Effect of the reactor model on steady detonation modeling

Abstract

The effect of employing four different reactor models, Zeldovich–von Neumann–Döring (ZND), constant volume (CV), constant pressure (CP), and the Shock routine from Chemkin, to perform detonation-relevant chemical modeling was assessed. The simulation results were compared in terms of characteristic length scales and chemical analyses with four representative mixtures: $\hbox {H}_{2}{-}\hbox {O}_{2}{-}\hbox {N}_{2}$, $\hbox {H}_{2}{-}\hbox {NO}_{2}/{\hbox {N}_{2}}{\hbox {O}_{4}}$, ${\hbox {C}_{3}}{\hbox {H}_{8}}{-}\hbox {O}_{2}{-}\hbox {N}_{2}$, and dimethyl ether (DME)–$\hbox {O}_{2}{-}\hbox {CO}_{2}$. The following conclusions were drawn: (i) CV and CP reactor models shorten the induction zone length and strengthen the energy release rate in most mixtures. In terms of chemical kinetics, the impact of CP and CV reactor models is quite limited for $\hbox {H}_{2}{-}\hbox {O}_{2}{-}\hbox {N}_{2}$ and $\hbox {H}_{2}{-}\hbox {NO}_{2}/{\hbox {N}_{2}}{\hbox {O}_{4}}$ mixtures. However, the C2 branch is enhanced in CV and CP reactor models for ${\hbox {C}_{3}}{\hbox {H}_{8}}{-}\hbox {O}_{2}{-}\hbox {N}_{2}$ mixture. Moreover, both reactor models weaken the intermediate-temperature chemistry and promote the high-temperature chemistry for DME–$\hbox {O}_{2}{-}\hbox {CO}_{2}$ mixtures; (ii) the Shock module can be employed to perform detonation modeling, as it provided similar results to the ZND simulations for all investigated mixtures; and (iii) the ZND reactor model is preferred over the zero-dimensional reactor models, while the Shock module of ANSYS is equivalent to the ZND reactor.

Appendix: Governing equations

The present appendix describes how the CV, CP, and ZND reactor models can be established from first principles. This enables to clearly understand the differences between these three types of reactors, which are all adiabatic. From the first law of thermodynamics [43], $\delta Q = \delta E + \delta W$, where Q, E, and W are the heat, the internal energy, and the work, respectively. As $\delta Q =0$, we obtain

$$\begin{aligned} \text {d}e = -P \text {d}v, \end{aligned}$$

(2)

where e is the specific internal energy; P is the pressure; and v is the specific volume. Considering gaseous species which obey the perfect gas law, the total internal energy of a mixture is given by

$$\begin{aligned} e = \sum _{i=1}^{k} Y_{i} e_{i}, \end{aligned}$$

(3)

where k is the total number of chemical species and $Y_i$ and $e_{i}$ are the mass fraction and the internal energy of species i, respectively. From the internal energy definition (i.e., $\text {d}e = \text {c}_{v} \text {d}T$, where T is the temperature) and (3), the specific heat capacity at constant volume ($c_{v}$) of a gas mixture is obtained by

$$\begin{aligned} c_{v} = \frac{\text {d}e}{\text {d}T} = \sum _{i=1}^{k} Y_{i} \frac{\text {d}e_{i}}{\text {d}T}. \end{aligned}$$

(4)

Let us recall the perfect gas law

$$\begin{aligned} P v = \tilde{R} T, \end{aligned}$$

(5)

where $\tilde{R}$ is the universal gas constant divided by the molecular weight ($\tilde{R}=R{/}W$), as well as the mass fraction equation

$$\begin{aligned} \frac{\text {d}Y_{i}}{\text {d}t} = \frac{W_{i} \dot{\omega _{i}}}{\rho }, \end{aligned}$$

(6)

where t is the time; $\dot{\omega _{i}}$ is the chemical source term; and $\rho$ is the density. Substituting (3)–(6) into (2), one obtains

$$\begin{aligned} \frac{\text {d}T}{\text {d}t} = - \frac{1}{c_{v}} \sum _{i=1}^{k} e_{i} \frac{W_{i} \dot{\omega _{i}}}{\rho } - T \frac{\tilde{R}}{c_{v}} \frac{1}{v} \frac{\text {d}v}{\text {d}t}. \end{aligned}$$

(7)

If the first law of thermodynamics is expressed as a function of enthalpy (h)

$$\begin{aligned} \text {d}h = v \text {d}P, \end{aligned}$$

(8)

(7) can be expressed as

$$\begin{aligned} \frac{\text {d}T}{\text {d}t} = - \frac{1}{c_{P}} \sum _{i=1}^{k} h_{i} \frac{W_{i} \dot{\omega _{i}}}{\rho } + T \frac{\tilde{R}}{c_{P}} \frac{1}{P} \frac{\text {d}P}{\text {d}t}, \end{aligned}$$

(9)

where $c_{P}$ is the specific heat capacity at constant pressure.

Considering a constant volume process, (7) reduces to

$$\begin{aligned} \frac{\text {d}T}{\text {d}t} = - \frac{1}{c_{v}} \sum _{i=1}^{k} e_{i} \frac{W_{i} \dot{\omega _{i}}}{\rho }, \end{aligned}$$

(10)

where as considering a constant pressure process, (9) reduces to

$$\begin{aligned} \frac{\text {d}T}{\text {d}t} = - \frac{1}{c_{P}} \sum _{i=1}^{k} h_{i} \frac{W_{i} \dot{\omega _{i}}}{\rho }. \end{aligned}$$

(11)

The constant volume reactor model corresponds to (10) and (6), whereas the constant pressure reactor model corresponds to (11) and (6).

To establish the ZND model, let us recall the Euler equations

$$\begin{aligned}&\frac{\partial \rho }{\partial t} + \nabla \cdot \left( \rho u \right) = 0 , \end{aligned}$$

(12)

$$\begin{aligned}&\frac{\partial \rho u }{\partial t} + \nabla \cdot \left( \rho u u \right) + \nabla P = 0, \end{aligned}$$

(13)

$$\begin{aligned}&\frac{\partial \rho e_\mathrm{{t}}}{\partial t} + \nabla \cdot \left( \rho u \left( e_\mathrm{{t}} + \frac{P}{\rho } \right) \right) = 0, \end{aligned}$$

(14)

$$\begin{aligned}&\frac{\partial \rho Y_{i}}{\partial t} + \nabla \cdot \left( \rho u Y_{i} \right) = \dot{\omega _{i}}^{\prime } , \end{aligned}$$

(15)

where $\nabla$ is the nabla operator; u is the flow velocity; $\dot{\omega _{i}}^{\prime }=\dot{\omega _{i}}/W$; and $e_\mathrm{{t}}$ is the specific total energy and is given by

$$\begin{aligned} e_\mathrm{{t}}= e + \frac{u^{2}}{2}. \end{aligned}$$

(16)

Since the ZND model considers a steady planar shock wave followed by a reaction zone, (12), (13), and (15) reduce to

$$\begin{aligned}&u \frac{\text {d}\rho }{\text {d}x} + \rho \frac{\text {d}u}{\text {d}x} = 0 , \end{aligned}$$

(17)

$$\begin{aligned}&\frac{\text {d}P}{\text {d}x} + \rho u \frac{\text {d}u}{\text {d}x} = 0 , \end{aligned}$$

(18)

$$\begin{aligned}&\frac{\text {d}Y_{i}}{\text {d}x} = \frac{\dot{\omega _{i}}^{\prime }}{\rho u}. \end{aligned}$$

(19)

The energy equation (14) is replaced by the adiabatic change equation [14] given by

$$\begin{aligned} \frac{\text {d}P}{\text {d}x} = \frac{c^{2}}{u} \left( u \frac{\text {d}\rho }{\text {d}x} + \rho \dot{\sigma } \right) , \end{aligned}$$

(20)

where c is the speed of sound and $\dot{\sigma }$ is the thermicity. Substituting (17) and (18) into (20) and using the definition of the Mach number, $M=\frac{u}{c}$, we obtain

$$\begin{aligned}&\frac{\text {d}u}{\text {d}x} = \frac{\dot{\sigma }}{1 - M^{2}}, \end{aligned}$$

(21)

$$\begin{aligned}&\frac{\text {d}P}{\text {d}x} = - \rho u \frac{\dot{\sigma }}{1 - M^{2}}, \end{aligned}$$

(22)

$$\begin{aligned}&\frac{\text {d}\rho }{\text {d}x} = - \frac{\rho }{u} \frac{\dot{\sigma }}{1 - M^{2}}, \end{aligned}$$

(23)

$$\begin{aligned}&\frac{\text {d}Y_{i}}{\text {d}x} = \frac{\dot{\omega _{i}}^{\prime }}{\rho u}. \end{aligned}$$

(24)

Equations (21) to (23) correspond to the ZND model expressed as a function of distance. For numerically solving this system of equation, it is convenient to employ a Lagrangian description. Recall that for a steady process

$$\begin{aligned} \frac{\text {D}}{\text {D}t} = u \frac{\text {d}}{\text {d}x}, \end{aligned}$$

(25)

where $\text {D}$/$\text {D}t$ is the Lagrangian derivative. The ZND model becomes

$$\begin{aligned}&\frac{\text {D}u}{\text {D}t} = u \frac{\dot{\sigma }}{\eta }, \end{aligned}$$

(26)

$$\begin{aligned}&\frac{\text {D}P}{\text {D}t} = - \rho u^{2} \frac{\dot{\sigma }}{\eta }, \end{aligned}$$

(27)

$$\begin{aligned}&\frac{\text {D}\rho }{\text {D}t} = - \rho \frac{\dot{\sigma }}{\eta }, \end{aligned}$$

(28)

$$\begin{aligned}&\frac{\text {D}Y_{i}}{\text {D}t} = \frac{\dot{\omega _{i}}^{\prime }}{\rho }, \end{aligned}$$

(29)

where $\eta$ is the sonic parameter given by $\eta =1 - M^{2}$.

To enable further comparison with the CV and CP reactor models, it is useful to establish the temperature-gradient equation along the path of a Lagrangian particle for the ZND model. Taking the substantial derivative of (5) leads to

$$\begin{aligned} \frac{\text {D}T}{\text {D}t} = \frac{1}{\rho \tilde{R}} \frac{\text {D}P}{\text {D}t} - \frac{P}{\tilde{R} \rho ^{2}} \frac{\text {D}\rho }{\text {D}t} - \frac{P}{\tilde{R}^{2} \rho } \frac{\text {D}\tilde{R}}{\text {D}t}. \end{aligned}$$

(30)

Substituting (27) and (28) into (30) and using (29) as well as the definition of the mixture molecular weight, $W_{\text {mix}}=1/\sum _{i}^{k}(Y_i/W_i)$, we obtain

$$\begin{aligned} \frac{\eta }{T} \frac{\text {D}T}{\text {D}t} = \dot{\sigma } \left( 1 - \gamma M^{2} \right) - \frac{\eta }{\tilde{R} \rho } \sum _{i=1}^{k} \frac{\dot{\omega _{i}}^{\prime }}{W_{i}}. \end{aligned}$$

(31)