VIEWS: 31 PAGES: 33 POSTED ON: 1/14/2012 Public Domain
Fluid Mechanics EIT Review School of Civil and Monroe L. Weber-Shirk Environmental Engineering Shear Stress F N Tangential force per unit area m2 A du change in velocity with respect to distance dy rate of shear Manometers for High Pressures Find the gage pressure in the center of the sphere. The sphere contains 1 fluid with g1 and the manometer contains fluid with g2. g2 What do you know? _____ P1 = 0 h1 3 g1 ? Use statics to find other pressures. h2 2 P1 + h1g2 - h2g1 =P3 For small h1 use fluid with high density. Mercury! Differential Manometers p1 Water p2 h3 orifice h1 h2 Mercury Find the drop in pressure p1 + h1gw - h2gHg- h3gw = p2 between point 1 and point p1 - p2 = (h3-h1)gw + h2gHg 2. p1 - p2 = h2(gHg - gw) Forces on Plane Areas: Inclined Surfaces Free surface O q x FR ghc A hc A’ xc xR centroid B’ O The origin of the y yc axis is on the free center of pressure yR y surface Statics Fundamental Equations Sum of the forces = 0 Sum of the moments = 0 F pc A centroid of the area pc is the pressure at the __________________ Ix y2 A Ix yp y Line of action is below the centroid yA yA Properties of Areas b a ba 3 a A = ab yc = I xc = I xyc = 0 Ixc yc 2 12 a yc = a ab 3 ba 3 ba 2 Ixc A= I xc = I xyc = (b - 2d ) yc 2 b+d 36 72 xc = b 3 d R p R4 I xyc = 0 Ixc A = p R 2 yc = R I xc = yc 4 Properties of Areas Ixc yc p R2 4R p R4 I xyc = 0 R A= yc = I xc = 2 3p 8 b a yc = a p ba 3 I xyc = 0 Ixc A = p ab I xc = yc 4 yc p R2 4R p R4 R A= yc = I xc = 4 3p 16 Inclined Surface Summary The horizontal center of pressure and the I xy horizontal centroid ________ when the surface x p x yA coincide has either a horizontal or vertical axis of symmetry The center of pressure is always _______ the below Ix yp y centroid yA The vertical distance between the centroid and the center of pressure _________ as the surface decreases is lowered deeper into the liquid What do you do if there isn’t a free surface? Example using Moments An elliptical gate covers the end of a pipe 4 m in diameter. If the gate is hinged at the top, what normal force F applied at the bottom of the gate is required to open the gate when water is 8 m deep above the top of the pipe and the pipe is open to the atmosphere on the other side? Neglect the weight of the gate. Solution Scheme Magnitude of the force hinge applied by the water 8m water Location of the resultant force F 4m Find F using moments about hinge Magnitude of the Force hinge Fr pc A 8m water A ab Fr F 4m h = _____ 10 m Depth to the centroid gh pc = ___ Fr gh ab a = 2.5 m N Fr 9800 10 m π2.5 m 2 m 3 m ________ Fr= 1.54 MN b=2m Location of Resultant Force Ix hinge yp y 8m water yA Fr yh Slant distance F 4m y ________ 12.5 m to surface ba 3 ba 3 yp y Ix A ab 4 yab 4 a = 2.5 m a2 yp y 2.5 m 2 yp y cp 4y 412.5 m y p y 0.125 m _______ x p __ x b=2m Force Required to Open Gate hinge How do we find the 8m water required force? Fr Moments about the hinge F 4m M hinge 0 =Fltot - Frlcp Fr lcp F 2.5 m ltot lcp=2.625 m ltot F 1.54 x 10 N 2.625 m 6 cp 5 m F = ______ 809 kN b=2m Example: Forces on Curved Surfaces Find the resultant force (magnitude and location) on a 1 m wide section of the circular arc. F V = W1 + W 2 3m W1 = (3 m)(2 m)(1 m)g + p/4(2 m)2(1 m)g water 2m = 58.9 kN + 30.8 kN = 89.7 kN W2 2m FH = p A x = g(4 m)(2 m)(1 m) = 78.5 kN y Example: Forces on Curved Surfaces The vertical component line of action goes through the centroid of the volume of water above the surface. A Take moments about a vertical axis through A. 3m W1 4(2 m) xFV (1 m) W1 W2 water 2m 3 4(2 m) W2 (1 m)58.9 kN 30.8 kN 2m x 3 89.7 kN = 0.948 m (measured from A) with magnitude of 89.7 kN Example: Forces on Curved Surfaces The location of the line of action of the horizontal component is given by A Ix yp y yA b 3m W1 bh 3 water Ix h 2m 12 W2 I x (1 m)(2 m)3/12 = 0.667 m4 2m y 4m x 0.667 m 4 yp 4 m 4.083 m 4 m 2 m 1 m y Example: Forces on Curved Surfaces 0.948 m 78.5 kN horizontal 4.083 m 89.7 kN vertical 119.2 kN resultant Cylindrical Surface Force Check 0.948 m 89.7kN All pressure forces pass C through point C. The pressure force 1.083 m applies no moment about point C. The resultant must pass 78.5kN through point C. 0 (78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ Curved Surface Trick Find force F required to open the gate. A The pressure forces and force F 3m W1 pass through O. Thus the hinge force must pass through O! water 2m O All the horizontal force is F W2 carried by the hinge Hinge carries only horizontal forces! (F = ________) W1 + W2 11.23 Dimensionless parameters Vl R Reynolds Number V F Froude Number gl V 2 l Weber Number W Mach Number V M c 2Drag 2p Cd Pressure Coefficient Cp V 2 V 2 A (the dependent variable that we measure experimentally) Model Studies and Similitude: Scaling Requirements dynamic similitude geometric similitude alllinear dimensions must be scaled identically roughness must scale kinematic similitude constant ratio of dynamic pressures at corresponding points streamlines must be geometrically similar Mach Reynolds Froude Weber _______, __________, _________, and _________ numbers must be the same a Cp f M, R, F,W,geometry f V F gl Froude similarity Fm Fp Froude number the same in model and 2 Vm Vp2 prototype g mLm g pLp ________________________ difficult to change g 2 Vm Vp 2 define length ratio (usually larger than 1) Lm Lp Vr L r Lp velocity ratio Lr Lr Lm tr Lr time ratio Vr Qr Vr Ar L r L r L r L5 / 2 discharge ratio r 3 Lr force ratio Fr = M r a r = r r L r 2 = L3r tr 11.33 Control Volume Equations Mass LinearMomentum Moment of Momentum Energy Conservation of Mass If mass in cv 2 v dA t d is constant 1 cs cv v1 v cs1 1 1 dA1 cs 2 2 v 2 dA 2 0 A1 Area vector is normal to surface and pointed out of cv 1V1 A1 2V2 A2 0 V = spatial average of v 1V1 A1 2V2 A2 m [M/t] V1 A1 V2 A2 Q If density is constant [L3/t] Conservation of Momentum F M 1 2 M M1 = - ( r 1V1 A1 ) V1 = - ( r Q ) V1 af af M 2 2V2 A2 V2 Q V2 a a F QfV QfV 1 2 F Qa f V V 2 1 F W F p1 Fp Fss 2 Energy Equation p1 V12 p2 V22 z1 1 Hp z2 2 H t hl g1 2g g2 2g LV2 hf f 2 V hl K 2g D 2g laminar turbulent 64 f Moody Diagram R Example HGL and EGL velocity head V2 2g p pressure head g energy grade line hydraulic grade line z elevation pump z=0 datum 2 2 pin V in pout Vout + zin + a in + hP = + zout + a out + hT + hL g 2g g 2g Smooth, Transition, Rough LV2 h f Turbulent Flow f D 2g Hydraulically smooth 1 Re f 2 log pipe law (von Karman, f 2.51 1930) Rough pipe law (von 3.7 D 1 Karman, 1930) 2 log f Transition function for both smooth and rough D pipe laws (Colebrook) 1 2 log 2.51 3.7 f Re f (used to draw the Moody diagram) Moody Diagram 0.10 0.08 D f Cp 0.05 0.04 l 0.06 0.03 0.05 0.02 friction factor 0.015 0.04 0.01 0.008 0.006 0.03 0.004 D laminar 0.002 0.02 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 smooth 1E+03 1E+04 1E+05 1E+06 1E+07 1E+08 R Solution Techniques find head loss given (D, type of pipe, Q) 0.25 8 LQ 2 4Q f hf f 2 Re D LF MH 5.74 0.9 2 IO P g D5 log N 3.7 D Re find flow rate given (head, D, L, type of pipe) KQ F I gh logG G 178 J . gh J Q 2.22 D5/ 2 f L G D LJ H 3.7 D K 2/3 f find pipe size given (head, type of pipe,L, Q) L F I 2 4 .75 FL I O 5.2 0.04 D 0.66M G J Q G J P LQ 1.25 9 .4 MH K N gh f H KP gh Qf Power and Efficiencies Electrical power Motor losses Pelectric IE bearing losses Shaft power Pshaft Tw Impeller power pump losses impeller P Tw Fluid power Pwater gQHp Manning Formula 1 2/3 1/2 V R h So n The Manning n is a function of the boundary roughness as well as other geometric parameters in some unknown way... A Rh P Hydraulic radius for wide channels A bh P b 2h bh Rh b 2h Drag Coefficient on a Sphere 1000 Drag Coefficient 100 Stokes Law 10 1 0.1 0.1 1 10 102 103 104 105 106 107 U 2 CD = 24 Reynolds Number Fd Cd A Re 2