Fluid Mechanics EIT Review Monroe L. Weber-Shirk School of Civil and Environmental Engineering Shear Stress F A du dy Tangential force per unit area N m 2 change in.
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Fluid Mechanics EIT Review Monroe L. Weber-Shirk School of Civil and Environmental Engineering Shear Stress F A du dy Tangential force per unit area N m 2 change in velocity with respect to distance rate of shear Manometers for High Pressures Find the gage pressure in the center of the sphere. The sphere contains fluid with g1 and the manometer contains fluid with g2. P1 = 0 What do you know? _____ 3 ? g1 Use statics to find other pressures. P1 + h1g2 - h2g1 =P3 For small h1 use fluid with high density. 2 Mercury! 1 g2 h1 h2 Differential Manometers p1 Water p2 h3 orifice h1 h2 Mercury Find the drop in pressure between point 1 and point 2. p1 + h1gw - h2gHg- h3gw = p2 p1 - p2 = (h3-h1)gw + h2gHg p1 - p2 = h2(gHg - gw) Forces on Plane Areas: Inclined Surfaces Free surface O q FR ghc A hc A’ x xc xR centroid B’ O center of pressure yR y yc The origin of the y axis is on the free 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 yA yA Line of action is below the centroid Properties of Areas b Ixc Ixc a yc ab A = yc 2 a b d Ixc A = ab R yc a yc = 2 a yc = 3 b+d xc = 3 A = p R 2 yc = R ba 3 I xc = 12 I xyc = 0 3 ba 2 = (b - 2d ) 72 ba I xc = 36 I xyc p R4 I xc = 4 I xyc = 0 Properties of Areas Ixc yc 4R yc = 3p p R4 I xc = 8 I xyc = 0 A = p ab yc = a p ba 3 I xc = 4 I xyc = 0 p R2 A= 4 4R yc = 3p p R4 I xc = 16 p R2 A= 2 R b a yc Ixc R yc Inclined Surface Summary The horizontal center of pressure and the I xy coincide when the surface x p x horizontal centroid ________ yA has either a horizontal or vertical axis of symmetry The center of pressure is always _______ below the y I x y p yA centroid The vertical distance between the centroid and the center of pressure _________ decreases as the surface 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 applied by the water hinge 8m Location of the resultant force Find F using moments about hinge water F 4m Magnitude of the Force hinge Fr pc A 8m A ab h = _____ 10 m Depth to the centroid water Fr F gh pc = ___ Fr gh ab N 10 m π2.5 m 2 m Fr 9800 3 m a = 2.5 m Fr= 1.54 ________ MN b=2m 4m Location of Resultant Force hinge Ix yp y yA yh 12.5 m y ________ ba 3 yp y 4 yab 2 a yp y 4y m y p y 0.125 _______ 8m Slant distance to surface Ix ba 3 4 Fr F 4m A ab 2.5 m yp y 412.5 m 2 x x p __ water a = 2.5 m cp b=2m Force Required to Open Gate How do we find the required force? hinge 8m Fr F Moments about the hinge M hinge 0 =Fltot - Frlcp F water 4m Fr lcp ltot lcp=2.625 m 2.5 m 1.54 x 10 N 2.625 m F cp 6 5 m F = ______ 809 kN b=2m ltot 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 = (3 m)(2 m)(1 m)g + p/4(2 m)2(1 m)g water = 58.9 kN + 30.8 kN = 89.7 kN 2m W1 2m W2 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. Take moments about a vertical axis through A. 4(2 m) xFV (1 m) W1 W2 3 4(2 m) (1 m)58.9 kN 30.8 kN 3 x 89.7 kN A 3m water W1 2m 2m W2 = 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 bh 3 Ix 12 W1 3m b h I x (1 m)(2 m)3/12 = 0.667 m4 y 4m 0.667 m 4 yp 4 m 4.083 m 4 m 2 m 1 m water 2m W2 2m x Ix yp y yA A y 4.083 m 0.948 m Example: Forces on Curved Surfaces 78.5 kN horizontal 89.7 kN vertical 119.2 kN resultant Cylindrical Surface Force Check 0.948 m C 1.083 m 78.5kN 89.7kN All pressure forces pass through point C. The pressure force applies no moment about point C. The resultant must pass through point C. (78.5kN)(1.083m) - (89.7kN)(0.948m) = ___ 0 Curved Surface Trick Find force F required to open the gate. The pressure forces and force F pass through O. Thus the hinge force must pass through O! All the horizontal force is carried by the hinge Hinge carries only horizontal forces! (F = ________) W1 + W2 A water 3m W1 O F 2m W2 11.23 Dimensionless parameters Reynolds Froude Weber Number Number Number R Vl V F gl W V 2 l V c 2Drag 2p Cd C Pressure Coefficient 2 p 2 V A V (the dependent variable that we measure experimentally) Mach Number M Model Studies and Similitude: Scaling Requirements dynamic similitude geometric similitude all linear dimensions must be scaled identically roughness must scale kinematic constant similitude ratio of dynamic pressures at corresponding points streamlines must be geometrically similar _______, Mach __________, Reynolds _________, Froude and _________ Weber numbers must be the same Cp f M, R, F,W,geometry a f V F gl Froude similarity Froude number the same in model and prototype ________________________ difficult to change g define length ratio (usually larger than 1) velocity ratio Vr L r Lr tr Lr time ratio Vr Qr Vr Ar L r L r L r L5r / 2 discharge ratio 3 Lr 3 F = M a = r L = L r r r r r 2 r force ratio tr Fm Fp Vp2 Vm2 g mLm g pLp 2 Vm2 Vp Lm Lp Lr Lp Lm 11.33 Control Volume Equations Mass Linear Momentum Moment of Momentum Energy Conservation of Mass 2 If mass in cv v dA t d is constant 1 cs cv v 1 cs1 1 dA1 2 v 2 dA 2 0 A1 v1 cs 2 Area vector is normal to surface and pointed out of cv 1V1 A1 2V2 A2 0 1V1 A1 2V2 A2 m V = spatial average of v [M/t] V1 A1 V2 A2 Q If density is constant [L3/t] Conservation of Momentum F M 1 M 2 M1 = - ( r 1V1 A1 ) V1 = - ( r Q ) V1 a f af F aQfV aQfV F QaV V f M 2 2V2 A2 V2 Q V2 1 2 2 1 F W F p1 Fp Fss 2 Energy Equation V12 p2 V22 z1 1 Hp z2 2 H t hl g1 2g g2 2g p1 2 V hl K 2g hf f LV2 D 2g laminar turbulent 64 f R Moody Diagram Example HGL and EGL velocity head V2 2g pressure head p g energy grade line hydraulic grade line z elevation pump z=0 2 in datum 2 pin V pout Vout + zin + a in + hP = + zout + a out + hT + hL g 2g g 2g Smooth, Transition, Rough h f Turbulent Flow LV2 f Hydraulically smooth pipe law (von Karman, 1930) Rough pipe law (von Karman, 1930) Transition function for both smooth and rough pipe laws (Colebrook) D 2g Re f 2 log 2.51 f 1 3.7 D 2 log f 1 D 2.51 2 log f Re f 3.7 1 (used to draw the Moody diagram) Moody Diagram 0.10 0.08 D f Cp l 0.06 0.05 0.04 0.03 friction factor 0.05 0.02 0.015 0.04 0.01 0.008 0.006 0.004 0.03 laminar 0.002 0.02 0.001 0.0008 0.0004 0.0002 0.0001 0.00005 0.01 1E+03 smooth 1E+04 1E+05 R 1E+06 1E+07 1E+08 D Solution Techniques find head loss given (D, type of pipe, Q) 2 0.25 8 LQ 4Q f 2 hf f 2 Re 5 5 . 74 g D D log 0. 9 3.7 D Re find flow rate given (head, D, L, type of pipe) L F M NH IO KP Q F I gh G 178 . J Q 2.22 D logG gh J L 3.7 D G J D H L K find pipe size given (head, type of pipe,L, Q) L F F LQ I L I O D 0.66M G J Q G J P gh K gh KP H H M N Q f 5/ 2 2/3 2 1.25 5.2 0.04 4 .75 9 .4 f f f Power and Efficiencies Electrical power Motor losses Pelectric IE Shaft power Pshaft Impeller Tw power Pimpeller Tw Fluid bearing losses power Pwater gQHp pump losses Manning Formula 1 2/3 1/2 R h So n The Manning n is a function of the boundary roughness as well as other geometric parameters in some unknown way... V Hydraulic radius for wide channels A Rh P A bh P b 2h bh Rh b 2h Drag Coefficient on a Sphere Drag Coefficient 1000 100 Stokes Law 10 1 0.1 0.1 CD = 1 24 Re 10 102 103 104 Reynolds Number 105 106 Fd Cd A 107 U 2 2