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Introduction to Convection: Flow and Thermal Considerations

Introduction to Convection: Flow and Thermal Considerations. Chapter Six and Appendix D Sections 6.1 through 6.8 and D.1 through D.3. Lecture 11b. Dimensionless Parameters. For a small Re: Viscous force is important For a large Re: Viscous force is negligible. Dimensionless Parameters.

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Introduction to Convection: Flow and Thermal Considerations

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  1. Introduction to Convection:Flow and Thermal Considerations Chapter Six and Appendix D Sections 6.1 through 6.8 and D.1 through D.3 Lecture 11b

  2. Dimensionless Parameters For a small Re: Viscous force is important For a large Re: Viscous force is negligible

  3. Dimensionless Parameters Pr ≈ 1 for gases, Pr<<1 for liquid metals, Pr>>1 for oils

  4. Dimensionless Parameters

  5. Dimensionless Parameters

  6. Dimensionless Parameters Dimensionless surface shear stress Dimensionless T gradient at the surface Dimensionless C gradient at the surface

  7. Boundary Layer Equations Normalized boundary layer equations Velocity: Thermal: Concentration:

  8. Boundary Layer Analogies To establish relationship between Cf, Nu and Sh

  9. Boundary Layer Analogies From Observations:

  10. Reynolds Analogies If dp*/dx*=0, Pr=Sc=1, the boundary layer equations for fluid flow, heat transfer and mass transfer become same form. Pr = Sc =1

  11. Chilton-Colburn Analogies 0.6 < Pr < 60 0.6 < Sc < 3000 For Laminar flow, need dP*/dx* =0; For turbulent flow, doesn’t need dp*/dx*=0

  12. Evaporative Cooling • The term evaporative cooling originates from association of the latent energy • created by evaporation at a liquid interface with a reduction in the thermal • energy of the liquid. If evaporation occurs in the absence of other energy transfer • processes, the thermal energy, and hence the temperature of the liquid, must decrease. • If the liquid is to be maintained at a fixed temperature, energy loss due • to evaporation must be replenished by other means. Assuming convection • heat transfer at the interface to provide the only means of energy inflow to • the liquid, an energy balance yields

  13. Evaporative Cooling With radiation from the interface and heat addition by other means,

  14. Example 1 Analogy between heat and mass transfer

  15. Example 1 Known:Boundary layer temperature and heat flux at a location on a solid in an air stream given T and V Find:Water vapor concentration and flux associate with the same location on a larger surface Assumptions:Steady state, 2-D, incompressible boundary layer behavior, constant properties; boundary layer approximations are valid; molar fraction of water vapor is much less than unit.

  16. Example 1 Properties:Table A.4 air (50 C) v, k, Pr, Table A.6 saturated water vapor (50C), , Table A.8 Water vapor-air (50C), DAB = Analysis: T* = f(x*, y*, ReL, Pr, dp*/dx*) CA*= f(x*, y*, ReL, Sc, dp*/dx*)

  17. Example 1 For case 1 (L=1m): ReL,1 = 5.5x106, Pr=0.7 For case 2 (L=2m): ReL,2 = 5.5x106, Sc=0.7 ReL,1 =ReL,2, Pr= Sc, x1*=x2*, y1*=y2*

  18. Example 1 T*=(T-Ts)/(T –Ts) = f(x*, y*, ReL, Pr, dp*/dx*) CA*=(CA-CA,s)/(CA, - CA,s)=f(x*, y*, ReL, Sc, dp*/dx*) We expect T* = CA* = f ( …. ) We can calculate CA from T* and hm from h (Sh=Nu ?) And NA” = hm (CA,s-C A,)

  19. Example 2 Dry air at atmospheric pressure blows across a thermometer whose bulb has been covered with a dampened wick. This classic “wet-bulb” thermometer indicates a steady-state T reached by a small amount of liquid evaporating into a large amount of unsaturated vapor-gas mixture. The thermometer reads at 18.3 ºC. At this T, the following properties were evaluated: Vapor pressure of water: 0.021 bar, density of air: 1.22 kg/m3 Latent heatof water vaporization: 2458 J/kg, Pr: 0.72, Sc: 0.61 Specific heat, cp of air: 0.56 J/kg/c What is the Temperature of dry air ?

  20. q”evap T =? CH2O,  = 0 =1.22 kg/m3 cp=0.56 J/kg/C Pr=0.72 Sc=0.61 pH2O=0.021 bar Ts=18.3 C hfg=2458 J/kg q”conv Example 2 Known: Thermophysical properties and T of water Find: Temperature of dry air Schematic:

  21. Example 2 Assumption: Steady-state, constant properties Analysis: Energy balance: q”conv = q”evap The energy required to evaporate the water is supplied by convective heat transfer. q”conv = h(T – Ts) = hfg*MH2O*N”H2O

  22. Example 2 Where N”H2O is the molar flux of water transferred from thermometer to air N”H2O = hm*(CH2O,S-CH2O,) How do we determine hm/h and CH2O,S ?

  23. Example 2 Pr = 0.72, Sc = 0.61, Reynolds analogy is not accurate Chilton-Colburn analogy can be used to relate hm and h. JH = Jm or St*Pr2/3 = Stm*Sc2/3 Stm/St= (Pr/Sc)2/3

  24. Example 2 St = h/(Vcp), Stm = hm/V Stm/St= (hm/h)*(cp) = (Pr/Sc)2/3 hm/h = (Pr/Sc)2/3/(cp) From ideal gas law: PV = nRT CH2O,S = n/V = P/(R*Ts)

  25. Example 2 M H2O = 18 (kg/kmol), R = 0.08314 (m3bar/kmol/K) T = 18.3C + 2458 (J/kg)* 18 (kg/kmol) * (0.72*0.61)2/3 /(1.22 kg/m3*0.56J/kg/C)*(0.021 bar/0.08314 (m3bar/kmol/K) /(18.3+273.15)K = 18.3 + 32.4 = 50.7 C

  26. Example 3 As a means to prevent ice formation on the wings of small aircraft, it is proposed that electric resistance heating elements be installed within the wings. To determine the representative power requirements, considering normal flying conditions for which plane moves at 100 m/s in air that is at -23 ºC and has properties of k=0.022 W/mK, Pr=0.72, and ν=16.3x10-6 m2/s. If the characteristic length of the airfoil is L=2m and the wind tunnel measurements indicate an average friction coefficient of =0.0025 for the normal conditions, what is the average heat flux needed to maintain a surface temperature of Ts=5 C?

  27. Example 3 Known:Nominal operating conditions of aircraft, characteristic length and average friction coefficient of wing Find: Average heat flux needed to maintain prescribed surface temperature of wing Schematic:

  28. Example 3 Assumption: Steady-state, constant properties Analysis: The average heat flux that must be maintained over the surface of the air foil is where the average convection coefficient ( ) may be obtained from the modified Reynolds analogy (Chilton-Colburn Analogy).

  29. Example 3 Lecture 11b

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