不同形式的能量方程,关于反应放热源项,绝对焓输运返方程无此源项,显焓输运方程的源项是生成焓乘以反应速率,温度输运方程的源项是绝对焓乘以反应速率。
绝对焓的输运方程:∂ ρ h ∂ t + ρ u i h ∂ x i = D p D t + ∇ ⋅ J + τ i j ∂ u i ∂ x j \frac{\partial \rho h}{\partial t}+\frac{\rho u_i h}{\partial x_i}=\frac{D p}{D t}+\nabla\cdot J+\tau_{ij}\frac{\partial u_i}{\partial x_j} ∂ t ∂ ρ h + ∂ x i ρ u i h = D t D p + ∇ ⋅ J + τ i j ∂ x j ∂ u i
其中J = k ∇ T − ρ ∑ h s V k Y k − ρ ∑ Δ h f , k 0 V k Y k = J s − ρ ∑ Δ h f , k 0 V k Y k J=k\nabla T-\rho\sum h_sV_kY_k-\rho\sum\Delta h_{f,k}^0 V_kY_k=J_s-\rho\sum\Delta h_{f,k}^0 V_kY_k J = k ∇ T − ρ ∑ h s V k Y k − ρ ∑ Δ h f , k 0 V k Y k = J s − ρ ∑ Δ h f , k 0 V k Y k 即J s = k ∇ T − ρ ∑ h s V k Y k J_s=k\nabla T-\rho\sum h_sV_kY_k J s = k ∇ T − ρ ∑ h s V k Y k 显焓h s = h − ∑ k = 1 N Δ h f , k 0 Y k h_s=h-\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}} h s = h − ∑ k = 1 N Δ h f , k 0 Y k 代入绝对焓的输运方程,得到:
∂ ρ h s + ∑ k = 1 N Δ h f , k 0 Y k ∂ t + ∂ ρ u i ( h s + ∑ k = 1 N Δ h f , k 0 Y k ) ∂ x i = D p D t + ∇ ⋅ J + τ i j ∂ u i ∂ x j \partial\rho\frac{ h_s+\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}}}{\partial t}+\partial\frac{\rho u_i (h_s+\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}})}{\partial x_i} =\frac{D p}{ D t}+\nabla\cdot J+\tau_{ij}\frac{\partial u_i}{\partial x_j} ∂ ρ ∂ t h s + ∑ k = 1 N Δ h f , k 0 Y k + ∂ ∂ x i ρ u i ( h s + ∑ k = 1 N Δ h f , k 0 Y k ) = D t D p + ∇ ⋅ J + τ i j ∂ x j ∂ u i
移项,得到:
∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∇ ⋅ J + τ i j ∂ u i ∂ x j − ∂ ρ ∑ k = 1 N Δ h f , k 0 Y k ∂ t − ∂ ρ u i ∑ k = 1 N Δ h f , k 0 Y k ∂ x i = D p D t + ∇ ⋅ J s + τ i j ∂ u i ∂ x j − ∂ ρ ∑ k = 1 N Δ h f , k 0 Y k ∂ t − ∂ ρ u i ∑ k = 1 N Δ h f , k 0 Y k ∂ x i − ∂ ( ρ ∑ k = 1 N Δ h f , k 0 V k , i Y k ) ∂ x i = D p D t + ∇ ⋅ J s + τ i j ∂ u i ∂ x j − ∑ Δ h f , k 0 [ ∂ ρ Y k ∂ t + ∂ ρ ( u i + V k , i ) Y k ∂ x i ] = D p D t + ∇ ⋅ J s + τ i j ∂ u i ∂ x j − ∑ Δ h f , k 0 ω ˙ k = D p D t + ∇ ⋅ J s + τ i j ∂ u i ∂ x j + ω ˙ T \begin{array}{ll} &\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i}\\ &= \frac{D p}{ D t}+\nabla\cdot J + \tau_{ij}\frac{\partial u_i}{\partial x_j} - \partial\rho\frac{\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}}}{\partial t} -\partial\frac{\rho u_i\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}}}{\partial x_i}\\ &= \frac{D p}{ D t}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}-\partial\rho\frac{\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}}}{\partial t}-\partial\frac{\rho u_i\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}}}{\partial x_i} -\partial\frac{(\rho\sum_{k = 1}^N{\Delta h_{f,k}^0V_{k,i}{Y_k}})}{\partial x_i}\\ &= \frac{D p}{ D t}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}-\sum\Delta h_{f,k}^0[\frac{\partial\rho Y_k}{\partial t}+\partial\frac{\rho(u_i+V_{k,i})Y_k}{\partial x_i}]\\ &= \frac{D p}{ D t}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}-\sum\Delta h_{f,k}^0\dot\omega_k\\ &=\frac{D p}{ D t}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T \end{array} ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∇ ⋅ J + τ i j ∂ x j ∂ u i − ∂ ρ ∂ t ∑ k = 1 N Δ h f , k 0 Y k − ∂ ∂ x i ρ u i ∑ k = 1 N Δ h f , k 0 Y k = D t D p + ∇ ⋅ J s + τ i j ∂ x j ∂ u i − ∂ ρ ∂ t ∑ k = 1 N Δ h f , k 0 Y k − ∂ ∂ x i ρ u i ∑ k = 1 N Δ h f , k 0 Y k − ∂ ∂ x i ( ρ ∑ k = 1 N Δ h f , k 0 V k , i Y k ) = D t D p + ∇ ⋅ J s + τ i j ∂ x j ∂ u i − ∑ Δ h f , k 0 [ ∂ t ∂ ρ Y k + ∂ ∂ x i ρ ( u i + V k , i ) Y k ] = D t D p + ∇ ⋅ J s + τ i j ∂ x j ∂ u i − ∑ Δ h f , k 0 ω ˙ k = D t D p + ∇ ⋅ J s + τ i j ∂ x j ∂ u i + ω ˙ T
其中:
D p D t = ∂ p ∂ t + u i ∂ p x i \dfrac{D p}{D t}=\dfrac{\partial p}{\partial t}+u_i\dfrac{\partial p}{x_i} D t D p = ∂ t ∂ p + u i x i ∂ p
上边推导的最后一步,是由组分的输运方程得到的:
∂ ρ Y k ∂ t + ∂ ρ ( u i + V k , i ) Y k ∂ x i = ω ˙ k \frac{\partial\rho Y_k}{\partial t}+\partial\frac{\rho(u_i+V_{k,i})Y_k}{\partial x_i}=\dot\omega_k ∂ t ∂ ρ Y k + ∂ ∂ x i ρ ( u i + V k , i ) Y k = ω ˙ k
ω ˙ T = − ∑ Δ h f , k 0 ω ˙ k \dot\omega_T=-\sum\Delta h_{f,k}^0\dot\omega_k ω ˙ T = − ∑ Δ h f , k 0 ω ˙ k 这一项是燃烧热。
Δ h f , k 0 \Delta h_{f,k}^0 Δ h f , k 0 是组分 k 的化学焓(chemical),也叫生成焓(formation)。
viscous heating source term:τ i j ∂ u i ∂ x j \tau_{ij}\frac{\partial u_i}{\partial x_j} τ i j ∂ x j ∂ u i 这一项是粘性热。
显焓的输运方程:∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∇ ⋅ J s + τ i j ∂ u i ∂ x j + ω ˙ T \frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} =\frac{D p}{ D t}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∇ ⋅ J s + τ i j ∂ x j ∂ u i + ω ˙ T
J s = k ∇ T − ρ ∑ k = 1 N h s V k Y k J_s=k\nabla T-\rho\sum_{k = 1}^N h_sV_kY_k J s = k ∇ T − ρ k = 1 ∑ N h s V k Y k
∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∇ ⋅ ( k ∇ T − ρ ∑ k = 1 N h s V k Y k ) + τ i j ∂ u i ∂ x j + ω ˙ T \frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} =\frac{D p}{ D t}+\nabla\cdot (k\nabla T-\rho\sum_{k = 1}^N h_sV_kY_k)+\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∇ ⋅ ( k ∇ T − ρ k = 1 ∑ N h s V k Y k ) + τ i j ∂ x j ∂ u i + ω ˙ T
∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∂ ∂ x i ( k ∂ T ∂ x i ) − ∂ ρ ∑ h s , k V k , i Y k ∂ x i + τ i j ∂ u i ∂ x j + ω ˙ T \frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{D p}{ D t} + \frac{\partial}{\partial x_i}(k\frac{\partial T}{\partial x_i})-\partial\frac{\rho\sum h_{s,k}V_{k,i}Y_k}{\partial x_i} +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∂ x i ∂ ( k ∂ x i ∂ T ) − ∂ ∂ x i ρ ∑ h s , k V k , i Y k + τ i j ∂ x j ∂ u i + ω ˙ T
∂ ρ ∑ h s , k V k , i Y k ∂ x i \partial\frac{\rho\sum h_{s,k}V_{k,i}Y_k}{\partial x_i} ∂ ∂ x i ρ ∑ h s , k V k , i Y k 经常被视为0 所以显焓的输运方程:
∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∂ ∂ x i ( κ ∂ T ∂ x i ) + τ i j ∂ u i ∂ x j + ω ˙ T \frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{D p}{ Dt} + \frac{\partial}{\partial x_i}(\kappa\frac{\partial T}{\partial x_i}) +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∂ x i ∂ ( κ ∂ x i ∂ T ) + τ i j ∂ x j ∂ u i + ω ˙ T
κ \kappa κ 就是λ \lambda λ ,导热系数。 由于温度和焓的关系:α = λ C p \alpha=\frac{\lambda}{C_p} α = C p λ , 代入上式,得:
∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∂ ∂ x i ( α ∂ h s ∂ x i ) + τ i j ∂ u i ∂ x j + ω ˙ T \frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{D p}{ D t} + \frac{\partial}{\partial x_i}(\alpha\frac{\partial h_s}{\partial x_i}) +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∂ x i ∂ ( α ∂ x i ∂ h s ) + τ i j ∂ x j ∂ u i + ω ˙ T
温度的输运方程:显焓的输运方程:
ρ d h s d t = ∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i = D p D t + ∂ ∂ x i ( k ∂ T ∂ x i ) − ∂ ρ ∑ h s , k V k , i Y k ∂ x i + τ i j ∂ u i ∂ x j + ω ˙ T \rho\frac{d h_s}{dt}=\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{D p}{ D t} + \frac{\partial}{\partial x_i}(k\frac{\partial T}{\partial x_i})-\partial\frac{\rho\sum h_{s,k}V_{k,i}Y_k}{\partial x_i} +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ρ d t d h s = ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s = D t D p + ∂ x i ∂ ( k ∂ x i ∂ T ) − ∂ ∂ x i ρ ∑ h s , k V k , i Y k + τ i j ∂ x j ∂ u i + ω ˙ T
h s = ∑ k = 1 N Y k h s , k h_s=\sum_{k = 1}^N Y_k{h_{s,k}} h s = k = 1 ∑ N Y k h s , k
代入显焓输运方程得到:
ρ d d t ∑ k = 1 N Y k h s , k = D p D t + ∂ ∂ x i ( k ∂ T ∂ x i ) − ∂ ρ ∑ k = 1 N h s , k V k , i Y k ∂ x i + τ i j ∂ u i ∂ x j + ω ˙ T \rho\frac{d }{dt}\sum_{k = 1}^N Y_k{h_{s,k}} = \frac{D p}{ D t} + \frac{\partial}{\partial x_i}(k\frac{\partial T}{\partial x_i})-\partial\frac{\rho\sum_{k = 1}^N h_{s,k}V_{k,i}Y_k}{\partial x_i} +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ρ d t d k = 1 ∑ N Y k h s , k = D t D p + ∂ x i ∂ ( k ∂ x i ∂ T ) − ∂ ∂ x i ρ ∑ k = 1 N h s , k V k , i Y k + τ i j ∂ x j ∂ u i + ω ˙ T
方程第一项:
ρ d d t ∑ k = 1 N ( Y k h s , k ) = ρ ∑ k = 1 N d d t ( h s , k Y k ) = ρ ∑ k = 1 N ( Y k d h s , k d t ) + ρ ∑ k = 1 N h s , k d Y k d t \rho\frac{d }{dt}\sum_{k = 1}^N (Y_k{h_{s,k}})=\rho\sum_{k = 1}^N \frac{d}{dt}(h_{s,k}Y_k)=\rho\sum_{k = 1}^N (Y_k\frac{dh_{s,k}}{dt})+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt} ρ d t d k = 1 ∑ N ( Y k h s , k ) = ρ k = 1 ∑ N d t d ( h s , k Y k ) = ρ k = 1 ∑ N ( Y k d t d h s , k ) + ρ k = 1 ∑ N h s , k d t d Y k
将h s , k = ∫ T 0 T C p , k d T h_{s,k}=\int_{T0}^TC_{p,k}dT h s , k = ∫ T 0 T C p , k d T 代入上式,得到:
= ρ ∑ k = 1 N Y k d d t ∫ T 0 T C p , k d T + ρ ∑ k = 1 N h s , k d Y k d t = ρ d ∫ T 0 T ∑ k = 1 N Y k C p , k d T d t + ρ ∑ k = 1 N h s , k d Y k d t =\rho\sum_{k = 1}^N Y_k\frac{d}{dt}\int_{T0}^TC_{p,k}dT+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt}=\rho\frac{d\int_{T0}^T\sum_{k = 1}^N Y_kC_{p,k}dT}{dt}+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt} = ρ k = 1 ∑ N Y k d t d ∫ T 0 T C p , k d T + ρ k = 1 ∑ N h s , k d t d Y k = ρ d t d ∫ T 0 T ∑ k = 1 N Y k C p , k d T + ρ k = 1 ∑ N h s , k d t d Y k
又有C p = ∑ Y k C p , k C_p=\sum Y_k{C_{p,k}} C p = ∑ Y k C p , k , 因此:
= ρ d ∫ T 0 T C p d T d t + ρ ∑ k = 1 N h s , k d Y k d t = ρ C p d T d t + ρ ∑ k = 1 N h s , k d Y k d t =\rho\frac{d\int_{T0}^TC_pdT}{dt}+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt}=\rho C_p\frac{dT}{dt}+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt} = ρ d t d ∫ T 0 T C p d T + ρ k = 1 ∑ N h s , k d t d Y k = ρ C p d t d T + ρ k = 1 ∑ N h s , k d t d Y k
接下来呢?
ρ C p d T d t + ρ ∑ k = 1 N h s , k d Y k d t = ∂ p ∂ t + ∂ ∂ x i ( k ∂ T ∂ x i ) − ∂ ρ ∑ k = 1 N h s , k V k , i Y k ∂ x i + τ i j ∂ u i ∂ x j + ω ˙ T \rho C_p\frac{dT}{dt}+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt}=\frac{\partial p}{ \partial t} + \frac{\partial}{\partial x_i}(k\frac{\partial T}{\partial x_i})-\partial\frac{\rho\sum_{k = 1}^N h_{s,k}V_{k,i}Y_k}{\partial x_i} +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ρ C p d t d T + ρ k = 1 ∑ N h s , k d t d Y k = ∂ t ∂ p + ∂ x i ∂ ( k ∂ x i ∂ T ) − ∂ ∂ x i ρ ∑ k = 1 N h s , k V k , i Y k + τ i j ∂ x j ∂ u i + ω ˙ T
把组分输运方程:ρ d Y k d t = − ∂ ∂ x i ρ V k , i Y k + ω ˙ k \rho\frac{dY_k}{dt}=-\frac{\partial}{\partial x_i}\rho V_{k,i}Y_k+\dot\omega_k ρ d t d Y k = − ∂ x i ∂ ρ V k , i Y k + ω ˙ k 代入上式左边第二项,得到:
ρ C p d T d t + ∑ k = 1 N h s , k ω ˙ k = ∂ p ∂ t + ∂ ∂ x i ( k ∂ T ∂ x i ) + τ i j ∂ u i ∂ x j + ω ˙ T \rho C_p\frac{dT}{dt}+\sum_{k = 1}^N h_{s,k}\dot\omega_k=\frac{\partial p}{ \partial t} + \frac{\partial}{\partial x_i}(k\frac{\partial T}{\partial x_i}) +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T ρ C p d t d T + k = 1 ∑ N h s , k ω ˙ k = ∂ t ∂ p + ∂ x i ∂ ( k ∂ x i ∂ T ) + τ i j ∂ x j ∂ u i + ω ˙ T
移项,得:
ρ C p d T d t = D p D t + ∂ ∂ x i ( κ ∂ T ∂ x i ) + τ i j ∂ u i ∂ x j + ω ˙ T ′ \rho C_p\frac{dT}{dt}=\frac{D p}{ D t}+\frac{\partial}{\partial x_i}(\kappa\frac{\partial T}{\partial x_i}) +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega'_T ρ C p d t d T = D t D p + ∂ x i ∂ ( κ ∂ x i ∂ T ) + τ i j ∂ x j ∂ u i + ω ˙ T ′
ω ˙ T ′ = ω ˙ T − ∑ k = 1 N h s , k ω ˙ k = − ∑ k = 1 N Δ h f , k ω ˙ k − ∑ k = 1 N h s , k ω ˙ k = − ∑ k = 1 N h k ω ˙ k \dot\omega^\prime_T =\dot\omega_T-\sum_{k = 1}^N h_{s,k}\dot\omega_k =-\sum_{k = 1}^N\Delta h_{f,k}\dot\omega_k-\sum_{k = 1}^N h_{s,k}\dot\omega_k =-\sum_{k = 1}^N h_k\dot \omega_k ω ˙ T ′ = ω ˙ T − k = 1 ∑ N h s , k ω ˙ k = − k = 1 ∑ N Δ h f , k ω ˙ k − k = 1 ∑ N h s , k ω ˙ k = − k = 1 ∑ N h k ω ˙ k
ω ˙ T \dot\omega_T ω ˙ T 和ω ˙ T ′ \dot\omega^\prime_T ω ˙ T ′ 的关系,来自 Poinsot 的书: These two terms are both called heat release so that different authors use the same term for different quantities. They differ by a small amount due to the contribution of sensible enthalpy termsh s , k h_{s,k} h s , k . Section 1.2.2 shows that they are equal when the heat capacitiesC p , k C_{p,k} C p , k are supposed equal for all species.
ω ˙ T ′ = − ∑ k = 1 N h k ω ˙ k = − ∑ k = 1 N h s , k ω ˙ k − ∑ k = 1 N Δ h f , k 0 ω ˙ k = − h s ∑ k = 1 N ω ˙ k − ∑ k = 1 N Δ h f , k 0 ω ˙ k = ω ˙ T \dot\omega^\prime_T=-\sum_{k = 1}^N h_k\dot\omega_k=-\sum_{k = 1}^N h_{s,k}\dot\omega_k-\sum_{k = 1}^N\Delta h_{f,k}^0\dot\omega_k=-h_s\sum_{k = 1}^N\dot\omega_k-\sum_{k = 1}^N \Delta h_{f,k}^0\dot\omega_k=\dot\omega_T\\ ω ˙ T ′ = − k = 1 ∑ N h k ω ˙ k = − k = 1 ∑ N h s , k ω ˙ k − k = 1 ∑ N Δ h f , k 0 ω ˙ k = − h s k = 1 ∑ N ω ˙ k − k = 1 ∑ N Δ h f , k 0 ω ˙ k = ω ˙ T
becauseh s , k = ∫ T 0 T C p , k d T h_{s,k}=\int_{T_0}^T C_{p,k}dT h s , k = ∫ T 0 T C p , k d T .
ifC p , k C_{p,k} C p , k are supposed equal for all species, then so doh s , k h_{s,k} h s , k , then∑ k = 1 N h s , k ω ˙ k = h s ∑ k = 1 N ω ˙ k \sum_{k = 1}^N h_{s,k}\dot\omega_k=h_s\sum_{k = 1}^N \dot\omega_k ∑ k = 1 N h s , k ω ˙ k = h s ∑ k = 1 N ω ˙ k and∑ k = 1 N ω ˙ k = 0 \sum_{k = 1}^N\dot\omega_k =0 ∑ k = 1 N ω ˙ k = 0 , thereforeh s ∑ k = 1 N ω ˙ k = 0 h_s\sum_{k = 1}^N \dot\omega_k\ =0 h s ∑ k = 1 N ω ˙ k = 0
reactingFoam 中的能量方程fvScalarMatrix EEqn ( fvm::ddt(rho, he) + mvConvection->fvmDiv(phi, he) + fvc::ddt(rho, K) + fvc::div(phi, K) + ( he.name() == "e" ? fvc::div ( fvc::absolute(phi/fvc::interpolate(rho), U), p, "div(phiv,p)" ) : -dpdt ) - fvm::laplacian(turbulence->alphaEff(), he) == Qdot + fvOptions(rho, he) );
其中 Qdot = reaction->Qdot();
,然后在燃烧模型中:Qdot.ref() = this->chemistryPtr_->Qdot();
。
template<class ReactionThermo, class ThermoType> Foam::tmp<Foam::volScalarField> Foam::StandardChemistryModel<ReactionThermo, ThermoType>::Qdot() const { tmp<volScalarField> tQdot ( new volScalarField ( IOobject ( ...... ), this->mesh_, dimensionedScalar("zero", dimEnergy/dimVolume/dimTime, 0) ) ); if (this->chemistry_) { scalarField& Qdot = tQdot.ref(); forAll(Y_, i) { forAll(Qdot, celli) { const scalar hi = specieThermo_[i].Hc(); Qdot[celli] -= hi*RR_[i][celli]; } } } return tQdot; }
动能输运方程:
∂ ρ K ∂ t + ∂ ρ u i K ∂ x i = u j ∂ σ i j ∂ x i \frac{\partial \rho K}{\partial t}+\partial\frac{\rho u_i K}{\partial x_i}=u_j\frac{\partial \sigma_{ij}}{\partial x_i} ∂ t ∂ ρ K + ∂ ∂ x i ρ u i K = u j ∂ x i ∂ σ i j
动能K = 1 2 u k u k K=\dfrac{1}{2}u_ku_k K = 2 1 u k u k 。
显焓+动能的输运方程推导如下,动能方程右边这一项可以表示为
u j ∂ σ i j ∂ x i = u j ∂ τ i j ∂ x i − u i ∂ p ∂ x i u_j\dfrac{\partial \sigma_{ij}}{\partial x_i}=u_j\dfrac{\partial \tau_{ij}}{\partial x_i}-u_i\dfrac{\partial p}{\partial x_i} u j ∂ x i ∂ σ i j = u j ∂ x i ∂ τ i j − u i ∂ x i ∂ p
显焓方程中的D p D t = ∂ p ∂ t + u i ∂ p x i \dfrac{D p}{D t}=\dfrac{\partial p}{\partial t}+u_i\dfrac{\partial p}{x_i} D t D p = ∂ t ∂ p + u i x i ∂ p ,这里的u i ∂ p x i u_i\dfrac{\partial p}{x_i} u i x i ∂ p 一正一负抵消了。
因此显焓+动能的输运方程:
∂ ρ h s ∂ t + ∂ ρ u i h s ∂ x i + ∂ ρ K ∂ t + ∂ ρ u i K ∂ x i = ∂ p ∂ t + ∂ ∂ x i ( α ∂ h s ∂ x i ) + ∂ τ i j u i ∂ x j + ω ˙ T \frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} +\frac{\partial \rho K}{\partial t}+\partial\frac{\rho u_i K}{\partial x_i} = \frac{\partial p}{ \partial t} + \frac{\partial}{\partial x_i}(\alpha\frac{\partial h_s}{\partial x_i}) +\frac{\partial \tau_{ij} u_i}{\partial x_j}+\dot\omega_T ∂ t ∂ ρ h s + ∂ ∂ x i ρ u i h s + ∂ t ∂ ρ K + ∂ ∂ x i ρ u i K = ∂ t ∂ p + ∂ x i ∂ ( α ∂ x i ∂ h s ) + ∂ x j ∂ τ i j u i + ω ˙ T
OF 代码中被忽略的部分是∂ τ i j u i ∂ x j \frac{\partial \tau_{ij} u_i}{\partial x_j} ∂ x j ∂ τ i j u i 。