Reynolds Transport Theorem
The Reynolds Transport Theorem (RTT) is given by
$$\frac{B_{sys}}{dt} = \frac{d}{dt}\int_{CV}\rho\beta dV + \int_{CS}\rho\beta\vec{v}_r\cdot\vec{n}dA.$$
The RTT relates a Lagrangian system on the left to an Eulerian control volume (CV) with control suface (CS) on the right. The system consists of a marked mass. The term marked mass means we define our system in terms of the fixed masses that constitute it. Those masses don’t have to be contiguous.
In the RTT, $B_{sys}$ is an extensive property of the system, typically mass, momentum, energy, or the mass of a given molecule or element. $\beta = B_{sys}/m$ is the corresponding intensive quantity. $\vec{v}_r$ is the relative velocity between the system and the CS, that is $\vec{v}_r = \vec{v}_{sys} - \vec{v}_{CS}$. We allow the control volume to move and deform, though we more often assume it is fixed.
The vector $\vec{n}$ is a unit normal vector on the surface pointing outward.
The RTT is a more general and flexible form of the familiar conservation expression “accumulation equals in minus out plus generation.” The left side of the RTT is the generation term. The first term on the right is the accumulation term, and the second term on the right is the transport term, out minus in.
The purpose of the RTT is to connect the Lagrangian system to the Eulerian control volume. We normally want conservation equations for Eulerian systems, but conservation laws are written for Lagrangian systems. Then the RTT connects the Lagrangian conservation law to the Eulerian control volume. For example, the Lagrangian conservation law for a system of a given marked mass is that the mass is conserved and neither created or destroyed. Then $dB_{sys}/dt = dm_{sys}/dt = 0.$ Mass is not generally conserved in a control volume, say a room. When things are moved in or out of the room the mass changes. The RTT connects the two.
Mass balance
Apply the RTT to mass conservation. Assume fixed $CV$ and $CS$.
Step 1
Define $B_{sys}=m$, $\beta = B_{sys}/m = 1$. The velocity is $\vec{v}_r = \vec{v}$, that is, the fluid velocity, since that is the velocity of the system and the control volume is stationary.
Step 2
Write the conservation law. In this case $dB_{sys}/dt = 0$.
Step 3
Insert in the RTT:
$$0 = \frac{d}{dt}\int_{CV}\rho dV + \int_{CS}\rho\vec{v}dA.$$
Step 4
Apply the Gauss Divergence Theorem (GDT). The GDT is
$$\int_{CS}\vec{f}\cdot\vec{n}dA = \int_{CV}\nabla\cdot\vec{f}dV.$$
This allows conversion between surface and volume integrals. Apply to the RTT, and rearrange to give
$$\frac{d}{dt}\int_{CV}\rho dV + \int_{CV}\nabla\cdot(\rho\vec{v})dV = 0.$$
Step 5
Finally, since the CV is fixed, we can move the $d/dt$ in the first term inside the integral, but since $\rho$ is a function of position, we use a partial derivative. Then combine the two integral terms:
$$\int_{CV}\left[\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec{v})\right]dV = 0.$$
This holds for arbitrary control volumes only if the term in brackets itself is zero, hence
$$\frac{\partial\rho}{\partial t} + \nabla\cdot(\rho\vec{v}) = 0.$$Species
Reynolds Transport Theorem (RTT):
$$\frac{dB_{sys}}{dt} = \int_V\frac{\partial\rho\beta}{\partial t}dV + \int_A\rho\beta\vec{v}_R\cdot\vec{n}dA.$$
Let $B=m_i$, and $\beta=B/m = y_i$. The conservation law is $(dm_i/dt)_{sys} = \int_V\dot{m}_i^{\prime\prime\prime}dV,$ and $\vec{v}_R=\vec{v}_i=\vec{v}+\vec{v}_i^{D}$. Inserting these give
$$\int_V\frac{\partial\rho y_i}{\partial t}dV + \int_A\rho y_i\vec{v}\cdot\vec{n}dA + \int_A\rho y_i\vec{v}_i^D\cdot\vec{n}dA = \int_V\dot{m}_i^{\prime\prime\prime}dV.$$
Use the Gauss Divergence Theorem to convert integrals over area to integrals over volume.
$$\int_V\frac{\partial\rho y_i}{\partial t}dV + \int_V\nabla\cdot(\rho y_i\vec{v})dV + \int_V\nabla\cdot(\rho y_i\vec{v}_i^D)dV = \int_V\dot{m}_i^{\prime\prime\prime}dV.$$
Now, use $\vec{j}_i=\rho y_i\vec{v}_i^D$ and collect terms under a single volume integral on the left hand side so that $\int_V(\cdot)dV = 0$. The resulting equation holds for any integration volume, hence $(\cdot)=0$, giving
$$\frac{\partial\rho y_i}{\partial t} + \nabla\cdot(\rho\vec{v}y_i) + \nabla\cdot\vec{j}_i=\dot{m}_i^{\prime\prime\prime}.$$Momentum
For the momentum equation, we have $B_{sys}$ is momentum $\mathcal{M}$ and $\beta=\vec{v}$ (momentum per mass). The conservation law follows from Newton’s second law, the rate of change of momentum is the sum of the external forces, which are pressure, viscous, and gravitational,
$$\frac{d\mathcal{M}}{dt} = -\int_A\tau\cdot\vec{n}dA -\int_AP\delta\cdot\vec{n}dA - \int_V\rho\vec{g}dV.$$
Here, $\tau$ is the stress tensor, $\delta$ is the unit dyad tensor, $\vec{g}$ is the gravitational acceleration vector, and $P$ is pressure. Note, $-(\tau+P\delta)\cdot\vec{n}$ is the force vector per unit surface area acting on the surface due to pressure and viscosity. Inserting these into the RTT, using the Gauss divergence theorem, and collecting integrals gives
$$\frac{\partial\rho\vec{v}}{\partial t} + \nabla\cdot{\rho\vec{v}\vec{v}} = -\nabla\cdot\tau - \nabla P + \rho\vec{g}.$$Here, we have used $\nabla\cdot(P\delta) = \nabla P$.
Energy
The total energy is denoted $E$ and consists of internal and kinetic energies. We use $B_{sys}=E$ and $\beta=e$, where
$$e = u + \frac{1}{2}\vec{v}\cdot\vec{v},$$ and $u=h-\frac{P}{\rho}$ is enthalpy.
The conservation law is the first law of thermodynamics, where the rate of change of energy is the sum of the heat and work on the system. The heat is due to heat flux $\vec{q}$ over the surface, and work is pressure and viscous on the surface and gravitational over the volume. The rate of work is the force vector dotted with velocity, and we use $\vec{n}\cdot\vec{v}=\vec{v}\cdot\vec{n}$.
$$\frac{dE_{sys}}{dt} = -\int_A\vec{q}\cdot\vec{n}dA -\int_AP\vec{v}\cdot\vec{n}dA -\int_A\tau\cdot\vec{v}\cdot\vec{n}dA+ \int_V\rho\vec{g}\cdot\vec{v}dV.$$ Insert this into the RTT and apply the Gauss divergence theorem, collecting terms; the integral argument becomes
$$\frac{\partial\rho e}{\partial t} + \nabla\cdot(\rho e\vec{v}) = -\nabla\cdot\vec{q}-\nabla\cdot(\tau\cdot\vec{v}) - \nabla\cdot(P\vec{v}) + \rho\vec{g}\cdot\vec{v}.$$The heat flux has contributions from thermal conduction, mass diffusion, and radiation
$$\vec{q} = -k\nabla T + \sum_ih_i\vec{j}_i + \vec{q}_{rad}.$$
Shvab-Zeldovich Energy Equation
Start with the energy transport equation, insert $e=h-\frac{P}{\rho} + \frac{1}{2}\vec{v}\cdot\vec{v}$, and assume
- steady state,
- neglect kinetic energy,
- neglect potential energy,
- neglect radiation,
- neglect viscous heating. The result is
where the $P\vec{v}$ work term has canceled.
Heat flux
The heat flux is given by
$$\vec{q} = -k\nabla T - \sum_ih_i\vec{j}_i.$$
Assume $j_i=-\rho D_i\nabla y_i$, and that all $D_i=\alpha$, where $\alpha=k/\rho c_p$ is the thermal diffusivity, that is, assume $Le_i = \alpha/D_i = 1$. Then we have
$$q = -k\nabla T - \sum_i\rho Dh_i\nabla y_i.$$
Use $h_i\nabla y_i = \nabla(y_ih_i) + y_i\nabla h_i$:
$$q = -k\nabla T - \sum_i\rho D\nabla(y_ih_i) + \sum_i\rho Dy_i\nabla h_i.$$
The summation commutes with $\nabla$, so the first term becomes $-\rho D\nabla h$, where $h=\sum_iy_ih_i$. In the second sum, $\nabla h_i=c_{p,i}\nabla T$ and we use $c_p = \sum_iy_ic_{p,i}$. Using this, along with $\rho Dc_p=k$ for $D=\alpha$ gives
$$\vec{q} = -\rho D\nabla h.$$The enthalpy equation with this expression for $\vec{q}$ is
$$\nabla\cdot(\rho\vec{v}h) = \nabla\cdot(\rho D\nabla h).$$Enthalpy is simply a conserved scalar subject to transport by advection and diffusion under the given assumptions.
Sensible enthalpy
We can write the enthalpy equation in terms of sensible enthalpy using $$h = h_f^o + h_s,$$ where $h_f^o$ is the heat of formation at a reference temperature of 298 K, and $h_s=\int_{T_r}^Tc_pdT$ is the sensible enthalpy. This is inserted into the enthalpy equation, which has two terms, to give a result with four terms, rearranged to have the sensible enthalpy on the left, and the reference formation enthalpy on the right:
$$\nabla\cdot(\rho\vec{v}h_s) - \nabla\cdot(\rho D\nabla h_s) = -\nabla\cdot(\rho\vec{v}h_f^o) +\nabla\cdot(\rho D\nabla h_f^o).$$
Now, use $h_f^o=\sum_iy_ih_{f,i}^o$, where $h_{f,i}^o$ is the heat of formation of species $i$ at $T_r$, which is a constant. Inserting this gives
$$\nabla\cdot(\rho\vec{v}h_s) - \nabla\cdot(\rho D\nabla h_s) = -\sum_ih_{f,i}^o\left(\nabla\cdot(\rho\vec{v}y_i) - \nabla\cdot(\rho D\nabla y_i)\right).$$
The term in the outer parentheses on the right-hand-side is $\dot{m}_i^{\prime\prime\prime}$ from the steady state species equation, with $\vec{j}_i=-\rho D\nabla y_i$. The final equation is
$$\nabla\cdot(\rho\vec{v}h_s) - \nabla\cdot(\rho D\nabla h_s) = -\sum_ih_{f,i}^o\dot{m}_i^{\prime\prime\prime}.$$Sensible enthalpy transported by advection and diffusion, but also created by the heat release rate (the quantity on the right-hand-side of the equation), which acts as a source term representing conversion of internal energy to sensible energy.
Temperature equation
Use the continuity equation $\nabla\cdot(\rho\vec{v}) = 0$, to convert the first term of the sensible energy equation as
$$\rho\vec{v}\cdot\nabla h_s - \nabla\cdot(\rho D\nabla h_s) = -\sum_ih_{f,i}^o\dot{m}_i^{\prime\prime\prime}.$$
Now, use $dh_s = c_pdT$, hence $\nabla h_s=c_p\nabla T$, and $k=\rho Dc_p$ when $D=\alpha$ to give
$$\rho c_p\vec{v}\cdot\nabla T - \nabla\cdot(k\nabla T) = -\sum_ih_{f,i}^o\dot{m}_i^{\prime\prime\prime}.$$Mixture fraction transport equation
A transport equation for mixture fraction can be computed starting with the species transport equation using $\vec{j}_i = -\rho D\nabla y_i$, and assuming $D_i=D$. Apply unsteady continuity $\frac{\partial\rho}{\partial t}\nabla\cdot(\rho\vec{v})$ to the species equation to give
$$\rho\frac{\partial y_i}{\partial t} + \rho\vec{v}\cdot\nabla y_i + \nabla\cdot\vec{j}_i=\dot{m}_i^{\prime\prime\prime}.$$
Now, use $\vec{j}=-\rho D\nabla y_i$, and the following relations defining mixture fraction $\xi$,
$$\xi = \frac{\beta-\beta_0}{\beta_1-\beta_0},$$ $$\beta = \sum_k\gamma_ky_k,$$ $$y_k = \sum_i\alpha_iy_i,$$
where $\beta$ is a linear combination of elemental mass fractions $y_k$ for element index $k$ (conserved scalars), which are linearly related to species mass fractions $y_i$ for species $i$. Here, $\gamma_k$ and $\alpha_i$ are some constant coefficients ($\alpha_i=a_{k,i}M_k/M_i$, where $a_{k,i}$ is the moles of element $k$ in species $i$, and $M$ are molecular weights).
Multiply the species equation by $\alpha_i$, and sum over all species. The sum commutes with derivatives. This gives an element transport equation, with $\sum_i\alpha_i\dot{m}_i^{\prime\prime\prime}=0$ since elements are conserved;
$$\rho\frac{\partial y_k}{\partial t} + \rho\vec{v}\cdot\nabla y_k - \nabla\cdot(\rho D\nabla y_k)=0.$$
Now, multiply by $\gamma_k$ and sum over elements to give
$$\rho\frac{\partial \beta}{\partial t} + \rho\vec{v}\cdot\nabla \beta - \nabla\cdot(\rho D\nabla \beta)=0.$$
Now, $d\beta = d(\beta-\beta_0)$ since $\beta_0$ is constant, so insert this and then divide through by $\beta_1-\beta_0$, which is also constant. This gives the same equation as $\beta$ but with $\xi$ replacing $. Use the continuity equation to get the conservation form of the equation
$$\frac{\partial \rho\xi}{\partial t} + \nabla\cdot(\rho\vec{v}\xi) - \nabla\cdot(\rho D\nabla \xi)=0.$$