Combustion Chemistry

We are now in a position to use the thermodynamic properties calculated in the previous section to predict the chemical composition and flame temperature of the exhaust gas mixture produced in the SSME pre-burners and main combustion chamber. The model is based on a fuel-rich combustion process utilizing the following set of reactions, as described by Sozen.[37] The combustion processes begins with $x$ moles of diatomic hydrogen and $y$ moles of diatomic oxygen as reactants (Equation 3.1.a). But as the reaction continues, $H_2$, $O_2$, and $H_2O$ disassociate and recombine according to Equations 3.1.c -3.1.d. When the system reaches equilibrium, all six species in these equations will be present.

(3.1.a)
$$x\,H_{2} + y\,O_{2} \rightleftharpoons \left(2\,y\right)H_{2}O +\left(x-2y\right)H_2 $$
(3.1.b)
$$H_{2} \rightleftharpoons 2\,H $$
(3.1.c)
$$O_{2} \rightleftharpoons 2\,O $$
(3.1.d)
$$ 2\,H_{2}O \rightleftharpoons H_2 +2\,OH $$

In a perfectly stoichiometric reaction the coefficients $x$ and $y$ would be 2 and 1, respectively. If the reactants are mixed in exactly this ratio there will be no excess $H_2$ on the right-hand side of Equation 3.1.a. All of the reactants will be turned into the product, $H_2O$. This can be verified by noting that the $(x-2y)$ coefficient on the excess hydrogen term in Equation 3.1.a vanishes for $x = 2$ and $y = 1$.

Again, this model is designed to analyze a fuel-rich combustion process. Fuel-rich combustion contains excess hydrogen as product. The term “fuel-rich” refers to the reactant mixture ratio, $M$:

(3.1.1)
$$ M = \frac{\dot m_{O_2}}{\dot m_{H_2}} $$

where $\dot m$ is the propellant mass flow rate into the combustion chamber in kg/sec. To calculate the mixture ratio, we need to multiply the coefficients $x$ and $y$ by the molecular mass of diatomic hydrogen (2 g/mol) and diatomic oxygen (32 g/mol). It now becomes evident that:

(3.1.2)
$$ M = \frac{\dot m_{O_2}}{\dot m_{H_2}} = \frac{32\,y}{2\,x} $$

In the stoichiometric case, this equation reduces to:

$$ M = \frac{\dot m_{O_2}}{\dot m_{H_2}} = \frac{32\left(1\right)}{2\left(2\right)} = 8 $$

This means that for fuel-rich combustion to occur, the mixture ratio must be less than 8. Plugging this value into Equation 3.1.2 and solving for $x$ reveals that in the case of fuel-rich combustion, $x > 2y$. This analysis ignores oxygen-rich combustion (mixture ratios greater than 8). As will be shown in following sections, for $H_2$ / $O_2$ combustion, specific impulse is maximized with a fuel-rich mixture ratio.[38]

Now we will examine the relationships between Equations 3.1 a-d, starting with Equation 3.1.a as an example. The analysis begins by multiplying each term in Equation 3.1.a by a factor of “$a$” to distinguish it from the terms in Equations 3.1 b-d:

(3.1.a)
$$a\,x\,H_{2} + a\,y\,O_{2} \implies a\,\left(2\,y\right)H_{2}O +a\,\left(x-2y\right)H_2 $$

Next, we need to determine how the coefficients for each species change between the left and right sides of the equation. This is accomplished by subtracting the left-hand coefficient for each species from the corresponding right-hand coefficient. For $H_2$ in Equation 3.1.a the results are:

$$ a\,\left(x-2\,y\right) -a\,x =-2\,a\,y $$

For $O_2$:

$$ 0-y\,a=-y\,a $$

Finally, for $H_2O$:

$$ 2\,y\,a-0 = 2\,y\,a $$

These quantities represent the change in number of moles of each individual species between when reaction 3.1.a began and the final composition it reaches at equilibrium. Repeating this process for Equations 3.1 b-d gives the following changes in composition, $\Delta_C$:

Equation $ H_2$ $ O_2$ $ H_2O$ $ OH$ $ O$ $ H$
3.1.a -2ya -ya 2ya 0 0 0
3.1.b -b 0 0 0 0 2b
3.1.c 0 -c 0 0 2c 0
3.1.d d 0 -2d 2d 0 0
$\sum\Delta_C$ -2ya-b+d -ya-c 2ya-2d 2d 2c 2b

Table 3.1.1: Composition Changes, $\Delta_C$

Now we examine the reaction modeled by Equations 3.1.a-d as a whole, rather than individually. Taking a summation of the $\Delta_C$ values in each column of Table 3.1.1 reveals the total change in number of moles of each species for the overall reaction. This summation is tabulated in the bottom row of Table 3.1.1.

We assume the reaction began with $x$ moles of $H_2$ and $y$ moles of $O_2$ as reactants. When the reaction reaches chemical equilibrium, all six species listed in Table 3.1.1 will be present in some unknown quantity, $C_i$:

(3.1.3)
$$x\,H_{2} + y\,O_{2} \implies C_1\,H_{2}O +C_2\,H_2 + C_3\,O_2+C_4\,OH+C_5\,H+C_6\,O $$

The challenge is now to determine the $C_i$ coefficients. The $\sum\Delta_C$ values in Table 3.1.1 provide an expression for how much the $C_i$ coefficients change from the start of the reaction all the way to its conclusion. The $C_i$ coefficients represent the final quantities of each species present at equilibrium. Therefore if we take the the quantity of each species present at the start of the reaction $\left(N_i\right)$ and add to it the expression for how the quantity changes, we will know the amount of each species present at equilibrium. This is expressed mathematically as follows:

(3.1.4)
$$C_i = N_i +\sum\Delta_{C_i} $$

In Equation 3.1.4, $N_{i}$ is equal to $x$ for $H_2$, $y$ for $O_2$, and 0 for all other species. This allows us to write an expression for the overall chemical equation:[39]

(3.1.5)
$$x\,H_{2} + y\,O_{2} \implies \left(2\,y\,a-2\,d\right)H_{2}O +\left(x-2y\,a-b+d\right)H_2 + \left(y-y\,a-c\right)O_2+2\,d\,OH+2\,b\,H+2\,c\,O $$

Equation 3.1.5 shows that combustion began with $(x+y)$ moles of reactants. The coefficients on the right-hand side of the equation represent the number of moles of each species the system comes to at equilibrium. Taking the sum of all the right-hand coefficients gives the total number of moles of products:

(3.1.6)
$$N_{prod} =\sum\,C_i =-ya+x+y+b+c+d $$

If we divide the right-hand coefficients by the summation in Equation 3.1.6 we obtain the mole fractions, $Z$, of each product in the resulting mixture.

(3.1.7)
$$Z_{O_2} =\frac{y-ya-c}{-ya+x+y+b+c+d} $$

(3.1.8)
$$Z_{H_2O} =\frac{2ya-2d}{-ya+x+y+b+c+d} $$

(3.1.9)
$$Z_{H_2} =\frac{x-2ya-b+d}{-ya+x+y+b+c+d} $$

(3.1.10)
$$Z_{OH} =\frac{2d}{-ya+x+y+b+c+d} $$

(3.1.11)
$$Z_{H} =\frac{2b}{-ya+x+y+b+c+d} $$

(3.1.12)
$$Z_{O} =\frac{2c}{-ya+x+y+b+c+d} $$

The $C_i$ coefficients can be used to calculate the mass generation rate, $\dot{N}_i$, of each product in terms of moles per second:

(3.1.13)
$$ \dot{N}_i = \frac{C_i}{x+y}\left(\dot{m}_{H_2}+\dot{m}_{O_2}\right) $$

The $ \dot{m}$ terms in the above equation represent the mass flow rates of hydrogen and oxygen reactants being pumped in to the combustion chamber in units of moles per second.

We now have an overdetermined system of six non-linear equations that are functions of the four variables a, b, c, and d. To complicate matters even further, this system does not include any mechanism to account for how the mole fractions vary as a function of temperature or pressure. To overcome this hurdle we must introduce the idea of an equilibrium constant.

Equilibrium Constants

The equilibrium constant, K, is a quantity that describes a chemical reaction’s tendency to approach completion. The equilibrium constant can be defined in terms of either the partial pressures of the species in the gas $\left(K_P\right)$ or in terms of mole fractions $\left(K_n\right)$.[40] Generally speaking, if $K_n$ values are small (less than 0.0001) the final composition the gas reaches at equilibrium will contain mostly reactants (terms on the left hand side of Equations 3.1 a-d). If $K_n$ is large (greater than 1000), the reaction tends to favor the products. Between these two values a more balanced mixture of reactants and products will be present.[41]

The equilibrium constant varies as function of temperature. This means that a chemical reaction could tend to favor reactants at low temperatures while leaning more toward products at high temperatures. The challenge is to develop the equilibrium constant for each reaction listed in Equations 3.1 a-d and use this information to determine the combustion temperature of the overall reaction described by Equation 3.1.5.

We begin with a model for $K_P$:[17]

(3.1.14)
$$ K_P = \text{exp}\left(\frac{-\Delta G}{R_uT}\right) $$

In the above equation, $G$ represents Gibbs free energy. It is a thermodynamic property of the gas related to the amount of energy available to drive chemical reactions. Fortunately this property does not require any models or curve fits beyond what has already discussed on this website. The functions for enthalpy and entropy established in Section 2.4 can be combined in order to calculate the Gibbs energy of an individual species:[17]

(3.1.15)
$$ G(T)= h(T)-T \times S(T) $$

where $h(T)$ and $S(T)$ are modeled by Equations 2.4.2 and 2.4.3, respectively. $\Delta G$ refers to the change in Gibbs in energy between the products and reactants of an entire chemical equation:

(3.1.16)
$$ \Delta G= \sum_{products} C_i \times G_i -\sum_{reactants} C_i \times G_i$$

$C_i$ represents the stoichiometric coefficients of the chemical reaction in question. In the case of Equation 3.1.a this means $x=2$ and $y=1$.

In other words, Equations 3.1.15 and 3.1.16 can be used to find $K_p$ as a function of temperature for each reaction shown in Equations 3.1.a-d. Results are shown below in Figure 3.1.1.

Fig 3.1.1

Figure 3.1.1

It is interesting to note that the equilibrium constant is positive for reaction 3.1.a and negative for reactions 3.1.b-3.1.d. The sign of the equilibrium constant reflects the reaction’s tendency to produce or absorb heat. Exothermic reactions produce heat and have positive equilibrium constants. Endothermic reactions absorb heat from the surroundings and have negative equilibrium constants.[42] In both cases, as temperature increases the absolute value of the equilibrium constants approach zero.

Now we need to link $K_p$ to the mole fractions described by Equations 3.1.7 - 3.1.12. We begin by defining the equilibrium constant based on mole fraction, $K_n$, as follows:[40]

(3.1.17)
$$ K_n=\frac{\prod_{prod} Z_i^{C_i}}{\prod_{react} Z_i^{C_i}} $$

In Equation 3.1.17, $C_i$ represents the stoichiometric coefficients of the chemical reaction in question (i.e. x =2 y =1). Using Equation 3.1.a as an example:

$$ K_n = \frac{Z_{H_2O}^2}{Z_{H_2}^2 Z_{O_2}} $$

We can now link $K_n$ and $K_p$ through the following relation:[40]

(3.1.18)
$$ K_P=K_n\left[P^{\sum_{prod}C_i-\sum_{react}C_i}\right] $$

In Equation 3.1.18, $C_i$ represents the not necessarily stoichiometric coefficients of the chemical reaction in question (i.e. leave the coefficients in terms of x and y) . $P$ represents the ratio of pressure in the rocket combustion chamber to atmospheric pressure. Using Equation 3.1.a as an example:

$$ K_P=\left[\frac{Z_{H_2O}^2}{Z_{H_2}^2 Z_{O_2}}\right]P^{-y} $$

Completing this process for each of Equations 3.1. a-d establishes the following system of 4 non-linear equations:

(3.1.19)
$$ 0=\left[\frac{Z_{H_2O}^2}{Z_{H_2}^2 Z_{O_2}}\right]P^{-y}-K_{P,a} $$
(3.1.20)
$$ 0=\left[\frac{Z_{H}^2}{Z_{H_2}}\right]P-K_{P,b} $$
(3.1.21)
$$ 0=\left[\frac{Z_{O}^2}{Z_{O_2}}\right]P-K_{P,c} $$
(3.1.22)
$$ 0=\left[\frac{Z_{H_2}Z_{OH}^2}{Z_{H_2O}^2}\right]P-K_{P,d} $$

It is important to remember that the $K_P$ terms in the above equations are not identical to each other. A unique value of $K_P$ must be determined for each chemical reaction described by Equations 3.1 a-d. This can be accomplished through the use of Equations 3.1.14 through 3.1.16. Another key point is that the mole fractions listed above, $Z$, are functions of the the variables a, b, c, and d. $K_P$ is a function of temperature. Assuming a known combustion chamber pressure ratio (P) and a known mixture ratio (set by x and y), we now have a system of 4 non-linear equations as a function of 5 unknown variables. One more equation is required to close the system: the first law of thermodynamics.

Previous Next

Top