SCHEMATIC:T∞, u∞hlam=Clamx-0.5hturb=Cturbx-0.2TurbulentLaminarxxcT∞, u∞hlam=Clamx-0.5hturb=Cturbx-0.2TurbulentLaminarxxcASSUMPTIONS:(1) Steady-state conditions, (2) Constant properties.PROPERTIES:Table A.4, air (T = 350 K): k = 0.030 W/m⋅K,ν= 20.92×10-6m2/s, Pr = 0.700.ANALYSIS:(a) Using air properties evaluated at 350 K with xc= 0.5 m,5cx,cuxRe510ν∞==×556c2u510x51020.9210m/ s 0.5m20.9m s−∞=×=×××=ν<(b) From Eq. 6.13, the average coefficient in the laminar region, 0≤x≤xc, is()()lamlamxx-0.50.5CClamlamlamlam0o111-0.5hx=hxdx =xdx =x2C2xxx=xh(x)=∫∫(1)<(c) The average coefficient in the turbulent region, xc≤x≤L, is()()()ccccxx0.50.8xxturblamturblamturb0x0x1xxhxhx dxhx dxCCx0.50.8=+=+∫∫Continued…

PROBLEM 6.15 (Cont.)()()0.50.80.8turblamcturbc1hx2Cx1.25Cxxx=+−(2)<(d) The local and average coefficients, Eqs. (1) and (2) are plotted below as a function of x for therange 0≤x≤L.00.51Distance from leading edge, x (m)050100150Convection coefficient (W/m^2.K)Local - laminar, x <= xcLocal - turbulent, x => xcAverage - laminar, x <= xcAverage - turbulent, x => xc

PROBLEM 6.16KNOWN:Air speed and temperature in a wind tunnel.FIND:(a) Minimum plate length to achieve a Reynolds number of 107, (b) Distance fromleading edge at which transition would occur.SCHEMATIC:Air,u∞= 30 m/sT∞= 23°CASSUMPTIONS:(1) Isothermal conditions, Ts= T∞.PROPERTIES:Table A-4, Air (23°C = 296K):ν= 15.53×10-6m2/s.ANALYSIS:(a) The Reynolds number isxuxuxRe.ρµν∞∞=To achieve a Reynolds number of 1×107, the minimum plate length is then()762xmin1 1015.53 10m/ sReLu30 m/sν−∞××==minL5.18 m.=<(b) For a transition Reynolds number of 5×105()5-62x,cc510 15.53 10m/ sRexu30 m/sν∞××==cx0.259 m.=<COMMENTS:Note thatx,ccLRexLRe=This expression may be used to quickly establish the location of transition from knowledge ofRe.x,cLand Re

PROBLEM 6.17KNOWN:Velocity and temperature of water flowing over a flat plate.Length of plate.Variation oflocal convection coefficient withxfor laminar and turbulent flow.FIND:Minimum and maximum average convection coefficient for roughness applied over the rangexr≤x≤L.Temperature at which extreme values of average convection coefficient occur andcorresponding values ofxr.SCHEMATIC:T= 300 K or 350 Ku∞L= 0.6 mxxrRoughnessASSUMPTIONS:(1) Steady-state conditions, (2) Constant properties, (3) Transition occurs at acritical Reynolds number of 5 × 105for the smooth plate, (4) Incompressible flow.PROPERTIES:Table A.6, Liquid water (T= 300 K):ρ=vf-1= 997kg/m3,µ= 855 × 10-6N⋅s/m2;Liquid water (T= 350 K):ρ=vf-1= 974kg/m3,µ= 3655 × 10-6N⋅s/m2.ANALYSIS:The smooth plate transition location,xc, was found in Example 6.4 to be 0.43 m and0.19 m, forT= 300 K and 350 K, respectively.For roughness applied over the range 0 ≤xr≤xc,transition occurs atxr. From Eq. 6.14,()0.50.8lamturblamturb0000.50.80.8lamturb1110.50.810.50.8rrrrxLLxLxxrrCChhdxhdxhdxxxLLLCCxLxL==+=+=+−∫∫∫(1)Roughness applied over the rangexr>xchas no effect on the transition since the transition occurs atxcfor the smooth plate. Thus, from Example 6.4,21620 W/mKh=⋅or23710 W/mK⋅forT= 300 K or350 K, respectively forxr>xc.

Upload your study docs or become a

Course Hero member to access this document

Upload your study docs or become a

Course Hero member to access this document

End of preview. Want to read all 110 pages?

Upload your study docs or become a

Course Hero member to access this document