Review for Exam Hyunse Yoon, Ph.D. Adjunct Assistant Professor Department of Mechanical Engineering, University of Iowa

Transcription

1 Review for Exam Hyunse Yoon, Ph.D. Adjunct Assistant Professor Department of Mechanical Engineering, University of Iowa Assistant Research Scientist IIHR-Hydroscience & Engineering, University of Iowa

2 57:020 Fluids Mechanics Fall 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.

3 57:020 Fluids Mechanics Fall 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

4 57:020 Fluids Mechanics Fall 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

5 57:020 Fluids Mechanics Fall 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

6 57:020 Fluids Mechanics Fall 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.

7 57:020 Fluids Mechanics Fall 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

8 57:020 Fluids Mechanics Fall 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.

9 57:020 Fluids Mechanics Fall 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.

10 57:020 Fluids Mechanics Fall 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

11 57:020 Fluids Mechanics Fall 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

12 57:020 Fluids Mechanics Fall 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 = = 68 m Pump power, WW pp = γγγγh pp = = 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.

13 57:020 Fluids Mechanics Fall 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).

14 57:020 Fluids Mechanics Fall Navier-Stokes Equations For incompressible, Newtonian fluids, Continuity: Momentum: + + = 0 ρρ + uu + vv + ww = + ρρg xx + μμ 2 uu xx uu yy uu zz 2 ρρ + uu + vv + ww = + ρρg yy + μμ 2 vv xx vv yy vv zz 2 ρρ + uu + vv + ww = + ρρg zz + μμ 2 ww xx ww yy ww zz 2

15 57:020 Fluids Mechanics Fall 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

16 57:020 Fluids Mechanics Fall 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)

17 57:020 Fluids Mechanics Fall 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.

18 57:020 Fluids Mechanics Fall 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 CC 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.

19 57:020 Fluids Mechanics Fall 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 CC 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.

20 57:020 Fluids Mechanics Fall 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

21 57:020 Fluids Mechanics Fall 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

22 57:020 Fluids Mechanics Fall 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.

23 57:020 Fluids Mechanics Fall 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.

24 57:020 Fluids Mechanics Fall 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.

25 57:020 Fluids Mechanics Fall Example Contd. Model (in water) DD mm = 0.1 m ρρ mm = 998 kg/m 3 μμ mm = 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 = 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 = mm ss Prediction equation: Δpp l pp = DD mm DD pp ρρ pp ρρ mm VV pp VV mm 2 Δpp l mm = = PPPP mm This slide contains an example problem and its contents (except for general formula) should NOT be included in your cheat-sheet.

26 57:020 Fluids Mechanics Fall 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

27 57:020 Fluids Mechanics Fall 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?

28 57:020 Fluids Mechanics Fall 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

29 57:020 Fluids Mechanics Fall 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

30 57:020 Fluids Mechanics Fall Turbulent Pipe Flow From a dimensional analysis ff = φφ Re, εε DD Moody chart: Empirical functional dependency of ff on Re and εε DD

31 57:020 Fluids Mechanics Fall Turbulent Pipe Flow Cond. Colebrook equation (difficult in its use as implicit) 1ff = 2 log εε DD RRRR ff Haaland equation (easier to use as explicit but approximation) 1 ff = 1.8 log εε DD RRRR

32 57:020 Fluids Mechanics Fall Example (pipe flow) VV = QQ AA = 25 ππ = 14.1 ft s Re = VVVV νν = (14.1)(1.5) = (turbulent) εε DD = = 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 h ff = ff LL DD VV 2 (1700) = g (1.5) If D = 1.5 ft and Q = 25 ft 3 /s, z = z 1 z 2? Neglect minor losses. Δzz = 5555 ffff ff = = 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.