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Prof Andrew Tay ME3122 Heat Transfer 1 Now, the volumetric thermal expansion coefficient T T T T p 1 1 1 Substituting for in eqn (11.3), we obtain 2 2 y u T T g y u v x u u T RT p T p 1 1 1 2 Note that for an ideal gas, RT / p (11.7) (11.6) (11.5) (11.4) We note that eqn (11.5) now contains T which means that u cannot be solved independently of T and the two problems are coupled. Although fluid motion is caused by density variations, these variations are very small and a satisfactory solution can be obtained by assuming incompressible flow, i.e. = constant , . This simplification is called the Boussinesq approximation.

CV5 pg 4&5

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Prof Andrew TayME3122 Heat Transfer1 Now, the volumetric thermal expansion coefficient ( )( ) T T T Tp=AA ~|.|

\|cc = |1 1 1Substituting forin eqn (11.3), we obtain( ) ( )22yuT T gyuvxuucc+ =cc+ccv |T RTpTp1 1 12 = ~|.|

\|cc = |Note that for an ideal gas, RT / p = (11.7) (11.6) (11.5) (11.4) We note that eqn (11.5) now contains T which means that u cannot be solved independently of T and the two problems are coupled. Although fluid motion is caused by density variations, these variations are very small and a satisfactory solution can be obtained by assuming incompressible flow,i.e. = constant,. This simplification is called the Boussinesq approximation. Prof Andrew TayME3122 Heat Transfer2 (11.8) (11.10) Hence we define the Grashof Number as (11.9) Substituting dimensionless variables into (11.5): ,T TT T* T ,uv* v ,uu* u ,Ly* y ,Lx* xs = = = = =0 0( )22201* y* uRe* TuL T T g* y* u* v* x* u* uL cc+=cc+cc|(11.7) We have ( )22yuT T gyuvxuucc+ =cc+ccv | (11.5) where L is a characteristic length and0 is a reference velocity, we get: ( ) ( ) | |2 12 3020 Then Let v | v | / L T T g / L u Re . L T T g us L s = = =( )23v| L T T gGrsL=We expect Gr to play the same role in free convection as Re plays in forced convection. It can be shown that forces viscousforces buoyancy =LGr