Final Equations for Fluid Mechanics | ENSC 3233, Exams of Fluid Mechanics

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Final Equations ENSC 3233 Fluid Mechanics 1 Units SI (MLT system) T = second (s) L = meter (m) M = kilogram (kg) The unit of force is: 1 Newton (N) = 1 kg m/s2 Work is: 1 joule (J) = 1 N m = 1 km m2/s2 0K = 273 + oC BG (FLT system) T = second (s) L = foot (ft) F = pound (lb) or (lbf) The unit of mass: 1 slug = 1 lb/(ft/s2) = 1 lb s2/ft Work is: 1 ft lb lbm (lb mass) = 1 lb / g = 1/32.2 slugs oR = 459.67 + oF Shear Stress dy duµτ = : ρ µυ = Drag force with laminar distribution Y UAAD wf µτ == Ideal gas law p = ρRT Bulk Modulus of Liquids VVd dpEv / −= if Ev = constant 12 1 2ln ppEv −=ρ ρ Compression of Gasses Isothermal: constant= ρ p Isentropic: constant=k p ρ v p c c k = Speed of Sound kRTkpc == ρ Surface Tension Capillary pressure across a meniscus: R ppp ei σ2 =−=∆ Capillary rise in tube: r h γ θσ cos2 = Archimedes' Principle: Wtsub= (ρbody - ρfluid) g V Statics ∫−=− 2 1 12 z z gdzpp ρ Constant density fluid p2-p1 =-ρg(z2-z1)= -γ(z2-z1) Gage pressure, h measured down from free water surface. p = ρgh = γh Force on flat surfaces ∫ ∫== A H R hdhbpdAF 0 γ Force on vertical rectangular surface, b wide. 2 2 1 bHFR γ= Acting at 2/3 H down from the surface. General plane surface AhAyF CCR γθγ == sin CyACy xcI Ry += CxACy xycI Rx += Final Equations ENSC 3233 Fluid Mechanics 2 Kinematics Velocity vector V = u(x,y,z,t)i + v(x,y,z,t)j + w(x,y,z,t)k 2-D Stream function ψ=− ∫∫ v dy u dx Acceleration z w y v x u t ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ = VVVVa Acceleration by components z uw y uv x uu t u xa ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ = z vw y vv x vu t v ya ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ = z ww y wv x wu t w za ∂ ∂ + ∂ ∂ + ∂ ∂ + ∂ ∂ = Continuity Steady compressible flow, no sources m& =ρQ = ρVA = constant Steady incompressible flow, no sources Q = VA = constant Momentum ∫∑ •= CS dAnVVForces )ˆ(ρ For discrete surfaces and Q = VA ∑∑ •= )ˆ( nVQForces ρ Forces ΣF = Fluid pressure on CV, Weight of fluid inside CV & External Forces on CV Dot Product Special Cases Outflow normal to surface QVnVQ ρρ =• )ˆ( Inflow normal to surface QVnVQ ρρ −=• )ˆ( X-component, one outflow, one inflow )( inxoutxx VVQF −=∑ ρ Moving CV CV moving with a velocity u, Define velocity of fluid, v relative to the CV v1 = V - u where V is the velocity relative to a fixed reference. Define flow relative to the CV, Q’ = v * A Bernoulli’s; No energy losses or gains 2 2 2 2 1 1 2 1 22 z g p g Vz g p g V ++=++ ρρ (head form) (or pressure form) 22 2 2 11 2 1 22 gzpVgzpV ρρρρ ++=++ Jet Propulsion Air speed u to the left; wind speed = W to the right. Vexhaust measured relative to the ground vin = W-(-u) = W+ u vexhaust = Vexhaust – (-u) = V2 + u inAinvinQairofmassinm ρρ ===& fminmfuelofmassairofmassexhaustm &&& +=+= invinmexhaustvfminmthurstF &&& −+= )( Bernoulli’s; No energy losses or gains 2 2 2 2 1 1 2 1 22 z g p g Vz g p g V ++=++ ρρ (head form) (or pressure form) 22 2 2 11 2 1 22 gzpVgzpV ρρρρ ++=++ Pitot-Darcy tube ( ) ρ/2 DpPpV −=∞ Energy equation For steady flow and along a streamline =+ CVinshaftWinnetQ )( &&         +++−+++ in gzVpu out gzVpum ) 2 2 () 2 2 ( ρρ (( & At constant pressure Tcu p= ( Neglecting heat transport lossgzp V Q inshaftWgzp V +++=+++ 222 2 2 112 2 1 ρρρρ & Lhzg p g V gQ inshaftWz g p g V +++=+++ 222 2 2112 2 1 ρρρ & Lhzg p g V phzg p g V +++=+++ 222 2 2112 2 1 ρρ Final Equations ENSC 3233 Fluid Mechanics 5 Hydraulic Jump (rectangular channel) 1 2 22 11 2 2 42 ygb Qyyy ++−= (Sections 1 and 2 may be exchanged up and down stream) Energy loss in Hydraulic jump       ++−      ++= 11 2 1 22 2 2 22 zy g Vzy g VhL Weirs Rectangular sharp edged Q = Cwr 2/3 b(2g)1/2 H3/2 For rectangular weirs with no end contraction, Cwr = 0.611 + 0.075(H/Pw) Broad crested rectangular weirs Q = Cwb bg1/2 (2/3 H)3/2 Cwb = 0.65/(1+H/Pw)1/2 Manning n Man made channels n Glass 0.010 brass 0.011 steel, smooth 0.012 steel, painted 0.014 steel, riveted 0.015 cast iron 0.013 concrete, finished 0.012 concrete, unfinished 0.014 wood, planed 0.012 clay tile 0.014 brick, neat 0.015 asphalt 0.016 corrugated metal 0.022 rubble masonry 0.025 Natural channels clean and straight 0.030 vegetation and bends 0.040 large rivers 0.035 cobble stream 0.035 PUMPS Energy equation with pumps: La hzg p g Vhz g p g V +++=+++ 2 2 2 2 1 1 2 1 22 ρρ Solving for ha:: La hzzg pp g VVh +−+−+−= )( 2 12 12 2 1 2 2 ρ The power gained by the fluid is: af QhP γ= In English horsepower units: 550 a f QhP γ= (hp) The power delivered to the pump shaft by the motor or engine must be greater than this. Wshaft > Pf The pump efficiency is thus, shaft f W P =η In English units: bhp Qha 550/γη = where bhp is the brake horsepower. bhp = Wshaft / 550 (when Wshaf = ft lb/s) Specific Speed ( ) 4/3gH Q Ns ω = English units ( ) 4/3)ft( (gpm)(rpm) H Q Nsd ω = NPSH R vss A NSPH p g VpNPSH ≥−+= γγ 2 2 Note ps must be in absolute pressure units. ps = pgage + patm Pump Similarity Laws (geometrically similar pumps) 3D Q ω = constant 22D gha ω = constant 53D Wshaft ρω = constant η = constant (when 3D Q ω = constant) Final Equations ENSC 3233 Fluid Mechanics 6 COMPRESSIBLE FLOW Mach Number c vMa = RTkc = Gas Equations of State RTp ρ= vp ccR −= )(ˆˆ 1212 TTcuu v −=− ρ puh +≡ ˆ )( 1212 TTchh p −=− v p c c k = 1− = k Rkcp 1− = k Rcv 1 2 1 2 12 lnln p pR T Tcss p −=− Isentropic flow, s2 = s1 1 2 1 2 1 1 2 p p T T k k k =      =      − ρ ρ Isentropic, Converging Nozzle 1 2 2 )1(1 1 −                 −+ = k k o Makp p 1 1 2* −       + = k k kp p o 528.0* 4.1 =      =ko p p 1 2* + = kT T o 833.0* 4.1 =      =ko T T 1 1 * 2 )1(1 1 −                 −+ = k o kρ ρ 634.0* 4.1 =      =ko ρ ρ                 − − = − k k oo o p p k kpV 1 2 1 )1( 2 ρ               −= 286.0 2 17 oo o p ppV ρ (k=1.4)           −      − = − 1 1 2 1 2 k k o p p k Ma         −      = 15 268.0 2 p pMa o (k = 1.4) )1)(1(5.02 * )1(5.0 )1(5.011 −+       + −+ = kk k Mak MaA A ( )32 * 73.1 2.011 Ma MaA A + = (k=1.4) V V A A ρ ρ ** * =                 −      − == − k k oo oo p p p p k pkAAVm k 1 1 1 2 2 ρρ&               −      = 286.043.1 17 oo oo p p p ppAm ρ& (With k = 1.4) Final Equations ENSC 3233 Fluid Mechanics Asba a 2 « a ding’ o—x he= ype | an 2 ed? 1e= fab 2» a pt (a) Rectangle (b) Circle = ER = ab = ba? a=B ed} aed I= OE Ty, = 0.1098R4 = by 94 he | Ine = BE MO ~ 2a a ¢ 1, = 0.3927R4 o—x rar 2 at 43) og 3 |-—_p—+}—_ rp —+] “ b+ d_,| 3 I b (0) Semicircle (a) Triangle = aR A= 4 1,,= 1, = 0.05488" Inyo = -0.01647R* (c) Quarter circle Final Equations ENSC 3233 Fluid Mechanics 10 Cylinder and Sphere Drag Coefficients. 2-D Shape drag Coefficients. Final Equations ENSC 3233 Fluid Mechanics 11 Drag Coeficients for object of Interest. Final Equations ENSC 3233 Fluid Mechanics 12 Lift and drag for NACA 64(1) airfoil.