Review for Exam2 11. 13. 2015 Hyunse Yoon, Ph.D. Adjunct Assistant Professor Department of Mechanical Engineering, University of Iowa Assistant Research Scientist IIHR-Hydroscience & Engineering, University of Iowa
57:020 Fluids Mechanics Fall2015 2 System vs. Control volume System: A collection of real matter of fixed identity. Control volume (CV): A geometric or an imaginary volume in space through which fluid may flow. A CV may move or deform.
57:020 Fluids Mechanics Fall2015 3 Laws of Mechanics for a System Laws of mechanics are written for a system, i.e., for a fixed amount of matter Conservation of mass DDDD DDDD = 0 Conservation of momentum DD mmvv DDDD = mmaa = FF Note: DD mmvv DDtt = DDmm DDtt =0 VV + mm DDVV DDtt =aa = mmaa Conservation of energy DDDD DDDD = Governing Differential Eq. (GDE): QQ WW DD DDDD mm, mmvv, EE system extensive properties, BB sys = RHS
57:020 Fluids Mechanics Fall2015 4 Reynolds Transport Theorem (RTT) In fluid mechanics, we are usually interested in a region of space, i.e., CV and not particular systems. Therefore, we need to transform GDE s from a system to a CV, which is accomplished through the use of RTT DDBB sys DDtt time rate ofchange of BB for a system = DD DDtt CV xx,tt ββββββvv time rate ofchange of BB in CV + CS xx,tt ββββvv RR ddaa net flux of BB across CS where, ββ = dddd dddd = 1, VV, ee for BB = (mm, mmvv, EE) Fixed CV, DDBB sys DDDD = tt + ββββvv ddaa CVββββββVV CS Note: BB CV = CV ββββββ = CV ββββββvv BB CS = CS ββββmm = ββρρvv ddaa CS
57:020 Fluids Mechanics Fall2015 5 Continuity Equation RTT with BB = mm and ββ = 1, Steady flow, Simplified form, tt ρρρρvv + CV CS ρρvv ddaa = 0 CS ρρvv ddaa = 0 mm out mm in = 0 Conduit flow with one inlet (1) and one outlet (2): Note: mm = ρρρρ = ρρρρρρ ρρ 2 VV 2 AA 2 ρρ 1 VV 1 AA 1 = 0 If ρρ = constant, VV 1 AA 1 = VV 2 AA 2
57:020 Fluids Mechanics Fall2015 6 Momentum Equation RTT with BB = mmvv and ββ = VV, Simplified form: tt VVρρρρVV + CV CS VVρρVV ddaa = FF or in component forms, mmvv out mmvv in = FF mmuu out mmvv out mmww out mmuu in = FF xx mmvv in = FF yy mmww in = FF zz Note: If VV = uuii + vvjj + ww kk is normal to CS, mm = ρρρρρρ, where VV = VV.
57:020 Fluids Mechanics Fall2015 7 Momentum Equation Contd. External forces: FF = FF body + FF surface + FF other o FF body = FF gravity FF gravity : gravity force (i.e., weight) o FF Surface = FF pressure + FF friction + FF other FF pressure : pressure forces normal to CS FF friction : viscous friction forces tangent to CS o FF other : anchoring forces or reaction forces Note: Surface forces arise as the CV is isolated from its surroundings, similarly to drawing a free-body diagram. A well-chosen CV exposes only the forces that are to be determined and a minimum number of other forces
57:020 Fluids Mechanics Fall2015 8 Example (Bend) uu 1 uu 2 Inlet (1): mm 1 = ρρvv 1 AA 1 uu 1 = VV 1 vv 1 = 0 Outlet (2): mm 2 = ρρvv 2 AA 2 uu 2 = VV 2 cos 45 vv 2 = VV 2 sin 45 vv2 mmuu out mmvv out mmuu in = ρρvv 2 AA 2 VV 2 cos 45 ρρvv 1 AA 1 VV 1 mmvv in = ρρvv 2 AA 2 VV 2 sin 45 ρρvv 1 AA 1 0 Since ρρvv 1 AA 1 = ρρvv 2 AA 2, mmuu out mmuu in = ρρvv 2 AA 2 VV 2 cos 45 + VV 1 mmvv out mmvv in = ρρvv 2 2 AA 2 sin 45 This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 9 Example Contd. pp 2 AA 2 sin 45 pp 2 AA 2 FF xx : 1) Body force = 0 2) Pressure force = pp 1 AA 1 + pp 2 AA 2 cos 45 3) Anchoring force = FF AAAA FF yy : 1) Body force = WW ww WW ee 2) Pressure force = pp 2 AA 2 sin 45 3) Anchoring force = FF AAzz pp 2 AA 2 cos 45 Thus, ρρvv 2 AA 2 VV 2 cos 45 + VV 1 = pp 1 AA 1 + pp 2 AA 2 cos 45 FF AAAA ρρvv 2 2 AA 2 sin 45 = γγvv ww WW ee + pp 2 AA 2 sin 45 FF AAAA FF AAAA = ρρvv 2 AA 2 VV 2 cos 45 + VV 1 + pp 1 AA 1 + pp 2 AA 2 cos 45 FF AAAA = ρρvv 2 2 AA 2 sin 45 γγvv ww WW ee + pp 2 AA 2 sin 45 This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 10 Energy Equation RTT with BB = EE and ββ = ee, Simplified form: tt eeρρρρvv + CV CS 2 pp in γγ + αα VV in in 2g + zz in + h pp = pp out γγ eeρρvv ddaa = + αα out VV in energy equation refers to average velocity VV QQ WW 2 VV out 2g + zz out + h tt + h LL 1 for uniform flow across CS αα : kinetic energy correction factor = 2 for laminar pipe flow 1 for turbulent pipe flow
57:020 Fluids Mechanics Fall2015 11 Energy Equation - Contd. Uniform flow across CS s: γγ + VV 2 1 2g + zz 1 + h pp = pp 2 γγ + VV 2 2 2g + zz 1 + h tt + h LL pp 1 WW Pump head h pp = pp = mmg WW Turbine head h tt = tt = mmg WW pp ρρρρg = WW tt ρρρρg = WW pp γγγγ WW pp = WW tt γγγγ WW tt = Head loss h LL = loss g = uu 2 uu 1 g QQ mmg > 0 mmgh pp = ρρgqqh pp = γγγγγ pp mmgh tt = ρρgqqh tt = γγγγγ tt
57:020 Fluids Mechanics Fall2015 12 Example (Pump) Energy equation: γγ + VV 2 1 2g + zz 1 + h pp = pp 2 γγ + VV 2 2 2g + zz 2 + h tt + h LL pp 1 With pp 1 = pp 2 = 0, VV 1 = VV 2 0, h tt = 0, and h LL = 23 m h pp = zz 2 zz 1 + h LL = 45 + 23 = 68 m Pump power, WW pp = γγγγh pp = 68 9790 0.03 746 = 80 hp (Note: 1 hp = 746 N m/s = 550 ft lbf/s) This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 13 Differential Analysis A microscopic description of fluid motions for a fluid particle by using differential equations*, Continuity equation Momentum equation + ρρvv = 0 ρρ VV + VV VV = ρρg pp + ττ iiii *CV analysis is a macroscopic description of fluid motions by using integral equations (RTT).
57:020 Fluids Mechanics Fall2015 14 Navier-Stokes Equations For incompressible, Newtonian fluids, Continuity: Momentum: + + = 0 ρρ + uu + vv + ww = + ρρg xx + μμ 2 uu xx 2 + 2 uu yy 2 + 2 uu zz 2 ρρ + uu + vv + ww = + ρρg yy + μμ 2 vv xx 2 + 2 vv yy 2 + 2 vv zz 2 ρρ + uu + vv + ww = + ρρg zz + μμ 2 ww xx 2 + 2 ww yy 2 + 2 ww zz 2
57:020 Fluids Mechanics Fall2015 15 Exact Solutions of NS Eqns. The flow of interest is assumed additionally (than incompressible & Newtonian), for example, 1) Steady (i.e., = 00 for any variable) 2) Parallel such that the yy-component of velocity is zero (i.e., vv = 00) 3) Purely two dimensional (i.e., ww = 00 and = 00 for any velocity component) 4) Fully developed (i.e., = 00 for any velocity component) e.g.) or 1) ρρ 4) + uu 3) 2) + vv + ww = + ρρg xx + μμ μμ dd2 uu ddyy 2 = ρρg xx 4) 2 uu xx 2 3) + 2 uu 2 yy 2 + uu zz 2
57:020 Fluids Mechanics Fall2015 16 Boundary Conditions Common BC s: No-slip condition (VV fluid = VV wall ; for a stationary wall VV fluid = 0) Interface boundary condition (VV AA = VV BB and ττ ss,aa = ττ ss,bb ) Free-surface boundary condition (pp liquid = pp gas and ττ ss,liquid = 0) Other BC s: Inlet/outlet boundary condition Symmetry boundary condition Initial condition (for unsteady flow problem)
57:020 Fluids Mechanics Fall2015 17 Example: No pressure gradient μμ dd2 uu ddyy 2 = 0 Integrate twice, uu yy = CC 1 yy + CC 2 B.C., uu 0 = (CC 1 ) 0 + CC 2 = 0 CC 2 = 0 uu bb = CC 1 bb + CC 2 = UU CC 1 = UU bb Analysis: ττ ww = μμ uu yy = UU bb yy dddd = μμ dddd yy=0 UU bb = μμμμ bb This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 18 Example: with Pressure Gradient Fixed plate Fixed plate Integrate twice, B.C., uu 0 = 1 2μμ dddd dddd uu bb = 1 2μμ μμ dd2 uu ddyy 2 = ddpp ddxx uu yy = 1 dddd 2μμ dddd yy2 + CC 1 yy + CC 2 dddd dddd 0 2 + CC 1 0 + CC 2 = 0 CC 2 = 0 bb 2 + CC 1 bb + CC 2 = 0 CC 1 = 1 2μμ dddd dddd bb Analysis: uu yy = 1 2μμ ddpp ddxx h qq = uuuuuu = bb3 h 12μμ yy 2 bbbb ττ ww = μμ dddd = bb dddd yy=0 2 This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 19 Example: Inclined wall Integrate twice, B.C., uu 0 = ρρg xx μμ dddd = dddd yy=h μμ dd2 uu ddyy 2 = ρρg xx uu yy = ρρg xx 2μμ yy2 + CC 1 yy + CC 2 ρρg xx μμ 0 2 + CC 1 0 + CC 2 = 0 CC 2 = 0 h + CC 1 = 0 CC 1 = ρρg xx μμ h Note: g = g xx ıı + g yy ȷȷ where, g xx = g sin θθ g yy = g cos θθ Analysis: uu yy = ρρg xx μμ hyy yy2 2 h qq = uuuuuu = ρρg xx h 3 0 μμ 3 ττ ww = μμ dddd = μμ dddd yy=0 ρρg xx μμ h = ρρg xxh This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 20 Buckingham Pi Theorem For any physically meaningful equation involving nn variables, such as uu 1 = ff uu 2, uu 3,, uu nn with minimum number of mm reference dimensions, the equation can be rearranged into product of rr dimensionless pi terms. Π 1 = φφ Π 2, Π 3,, Π rr where, rr = nn mm
57:020 Fluids Mechanics Fall2015 21 Similarity and Model Testing If all relevant dimensionless parameters have the same corresponding values for model and prototype, flow conditions for a model test are completely similar to those for prototype. For, Similarity requirements: Prediction equation: Π 1 = φφ Π 2,, Π nn Π 2,model = Π 2,prototype Π nn,model = Π nn,prototype Π 1,model = Π 1,prototype
57:020 Fluids Mechanics Fall2015 22 Example (Repeating Variable Method) Example: The pressure drop per unit length Δpp l in a pipe flow is a function of the pipe diameter DD and the fluid density ρρ, viscosity μμ, and velocity VV. Δpp l = ff DD, ρρ, μμ, VV Δpp l DD ρρ μμ VV MMLL 2 TT 2 LL MMMM 3 MMLL 1 TT 1 LLTT 1 rr = nn mm = 5 3 = 2 This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 23 Example Contd. Select mm = 3 repeating variables, DD, VV, ρρ for LL, TT, MM, then Π 1 = DD aa VV bb ρρ cc Δpp l = LL aa LLTT 1 bb MMLL 3 cc MMLL 2 TT 2 = MM 0 LL 0 TT 0 aa + bb 3cc 2 = 0 bb 2 = 0 aa = 1, bb = 2, cc = 1 cc + 1 = 0 Π 1 = DD 1 VV 2 ρρ 1 Δpp l = Δpp ldd ρρvv 2 Π 2 = DD aa VV bb ρρ cc μμ = LL aa LLTT 1 bb MMLL 3 cc MMLL 1 TT 1 = MM 0 LL 0 TT 0 aa + bb 3cc 1 = 0 bb 1 = 0 cc + 1 = 0 Π 2 = DD 1 VV 1 ρρ 1 μμ = Δpp ldd ρρρρρρ = φφ ρρvv2 μμ aa = 1, bb = 1, cc = 1 μμ DDDDDD This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 24 Example (Model Testing) Model Prototype Δpp l DD ρρρρρρ = φφ ρρvv2 μμ If, Then, ρρ mm VV mm DD mm μμ mm Δpp l mm DD mm ρρ mm VV mm 2 = ρρ ppvv pp DD pp μμ pp = Δpp lpp DD pp ρρ pp VV pp 2 similarity requirement (Prediction equation) This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 25 Example Contd. Model (in water) DD mm = 0.1 m ρρ mm = 998 kg/m 3 μμ mm = 1.12 10-3 N s/m 2 VV mm =? Δpp l mm = 27.6 Pa/m Prototype (in air) DD pp = 1 m ρρ pp = 1.23 kg/m 3 μμ pp = 1.79 10-5 N s/m 2 VV pp = 10 m/s Δpp l mm =? Similarity requirement: VV mm = ρρ pp ρρ mm μμ mm μμ pp DD pp VV DD pp = 1.23 mm 998 1.12 10 3 1.79 10 5 1 0.1 10 = 77. 7777 mm ss Prediction equation: Δpp l pp = DD mm DD pp ρρ pp ρρ mm VV pp VV mm 2 Δpp l mm = 0.1 1 1.23 998 10 7.71 2 27.6 = 55. 7777 1111 33 PPPP mm This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.
57:020 Fluids Mechanics Fall2015 26 Pipe Flow: Laminar vs. Turbulent Reynolds number regimes Re = ρρρρρρ μμ Re < Re crit 2,000 Re crit < Re < Re trans Re > RRee trans 4,000
57:020 Fluids Mechanics Fall2015 27 Flow in Pipes Basic piping problems: Given the desired flow rate, what pressure drop (e.g., pump power) is needed to drive the flow (i.e., to overcome the head loss through piping)? Given the pressure drop (e.g., pump power) available, what flow rate will ensue? Given the pressure drop and the flow rate desired, what pipe diameter is needed?
57:020 Fluids Mechanics Fall2015 28 Head Loss h LL = h LL major + h LL minor h LL major (or h ff ): Major loss, the loss due to viscous effects h LL minor : Minor loss, the loss in the various pipe components Darcy-Weisbach equation h ff = ff LL DD VV 2 2g ff = 8ττ ww ρρvv2: Friction factor LL: Pipe length DD: Pipe diameter VV: Average flow velocity across the pipe cross-section
57:020 Fluids Mechanics Fall2015 29 Laminar Pipe Flow Exact solution exists by solving the NS equation uu rr = VV max 1 rr RR 2, VVmax = 2VV Wall shear stress Friction factor Head loss h ff = ff LL DD ττ ww = μμ dddd = 8μμμμ dddd rr=rr DD ff = 8ττ ww ρρvv 2 = 64μμ ρρρρρρ = 64 Re VV 2 2g = 32μμμμμμ γγdd 2 = 128μμμμμμ ππππdd 4 Notes: QQ = VVVV AA = ππdd2 4
57:020 Fluids Mechanics Fall2015 30 Turbulent Pipe Flow From a dimensional analysis ff = φφ Re, εε DD Moody chart: Empirical functional dependency of ff on Re and εε DD
57:020 Fluids Mechanics Fall2015 31 Turbulent Pipe Flow Cond. Colebrook equation (difficult in its use as implicit) 1ff = 2 log εε DD 3.7 + 2.51 RRRR ff Haaland equation (easier to use as explicit but approximation) 1 ff = 1.8 log εε DD 3.7 1.1 + 6.9 RRRR
57:020 Fluids Mechanics Fall2015 32 Example (pipe flow) VV = QQ AA = 25 ππ 1.5 2 4 = 14.1 ft s Re = VVVV νν = (14.1)(1.5) 1.21 10 5 = 1.75 106 (turbulent) εε DD = 0.0005 1.63 = 0.00033 Energy equation: 2 pp 1 γγ + αα VV 1 1 2g + zz 1 = pp 2 2 γγ + αα VV 2 2 2g + zz 2 + h ff Since pp 1 = pp 2 = 0 and VV 1 = VV 2, h ff = zz 1 zz 2 = Δzz Friction factor, Head loss 1 ff = 1.8 log 0.00033 3.7 h ff = ff LL DD 1.1 + VV 2 (1700) = 0.0159 2g (1.5) If D = 1.5 ft and Q = 25 ft 3 /s, z = z 1 z 2? Neglect minor losses. Δzz = 5555 ffff 6.9 1.75 10 6 ff = 0.0159 14.1 2 = 56 ft 2 (32.2) This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.