Equations of State

The previous section tracked the enthalpy of the propellants as they flowed through the components of the engine. But what exactly is enthalpy? And what does it have to do with an equation of state? Enthalpy is not as intuitive to understand as the pressure and temperature of a fluid. Mathematically, enthalpy is defined as follows:

(2.1.1)
$ h=u+\frac{P}{\rho} $

$ h= $ enthalpy $($J/kg$)$

$ u= $ internal energy $($J/kg$)$

$ P= $ pressure $($Pa$)$

$ \rho= $ density $($kg/m3$)$

As shown by Equation 2.1.1, the enthalpy of a fluid is the sum of the fluid’s internal energy (due to intermolecular forces and temperature) and its flow energy (due to density and pressure).[3] It is mathematically convenient to combine both internal energy and flow energy into one property known as enthalpy. Knowing the enthalpy of the propellants as they flow through rocket engine aids greatly in estimating the amount of power required to drive the engine’s turbomachinery (see Section 4). But how can the enthalpy of a propellant be estimated? Pressure is measured easily enough by mounting a transducer on a pipeline, but it is not so easy to measure internal energy due to intermolecular forces. An equation of state must be used to help solve this problem. An equation of state relates the temperature, pressure, and density of a fluid [4]. . Perhaps the most well-known equation of state is the ideal gas law:

(2.1.2)
$ P=\rho\,R_s\,T $

$ R_s=$ specific gas constant

$ T= $ temperature $($K$)$

The following example illustrates the link between enthalpy and the equation of state.

Example 2.1

Consider liquid hydrogen flowing through a pipeline at a constant pressure of 1 atmosphere at an initial temperature of 20 K. After travelling a short distance it flows through a check valve, and on the other side of the valve the temperature has increased to 21 K while the pressure remains 1 atmosphere. How has this small temperature change affected the enthalpy of the flow?

The first item to bear in mind is that it may not be possible to use the ideal gas equation to predict the initial density of the flow, since it began as a liquid (where intermolecular forces cannot be neglected). Secondly, even though the temperature change across the check valve was only 1 K, this small change drastically altered the fluid. In fact, the hydrogen has now boiled and is no longer a liquid, but a vapor. Figures 2.1.1 and 2.1.2 plot the density and enthalpy of the hydrogen on each side of the valve. Note how the small temperature change caused the density to decrease by a factor of 50. The ideal gas equation of state is not capable of modeling phase changes like this.

Fig 2.1

Figure 2.1.1

Fig 2.2

Figure 2.1.2

How are phase changes modeled?

A traditional fluid phase diagram splits the pressure-temperature plane into four quadrants separated by the critical isobar $ P_c $ and critical isotherm $T_c$ [5] as shown below below in Figure 2.1.3. The intersection of the critical isobar and critical isotherm is known as the critical point. This is the location where liquid and vapor states become indistinguishable from each other. [4] The blue line shown on Figure 2.1.3 represents the saturation curve. Along this line, the fluid is boiling and the distinct liquid and vapor states in regions $I_L$ and $ I_V $ exist together in equilibrium. Discontinuities in the density and enthalpy fields arise when crossing this line (as shown above in Figures 2.1.1 and 2.1.2).

Fig 2.1.3

Figure 2.1.3 [6]

Depending on who you ask, each quadrant on the phase diagram corresponds to a different fluid state. [5] Some typical examples are listed here:


Engineering Toolbox [7] National Institute of Standards and Technology [8] Younglove, B.A. [9]
$I_L$ Liquid Liquid Liquid
$I_V$ Gaseous Vapor Fluid
$II$ Gaseous Vapor Fluid
$III$ Supercritical Supercritical Fluid
$IV$ Compressible Liquid Liquid Liquid

An equation of state that is more complex than the ideal gas law must be used to adequately model fluid properties in these different regions.

The Helmholtz Free Energy Equation of State

Helmholtz energy is used to describe the amount work that can be extracted from a closed system at constant temperature and volume. [10] In a similar manner to the ideal gas equation, two properties (in this case density and temperature) must be specified to calculate the Helmholtz energy of a fluid:

(2.1.3)
$ a=a(\rho, T) $

$a$ = Helmholtz energy $($J/kg$)$

Helmholtz energy is typically specified in reduced form:

(2.1.4)
$$\alpha =\frac{a(\delta, \tau)}{R_s} $$

$\alpha$ = reduced Helmholtz energy

$\delta$ = reduced density

(2.1.5)
$$\delta =\frac{\rho}{\rho_c} $$

$\tau$ = reduced reciprocal temperature

(2.1.6)
$$\tau =\frac{T_c}{T} $$

$T_c$ and $\rho_c$ represent temperature and density at the critical point (refer to Figure 2.1.3).

A unique characteristic of the Helmholtz free energy equation of state (as opposed to the ideal gas law) is that it can be split into two terms, one describing the ideal gas contribution, and the other describing the residual contribution due to intermolecular forces:[11]

(2.1.7)
$$\alpha =\alpha^0(\tau, \delta)+\alpha^r(\tau, \delta) $$

$\alpha^0$ = ideal gas compent of reduced Helmholtz energy

$\alpha^r$ = residual compent of reduced Helmholtz energy

Over the years, researchers tabulated thousands of thermodynamic data points and developed sets of non-linear regressions to model $\alpha^0$ and $\alpha^r$ as functions of $\tau$ and $\delta$ .

Finally: The Link Between Enthalpy and the Equation of State

The Helmholtz free energy equation of state provides a single tool capable of determining fluid properties in all four quadrants of Figure 2.1.3. A reduced reciprocal temperature ($\tau$) and a reduced density ($\delta$) are plugged in to the non-linear regression model (Equation 2.1.7). All of the data required to use this equation for SSME propellants is tabulated in Sections 1.1 and 1.2. The regression model returns the ideal gas and residual contributions of the Helmholtz energy, $\alpha^0$ and $\alpha^r.$ Taking derivatives of these contributions with respect to $\tau$ and $\delta$ allows many different thermodynamic properties to be calculated, including enthalpy.[11] Sections 2.2 and 2.3 and present a qualitative description of the results applicable to hydrogen and oxygen, as well as mathematical details on how to perform the calculations.

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