5.1 DETERMINATION OF PRESSURANT REQUIREMENTS

The physical and chemical processes which take place during the expulsion of a liquid propellant from a tank by a gas or gas mixture are numerous and difficult to analyze. Applicable experimental data for a selected system are often limited. Thus, the basis for the analytical approach is frequently narrow and uncertain. As a result, the initial design calculations of the quantity of pressuring gas required must be considered approximate until verified experimentally. The refinement of the analytical approach to minimize discrepancies between theoretical predictions and actual test results is an art requiring experience and thorough understanding of the physical processes.

Basic considerations and necessary procedures for the calculation of pressurant requirements are described below.

Required Engine System Data and Assumptions

Before starting calculations of pressurant requirements, the following significant engine operating parameters must be known or assumed: (1) Design operating temperature range of the propellants and the feed system including pressurant (2) Type of propellants, their weights and corresponding volumes at the extremes of the operating temperature range (3) Total tank volumes: Nominal value and tolerances (4) Initial tank ullage volume, percent of total tank volume at the temperature limits. (The term "ullage" denotes that portion of a propellant tank not occupied by the liquid propellants.) (5) Trapped propellant volumes, percent of total propellant load, at engine burnout (6) Operating tank pressure: Nominal value and tolerances (7) Operating duration of the engine systems: Nominal value and variations. To avoid later marginal conditions, calculations should assume that pressurization must be supplied for maximum systems operating duration even though some missions may require shorter durations. Realistic assumptions for the temperature of tanks and propellants at burnout must be made. If a mission calls for several system restarts and coasting periods, the environmental conditions during the coasting periods must also be given or assumed.

Factors Influencing Pressurant Requirements

Several important factors which will influence considerably the final state of a pressurizing gas or gases, and thus their required quantity, are discussed below.

1. Propellant vaporization.-Propellants evaporate to various degrees from the gas-liquid interface within the tank. The amount depends upon the volatility of the propellant, the temperature of the entering gas, the turbulence of the gas, the sloshing of the liquid, the tank geometry including internal structural members, and the rate of propellant expulsion. To whatever degree vaporization takes place, it lowers the temperature of the gas and adds propellant vapor as a component of the pressurizing gases. Also, as the liquid propellant level recedes, a film of liquid may be left on the tank wall surface, further contributing to propellant evaporation.
2. Tank wall temperature.-If the pressurizing gas is hotter than the tank walls, cooling of the gas and heating of the wall may result. On the other hand, aerodynamic heating of the tank walls in flight may cause heating of the pressurizing gas. It may also heat the propellant and thus increase vaporization effects and raise NPSH requirements in turbopump-fed systems.
3. Vapor condensation.-Certain components of the pressurizing gas, such as water vapor, may condense. Even if the bulk of the gas remains above the dew point for the condensible component, local condensation may occur at the tank walls, or at the propellant surface.
4. Solubility of the pressurizing gas.-If the pressurizing gas contains components which are soluble in the propellant, diffusion of these components into the propellant can occur. Solubility is generally affected by temperature and pressure conditions.
5. Ullage gas compression.-If, before start, the tank ullage space is filled with low-pressure gas, onset of pressurization will cause adiabatic compression. This can raise the ullage space temperature considerably during the initial few seconds of operation.
6. Chemical reaction.-If any components of the pressurizing gas are chemically reactive with the propellant, the reaction products may become a component of the gas.
7. Pressurizing gas turbulence.-The heat exchange between pressurizing gas and propellant would be extremely large if the gas were permitted to agitate the liquid propellant surface. This effect can be prevented through the use of a diffuser which spreads the gas in a gentle flow toward the top and sides of the tank.

Design Calculations of Pressurant Requirements

If the system operating duration is relatively short, or if the pressurant temperature is close to or lower than the propellant temperature, heatand mass-transfer effects can be neglected. The required pressurant weight can then be calculated by the perfect gas law:

$$\begin{equation*} W_{g}=\frac{P_{T} V_{T}}{R_{g} T_{g}} \tag{5-1} \end{equation*}$$

where $W_{g}=$ required pressurant weight in the tank, lb $P_{T}=$ propellant tank pressure, $\mathrm{lb} / \mathrm{ft}^{2}$ $V_{T}=$ total volume of the empty propellant tank, $\mathrm{ft}^{3}$ $R_{g}=$ gas constant of the pressurant, $\mathrm{ft}-\mathrm{lb} / \mathrm{lb}- \operatorname{deg} \mathrm{R}$ $T_{g}=$ mean temperature of entering pressurant, ${ }^{\circ} \mathrm{R}$ However, in cases where longer systems duration and higher pressurant temperatures are involved, the pressurant requirement can best be determined by the following procedure, keeping in mind the limitations set forth at the beginning of section 5.1.

Considering first a single-start operation (not requiring coast periods and restarts), and neglecting heat transfer from the tank walls, the total heat transferred from the pressurant gas to the vaporized propellant can be approximated by equation (5-2).

$$\begin{equation*} Q=H A t\left(T_{u}-T_{e}\right) \tag{5-2} \end{equation*}$$

where $Q=$ total heat transferred, Btu $H=$ experimentally determined heat transfer coefficient at the gas-liquid interface, $\mathrm{Btu} / \mathrm{sec}-\mathrm{ft}^{2}-\mathrm{deg} \mathrm{R}$ $A=$ area of the gas-liquid interface (in case of a cylindrical tank, the cross-section area of the tank), $\mathrm{ft}^{2}$ t =operating duration, sec $T_{u}=$ temperature of the gases at burnout, ${ }^{\circ} \mathrm{R}$ $T_{e}=$ temperature of the propellant, ${ }^{\circ} \mathrm{R}$ Both $T_{u}$ and $T_{e}$ are treated as constant values at the interface between liquid and gas.

This heat, $Q$, is assumed to have heated and vaporized the propellant according to the equation

$$\begin{equation*} Q=W_{V}\left[C_{p l}\left(T_{V}-T_{\mathrm{e}}\right)+h_{V}+C_{p V}\left(T_{\mathrm{u}}-T_{v}\right)\right] \tag{5-3} \end{equation*}$$

where $W_{V}=$ total weight of vaporized propellant, lb $C_{p l}=$ specific heat of the liquid propellant, Btu/lb-deg R $h_{V}=$ heat of vaporization of the propellant, Btu/lb $C_{p v}=$ specific heat of the propellant vapor, Btu/lb-deg R $T_{V}=$ vaporization temperature of the liquid propellant, ${ }^{\circ} \mathrm{R}$ The value of $W_{V}$ can now be obtained from equations (5-2) and (5-3) with an assumed value for $T_{u}$.

The partial volume occupied by the vaporized propellant is given by

$$\begin{equation*} V_{v}=\frac{W_{V} Z R_{p} T_{u}}{P_{T}} \tag{5-4} \end{equation*}$$

where $V_{V}=$ total volume occupied by the vaporized propellant, $\mathrm{ft}^{3}$ $Z=$ compressibility factor evaluated at the total pressure ( $P_{T}$ ) and the temperature ( $T_{u}$ ) of the gaseous mixture at burnout $R_{p}=$ gas constant of the propellant vapor, $\mathrm{ft}-\mathrm{lb} /$ lb-deg R The remaining tank volume at burnout, neglecting residual propellants, can be assumed to be occupied by the pressurant gas

$$\begin{equation*} V_{g}=V_{T}-V_{V} \tag{5-5} \end{equation*}$$

where $V_{g}=$ volume of pressurant gas at burnout,

$$\mathrm{ft}^{3}$$

The weight of pressurant is calculated by the perfect gas law

$$\begin{equation*} W_{g}=\frac{P_{T} V_{g}}{R_{g} T_{u}} \tag{5-6} \end{equation*}$$

In order to maintain the heat balance, the value for $Q$ should satisfy the following equation:

$$\begin{equation*} Q=W_{g} C_{p g}\left(T_{g}-T_{u}\right) \tag{5-7} \end{equation*}$$

where $C_{p g}=$ specific heat at constant pressure of pressurant gas, Btu/lb-deg R From equation (5-7) the required value of $T_{g}$ for the assumed $T_{u}$ can thus be calculated. If however, $T_{g}$ is a predetermined fixed value, then the values of $W_{g}, W_{v}$, and $T_{u}$ must satisfy the following as well as other correlated equations:

$$\begin{align*} & W_{g} C_{p g}\left(T_{g}-T_{u}\right) \\ & \quad=W_{v}\left[C_{p l}\left(T_{v}-T_{e}\right)+h_{v}+C_{p v}\left(T_{u}-T_{v}\right)\right] \tag{5-8} \end{align*}$$

Thus far, heat transfer from the tank walls has been neglected. However, if there is a considerable temperature differential between pressurizing gases, propellant, and tank walls, the total heat transferred between them during the mission must be taken into consideration for the determination of vaporized propellant at burnout.

Equation (5-3) can be rewritten as

$$\begin{align*} Q \pm & Q W_{1} \\ & =W_{v}\left[C_{p l}\left(T_{v}-T_{\mathrm{e}}\right)+h_{v}+C_{p v}\left(T_{u}-T_{v}\right)\right] \tag{5-9} \end{align*}$$

where $Q_{W_{1}}=$ total heat transferred between tank walls and liquid and gaseous propellant during the mission, Btu. The positive ( + ) or negative ( - ) sign indicates whether $Q_{W_{1}}$ is contributed by, or lost to, the tank walls. Furthermore, equation (5-7) becomes

$$\begin{equation*} Q=W_{g} C_{p g}\left(T_{g}-T_{u}\right) \pm Q_{W_{2}} \tag{5-10} \end{equation*}$$

where $Q_{W_{2}}=$ total heat transferred between pressurizing gases and tank walls during a mission, Btu. Again the positive ( + ) or negative ( - ) sign indicates whether heat is contributed by, or lost to, the tank walls. Combining equations (5-9) and (5-10), the heat balance considering heat transfer from the tank walls can be written as

$$\begin{array}{r} {\left[W_{g} C_{p g}\left(T_{g}-T_{u}\right)\right] \pm Q_{W_{2}}=W_{v}\left[C_{p l}\left(T_{v}-T_{e}\right)\right.} \\ \left.+h_{v}+C_{p v}\left(T_{u}-T_{v}\right)\right]-\left( \pm Q W_{1}\right) \tag{5-11} \end{array}$$

If the vehicle mission includes several powered flight and coasting periods, the calculation of the heat transfer across the gas-liquid interface should take the total mission time into consideration. Equation (5-2) can be rewritten as

$$\begin{equation*} Q=H A t_{m}\left(T_{m}-T_{e}\right) \tag{5-12} \end{equation*}$$

where $t_{m}=$ total mission time including powered flight and coasting period, sec $T_{m}=$ mean temperature of the pressurizing gases during the entire mission, ${ }^{\circ} \mathrm{R}$. This is a function of many factors such as length of coasting periods, heat transfer between gases and tank wall, etc. Other effects such as vapor condensation, solubility of the pressurizing gas in the propellant, and chemical reactions of the pressurizing gas with the propellant can be included, based on experimental data. However, no set of equations can be applied directly.

In some cases, the uncertainties in pressurizing system design can be reduced by providing adjustability of the pressurant temperature at the propellant tank inlet. In this approach, the temperature of the pressurizing gas at system burnout is assumed or targeted from the beginning. Based on this and other given or assumed data, the values of required pressurant quantity and inlet temperature can be calculated by equations (5-1) through (5-11). Certain correction factors such as pressurant solubility, etc., can be applied later. If the required pressurant quantity in experimental engine system evaluation deviates from the calculations, because of temperature discrepancies of the pressurizing gas at system burnout, an adjustment of the pressurant temperature at the propellant tank inlet can often be made to correct for this difference, such as through an adjustment of the pressurant supply from a heat exchanger, or from a gas generator. Because of the narrow safety margins employed in rocket vehicle design, the effect of varied gas temperatures on structural members must be carefully weighed.

The following is a sample calculation to demonstrate this design approach:

Sample Calculation 5-1

From table $3-5$, the following data are obtained for the oxidizer tank of the A-4 stage propulsion system:

Oxidizer, $\mathrm{N}_{2} \mathrm{O}_{4}$ Pressurant, gaseous He Tank volume, neglecting the volume of residual propellant, $\left(V_{T}\right)=119 \mathrm{ft}^{3}$ Average tank cross-section area, $(A)=20 \mathrm{ft}^{2}$ Tank pressure, $\left(P_{T}\right)=165 \mathrm{psia}$, or 23760 psfa Propellant temperature, $\left(T_{e}\right)=520^{\circ} \mathrm{R}$ Calculate the following: (a) The total pressurant weight ( $W_{g}$ ) and required temperature ( $T_{g}$ ) at the tank iniet, for a single operation time ( $t$ ) of 500 seconds, with an experimentally determined heat transfer coefficient ( $H$ ) at the gas-liquid interface of 0.002 $\mathrm{Btu} / \mathrm{sec}-\mathrm{ft}^{2}-\mathrm{deg} \mathrm{R}$. It is assumed the ullage gas temperature $T_{u}$ at burnout is $700^{\circ} \mathrm{R}$ and that there is no heat transferred at the tank wall surfaces. (b) The required pressurant weight ( $W_{g}$ ) and temperature ( $T_{g}$ ) at the tank inlet, for a mission consisting of several powered flight and coasting periods, with a total mission duration ( $t_{m}$ ) of 18000 seconds. The mean temperature of the pressurizing gases during the mission ( $T_{m}$ ) is $526^{\circ} \mathrm{R}$. The total heat transferred between propellant and tank walls ( $Q W_{1}$ ) is -2000 Btu. The total heat transferred between pressurizing gas and tank walls ( $Q_{W_{2}}$ ) is -600 Btu . The temperature ( $T_{u}$ ) of the ullage gases at final burnout is $660^{\circ} \mathrm{R}$.

Solution

(a) From standard propellant references, the following data are obtained for $\mathrm{N}_{2} \mathrm{O}_{4}$ at a pressure of 165 psia:

Vaporization temperature, $T_{V}=642^{\circ} \mathrm{R}$ Heat of vaporization, $h_{v}=178 \mathrm{Btu} / \mathrm{lb}$ Mean value of specific heat in liquid state, $C_{p l}=0.42 \mathrm{Btu} / \mathrm{lb} \operatorname{deg} \mathrm{F}$ Mean value of specific heat in vapor state, $C_{p v}=0.18 \mathrm{Btu} / \mathrm{lb}-\operatorname{deg} \mathrm{F}$ Compressibility factor, $Z=0.95$ Molecular weight $=92$ The specific heat of helium, $C_{p g}$, is 1.25 $\mathrm{Btu} / \mathrm{lb}-\operatorname{deg} \mathrm{F}$, and its molecular weight is 4 .

From equation (5-2), total heat transferred at the gas-liquid interface:

$$\begin{aligned} & Q=H A t\left(T_{u}-T_{e}\right) \\ & \quad=0.002 \times 20 \times 500(700-520)=3600 \mathrm{Btu} \end{aligned}$$

Substitute this into equation (5-3):

$$\begin{aligned} 3600 & =W_{V}\left[C_{p I}\left(T_{V}-T_{\mathrm{e}}\right)+h_{V}+C_{p V}\left(T_{u}-T_{v}\right)\right] \\ & =W_{v}[0.42(642-520)+178+0.18(700-642)] \\ & =W_{v} \times 239.6 \end{aligned}$$

Total weight of vaporized propellant, $W_{V}=15.1 \mathrm{lb}$ Substitute into equation (5-4), to obtain the volume occupied by the vaporized propellant:

$$\begin{aligned} V_{V} & =\frac{W_{V} Z R_{p} T_{u}}{P_{T}} \\ & =\frac{15.1 \times 0.95 \times \frac{1544}{92} \times 700}{23760} \\ & =7.13 \mathrm{ft}^{3} \end{aligned}$$

Substitute this into equation (5-5), to obtain the volume occupied by pressurant gas

$$V_{g}=V_{T}-V_{V}=119-7.13=111.87 \mathrm{ft}^{3}$$

From equation (5-6), the required pressurant weight results:

$$\begin{aligned} W_{g} & =\frac{P_{T} V_{g}}{R_{g} T_{u}} \\ & =\frac{23760 \times 111.87}{\left(\frac{1544}{4}\right) \times 700} \\ & =9.79 \mathrm{lb} \end{aligned}$$

Substitute results into equation (5-7),

$$\begin{aligned} Q & =9.79 C_{p g}\left(T_{g}-T_{u}\right) \\ 3600 & =9.79 \times 1.25\left(T_{g}-T_{u}\right) \end{aligned}$$

The required pressurant temperature at tank inlet is:

$$T_{g}=\frac{3600}{9.79 \times 1.25}+700=995^{\circ} \mathrm{R}$$

(b) From equation (5-12), the total heat transferred at the gas-liquid interface:

$$\begin{aligned} Q & =H A t_{m}\left(T_{m}-T_{e}\right) \\ & =0.002 \times 20 \times 18000(526-520) \\ & =4320 \mathrm{Btu} \end{aligned}$$

Substitute into equation (5-9):

$$\begin{aligned} 4320-2000 & =W_{V}[0.42(642-520) \\ \quad+178+0.18(660-642)] & \\ W_{V} & =10.0 \mathrm{lb} \end{aligned}$$

Substitute into equation (5-4):

$$\begin{aligned} V_{v} & =\frac{10.0 \times 0.95 \times\left(\frac{1544}{92}\right) \times 660}{23760} \\ & =4.45 \mathrm{ft}^{3} \end{aligned}$$

Substitute into equation (5-5):

$$V_{g}=119-4.45=114.5 \mathrm{ft}^{3}$$

Substitute into equation (5-6):

$$\begin{aligned} W_{g} & =\frac{23760 \times 110.76}{\left(\frac{1544}{4}\right) \times 660} \\ & =10.65 \mathrm{lb} \end{aligned}$$

Substitute into equation (5-10):

$$\begin{aligned} 4320 & =10.65 \times 1.25\left(T_{g}-660\right)-600 \\ T_{g} & =\frac{4320+600}{10.65 \times 1.25}+660=1030^{\circ} \mathrm{R} \end{aligned}$$