energy equations in combustion

several types of energy equations in combustion

绝对焓的输运方程:

ρht+ρuihxi=dpdt+J+τijuixj\frac{\partial \rho h}{\partial t}+\frac{\rho u_i h}{\partial x_i}=\frac{dp}{dt}+\nabla\cdot J+\tau_{ij}\frac{\partial u_i}{\partial x_j}

其中 J=kTρhsVkYkρΔhf,k0VkYk=JsρΔhf,k0VkYkJ=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
Js=kTρhsVkYkJ_s=k\nabla T-\rho\sum h_sV_kY_kJJ
显焓 hs=hk=1NΔhf,k0Ykh_s=h-\sum_{k = 1}^N {\Delta h_{f,k}^0{Y_k}} 代入绝对焓的输运方程,得到:

ρhs+k=1NΔhf,k0Ykt+ρui(hs+k=1NΔhf,k0Yk)xi=dpdt+J+τijuixj\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{dp}{dt}+\nabla\cdot J+\tau_{ij}\frac{\partial u_i}{\partial x_j}

移项,得到:

ρhst+ρuihsxi=dpdt+J+τijuixjρk=1NΔhf,k0Yktρuik=1NΔhf,k0Ykxi=dpdt+Js+τijuixjρk=1NΔhf,k0Yktρuik=1NΔhf,k0Ykxi(ρk=1NΔhf,k0Vk,iYk)xi=dpdt+Js+τijuixjΔhf,k0[ρYkt+ρ(ui+Vk,i)Ykxi]=dpdt+Js+τijuixjΔhf,k0ω˙k=dpdt+Js+τijuixj+ω˙T\begin{array}{ll} &\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i}\\ &= \frac{dp}{dt}+\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{dp}{dt}+\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{dp}{dt}+\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{dp}{dt}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}-\sum\Delta h_{f,k}^0\dot\omega_k\\ &=\frac{dp}{dt}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T \end{array}

上边推导的最后一步,是由组分的输运方程得到的:

ρYkt+ρ(ui+Vk,i)Ykxi=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

ω˙T=Δhf,k0ω˙k\dot\omega_T=-\sum\Delta h_{f,k}^0\dot\omega_k is heat release due to combustion.

显焓的输运方程:

ρhst+ρuihsxi=dpdt+Js+τijuixj+ω˙T\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} =\frac{dp}{dt}+\nabla\cdot J_s+\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T

Js=kTρk=1NhsVkYkJ_s=k\nabla T-\rho\sum_{k = 1}^N h_sV_kY_k

ρhst+ρuihsxi=dpdt+(kTρk=1NhsVkYk)+τijuixj+ω˙T\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} =\frac{dp}{dt}+\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

ρhst+ρuihsxi=dpdt+xi(kTxi)ρhs,kVk,iYkxi+τijuixj+ω˙T\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{dp}{dt} + \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

ρhs,kVk,iYkxi\partial\frac{\rho\sum h_{s,k}V_{k,i}Y_k}{\partial x_i} 经常被视为0
所以显焓的输运方程:

ρhst+ρuihsxi=dpdt+xi(κTxi)+τijuixj+ω˙T\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{dp}{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

κ\kappa 就是 λ\lambda,导热系数。
由于温度和焓的关系: α=λCp\alpha=\frac{\lambda}{C_p}, 代入上式,得:

ρhst+ρuihsxi=dpdt+xi(αhsxi)+τijuixj+ω˙T\frac{\partial \rho h_s}{\partial t}+\partial\frac{\rho u_i h_s}{\partial x_i} = \frac{dp}{dt} + \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

τijuixj\tau_{ij}\frac{\partial u_i}{\partial x_j} 这一项也被忽略了,因为粘性热远小于化学热???

温度的输运方程:

显焓的输运方程:

ρdhsdt=ρhst+ρuihsxi=dpdt+xi(kTxi)ρhs,kVk,iYkxi+τijuixj+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{dp}{dt} + \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

hs=k=1NYkhs,kh_s=\sum_{k = 1}^N Y_k{h_{s,k}}

代入显焓输运方程得到:

ρddtk=1NYkhs,k=dpdt+xi(kTxi)ρk=1Nhs,kVk,iYkxi+τijuixj+ω˙T\rho\frac{d }{dt}\sum_{k = 1}^N Y_k{h_{s,k}} = \frac{dp}{dt} + \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

方程第一项:

ρddtk=1N(Ykhs,k)=ρk=1Nddt(hs,kYk)=ρk=1N(Ykdhs,kdt)+ρk=1Nhs,kdYkdt\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}

hs,k=T0TCp,kdTh_{s,k}=\int_{T0}^TC_{p,k}dT 代入上式,得到:

=ρk=1NYkddtT0TCp,kdT+ρk=1Nhs,kdYkdt=ρdT0Tk=1NYkCp,kdTdt+ρk=1Nhs,kdYkdt=\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}

又有 Cp=YkCp,kC_p=\sum Y_k{C_{p,k}}, 因此:

=ρdT0TCpdTdt+ρk=1Nhs,kdYkdt=ρCpdTdt+ρk=1Nhs,kdYkdt=\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}

接下来呢?

ρCpdTdt+ρk=1Nhs,kdYkdt=dpdt+xi(kTxi)ρk=1Nhs,kVk,iYkxi+τijuixj+ω˙T\rho C_p\frac{dT}{dt}+\rho\sum_{k = 1}^N h_{s,k}\frac{dY_k}{dt}=\frac{dp}{dt} + \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

把组分输运方程: ρdYkdt=xiρVk,iYk+ω˙k\rho\frac{dY_k}{dt}=-\frac{\partial}{\partial x_i}\rho V_{k,i}Y_k+\dot\omega_k 代入上式左边第二项,得到:

ρCpdTdt+k=1Nhs,kω˙k=dpdt+xi(kTxi)+τijuixj+ω˙T\rho C_p\frac{dT}{dt}+\sum_{k = 1}^N h_{s,k}\dot\omega_k=\frac{dp}{dt} + \frac{\partial}{\partial x_i}(k\frac{\partial T}{\partial x_i}) +\tau_{ij}\frac{\partial u_i}{\partial x_j}+\dot\omega_T

移项,得:

ρCpdTdt=dpdt+xi(κTxi)+τijuixj+ω˙T\rho C_p\frac{dT}{dt}=\frac{dp}{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

ω˙T=ω˙Tk=1Nhs,kω˙k=k=1NΔhf,kω˙kk=1Nhs,kω˙k=k=1Nhkω˙k\dot\omega'_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

Author: Yan Zhang
Link: https://openfoam.top/hEqn/
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