Review for Exam3 12. 9. 2015 Hyunse Yoon, Ph.D. Adjunct Assistant Professor Department of Mechanical Engineering, University of Iowa Assistant Research Scientist IIHR-Hydroscience & Engineering, University of Iowa
2 Chapter 8 Flow in Conduits Internal flow: Confined by solid walls Basic piping problems: Given the desired flow rate, what pressure drop (e.g., pump power) is needed to drive the flow? Given the pressure drop (e.g., pump power) available, what flow rate will ensue?
3 Pipe Flow: Laminar vs. Turbulent Reynolds number regimes Re = ρρρρρρ μμ Re < Re crit 2,000 Re crit < Re < Re trans Re > Re trans 4,000
4 Entrance Region and Fully Developed Entrance Length, LL e : o Laminar flow: LL e DD = 0.06RRRR (LL e,max = 0.06Re crit 138DD) o Turbulent flow: LL e DD = 4.4Re 1 6 (20DD < LLe < 30DD for 10 4 < Re < 10 5 )
5 Pressure Drop and Shear Stress Pressure drop, Δpp = pp 1 pp 2, is needed to overcome viscous shear stress. Considering force balance, ππdd 2 pp 1 4 pp 2 ππdd 2 4 = ττ ww ππππππ Δpp = 4ττ ww LL DD
6 Head Loss and Friction Factor Energy equation h LL = pp 1 pp 2 γγ + αα 1VV 1 2 αα 2 VV 2 2 2gg + zz 1 zz 2 = Δpp γγ Δpp = 4ττ ww LL DD from force balance and γγ = ρρg h LL = LL 4ττ ww DD ρρg = 8ττ ww ρρvv 2 LL DD VV 2 VV 2 2g = ff LL DD 2g Friction factor Darcy Weisbach equation ff 8ττ ww ρρvv 2
7 Fully-developed Laminar Flow Exact solution, uu rr = VV c 1 rr RR 2 Wall sear stress where, VV = QQ AA ττ ww = μμ dddd = 8μμμμ ddrr rr=rr DD Friction factor, ff = 8ττ ww ρρvv 2 = 8 ρρvv 2 8μμμμ DD = 64 ρρρρρρ μμ = 64 Re
8 Fully-developed Turbulent Flow Dimensional analysis ττ ww = ff DD, VV, μμ, ρρ, εε kk rr = 6 3 = 3 Π ss ττ ww ρρvv 2 = φφ ρρρρρρ μμ, εε DD ff = φφ RRRR, εε DD
9 Moody Chart ff = φφ εε DD
10 Moody Chart Contd. Colebrook equation 1ff = 2 log εε DD 3.7 + 2.51 RRRR ff Haaland equation 1 ff = 1.8 log εε DD 3.7 1.1 + 6.9 RRRR
11 Minor Loss Loss of energy due to pipe system components (valves, bends, tees, and the like). Theoretical prediction is, as yet, almost impossible. Usually based on experimental data. KK LL : Loss coefficient h mm = KK LL VV 2 2g E.g.) Pipe entrance (sharp-edged), KK LL =0.8 (well-rounded), KK LL =0.04 Regular 90 elbows (flanged), KK LL =0.3 Pipe exit, KK LL =1.0
12 Pipe Flow Problems Energy equation for pipe flow: γγ + VV 2 1 2g + zz 1 + h pp = pp 2 γγ + VV 2 2 2g + zz 2 + h tt + h LL pp 1 h LL = h ff + h mm = ff LL DD + KK LL VV 2 2g Type I: Determine head loss h LL (or Δpp) Type II: Determine flow rate QQ (or VV) Type III: Determine pipe diameter DD Iteration is needed for types II and III
13 Type I Problem Typically, VV (or QQ) and DD are given Find the pump power WW pp required. For example, if pp 1 = pp 2, VV 1 = VV 2 and Δzz = zz 2 zz 1, 0 + 0 + zz 1 + h pp = 0 + 0 + zz 2 + ff LL DD + KK LL VV 2 2g ff = φφ ρρρρρρ μμ, εε DD Solve the energy equation for h pp, h pp = from Moody Chart ff LL DD + KK LL VV 2 2g Δzz Thus, WW pp = h pp γγγγ
14 Type II Problem QQ (thus VV) is unknown RRRR? Solve energy equation for VV as a function of f. For example, if pp 1 = pp 2, VV 1 = VV 2 and Δzz = zz 2 zz 1, 0 + 0 + zz 1 + h pp = 0 + 0 + zz 2 + ff LL DD + KK LL VV 2 2g VV = 2g h pp Δzz ff LL DD + KK LL Guess ff VV RRRR ff new ; Repeat until ff is converged VV
15 Type III Problem DD is unknown RRRR and εε DD? Solve energy equation for DD as a function of f. For example, if pp 1 = pp 2, VV 1 = VV 2, Δzz = zz 1 zz 2, and KK LL = 0 and using VV = QQ ππdd 2 4, 0 + 0 + zz 1 + h pp = 0 + 0 + zz 2 + ff LL DD + 0 1 2g QQ ππdd 2 4 2 DD = 8LLQQ 2 ff ππ 2 g h pp Δzz 1 5 Guess ff DD RRRR and εε DD ff new ; Repeat until ff is converged DD
16 Chapter 9 Flow over Immersed Bodies External flow: Unconfined, free to expand Complex body geometries require experimental data (dimensional analysis)
17 Drag Resultant force in the direction of the upstream velocity CC DD = DD 1 2 ρρvv2 AA = 1 1 2 ρρvv2 AA pp pp nn iidddd SS CC DDDD = Pressure drag (or Form drag) + ττ ww tt iidddd SS CC ff =Friction drag Streamlined body ( tt l << 1): CC ff >> CC DDDD, Boundary layer flow Bluff body ( tt l 1): CC DDDD >> CC ff where, tt is the thickness and l the length of the body
18 Boundary Layer High Reynolds number flow, Re xx = UU xx Viscous effects are confined to a thin layer, δδ uu = 0.99 at yy = δδ UU Inviscid region νν >> 1,000 Viscous region
19 Friction Coefficient Local friction coefficient cc ff xx = 2ττ ww xx ρρuu 2 Friction drag coefficient CC ff = DD ff 1 2 ρρuu2 AA DD ff = CC ff 1 2 ρρuu2 AA Note: Note: Darcy friction factor for pipe flow ff = 8ττ ww ρρvv 2 DD ff = ττ ww dddd = ττ ww bbbbbb AA 0 CC ff = 2 LL ρρuu 2 bbbb ττ ww bbbbbb 0 = 1 LL 0 LL LL cc ff xx dddd CC ff = cc ff xx
20 Reynolds Number Regime Transition Reynolds number RRee xx,tttt = 5 10 5
21 Laminar boundary layer Blasius introduced coordinate transformations Ψ ηη yy UU νννν ννννuu ff ηη Then, rewrote the BL equations as a simple ODE, ffff + 2ff = 0 From the solutions, δδ xx xx = 5 RRee xx ; cc ff xx = 0.664 RRee xx ; CC ff = 1.328 RRee LL
22 Turbulent boundary layer uu UU yy δδ 1 7 one seventh power law cc ff 0.02RRee δδ 1 6 power law fit δδ xx = 0.16 1 xx RRee7 xx ; cc ff xx = 0.027 1 RRee7 xx ; CC ff = 0.031 1 RRee7 LL Valid for a fully turbulent flow over a smooth flat plate from the leading edge. Better results for sufficiently large RRee LL > 10 7
23 Turbulent boundary layer Contd. Alternate forms by using an experimentally determined shear stress formula: ττ ww = 0.0225ρρUU 2 νν UUUU 1 4 δδ xx xx = 0.37RRee 1 5 xx ; cc ff xx = 0.058 1 RRee5 xx ; CC ff = 0.074 1 RRee5 LL Valid only in the range of the experimental data; RRee LL = 5 10 5 ~10 7 for smooth flat plate
24 Turbulent boundary layer Contd. Other formulas for smooth flat plates are by using the logarithmic velocity-profile instead of the 1/7-power law: δδ LL = cc ff 0.98 log RRee LL 0.732 cc ff = 2 log RRee xx 0.65 2.3 CC ff = 0.455 log 10 RRee LL 2.58 These formulas are valid in the whole range of RRee LL 10 9
25 Turbulent boundary layer Contd. Composite formulas (for flows initially laminar and subsequently turbulent with RRee tt = 5 10 5 ): CC ff = 0.031 1 7 RRee LL CC ff = 0.074 1 5 RRee LL 1440 RRee LL 1700 RRee LL CC ff = 0.455 log 10 RRee 2.58 1700 LL RRee LL
26 Turbulent boundary layer Contd.
27 Bluff Body Drag In general, DD = ff VV, LL, ρρ, μμ, cc, tt, εε, Drag coefficient: CC DD = DD 1 2 ρρvv2 AA = φφ AAAA, tt LL, RRRR, cc VV, εε LL, For bluff bodies experimental data are used to determine CC DD
28 Shape dependence The blunter the body, the larger the drag coefficient The amount of streamlining can have a considerable effect Blunter More streamlined
29 Separation Fluid stream detaches from a surface of a body at sufficiently high velocities. Laminar Turbulent Only appears in viscous flows. Inside a separation region: lowpressure, existence of recirculating /backflows; viscous and rotational effects are the most significant
30 Reynolds number dependence Very low RRRR flow (RRRR < 1) Inertia effects are negligible (creeping flow) CC DD ~ RRee 1 Streamlining can actually increase the drag (an increase in the area and shear force) Moderate RRRR flow (10 3 < RRRR < 10 5 ) For streamlined bodies, CC DD ~ RRee 1 2 For blunt bodies, CC DD ~ constant Very large RRRR flow (turbulent boundary layer) For streamlined bodies, CC DD increases For relatively blunt bodies, CC DD decreases when the flow becomes turbulent (10 5 < RRRR < 10 6 ) For extremely blunt bodies, CC DD ~ constant
31 Surface roughness For streamlined bodies, the drag increases with increasing surface roughness For blunt bodies, an increase in surface roughness can actually cause a decrease in the drag. For extremely blunt bodies, the drag is independent of the surface roughness
32 Lift Lift, LL: Resultant force normal to the upstream velocity CC LL = LL 1 2 ρρuu2 AA LL = CC LL 1 2 ρρuu2 AA
33 Magnus Effect Lift generation by spinning Breaking the symmetry causes a lift
34 Minimum Flight Velocity Total weight of an aircraft should be equal to the lift Thus, WW = FF LL = 1 2 CC 2 LL,mmmmmmρρVV mmmmmm AA VV mmmmmm = 2WW ρρcc LL,mmmmmm AA