First Law of Thermodynamics
The analysis begins with a thermodynamic system (perhaps a rocket combustion chamber, turbine, or pump) as shown below in Figure 3.2.1.
Figure 3.2.1
Fluid enters the system at Station 1 with an internal energy $u_1$, pressure $P_1$, temperature $T_1$, and density $\rho_1$. It flows through the device, exiting at Station 2 with internal energy $u_2$, pressure $P_2$, temperature $T_2$, and density $\rho_2$. Fluid enters at velocity $V_1$, and exits at velocity $V_2$. The mass flow rate through the device is represented by $\dot{m} $. $A_1$ and $A_2$ are the cross-sectional areas of the inlet and outlet. $Z$ represents the heights of the inlet and outlet above a common reference plane. $\dot{W}$ represents shaft work going into or out of the system. In the case of a pump, shaft work goes into the system in order to pressurize the fluid. In the case of a turbine, shaft work extracts energy out of the system. $\dot{Q}$ represents heat transfer through the walls of the system.
The first law of thermodynamics says that the amount of energy entering this system must be equal to the amount of energy leaving the system. This is described mathematically as follows:[3]
The terms inside the integral represent the different types of energy flowing through the system. $u+\frac{P}{\rho}$ is defined as the enthalpy of the fluid, $h$ (refer to Section 2.1). $\frac{V^2}{2}$ is the kinetic energy of the fluid, and $gZ$ is gravitational potential energy. Assuming fluid properties are constant over the cross section of both the inlet and outlet (but not necessarily constant between Stations 1 and 2), $\rho V \cdot \hat{n} dA$ becomes the mass flow rate of fluid across the control surface, $\dot m$. Assuming steady-state conditions, integrating between Stations 1 and 2 gives the following:
Equation 3.2.2 is a general result applicable to any steady-state thermodynamic system with one inlet and one outlet to satisfy conservation of mass. We will now make several assumptions to tailor this general result into a better model of a rocket combustion chamber. First, we assume that very little time elapses as the fluid travels from Station 1 to Station 2 and completes the combustion process. This means that the distance between Stations 1 and 2 is relatively small. We can therefore assume that the combustion process is complete before the fluid enters the converging section of the nozzle and begins to accelerate. This means that $V_1$ = $V_2$ and $A_1$ = $A_2$. We will also assume that the difference in height between the entrance and exit of the combustion chamber is negligible $\left(Z_1 = Z_2\right)$. There is no shaft work into or out of the combustion chamber, so both $W_1$ and $W_2$ are 0. For now, we assume there is no heat transfer into or out of the combustion chamber ($Q_1$ and $Q_2$ = 0). This reduces the equation to:
This simple equation gives a powerful result. The enthalpy of the fluid entering the combustion chamber, $h_1$, must be equal to the enthalpy of the fluid exiting the chamber after combustion is complete, $h_2$.
Enthalpy of the Reactants
Let’s begin by analyzing $h_1$, the enthalpy of the reactants entering the combustion chamber. $h_1$ is found by taking the sum of the enthalpies of the individual reactants:
In Section 2.5, a great deal of effort was spent showing that the enthalpy of an ideal gas does not depend on pressure. Then why does pressure show up as a variable in Equation 3.2.4? This occurs because the propellants entering the combustion chamber may be in either a liquid or gaseous state. The enthalpy of liquid propellants requires a “real” equation of state (such as the Helmholtz correlations in Sections 2.2 and 2.3) to account for the effects of pressure and intermolecular forces. The enthalpy of gaseous reactants can be modeled with the “ideal” correlations described in Section 2.4 that are independent of pressure.
Enthalpy of the Products (Ideal Gas Model)
Now we shift our attention to the right-hand side of Equation 3.1.5 and focus on the species produced as a result of combustion. The overall enthalpy of these combustion products, $h_2$, is the summation of the enthalpies of the individual species in the resulting gaseous mixture. For the sake of simplicity, we will will restrict our focus to only one of these compounds: $H_2O$. Since we are now dealing solely with gaseous products, we can utilize the ideal gas enthalpy model (Equation 2.4.2) and ignore any effects of pressure. The enthalpy magnitude $h(T)$ returned by this equation is really a summation of two different components: the heat of formation at the reference temperature ($h_f$) plus difference in enthalpy between the temperature of interest and the reference temperature ($\Delta h$):
The heat of formation, $h_f$ accounts for the energy that is released (exothermic reactions) or absorbed (endothermic relations) to form the compound of interest at the desired temperature, $T$. Heat of formation for a chemical compound is calculated as follows:[17]
Similar expressions can be developed for each species in the chemical model. It should be noted that as a consequence of Equation 3.2.6 the heat of formation for the reference elements $H_2$ and $O_2$ is zero at all temperatures.
We now turn our attention to the $\Delta h$ term on the right-hand side of Equation 3.2.5. From the discussion of enthalpy in Section 2.5, it is important to remember that enthalpy of a fluid is only meaningful as a difference in energy between two states. We can define the overall magnitude of enthalpy arbitrarily, as long as it is referenced to a standardized baseline (see Equation 2.5.12). The $\Delta h$ term accounts for variation from this baseline. This is accomplished by defining a reference temeperature, $T_{ref}$, from which to make enthalpy comparisons. The model developed here utilizes a reference temperature of 298.15 K. This concept is described mathematically by Equation 3.2.8:
Plugging this expression for $\Delta h$ into Equation 3.2.5 provides an expression for the enthalpy of an individual gaseous product:
It is worth taking a moment to examine this equation in greater in detail. The term $h_{2,i}(T)$ corresponds to the value returned by Equation 2.4.2 at a given temperature. $h_{f,i}(T)$ is calculated with Equation 3.2.6, in conjunction with Equation 2.4.2 and the polynomial coefficients for each relevant species in the formation reaction. The final two $h_{i}(T)$ terms in Equation 3.2.9 are also calculated with Equation 2.4.2, utilizing the temperature of interest ($T$) and the reference temperature (298.15 K). Let’s use these expressions to tabulate the values of $h_{2,i}$ and $h_f$ for $H_2O$ at several temperatures.
| $T $ | $h(T)$ (Eqn. 2.4.2) | $h_f(T)$ (Eqn 3.2.8) |
|---|---|---|
| 298.15 | -241826 | -241826 |
| 400 | -238373 | -242852 |
| 500 | -234901 | -243839 |
Table 3.2.1: $h$ and $h_f$ for $H_2O$ at $T_{ref}=298.15$ [J/mol]
Now let’s plug the data from Table 3.2.1 in to each term of equation 3.2.9 to see how Equation 2.4.2 calculates the enthalpy of $H_2O$ at 500 K utilizing the standard reference temperature of 298.15 K.
It is interesting to note that at the reference temperature of 298.15 K, the heat of formation is identical to the enthalpy value returned by Equation 2.4.2. Another noteworthy point arises when examining enthalpies of the reference elements used to form a water molecule: $O_2$ and $H_2$.
| $O_2$ | $H_2$ | $H_2O$ | |
|---|---|---|---|
| $T=298.15$ | 0 | 0 | -241826 |
Table 3.2.2: Enthalpy at $T=T_{ref}=298.15$ [J/mol]
This table shows how the polynomial curve fit described by Equation 2.4.2 is built in such a way that the enthalpy of the reference elements is defined to be zero at the reference temperature of 298.15 K. An important distinction to be aware of is that $H_2O$ is not one of the “reference elements”[16] used by the ideal gas enthalpy model of Section 2.4. This means that the magnitude of $h_{H_2O}\left(298.15\right)$ returned by Equation 2.4.2 is non-zero. This makes physical sense, because at standard temperature and pressure, water is a liquid, rather than an ideal gas.
Back to the First Law
Now we return to where we left off on our discussion of the first law of thermodynamics. Referring to back to Equation 3.2.3, we were searching for $h_2$, the enthalpy of the gaseous mixture exiting the combustion chamber. All the necessary tools are now in place. The enthalpy $h_2$ is the sum of the enthalpies of the individual species, $h_{2,i}$. These values are computed directly with Equation 2.4.2, which in expanded form is identical to equation 3.2.9. to obtain the total enthalpy of the fluid exiting the combustion chamber, this equation must be applied to each of the six species present on the right-hand side of of Equation 3.1.5 (the total chemical reaction). This is represented by the following summation:
Now the expressions for $h_1$ and $h_2$ can be plugged in to the simplified version of the First Law, Equation 3.2.3, to reach the following conclusion:
Only one step remains! Equation 3.2.12 gives enthalpy in terms of $\frac{J}{mol}$. We want to know how much power this corresponds to in terms of watts. We accomplish this by multiplying the each term in Equation 3.2.12 by its respective mass flow rate $\dot{N}_i$, in terms of $\frac{mol}{sec}$. Equation 3.1.13 contains an expression for $\dot{N}_i$.
This is an excting result! Analyzing Equation 3.2.12 in conjunction with Equations 3.1.19 - 3.1.22 provides a closed system of five non-linear equations that are functions of the five variables a, b, c, d, and T. This system can be solved to determine the resulting equilibrium mole fractions and flame temperature for a given combustion chamber pressure, mixture ratio, and reactant input conditions. Read on for the solution procedure!