Chapter 7 pp.14-23In potential situation, the flow goes through different region when px goes from xxx to advance with separation points dpending on magnitude of pxMomentum Integral EquationTo obtain general momentum integral relation which is valid for both laminar and turbulent flow: dyvuy0continuityequation momentum of form BL For flat plate equation0dxdU 0**0211221dyUuHdyUuUudxdUUHdxdCUfwMomentum: ypxvuuuyx1const.212 Up00221xxUUpxxUUp yxyxyxUvUuuvuuvuUu , continuityyxvu vuvUyUuUuuUxUvUuuUUvuuuxyxyxyxy221 0 0001vuvUdyuUUdyuUuxdyxy,0 vuvU *200221xxxxwUUUUdyuUUdyUuUuUx dxdUUdxdCf122* dxdUUHdxdCf 22, *H xxfwUUHCU 2212Laminar flow One parameter SolutionsProblem method:Assume 432dcbafUu, and let ya, b, c, d determined from boundary conditions1) 0yxyyUUu2) yUu , 0yu, 0yyuNo slip is automatically satisfied. 3431622GF GFUu, 1212 221UpUxx, ww,,**1262,907274531537,120102*UwThus, by assuming the form of the profile we have expressed all the momentum integral xxx in terms of i.e.(MISSING THE NOTES)To determineand compute the solution we make use of momentum integral relation.1st multiply momentum integral equation by U HdxdUdxdUUw 22, dxdUdxdUdxdUUSw22,21,Define a new parameter 222dxdU, and seek solution in terms of ;i.e. HHSUw ,Define 2z,dxdUz FHSdxdzU 22 645.0expression dcomplicate F(Th waits based on xxx xxx of experimental data)dxdUzdxdzU 645.0645.0 45.06 dxdUzdxdzUi.e. 45.0165zUdxdUCdxUzUx05645.0, 2zxdxUU05620245.0, 0)0(0xCompute solutions: dxdU2 SUwcomputed expression= 62.009.0 (This was based on curve fit to data.)Note: S=0 for 09.0i.e. 0wwhich is the condition for xxxxxx: 09.0To complete solution; that is, to evaluate velocity profile at at H we must determinefrom 9072745315372Assuming: evaluation : 10%Velocity profile: just OK but can be improved by using better approximation to the xxx xxx polynomial.Example 7.5: xxx xxx xxx flowa) Compute Xsepb) 1.0LxCfa) Sollution11075.01195.06005506602LxULdxLxULxUxcan be evaluated for given L, ReL (Lxx ,00)11075.062LxdxdU123.009.0 LXsepsep: 3% higher than exact solution (=0.1199)b) Solution 1.0LxCfi.e. just before separation 77.0257.0099.02,Re257.0099.02Re257.0ReReRe257.0Re0661.00661.00661.011.01075.0Re)099.0(2Re21099.00661.02121210220602LfLLLLffCLLULLULULCCSTo compute solution, i.e. get *,Uu, need to determine .Consider the complex potential ierazazF22222 2cos2Re2razF 2sin2Im2razF 2sin2cosˆ1ˆarvarvereVrrr,02,ˆcosˆsinˆ,ˆsinˆcosˆjiejier jijijiVrrrˆcossinˆsincosˆcosˆsinˆsinˆcosPotential flow xxx along surface: (consider 90)1) determine a such that 0Uat r=L, 90 aLU 0, i.e. LUa02) let xUat x=L-r LxUxLLUxU 100Chapter 7: pp.26-30Integral Methods:Although xxxx, it is usefull to read section 6-8 of the text and study xxxx 6-8.The momentum-integral equation represents one equation:2)2(fccCdxdUUHdxdIn three unknowns:fCH ,, therefore xxxx xxxx the addition of at least two xxxx. The usual approach is to use a xxxx-shear item correction.),(HCCffAnd then an additional relation associated with the xxxx velocity xxxx xxxx xxxxx xxxx eg )( + xxxx’s xxxx log-law (eg (6-76)+(6-77)). Such a method is described in section 6-8.3. two alternative approaches are discussed in sections 6-8.4 and 6-8.5. here we shall xxxx xxxx xxxx method which has been xxxx as part 2 of HW#3.Turbulent flows*)2(21HdxdUUHdxdCfJust as for the laminar flow case we would like to xxxx)..,()()()(gexxxxUdxdUdxdUdxdUHHdxdUCCxttFor laminar flow all xxxx correlated xxxx terms xxxx xxxx dxdUr2However for turbulent flow (,, HCf) cannot be correlated in terms a single parameter. Additional parameters and relationships are required that model the influenceof the turbulent fluctuations. There are many xxxx all of which require a curtain amount of empirical data. As an example we will review the method of xxxx.Xxxx methodClosure of momentum integral equations for turbulent flow is accomplished through the use of the entrainment-integral xxxx.jixJxyxxxxxyˆˆ))(()(Continuity:
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