, ﬂuid density p, and body size (indicated by frontal area A) and shape (indicated by drag coefﬁ—
cient CD). What are the dimensions of CD? Given: Equation for drag on a body. Find: Dimensions of CD. Solution: Use the drag equation. Then "solve" for CD and use dimensions. The drag equation is "Solving" for CD, and using dimensions But, From Newton's 2nd law Henee The drag coefﬁcient is dimensionless. i 2
F33 = E'P'V 'A'CD
2F
03: D
' 2
pV A
Cj— F
' 2
E3><[£] ><L2
L
Force = MassAcceleration or
F ML L3 t2 1
CD‘ 2 = 2 ><—><—2><—2=0
M (1.] 2 t M L L
—3>< — XL
L Problem 4 For the velocity ﬁelds given below, determine: a. whether the ﬂow ﬁeld is one—, two—, or three—
dimensional, and why. 13. whether the ﬂow is steady or unsteady, and why. 0. (The quantities a and b are constants.)
m 17'=[ayze_“]f (2) V=axzf+bxj+cic
(3) 12" = my? — 532:; (4) 12" = art— sﬁ+ art
(5) I? = [ac—”qft m2) (6) i" = an? + 32)“? ”33):“.
(7) I} = [ax + r}?— by?) (3) I? = ar2f+ bxzj+ cyE Given: Velocity fields Find: "Whetherflows are 1, 2 or 3D, steady or unsteady. Solution: 9 —> a —> (1) V = V (y) ID V = V () Unsteady
a —> a —> (2) V = V (X) 1D V i V () Steady
a —> a —> (3) V = V (X,y) 2D V = V () Unsteady
a —> a —> (4) V = V (X,y) 2D V = V () Unsteady
a —> a —> (5) V = V (X) ID V = V () Unsteady
a —> a —> (6) V = V (X,y,z) 3D V i V () Steady
a —> a —> (7") V = V (X,y) 2D V = V () Unsteady
a —> a —> (8) V = V (X,y,z) 3D V i V () Steady Pr. 5 I._R_.l Solution: For any radius r S R, the liquid gap is h = r tanﬁ. Then d(Torque)=dM= IdAWr={,u Qr j[27zr dr )r, or r tanﬂ cos6
:RZ7zQ/r Jrz dr— _ 2729th 9 or: ,u = 3M s1n0 Ans.
sin 6 3 sm 6 27zQR3 Pr. 6 A block of weight W slides down an
inclined plane on a thin ﬁlm of oil, as in
Fig. ' at right. The ﬁlm contact area
is A and its thickness h. Assuming a linear
velocity distribution in the ﬁlm, derive an
analytic expression for the terminal velocity V of the block. Oil ﬁlm,
thickness h Solution: Let “x” be down the incline, in the direction of V. By “terminal” velocity we
mean that there is no acceleration. Assume a linear viscous velocity distribution in the
ﬁlm below the block. Then a force balance in the X direction gives: ZFX = W sin6— TA 2 W sin6—[y%] A = max = 0, hW sin 0 erminal : Al’lS.
yA or: Vt Problem 7 The ﬁovtr ﬁeld for an atmospheric ﬂow is given by Kv a Kr ., f’ = — —‘1 + —
lain:2 + 3:2} anx" + y?) ”r where K = 5 X 104 mats and the x and y coordinates are parallel
to the local latitude and longitude. Plot the velocity magnitude
along the x axis, along the y axis, and along the line 3! = I. For
each plot use the range — 10 km 5 x or y 5 10 km. excluding LII or iv! 5 100 In. Find the equation for the streamlines and sketch
several of them. What does Utis ﬂow ﬁeld model? Given: Flow field
Find: Plot ofvelocity magnitude along axes, andy =x; Equation of streamlines Solution: Onthex axis,y=0, so Plotting v( mfs) Xﬂém) The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero. This can also be plottedin Excel. Onthey axis,x=0, so Plotting 11 (1114’s) r (km) The velocity is perpendicular to the axis, is very high close to the origin, and falls off to zero. This can also be plotted in Excel. . Kx K KX K
Onthey:xax13 u:_ﬁ:_ VZﬁ:
21T(X +X) 41TX 27T(X +x) 41TX
The ﬂow is perpendicular to line y = X: Slope of line y = X: l Slope of trajectory of motion: 2 : —1
v vave define the radial position: r : \I x2 + y2 then along y : X r : a" x2 + x2 : JEX
Then the magnitude of the velocity along y : X is V : \l u2 + v2 : i i + i : L : K
4'“ X2 X2 2.1THJE.X 27T1' Plotting V(m/s) r (km)
This can also be plotted in Excel.
Kx
2
For streamlines X : ﬂ : M : if
u dx Ky y
2 2
2 11' X + y
So, separating variables ydy : —Xdx
y2 X2
Inte rat' _ : ,_+ c
g mg 2 2
The solution is X2 + y2 : C which is the equation of a circle. Streamlines form a set of concentric circles. This ﬂow models a vortex ﬂow. See Example 5.6 for streamline plots. Streamlines are circular, and the velocity approaches infinity as we
approach the center. In Problem 2.1 l, we see that the streamlines are also circular. In a real tornado, at large distances from the center, the
velocities behave as in this problem; close to the center, they behave as in Problem 2.11. Problem 3 The velocity distribution for laminar ﬂow between parallel
plates is given by where h is the distance separating the plates and the origin is
placed midway between the plates. Consider a ﬂow of water at 153C. with arm“ = 0.10 [tits and h = 0.1 mm. Calculate the shear
stress on the upper plate and give its direction. Sketch the vari— ation of shear stress across the channel. Given: 1Velocity distribution between flat plates
Find: Shear stress on upper plate; Sketch stress distribution
Solution:
2— 811 .
. . du du d 2y 4 many
Basic equation Tyx = Hid—y (1—3? = d—yumaX[l _ [T] _ = “max' [—FJIZIY = _T
8' l‘l’untarr'y
Tyx = — 2
h
h — N
Atthe upper surface 3: = E and h = [llmm umax = 0.1? it = 1.14><10 3—: (Table AB)
in
—3 N5 m 01 lrn 1 lUUUmm 2
Hence TX=—8Xl.l4XlU —><0.1—><+mmx—>< x—
y m2 s 2 1000mm [llmm lm
N
Tyx = —4.56—2
m The upper plate is a minus y surface. Since cw ‘1 U, the shear stress on the upper plate must act in the plus x direction. . . . 8' l‘l’umax
The shear stress varies linearly With 3r Tyxﬁl) = — — .3; h2 ,_.
i:pi:pi:i:i:p 3’ (min) D
D
— [J]
D
D Shear Stress (Pa) Problem ? i through the same point at the instants r = 0, l, and ‘2 5. Given: Velocityﬁeld
Find: Plotpajhlinesmdstremlines
Solution:
. . dx
Palhhnes are given by E = u = axt
. . dx
So, sepaianngvmables — = atdt
x
. l 2
Integrating 111(x) = Eat + (:1
For initialposiﬁon (10,30) :1: = xﬂe Usmgmegimdma, deC£xg=Y0)=(Ll)att=ﬂ 0.0542
x = e
Shemlioesaiegivenby : = E — —by
1:1 dx axt
. . b dx
So, separehogvanables ﬂ = —— —
y a t
. b
Integtajmg 111(y) = —— ]n(x) + C
at
_i
Thesohitionis y = Cx at
Forsheamlioeai(l,1)otr=l}s xzc
. —10
Forsheamhoeai(l,l)otr=ls yzx Forsheamlioeai(l,1)otr=2s yzx _ A velocity ﬁeld is given by i7 = an? — bﬁ, when: a = 0.1
3—2 and b = 1 s_'. For the particle that passes through the point
(x, y) = (l, l) at instant I‘ = 0 s, plot the pathl'me during the inter—
val from t = 0 to i = 3 5. Compare with the streamlines plotted Pathline ,_.
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Streamline (t : 0) —Stream1ine (t : 1 s) — Streamline (t : 2 s) 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 Problem [0 Tiny hydrogen bubbles are being used as tracers to visuale
ize a flow. All the bubbles are generated at the origin (.1: = 0,
y = 0). The velocity ﬁeld is unsteady and obeys the equations: u=1mfs u=2mfs Dir<25
u=0 U=—1sz 05:54:; Plot the pathlines of bubbles that leave the origin at t = 0. l. 2. 3.
and 4 5. Mark the locations of these ﬁve bubbles at t = 4 5. Use a
dashed line to indicate the position of a streakline at t = 4 5. Solution The particle starting att : 3 5 follows the particle starting at t : 2 s;
The particle starting att = 4 5 doesn't move! Pathlines: Starting at t = 0 Starting at t = 1 5 Starting at t = 2 s Streakline at t = 4 s H .2‘53535353
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 Fall '10
 AlisonMarsden

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