AEROFROSH: a shock condition calculator for multi-component fuel aerosol-laden flows

Abstract

This article introduces an algorithm that determines the thermodynamic conditions behind incident and reflected shocks in aerosol-laden flows. Importantly, the algorithm accounts for the effects of droplet evaporation on post-shock properties. Additionally, this article describes an algorithm for resolving the effects of multiple-component-fuel droplets. This article presents the solution methodology and compares the results to those of another similar shock calculator. It also provides examples to show the impact of droplets on post-shock properties and the impact that multi-component fuel droplets have on shock experimental parameters. Finally, this paper presents a detailed uncertainty analysis of this algorithm’s calculations given typical experimental uncertainties.

Appendix 1: Derivation of select equations

1.1 1.1 Derivation of equation (20)

The total number of moles in the gas phase following evaporation $N_{2,\mathrm{g}}$ is equal to the number of moles of carrier gas (the carrier gas consists of bath gas plus fuel vapor) in Region 1 $N_\mathrm{1,g}$ plus the number of moles in the droplets in Region 1 $N_{1,\mathrm{d}}$:

$$\begin{aligned} N_\mathrm{2,g} = N_\mathrm{1,g} + N_\mathrm{1,d} \end{aligned}$$

(35)

For Region 1, the number of moles of bath gas (non-fuel vapor components) in Region 1 $N_\mathrm{1,bath}$ can be obtained using the total number of moles of carrier gas in Region 1 and the mole fraction of bath gas in Region 1 $x_\mathrm{1,bath}$:

$$\begin{aligned} N_\mathrm{1,bath} = N_\mathrm{1,g} x_\mathrm{1,bath} \end{aligned}$$

(36)

The same can be accomplished for Region 2:

$$\begin{aligned} N_\mathrm{2,bath} = N_\mathrm{2,g} x_\mathrm{2,bath} \end{aligned}$$

(37)

Also, mass is conserved in the bath gas components between Regions 1 and 2:

$$\begin{aligned} N_\mathrm{1,bath} = N_\mathrm{2,bath} \end{aligned}$$

(38)

Equations (35)–(38) can be combined and rearranged, yielding

$$\begin{aligned} \frac{N_\mathrm{1,g} x_\mathrm{1,bath}}{x_\mathrm{2,bath}} = N_{1,\mathrm{g}} + N_{1,\mathrm{d}} \end{aligned}$$

(39)

Finally, the ratio $\frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}}$ can be isolated to achieve the desired result:

$$\begin{aligned} \frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}} = \frac{x_\mathrm{1,bath}-x_\mathrm{2,bath}}{x_\mathrm{2,bath}} \end{aligned}$$

(20)

1.2 1.2 Derivation of equation (21)

The mole fraction of bath gas component i in Region 2 following complete evaporation is given by the molar ratio

$$\begin{aligned} x_{\mathrm{2,bath},i} = \frac{N_{\mathrm{2,bath},i}}{N_{2,\mathrm{g}}} \end{aligned}$$

(40)

Mass conservation of bath gas components across Regions 1 and 2 gives

$$\begin{aligned} N_\mathrm{1,bath} = N_\mathrm{2,bath} \end{aligned}$$

(41)

and hence

$$\begin{aligned} x_{\mathrm{2,bath},i} = \frac{N_{\mathrm{1,bath},i}}{N_\mathrm{2,g}} \end{aligned}$$

(42)

The following equation relates the number of moles of the i ^th bath gas component $N_{1,\mathrm{bath},{i}}$ to that component’s mole fraction $x_{1,\mathrm{bath},{i}}$:

$$\begin{aligned} N_{\mathrm{1,bath},i} = N_\mathrm{1,g} x_{\mathrm{1,bath},i} \end{aligned}$$

(43)

This can be used in (42), together with (35), to write

$$\begin{aligned} x_{\mathrm{2,bath},i} = \frac{N_\mathrm{1,g} x_{\mathrm{1,bath},i}}{N_\mathrm{1,g} + N_{1,\mathrm{d}}} \end{aligned}$$

(44)

Finally, rearranging yields

$$\begin{aligned} x_{\mathrm{2,bath},i} = \frac{x_{\mathrm{1,bath},i}}{1 + \frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}}} \end{aligned}$$

(21)

1.3 1.3 Derivation of equation (22)

The gas-phase mole fraction of fuel in Region 2 following complete evaporation is equal to the molar ratio

$$\begin{aligned} x_\mathrm{2,fuel} = \frac{N_\mathrm{2,fuel}}{N_\mathrm{2,g}} \end{aligned}$$

(45)

Mass conservation for the fuel molecules dictates that the total number of moles of fully evaporated fuel in Region 2 $N_\mathrm{2,fuel}$ is the sum of the moles of fuel vapor $N_\mathrm{1,fuel,vap}$ in Region 1 and the moles of droplets $N_{1,\mathrm{d}}$ in Region 1:

$$\begin{aligned} N_\mathrm{2,fuel} = N_\mathrm{1,fuel,vap} + N_{1,\mathrm{d}} \end{aligned}$$

(46)

Combining (35) and (46) in Equation (45) produces

$$\begin{aligned} x_\mathrm{2,fuel} = \frac{N_\mathrm{1,fuel,vap} + N_{1,\mathrm{d}}}{N_\mathrm{1,g} + N_{1,\mathrm{d}}} \end{aligned}$$

(47)

The quantity $N_\mathrm{1,fuel,vap}$ can be rewritten as

$$\begin{aligned} N_\mathrm{1,fuel,vap} = N_\mathrm{1,g} x_\mathrm{1,fuel} \end{aligned}$$

(48)

yielding

$$\begin{aligned} x_\mathrm{2,fuel} = \frac{N_\mathrm{1,g} x_\mathrm{1,fuel} + N_{1,\mathrm{d}}}{N_{1,\mathrm{g}} + N_{1,\mathrm{d}}} \end{aligned}$$

(49)

Finally, rearranging gives

$$\begin{aligned} x_\mathrm{2,fuel} = \frac{x_\mathrm{1,fuel} + \frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}}}{1 + \frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}}} \end{aligned}$$

(22)

1.4 1.4 Derivation of equation (31)

Since absorbance values are additive, the total Region 2 absorbance value $\alpha _2$ for a mixture of absorbing fuels [see also (11)] is given by

$$\begin{aligned} \alpha _2 = \displaystyle \sum _{j=1}^{C_{\text {fuel}}} n_{\mathrm{2,fuel},j} \sigma _{\mathrm{2,fuel},j} L \end{aligned}$$

(50)

Also, the total absorbance, computed using the total fuel concentration $n_\mathrm{2,fuel}$ and an average cross section $\sigma _\mathrm{2,fuel,avg}$ can be defined by

$$\begin{aligned} \alpha _2 = n_\mathrm{2,fuel} \sigma _\mathrm{2,fuel,avg} L \end{aligned}$$

(51)

Equating (50) and (51) and canceling the path length L gives

$$\begin{aligned} \displaystyle \sum _{j=1}^{C_{\text {fuel}}} n_{\mathrm{2,fuel},j} \sigma _{\mathrm{2,fuel},j} = n_\mathrm{2,fuel} \sigma _\mathrm{2,fuel,avg} \end{aligned}$$

(52)

The concentration of fuel component j can be written

$$\begin{aligned} n_{\mathrm{2,fuel},j} = x_{\mathrm{2,fuel},j} n_\mathrm{2,total} \end{aligned}$$

(53)

and likewise $n_\mathrm{2,fuel}$ can be written

$$\begin{aligned} n_\mathrm{2,fuel} = x_\mathrm{2,fuel} n_\mathrm{2,total} \end{aligned}$$

(54)

where $n_{2,\mathrm{total}}$ is the total gas-phase molar concentration following complete evaporation. Substituting these relations into (52) and canceling $n_{2,\mathrm{total}}$ gives

$$\begin{aligned} \displaystyle \sum _{j=1}^{C_{\text {fuel}}} x_{\mathrm{2,fuel},j} \sigma _{\mathrm{2,fuel},j} = x_\mathrm{2,fuel} \sigma _\mathrm{2,fuel,avg} \end{aligned}$$

(55)

Finally, solving for $\sigma _\mathrm{2,fuel,avg}$ yields

$$\begin{aligned} \sigma _\mathrm{2,fuel,avg} = \displaystyle \sum _{j=1}^{C_{\text {fuel}}} \frac{x_{\mathrm{2,fuel},j}}{x_\mathrm{2,fuel}} \sigma _{\mathrm{2,fuel},j} \end{aligned}$$

(31)

1.5 1.5 Derivation of equation (32)

The gas-phase mole fraction of fuel component j in Region 2 following complete evaporation is equal to the molar ratio

$$\begin{aligned} x_{\mathrm{2,fuel},j} = \frac{N_{\mathrm{2,fuel},j}}{N_\mathrm{2,g}} \end{aligned}$$

(56)

Mass conservation for the fuel component j dictates that the total number of moles of fully evaporated component j in Region 2 $N_{\mathrm{2,fuel},j}$ is the sum of the moles of fuel component j vapor $N_{\mathrm{1,fuel,vap},j}$ in Region 1 and the moles of component j in the droplets $N_{1,{\mathrm{d},j}}$ in Region 1:

$$\begin{aligned} N_{\mathrm{2,fuel},j} = N_{\mathrm{1,fuel,vap},j} + N_{\mathrm{1,d},j} \end{aligned}$$

(57)

Using (57) and also (35) in (56) produces

$$\begin{aligned} x_{\mathrm{2,fuel},j} = \frac{N_{\mathrm{1,fuel,vap},j} + N_{1,{\mathrm{d},j}}}{N_\mathrm{1,g} + N_{1,\mathrm{d}}} \end{aligned}$$

(58)

The quantity $N_{\mathrm{1,fuel,vap},j}$ can be rewritten as

$$\begin{aligned} N_{\mathrm{1,fuel,vap},j} = N_\mathrm{1,g} x_{\mathrm{1,fuel},j} \end{aligned}$$

(59)

Also, recalling that y represents liquid mole fractions, the number of moles of fuel component j in the drops can be written

$$\begin{aligned} N_{\mathrm{1,d},j} = N_{1,\mathrm{d}} y_{\mathrm{fuel,droplet},j} \end{aligned}$$

(60)

Equations (59) and (60) can be used to simplify (58), yielding

$$\begin{aligned} x_{\mathrm{2,fuel},j} = \frac{N_\mathrm{1,g} x_{\mathrm{1,fuel},j} + N_{1,\mathrm{d}} y_{\mathrm{fuel,droplet},j}}{N_\mathrm{1,g} + N_{1,\mathrm{d}}} \end{aligned}$$

(61)

Assuming that the mole fractions of the fuel components in the droplets are equivalent to those prepared by the supplier (i.e., $y_{\mathrm{fuel,droplet},j} = y_{\mathrm{fuel,supplier},j}$, as discussed in Sect. 8.5) allows further simplification:

$$\begin{aligned} x_{\mathrm{2,fuel},j} = \frac{N_{1,\mathrm{g}} x_{\mathrm{1,fuel},j} + N_{1,\mathrm{d}} y_{\mathrm{fuel,supplier},j}}{N_\mathrm{1,g} + N_{1,\mathrm{d}}} \end{aligned}$$

(62)

Finally, rearranging produces

$$\begin{aligned} x_{\mathrm{2,fuel},j} = \frac{x_{\mathrm{1,fuel},j} + \frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}} y_{\mathrm{fuel,supplier},j}}{1 + \frac{N_{1,\mathrm{d}}}{N_\mathrm{1,g}}} \end{aligned}$$

(32)

1.6 1.6 Derivation of equation (33)

Consider a volume of aerosol in Region 1 that contains $N_\mathrm{total}$ moles in the gas and liquid phases. $N_\mathrm{total}$ can be computed from the sum of the number of moles of gas in that volume $N_\mathrm{1,g}$ and the number of moles comprising the droplets in that volume $N_{1,\mathrm{d}}$:

$$\begin{aligned} N_\mathrm{total} = N_\mathrm{1,g} + N_{1,\mathrm{d}} \end{aligned}$$

(63)

Prior to evaporation, the fraction of moles in the liquid phase within this volume is given by $C_\mathrm{m}$

$$\begin{aligned} C_\mathrm{m} = \frac{N_{1,\mathrm{d}}}{N_{1,\mathrm{d}} + N_\mathrm{1,g}} \end{aligned}$$

(5)

such that $N_{1,\mathrm{d}}$ can be expressed as

$$\begin{aligned} N_{1,\mathrm{d}} = C_\mathrm{m} N_\mathrm{total} \end{aligned}$$

(64)

Following evaporation at the end of Region 2, all droplets are in the gas phase; thus

$$\begin{aligned} N_\mathrm{total} = N_\mathrm{2,g} \end{aligned}$$

(65)

Also, the number of moles of fuel in the gas phase following evaporation $N_\mathrm{2,fuel}$ is

$$\begin{aligned} N_\mathrm{2,fuel} = x_\mathrm{2,fuel} N_\mathrm{2,g} \end{aligned}$$

(66)

The fraction of fuel loading attributable to the droplets is equal to

$$\begin{aligned} \Upsilon _\mathrm{d} = \frac{N_\mathrm{1,d}}{N_\mathrm{2,fuel}} \end{aligned}$$

(67)

Finally, substituting (64)–(66) into (67) and canceling $N_\mathrm{total}$ yields

$$\begin{aligned} \Upsilon _\mathrm{d}=\frac{C_\mathrm{m}}{x_\mathrm{2,fuel}} \end{aligned}$$

(33)