3.4 The Differential Equations on Non-Conservation FormNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
The Substantial Derivative ¶ The substantial derivative operator is defined as
D D t = ∂ ∂ t + v ⋅ ∇ \frac{D}{Dt}=\frac{\partial}{\partial t}+\mathbf{v}\cdot\nabla D t D = ∂ t ∂ + v ⋅ ∇ where the first term of the right hand side is the local derivative and the second term is the convective derivative.
Conservation of Mass ¶ If we apply the substantial derivative operator to density we get
D ρ D t = ∂ ρ ∂ t + v ⋅ ∇ ρ \frac{D\rho}{Dt}=\frac{\partial \rho}{\partial t}+\mathbf{v}\cdot\nabla\rho D t D ρ = ∂ t ∂ ρ + v ⋅ ∇ ρ From before we have the continuity equation on differential form as
∂ ρ ∂ t + ∇ ⋅ ( ρ v ) = 0 \frac{\partial \rho}{\partial t} + \nabla\cdot(\rho\mathbf{v})=0 ∂ t ∂ ρ + ∇ ⋅ ( ρ v ) = 0 which can be rewritten as
∂ ρ ∂ t + ρ ( ∇ ⋅ v ) + v ⋅ ∇ ρ = 0 \frac{\partial \rho}{\partial t} + \rho(\nabla\cdot\mathbf{v}) + \mathbf{v}\cdot\nabla\rho=0 ∂ t ∂ ρ + ρ ( ∇ ⋅ v ) + v ⋅ ∇ ρ = 0 and thus
D ρ D t + ρ ( ∇ ⋅ v ) = 0 \frac{D\rho}{Dt}+\rho(\nabla\cdot\mathbf{v})=0 D t D ρ + ρ ( ∇ ⋅ v ) = 0 Eqn. (3.47)
Conservation of Momentum ¶ We start from the momentum equation on differential form derived above
∂ ∂ t ( ρ v ) + ∇ ⋅ ( ρ v v ) + ∇ p = ρ f \frac{\partial}{\partial t}(\rho \mathbf{v}) + \nabla\cdot(\rho \mathbf{v}\mathbf{v}) + \nabla p = \rho \mathbf{f} ∂ t ∂ ( ρ v ) + ∇ ⋅ ( ρ vv ) + ∇ p = ρ f Expanding the first and the second terms gives
ρ ∂ v ∂ t + v ∂ ρ ∂ t + ρ v ⋅ ∇ v + v ( ∇ ⋅ ρ v ) + ∇ p = ρ f \rho\frac{\partial \mathbf{v}}{\partial t} + \mathbf{v}\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla\mathbf{v} + \mathbf{v}(\nabla\cdot\rho\mathbf{v}) + \nabla p = \rho \mathbf{f} ρ ∂ t ∂ v + v ∂ t ∂ ρ + ρ v ⋅ ∇ v + v ( ∇ ⋅ ρ v ) + ∇ p = ρ f Collecting terms, we can identify the substantial derivative operator applied to the velocity vector and the continuity equation.
ρ [ ∂ v ∂ t + v ⋅ ∇ v ] ⏟ = D v D t + v [ ∂ ρ ∂ t + ∇ ⋅ ρ v ] ⏟ = 0 + ∇ p = ρ f \rho\underbrace{\left[\frac{\partial \mathbf{v}}{\partial t}+\mathbf{v}\cdot\nabla\mathbf{v}\right]}_{=\frac{D\mathbf{v}}{Dt}}+\mathbf{v}\underbrace{\left[\frac{\partial \rho}{\partial t}+\nabla\cdot\rho\mathbf{v}\right]}_{=0}+ \nabla p = \rho \mathbf{f} ρ = D t D v [ ∂ t ∂ v + v ⋅ ∇ v ] + v = 0 [ ∂ t ∂ ρ + ∇ ⋅ ρ v ] + ∇ p = ρ f which gives us the non-conservation form of the momentum equation
D v D t + 1 ρ ∇ p = f \frac{D\mathbf{v}}{Dt}+\frac{1}{\rho}\nabla p = \mathbf{f} D t D v + ρ 1 ∇ p = f Conservation of Energy ¶ The last equation on non-conservation differential form is the energy equation. We start by rewriting the energy equation on differential form (Eqn. (3.42)
∂ ∂ t ( ρ e o ) + ∇ ⋅ ( ρ h o v ) = ρ f ⋅ v + q ˙ ρ \frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho h_o\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho ∂ t ∂ ( ρ e o ) + ∇ ⋅ ( ρ h o v ) = ρ f ⋅ v + q ˙ ρ Total enthalpy, h o h_o h o e o e_o e o
h o = e o + p ρ h_o=e_o+\frac{p}{\rho} h o = e o + ρ p which gives
∂ ∂ t ( ρ e o ) + ∇ ⋅ ( ρ e o v ) + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ \frac{\partial}{\partial t}(\rho e_o) + \nabla\cdot(\rho e_o\mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho ∂ t ∂ ( ρ e o ) + ∇ ⋅ ( ρ e o v ) + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ Expanding the two first terms as
ρ ∂ e o ∂ t + e o ∂ ρ ∂ t + ρ v ⋅ ∇ e o + e o ∇ ⋅ ( ρ v ) + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ \rho\frac{\partial e_o}{\partial t} + e_o\frac{\partial \rho}{\partial t} + \rho\mathbf{v}\cdot\nabla e_o + e_o\nabla\cdot(\rho \mathbf{v}) + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho ρ ∂ t ∂ e o + e o ∂ t ∂ ρ + ρ v ⋅ ∇ e o + e o ∇ ⋅ ( ρ v ) + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ Collecting terms, we can identify the substantial derivative operator applied on total energy, D e o / D t De_o/Dt D e o / D t
ρ [ ∂ e o ∂ t + v ⋅ ∇ e o ] ⏟ = D e o D t + e o [ ∂ ρ ∂ t + ∇ ⋅ ( ρ v ) ] ⏟ = 0 + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ \rho\underbrace{\left[ \frac{\partial e_o}{\partial t} + \mathbf{v}\cdot\nabla e_o \right]}_{=\frac{De_o}{Dt}} + e_o\underbrace{\left[\frac{\partial \rho}{\partial t} + \nabla\cdot(\rho \mathbf{v}) \right]}_{=0} + \nabla\cdot(p\mathbf{v})= \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho ρ = D t D e o [ ∂ t ∂ e o + v ⋅ ∇ e o ] + e o = 0 [ ∂ t ∂ ρ + ∇ ⋅ ( ρ v ) ] + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ and thus we end up with the energy equation on non-conservation differential form
ρ D e o D t + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ \rho\frac{De_o}{Dt} + \nabla\cdot(p\mathbf{v}) = \rho\mathbf{f}\cdot\mathbf{v} + \dot{q}\rho ρ D t D e o + ∇ ⋅ ( p v ) = ρ f ⋅ v + q ˙ ρ