CHEMICAL ROCKET PROPELLANT PERFORMANCE ANALYSIS
5.2. ANALYSIS OF CHAMBER OR MOTOR CASE CONDITIONS
k k
k k 162 CHEMICAL ROCKET PROPELLANT PERFORMANCE ANALYSIS
(e.g., propellants, their ingredients and proportions, desired chamber pressure, and all likely reaction products). Although combustion processes really consist of a series of different chemical reactions that occur almost simultaneously and include the break-down of chemical compounds into intermediate and subsequently into final products, here we are only concerned with initial and final conditions, before and after com-bustion. We will mention several approaches to analyzing chamber conditions; in this section we first give some definitions of key terms and introduce relevant principles.
The first principle concerns the conservation of energy. The heat created by the combustion is set equal to the heat necessary to adiabatically raise the resulting gases to their final combustion temperature. The heat of reaction of the combustion ΔrH has to equal the enthalpy change ΔH of the reaction product gases. Energy balances may be thought of as a two-step process: the chemical reaction process occurs instan-taneously but isothermally at the reference temperature, and then the resulting energy release heats the gases from this reference temperature to the final combustion tem-perature. The heat of reaction, Equation 5–9, becomes
ΔrH =
∑m 1
nj
∫
T1 Tref
CpjdT =
∑m 1
njΔhj||TT1ref (5–18) Here, the Δhj, the increase in enthalpy for each species, is multiplied by its molar concentration njand Cpjis the species molar specific heat at constant pressure.
The second principle is the conservation of mass. The mass of any atomic species present in the reactants before the chemical reaction must equal that of the same species in the products. This can be better illustrated with a more general case of the reaction shown in Equation 5–8 when the reactants are not in stoichiometric proportion.
The combustion of hydrogen with oxygen is used below as an example. It may yield six possible products: water, hydrogen, oxygen, hydroxyl, atomic oxygen, and atomic hydrogen. Here, all reactants and products are gaseous. Theoretically, there could be two additional products: ozone O3and hydrogen peroxide H2O2; however, these are unstable compounds that do not exist for long at high temperatures and can be ignored. In chemical notation the mass balance may be stated as
aH2+ bO2 → nH2OH2O + nH
2H2+ nO
2O2+ nOO + nHH + nOHOH (5–19) The left side shows the condition before the reaction and the right side the condition after. Since H2and O2are found on both sides, it means that not all of these species are consumed and a portion, namely, nH2 and nO2, will remain unreacted. At any particular temperature and pressure, the molar concentrations on the right side will remain fixed when chemical equilibrium prevails. Here, a, b, nH2O, nH2, nO2, nO, nH, and nOHare the respective molar concentrations of these substances before and after the reaction, these are expressed in kg-mol per kilogram of propellant reaction prod-ucts or of mixture; initial proportions of a and b are usually known. The number of kg-mol per kilogram of mixture of each element can be established from this initial
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5.2. ANALYSIS OF CHAMBER OR MOTOR CASE CONDITIONS 163 mix of oxidizer and fuel ingredients. For the hydrogen–oxygen relation above, the mass balances would be
for hydrogen∶ 2a = 2nH2O+ 2nH2+ nH+ nOH
for oxygen∶ 2b = nH2O+ 2nO2 + nO+ nOH (5–20) The mass balances of Eq. 5–20 provide two more equations for this reaction (one for each atomic species) in addition to the energy balance equation. There are six unknown product percentages and an unknown combustion or equilibrium tempera-ture. However, three equations can only solve for three unknowns, say the combustion temperature and the molar fractions of two of the species. When, for example, it is known that the initial mass mixture ratio of b/a is fuel rich, so that the combustion temperature will be relatively low, the percentage of remaining O2 and the percent-age of the dissociation products (O, H, and OH) would all be very low and may be neglected. Thus, nO, nH, nOH, and nO
2 are set to be zero. The solution requires knowledge of the enthalpy change of each of the species, and that information can be obtained from existing tables, such as Table 5–2 or Refs. 5–8 and 5–9.
In more general form, the mass for any given element must be the same before and after the reaction. The number of kg-mol of a given element per kilogram of reactants and product is equal, or their difference is zero. For each atomic species, such as the H or the O in Eq. 5–20,
[m
∑
j=1
aijnj ]
products
− [ r
∑
j=1
aijnj ]
reactants
= 0 (5–21)
Here, the atomic coefficients aijare the number of kilogram atoms of element i per kg-mol of species j, and m and r are indices as defined above. The average molecular mass for the products, using Eqs. 5–5 and 5–19, becomes
𝔐 = 2nH
2 + 32nO
2+ 18nH
2O+ 16nO+ nH+ 17nOH nH
2+ nO
2+ nH
2O+ nO+ nH+ nOH (5–22) The computational approach used in Ref. 5–13 is the one commonly used today for thermochemical analyses. It relies on the minimization of the Gibbs free energy and on the mass balance and energy balance equations. As was indicated in Eqs. 5–12 and 5–13, the change in the Gibbs free energy function is zero at equilibrium; here, the chemical potential of the gaseous propellants has to equal that of the gaseous reaction products, which is Eq. 5–12:
ΔG =∑
(njΔGj)products−∑
(njΔGj)reactants= 0 (5–23) To assist in solving this equation a “Lagrangian multiplier,” a factor representing the degree of the completion of the reaction, is often used. An alternative older method
k k 164 CHEMICAL ROCKET PROPELLANT PERFORMANCE ANALYSIS
in solving for gas composition, temperature, and gas properties is to use the energy balance (Eq. 5–18) together with several mass balances (Eq. 5–21) and certain equi-librium constant relationships (see for example Ref. 5–16).
After assuming a chamber pressure and setting up the energy balance, mass bal-ances, and equilibrium relations, another method of solving all the equations is to estimate a combustion temperature and then solve for the various values of nj. Then, check to see if a balance has been achieved between the heat of reaction ΔrH0 and the heat absorbed by the gases, HT0− H00, going from the reference temperature to the combustion temperature. If they do not balance, the value of the combustion tem-perature is iterated until there is convergence and the energy balances.
The energy release efficiency, sometimes called the combustion efficiency, can be defined here as the ratio of the actual change in enthalpy per unit propellant mixture to the calculated change in enthalpy necessary to transform the reactants from the initial conditions to the products at the chamber temperature and pressure. The actual enthalpy change is evaluated when the initial propellant conditions and the actual compositions and the temperatures of the combustion gases are measured. Mea-surements of combustion temperature and gas composition are difficult to perform accurately, and combustion efficiency is therefore only experimentally evaluated in rare instances (such as in some R & D programs). Combustion efficiencies in liquid propellant rocket thrust chambers also depend on the method of injection and mixing and increases with increasing combustion temperature. In solid propellants the combustion efficiency becomes a function of grain design, propellant composition, and degree of uniform mixing among the several solid constituents. In well-designed rocket propulsion systems, actual measurements yield energy release efficiencies from 94 to 99%. These high efficiencies indicate that combustion is essentially complete, that is, that negligible amounts of unreacted propellant remain and that chemical equilibrium is indeed closely established.
The number of compounds or species in combustion exhausts can be large, up to 40 or more with solid propellants or with liquid propellants that have certain additives.
The number of nearly simultaneous chemical reactions that take place may easily exceed 150. Fortunately, many of these chemical species are present only in relatively small amounts and may usually be neglected.
Example 5–1. Hydrogen peroxide is used both as a monopropellant and as an oxidizer in bipropellant systems (see Chapter 7). It is stored in liquid form and available in various degrees of dilution with liquid water. For rocket applications, concentrations (70 to 98+%), known as high-test peroxide (HTP) are used. For a monopropellant application, calculate the adiabatic flame or dissociation temperature as a function of water content based on an initial mixture temperature of 298.15 K (the standard condition).
SOLUTION. In this application, one kilogram hydrogen peroxide dissociates while passing through a catalyst and releases energy, which only goes to increase the propellant temperature in the absence of any heat transfer losses. But some of this heat will be required to evaporate the diluent water. The mass balance, Eq. 5–21, is satisfied by 2 mol of hydrogen peroxide in
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5.2. ANALYSIS OF CHAMBER OR MOTOR CASE CONDITIONS 165 n mols of liquid water, producing n + 2 mol of water vapor plus 1 mol of oxygen gas. Since the reaction goes to completion, no equilibrium constant is needed:
2H2O2(l) + nH2O(l)→ (n + 2)H2O(g) + O2(g)
The symbols (l) and (g) refer to the liquid state and the gaseous state, respectively. The heats of formation from the standard state ΔfH0and molar specific heats Cpare shown below (see Table 5–1 and other common sources such as the NIST Chemistry Web-Book, http://webbook .nist.gov/chemistry/). For these calculations the heat of mixing may be ignored.
Species ΔfH0(kJ/kg-mol) Cp(J/kg-mol-K) 𝔐, (kg/kg-mol)
H2O2(l) −187.69 34.015
H2O(l) –285.83 18.015
H2O(g) −241.83 0.03359 18.015
O2(g) 0 0.02938 31.999
The energy balance, Eq. 5–9, for 2 mol of decomposing hydrogen peroxide becomes ΔrH0= [nΔfH0]H
2O− [nΔfH0]H
2O2
= 2 × (−241.83) − 2 × (−187.69)
= −108.28 kJ
The reaction is exothermic but, as stated, some of this energy is used up in vaporizing the diluent liquid water, namely, 285.83 − 241.83 = 44.0 kJ∕kg-mol (at standard conditions). The net available heat release thus becomes 108.28 − 44.0n (kJ). In order to calculate the adiabatic temperature, we assume ideal-gas heating at constant pressure, Eq. 5–18 (values for the molar specific heats are from Table 5–1 and are taken as constant):
∫ (nH
2OCpH
2O+ nO
2CpO
2)dT = [(2 + n)CpH
2O+ CpO
2]ΔT = 108.28 − 44.0n It will be more convenient to give results in terms of z, a mass fraction of the diluent water in the original mixture, and for this molecular masses need to be inserted (also shown in Table 5–1):
z = mH
2O∕(mH
2O2+ mH
2O) = n𝔐H2O∕(2𝔐H2O2+ n𝔐H2O)
= 18.015n∕(2 × 34.015 + 18.015n)
Now solve for n, n = 3.78z∕(1 − z) n = 3.78z∕(1 − z), and substitute it in the relation for the temperature. The resulting value (the adiabatic temperature, Tad) may then be plotted (see figure next page) in terms of the mass fraction z with the initial temperature as given:
Tad= 298.15 + 108.28 − 44 × 3.78z∕(1 − z) 0.03359 × [2 + 3.78z∕(1 − z)] + 0.02838
k k 166 CHEMICAL ROCKET PROPELLANT PERFORMANCE ANALYSIS
The figure also displays the values of c∗(where T1= Tad),𝔐, and k, which are calculated according to Eqs. 3–32 and 5–5 and 5–7.
40 50
30 20 10 0
5000 4000 3000 2000 1000
0
0 0.05 0.1 0.15 0.2 0.25 0.3
c* (characteristic velocity)
(molecular mass) k (ratio of specific heats)
z (H2O mass fraction)
Tad (adiabatic temperature)
Tad (K) and c* (m/sec) 2.0
1.5 k 1.0
(kg/kg–mol)