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Lecture_Notes_10_06

# Lecture_Notes_10_06 - Flows with Friction Simple Shear...

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Unformatted text preview: Flows with Friction Simple Shear Flows Shear stress = tangent force per unit area at a point lim AFt Shear stress is caused by slower moving molecules in one ﬂuid layer getting into a faster moving adjacent layer and tending to slow it down. Conversely, molecules in the faster moving layer get into the slower moving layer and tend to speed it up. There is a net momentum exchange across the surface which gives rise to a shear stress. Shear Flow betweenaParallel Plates u(0) = 0, u(h) = U Wall and plate are long — ﬂow is parallel and no pressure forces ZFx = (2' + 22:a’y)dxa’z — rdxdz : O 5y —— = 0‘ ' z" = constant. Strain measured by change in angle @ dydt 6y — @ dt dy 5y tandé’ E d6 2 Strain rate = ﬁ = 6—“ dz‘ 8y Assume stress proportional to strain rate = Newtonian ﬂuid z- -— 6—14 I“ (3)} ,u = Viscosity, a property of the ﬂuid 7 N / m2 NS 'u : du/ay N (m/S)/m : m2 Air: ,u =1.80x10‘5Ns/m2 u = l.47x10_5m2 /s Water: # =l.00xlO_3Ns/m2 u = 1.00x10_6m2 /s SAE 30 Oil: ,u = 2.30x10_le/m2 U = 3.16x10_4m2 /s Mercury: ,u =1.56x10_3NS/m2 u =1.15x10‘7m2 /s ,u = absolute viscosity 0 = ,u/ p =‘ kinematic Viscosity = property of a ﬂuid 0 ~ m2 /s = kinematic units 2' = constant for the moving plate problem 6—712: uziy Whereu=0aty=0 5y # # A u=U for y=h Then I = ,uU/ h Example A rotary viscometer is a devise for measuring the Viscosity of a ﬂuid. The sheared ﬂuid between the cylinders puts a torque on the outer cylinder which is measured. What is the Viscosity of the ﬂuid as determined from the measured torque? Data: Radius of inner cylinder = R : 5cm Angular velocity inner cylinder = (lOOOrpm) Clearance between the cylinders = e 2 2mm Length of the cylinders = L 2 15cm Measured torque on outer cylinder =' T: 2.5x10_3 Nm U. er 7:”? or “:36 a) = 2%(1000)/60 :104;7rad/S T = (R + e)r[A] : (R + e)2'[27r(R + e)L] T 2.5x10‘3 ‘ = 2 = 2 =0.981N/m2 27r(R+e) L 2%(0052) (0.15) Z. _ (0.002)(0.981) " (0.05)(104.7) y = 3.75x10“‘Ns/m2 Parallel ﬂows involving pressure and friction. Fully Developed ﬂow in a parallel plate channel Fully developed ﬂow: x 2 Le Where Le: entry length Parallel ﬂow, no acceleration, balance between pressure and friction force ZFx =polde—(p+gadxyz'ydz~dedz+(r+%dy)dxdz=0 x y ﬁ_a_P_d_P 8y 8x dx dp an dp - ‘ 72—— 7: —=— Wherea/ =0f =0 dxy ﬂay» dxy u 8y ory 2 2 u=ié€y—+C Where C=—Z1—~d—pv 11:0 for y=ih/2 ,udx 2 8;! dx 2 2 2 u:_ﬁ_d_p(1_4_yi):u0(1_4i 8y dx 2 h2 2 u0 = —h—gg = centerline velocity 8,u dx 2 , q' = ﬁhw = 3/22 uwa’y = r2u0wK/Zﬂ — gig—)0} = guohw MO q,_ 2u0wh H_h3w dp _ h3Wp1—p2 3 12;; dx 12;! L General Viscous Flows Stresses: normal and shear lim AF” 0'” and 0' :AA—)0AA SzAA—>0AA Normal stress is unique Shear given by a stress and angle or two components State of stress On an area is given by three components, one normal and two shears axx, axy, 0x2 = X,y,z components of stress on the X area. Three dimensions - three surface orientations which are perpendicular to the X,y, z axes._ State of stress - nine components, three components on each of three surfaces Stress Matrix 0‘ 0' Three normal stresses axx, a 022 and siX shear stresses, yy 9 however, it can be shown using the moment of moment equation that 0' a = 0' and 0' 02y. Thus 3/2 2 there are siX independent components of stress. xy :O-yxa xz 2x: (7 is the total stress which includes the pressure. _ p + Txx Txy sz ryx — p + Tyy ryz sz sz _ p + 2‘22 -p is the normal stress due to pressure r components are due to Viscosity. Stresses in the X—Direction .V a ‘2‘... Newton’s Second Law for the X-direction D(updV)__ _ _ 5(-P+Txx) 7—2Fx—K p+7xx)'+ ax ' 6 dx]dydz — (—p + Txx )dydz Tyx , 5y +[a(_p + 2306) + 87)“ + 672x 8x (33/ 82 '+ +[ dy]dxdz — Tyx 0’de + [sz + 8 Tyx sz dz]dxdy — 72x dxdy 82 Equation of Motion For the y and 2 directions at 61 62 pﬂ:_§lj+ xy+ yy+ zy Dz‘ 6y 6x 8y 62 Dw _ 6p + aer aTyz + 6722 ,0 Z + Dt 82 6x 6y 62 Strains _ 5:3 1]— - " ,~ Aim, Q5: _‘ E I m”; t La“ “Li . s i? I Strains: a) Normal, b) Shear Strain due to the normal stress rxx- elongation in X—direction (strain) xx = M (51‘ )r Shear stresses change the corner angles (strain) xy = (91 + 6’2 Relate strains to velocity ﬁeld 17(x + 5x, y, 2, 2‘) = 17(x, y, 2, t) + Wéx # 6x [7(x’y + 632’2’1.) : 17(x9yaizat) +an—xéyz—Z—2—Qéj} y I7(x,y,z+6z,t) = I7<x,y,z,t)+W§Z 82 Relative Velocities and Rotation of Coordinate Elements (— am ax dug d = — d gxx 5x 6x Strain rate dam _ é _ ﬁg dz‘ xx 6x - - . 5v . 6w Slmilarly, 5W = b}— and 5Z2 2 EZ— Velocity component (av/ 6x)§x rotates the segment 5x around the z axis 8v (— 520a? tandé’l E d61 2L :th and £29129 5x 8x dz‘ 8x (3M (5— 5wa au d6? au Similarly, tan c1492 ; daz = —y—-— = ﬂair and _2 = 9'2 2 _ éj/ By 611‘ 6y Shear strain rate = average of the two angles 1 6v ﬂu 6‘ — - (—— + —) = a Rotation of the ﬂuid element = average rotation of two a perpendicular line Segments wz : l(6'1 — 6'2) 2 l(@ — %) = rate of rotation (rad/s) 2 2 8x 6y ’ of a ﬂuid element about the z axis N avier-Stokes Equations for two dimensions Newtonian ﬂuid — stress is proportional to strain 8L1 [ av 82;] 2#-—- #-—-+-* [Txx racy] : 2ﬂ[éxx ‘5“ny : 6x 8x 8y Tyx Tyy éyx éyy [u ‘1“ + Q 2” 93 (33/ 8x 8y + + Dz‘ 6x 8x 6y 6x 8x 8y (9y 6x 82' pDu:_6p+6Txx+ yx:_§£+ﬂ[a(26u) 6(6u 812)] With the continuity equation au/ 5x + 612/ 6y 2 0 &__.__62+ th 6x [a 6x2 6))2 Where the Viscosity was taken to be constant. [Bu au 8a) ap [azu ﬁzu] ,0 ——+u——+v— =———+,u —+— 81‘ 6x 6x 8x2 I ayz 6v 8v 8v 81) (32v 62v ,0 —+u—-——+v—— =——+,u ~—+—— at 8x 6y 8y 6x2 (33/2 With the continuity equation an / 6x + 812/ 6y 2 0 there are three equations to determine u,V and p. ...
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Lecture_Notes_10_06 - Flows with Friction Simple Shear...

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