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mae_101a_hw1s - M45 101 4 Y/omawohé l So/HZ'IOAS Problem 1...

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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 p-V A C-j— F ' 2 E3><[£] ><L2 L Force = Mass-Acceleration 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—”qf-t- m2) (6) i" = an? + 3-2)“? ”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. . K-x K K-X K Onthey:xax13 u:_fi:_ VZfi: 2-1T-(X +X) 41TX 2-7T-(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 : JE-X 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 2-7T-1' Plotting V(m/s) r (km) This can also be plotted in Excel. K-x 2 For streamlines X : fl : M : if u dx K-y y 2 2 2 11'- X + y So, separating variables y-dy : —X-dx 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— 8-11 . . . 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 = [ll-mm umax = 0.1-? it = 1.14><10 3—: (Table AB) in —3 N5 m 01 l-rn 1 lUUU-mm 2 Hence TX=—8Xl.l4XlU -—><0.1-—><+-mmx—>< x— y m2 s 2 1000mm [ll-mm l-m 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 = a-x-t . . dx So, sepaianngvmables — = a-t-dt x . l 2 Integrating 111(x) = E-a-t + (:1 For initialposifion (10,30) :1: = xfl-e Usmgmegimdma, deC£xg=Y0)=(Ll)att=fl 0.0542 x = e Shemlioesaiegivenby : = E — —b-y 1:1 dx a-x-t . . b dx So, separehogvanables fl = —— — y a t . b Integtajmg 111(y) = —— -]n(x) + C a-t _i Thesohitionis y = C-x 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 ,_. -.] L11 EHEQQQ {41030qu Omomom ,_.,_.,_.,_.,_.,_.,_. 1—‘CDCDCDCDCDCD [\JOOU‘IUJHOO 5.] [\J O O ,_. [\J [\J .N . [\J U] M \O .N . «.4 U1 ,_n A O‘\ L...) O O ,_. U1 -..] P“ o o m . éeeéww. qmwoqm momomo N L11 5.] O ,_I DO U] Q [\J WWNNM 4205»th \O‘OUIKIUJ Streamlines t = 0 I. - -mm -m ,_.,_n 00 co _. ©< c: . 0.78 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00 m 1.00 1.00 1.00 1.00 0.02 1.00 1.00 1.00 1.00 1.00 0.01 -IEI -IEI -m 0000 0000 b—ib—ID—‘M 1.0— 0.8 7 0.6 — 0.4 7 0.2 - 0.0 Pathline and Streamline Plots 0 Pathline 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 Doom-lbw ooooo Hoooo coca-hm ooooo NEE—‘99 cameo-l: ‘< ooooo .E‘E—‘E—‘E‘E—‘N ou-Loxooo oooooo QQE‘E—‘P Loomoxo": ooooo ,_. N o ._. N o to .1; o o to o o 4; o o oo o . .0 4;. o ,_. 4; O l—l .1; O P . 00 O O 4: o o 00 o C ox o I 5:: 00 O ,_. O\ O |—l O\ O U.) N O O O\ O ,_. N O O J}. O I |—l N O I 1“ . O\ O [\J o o N o o :5: o o ._. o o m o o . N o o I I 1“ l“ . O\ 00 O O I E“ 4:. O I H . O O P’P’PPPP. Noooox-ILN oooooo HPPPNN. oooooo oooooo Nmmmmm mom-boon oooooo DJ 4; o P o o to Ch 0 ._. o o I—AI—tI—tI—II—II—t OOOOOO 000000 CD QI—-I—-I—-I—-I—- 45- WON-5000 0 000000 Ln 0\ O P C) O [\J .b. O l—‘ O O I I .C' Q . J:- 00 O O LN 00 o N O o .N . m o ._. o o O N o . o N o mm mm mm mm mm- mm [MI W W] mm W] .b. O O N C) O N C) O l—‘ O O Pathline and Streakline Plots 4 a o o n . s 3 O o a a o o 2 I 0 I i I : / 1 I : I I : f f Y I I .0 5 0.5 I 1.0 1.5 2.0 2.5 / _ _ 1 / o Pathllne starting at t= 0 z I Pathline starting at t= 1 s l\J O Pathline starting at t= 2 s 'Streakline at t = 4 s ...
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