AE/ME 339 Computational Fluid Dynamics (CFD) 9//005 Topic7_NS_ F0 1 Momentum equation 9//005 Topic7_NS_ F0 1
Consider the moving fluid element model shown in Figure.b Basis is Newton s nd Law which says F = m a Note that this is a vector equation. It can be written in terms of the three cartesian scalar components, the first of which becomes F x = m a x Since we are considering a fluid element moving with the fluid, its mass, m, is fixed. The momentum equation will be obtained by writing expressions for the externally applied force, F x, on the fluid element and the acceleration, a x, of the fluid element. The externally applied forces can be divided into two types: 9//005 Topic7_NS_ F0 3 1. Body forces: Distributed throughout the control volume. Therefore, this is proportional to the volume. Examples: gravitational forces, magnetic forces, electrostatic forces.. Surface forces: Distributed at the control volume surface. Proportional to the surface area. Examples: forces due to surface and normal stresses. These can be calculated from stress-strain rate relations. Body force on the fluid element = f x dm = f x ρ dx dy dz where f x is the body force per unit mass in the x-direction The shear and normal stresses arise from the deformation of the fluid element as it flows along. The shape as well as the volume of the fluid element could change and the associated normal and tangential stresses give rise to the surface stresses. 9//005 Topic7_NS_ F0 4
The relation between stress and rate of strain in a fluid is known from the type of fluid we are dealing with. Most of our discussion will relate to Newtonian fluids for which Stress is proportional to the rate of strain For non-newtonian fluids more complex relationships should be used. Notation: stress τij indicates stress acting on a plane perpendicular to the i-direction (x-axis) and the stress acts in the direction, j, (y-axis). The stresses on the various faces of the fluid element can written as shown in Figure.8. Note the use of Taylor series to write the stress components. 9//005 Topic7_NS_ F0 5 Strictly, it should be τ yx on the LHS 9//005 Topic7_NS_ F0 6 3
The normal stresses also has the pressure term. Net surface force acting in x direction = p p p+ dx dydz x xx yx + τxx + dx τxx dydz + τ yx + dy τ yx dxdz x y zx + τzx + dz τzx dxdy...(.46) z 9//005 Topic7_NS_ F0 7 p xx yx zx Fx = + + + dxdydz + ρ f xdxdydz...(.47) x x y z m = ρdxdydz...(.48) Du ax =...(.49) Dt 9//005 Topic7_NS_ F0 8 4
Du p xx yx zx ρ = + + + + ρ f x...(.50 a) Dt x x y z Dv p xy yy zy ρ = + + + + ρ f y...(.50 b) Dt y x y z Dw p xz yz zz ρ = + + + + ρ f z...(.50 c) Dt z x y z 9//005 Topic7_NS_ F0 9 ρ Du u V u...(.51) Dt ρ = + t ρ u ρ = ρ + u t t t u ρ ρ = u...(.5) t t t ( ) ( )...(.53) ρv u= ρuv u ρv Du ρ ρ = u u ρv + ρuv Dt t t ρ = u + V + uv t t ( ) ( ) ( ρ ) ( ρ )...(.54) 9//005 Topic7_NS_ F0 10 5
The term in the brackets is zero (continuity equation) The above equation simplifies to Du ρ = + ( ρuv )...(.55) Dt t Substitute Eq. (.55) into Eq. (.50a) shows how the following equations can be obtained. p τ xx yx τzx + ( ρuv ) = + + + + ρ f x...(.56 a) t x x y z ( ρv) p + = + + + + t y x y z xy yy zy ( ρvv ) ρ f y...(.56 b) ( ρw) p τ xz yz τzz + ( ρwv ) = + + + + ρ f z...(.56 c) t z x y z 9//005 Topic7_NS_ F0 11 For Newtonian fluids the stresses can be expressed as follows u τxx = λ + µ x ( V)...(.57 a) τ v yy = λ + µ y ( V )...(.57 b) τ w zz = λ + µ z ( V )...(.57 c) τ v u xy = τ yx = µ +...(.57 d ) x y τ u w xz = τ zx = µ +...(.57 e) z x τ w v yz = τ zy = µ +...(.57 f ) y z 9//005 Topic7_NS_ F0 1 6
The above are the Navier-Stokes equations in conservation form. 9//005 Topic7_NS_ F0 13 In the above µ is the coefficient of dynamic viscosity and λ is the second viscosity coefficient. Stokes hypothesis given below can be used to relate the above two coefficients λ= - /3 µ The above can be used to get the Navier-Stokes equations in the following familiar form 9//005 Topic7_NS_ F0 14 7
( ρ ) ( ρu ) ( ρ ) ( ρ ) u uv uw p u + + + = + λ V + µ t x y z x x x v u u w + µ + + µ + + ρ f...(.58 ) x a y x y z z x ( ρ ) ( ρv ) ( ρ ) ( ρ ) v uv vw p v + + + = + λ V + µ t y x z y y y v u v w + µ + + µ + + ρ f...(.58 ) y b x x y z z y 9//005 Topic7_NS_ F0 15 ( ρ ) ( ρw ) ( ρ ) ( ρ ) w uw vw p w + + + = + λ V + µ t z x y z z z w v u w + µ + + µ + + ρ f...(.58 ) z c y y z x z x The energy equation can also be derived in a similar manner. Read Section.7 9//005 Topic7_NS_ F0 16 8
For a summary of the equations in conservation and non-conservation forms see Anderson, pages 76 and 77. The above equations can be simplified for inviscid flows by dropping the terms involving viscosity. (read Section.8.) 9//005 Topic7_NS_ F0 17 Navier Stokes Equations ρ + ( ρv )...(.55) t Du p u v w u v w u ρ = ρ f x + µ + µ + + µ + Dt x x 3 x y z y y x z x z Dv p v u v u w v w ρ = ρ f y + µ + + µ + µ + Dt y x x y y 3 y x z z z y Dw p w u v w w u v ρ = ρ f z + µ + ++ µ + + µ Dt z x x z y z y z 3 z x y 9//005 Topic7_NS_ F0 18 9
Navier Stokes Equations (continued) Energy Equation De ρ = ρq + k T +Φ Dt u v w v u w v u w u v w Φ= µ + + + + + + + + + + x y z x y y z z x 3 x y z e - internal energy q - heat generation per unit volume Φ viscous dissipation term 9//005 Topic7_NS_ F0 19 Navier Stokes Equations ρ + ( ρv )...(.55) t Du p fx u v w ρ ρ µ µ u v µ w u = + + + + + Dt x x 3 x y z y y x z x z Dv p v u v u w v w ρ = ρ f y + µ + + µ + µ + Dt y x x y y 3 y x z z z y Dw p w u v w w u v ρ = ρ fz + µ + ++ µ + + µ Dt z x x z y z y z 3 z x y 9//005 Topic7_NS_ F0 0 10
Energy Equation Navier Stokes Equations (continued) De ρ = ρq + k T +Φ Dt u v w v u w v u w u v w Φ= µ + + + + + + + + + + x y z x y y z z x 3 x y z e - internal energy q - heat generation per unit volume Φ viscous dissipation term 9//005 Topic7_NS_ F0 1 11