Derivations

Equation 7 in Bhagatwala, 2015 is the following

∂ρ∂t=−∂ρui∂xi+m˙∂ρui∂t=−∂ρuiuj∂xi−∂P∂xi+∂τij∂xi+m˙ui∂ρet∂t=−∂ρetuj∂xi−∂Puj∂xj+∂τijui∂xj−∂qj∂xj+m˙et∂ρYk∂t=−∂ρYkuj∂xi−∂Jk,j∂xj+ωk+m˙Yk\begin{aligned} \dfrac{\partial \rho}{\partial t} & = - \dfrac{\partial \rho u_i}{\partial x_i} + \dot{m} \\ \dfrac{\partial \rho u_i}{\partial t} & = - \dfrac{\partial \rho u_i u_j}{\partial x_i} - \dfrac{\partial P}{\partial x_i} + \dfrac{\partial \tau_{ij}}{\partial x_i} + \dot{m}u_i \\ \dfrac{\partial \rho e_t}{\partial t} & = - \dfrac{\partial \rho e_t u_j}{\partial x_i} - \dfrac{\partial P u_j}{\partial x_j} + \dfrac{\partial \tau_{ij}u_i}{\partial x_j} - \dfrac{\partial q_j}{\partial x_j} + \dot{m}e_t \\ \dfrac{\partial \rho Y_k}{\partial t} & = - \dfrac{\partial \rho Y_k u_j}{\partial x_i} - \dfrac{\partial J_{k,j}}{\partial x_j} + \omega_k + \dot{m}Y_k \\ \end{aligned}

and equation 8 is the following

P(x,t)=Pt(t)+p(x,t)=ρRTP(\mathbf{x}, t) = P_t(t) + p(\mathbf{x}, t) = \rho \mathrm{R} T

According to Bhagatwala, 2015, upon differentiating equation 8 and using equation 7, the rate of change of pressure is determined to be

∂P∂t=(γ−1)(∂ρet∂t−ui∂ρui∂t+ui22∂ρ∂t−∑khk∂ρYk∂t)+γRuT∑k1Wk∂ρYk∂t\dfrac{\partial P}{\partial t} = (\gamma - 1) \bigg( \dfrac{\partial \rho e_t}{\partial t} - u_i\dfrac{\partial \rho u_i}{\partial t} + \dfrac{u_i^2}{2}\dfrac{\partial \rho}{\partial t} - \sum_{k} h_k \dfrac{\partial \rho Y_k}{\partial t} \bigg) + \gamma \mathrm{R_u} T \sum_{k} \dfrac{1}{W_k} \dfrac{\partial \rho Y_k}{\partial t}

which is equation 9 in his paper. In this section I will derive the equation 9 from equation 7 and equaiton 8. Let’s start with the equation of state.

P=ρRTP = \rho \mathrm{R} T

Let’s write temperature in terms of internal energy

T=eCvP=ρReCvP=ρeRCv=ρeCp−CvCv=ρe(CpCv−1)=ρe(γ−1)\begin{aligned} T & = \dfrac{e}{C_\mathrm{v}} \\ P & = \rho \mathrm{R} \dfrac{e}{C_\mathrm{v}} \\ P & = \rho e \dfrac{\mathrm{R}}{C_\mathrm{v}} \\ & = \rho e \dfrac{C_\mathrm{p}-C_\mathrm{v}}{C_\mathrm{v}} \\ & = \rho e \bigg(\dfrac{C_\mathrm{p}}{C_\mathrm{v}} - 1\bigg) \\ & = \rho e \big(\gamma-1\big) \\ \end{aligned}

Differentiating both sides.

∂P∂t=(γ−1)(ρ∂e∂t+e∂ρ∂t)+ρe∂∂t(γ−1)\dfrac{\partial P}{\partial t} = \big(\gamma-1\big) \bigg( \rho\dfrac{\partial e}{\partial t} + e\dfrac{\partial \rho}{\partial t} \bigg) + \rho e\dfrac{\partial}{\partial t} \big(\gamma-1\big)

Now lets express internal energy in terms of total internal energy and kinetic energy

e=et−uiui2e=e_\mathrm{t} - \frac{u_i u_i}{2} ∂P∂t=(γ−1)(et∂ρ∂t−uiui2∂ρ∂t+ρ∂et∂t−ρ∂uiui/2∂t)+ρe∂γ∂t=(γ−1)(et∂ρ∂t+ρ∂et∂t−uiui2∂ρ∂t−ρ∂uiui/2∂t)+ρe∂γ∂t=(γ−1)(∂ρet∂t−uiui2∂ρ∂t−ρ∂uiui/2∂t)+ρe∂γ∂t\begin{aligned} \dfrac{\partial P}{\partial t} & = \big(\gamma-1\big) \bigg(e_\mathrm{t}\dfrac{\partial \rho}{\partial t} - \frac{u_i u_i}{2}\dfrac{\partial \rho}{\partial t} + \rho\dfrac{\partial e_\mathrm{t}}{\partial t} - \rho\dfrac{\partial u_i u_i/2}{\partial t} \bigg) + \rho e\dfrac{\partial \gamma}{\partial t} \\ & = \big(\gamma-1\big) \bigg(e_\mathrm{t}\dfrac{\partial \rho}{\partial t} + \rho\dfrac{\partial e_\mathrm{t}}{\partial t} - \frac{u_i u_i}{2}\dfrac{\partial \rho}{\partial t} - \rho\dfrac{\partial u_i u_i/2}{\partial t} \bigg) + \rho e\dfrac{\partial \gamma}{\partial t} \\ & = \big(\gamma-1\big) \bigg(\dfrac{\partial \rho e_\mathrm{t}}{\partial t} - \frac{u_i u_i}{2}\dfrac{\partial \rho}{\partial t} - \rho\dfrac{\partial u_i u_i/2}{\partial t} \bigg) + \rho e\dfrac{\partial \gamma}{\partial t} \\ \end{aligned}

References

Bhagatwala, A. (2015). On the representation of chemistry in low Mach number reactive flows. Combustion and Flame, 162(6), 2108-2123. https://doi.org/10.1016/j.combustflame.2015.06.005