8.3 STRUCTURAL DESIGN OF PROPELLANT TANKS

As a rule, the wall thickness of propellant tanks is first calculated from stresses caused by internal pressure loads and discontinuities. Then the design is checked for other loads. If the wall thickness of a pressure vessel is small compared to the radii of wall curvature $(t / r \leq 1 / 15)$, and offers no resistance to bending, the wall is subjected only to direct or hoop-membrane stresses, assumed to be uniformly distributed over the thickness. However, if any discontinuity is present along the wall, such as an abrupt change in radius of curvature or wall thickness, discontinuity and bending stresses are added. At a sufficient distance from the juncture between tank ends (sphere or ellipsoid) and cylindrical shell (where interaction does not occur), the maximum stress in the tank wall due to internal pressure is calculated using the hoopmembrane stress formula only.

Figure 8-6 identifies the major tank elements. In an optimum tank design, the wall thickness varies according to a combination of local membrane, bending, and discontinuity stresses. This is especially true for the spherical and ellipsoidal tank ends. The structural calculation methods for volume, wall thickness, wall surface area, and weight of various tank shapes will now be discussed. The following general terminology is used:

$$\begin{aligned} p_{t}= & \text { maximum tank working pressure, psig } \\ S_{W}= & \begin{array}{r} \text { maximum allowable working stress of the } \\ \quad \text { tank construction material, psi } \end{array} \\ \rho= & \text { density of the tank construction material, } \\ & \quad \mathrm{lb} / \mathrm{in}^{3} \\ E= & \text { modulus of elasticity, psi } \\ V= & \text { Poisson's ratio } \\ e_{W}= & \text { weld efficiency } \end{aligned}$$

Spherical Tanks

(1) Volume, $i n^{3}$ :

$$\begin{equation*} V_{s}=\frac{4 \pi \mathrm{a}^{3}}{3} \tag{8-10} \end{equation*}$$

Figure 8-6.-Nomenclature of principal tank elements.

where $\mathbf{a}=$ nominal radius of the tank, in (2) Wall thickness, in, required to withstand membrane stresses from internal tank pressure:

$$\begin{equation*} t_{S}=\frac{p_{t} a}{2 S_{w} e_{w}} \tag{8-11} \end{equation*}$$

(3) Wall surface area, in ${ }^{2}$ :

$$\begin{equation*} A_{S}=4 \pi a^{2} \tag{8-12} \end{equation*}$$

(4) Weight, Ib:

$$\begin{equation*} W_{S}=4 \pi a^{2} t_{S \rho} \tag{8-13} \end{equation*}$$

(5) Critical pressure due to external loading, psi . When the external pressure is higher than the internal tank pressure, the pressure differential across the tank wall may cause the tank to buckle.

$$\begin{equation*} p_{\mathrm{cr}}=\frac{2 E t_{\mathrm{s}}{ }^{2}}{\mathrm{a}^{2}} \sqrt{3\left(1-v^{2}\right)} \tag{8-14} \end{equation*}$$

Ellipsoidal and Spherical Tank Ends (fig. 8-6)

Note that the spherical end is a special case of ellipsoidal end, in which the major halfdiameter, a, equals the minor half-diameter, $b$. (1) Volume:

Ellipsoidal tank end volume, in ${ }^{3}$ :

$$\begin{equation*} V_{e}=\frac{2 \pi a^{2} b}{3} \tag{8-15} \end{equation*}$$

Spherical tank end volume, in ${ }^{3}$ :

$$\begin{equation*} V_{s}=\frac{2 \pi a^{3}}{3} \tag{8-16} \end{equation*}$$

where $a=$ elliptical tank end major half-diameter, in = radius of the cylindrical tank section $b=$ elliptical tank end minor half-diameter, in (2) Wall thicknesses considering combined membrane, discontinuity, and local bending stresses caused by internal tank pressure $p_{t}$. An equivalent wall thickness, which is an average value of knuckle-and-crown thickness, may be used to calculate the weight of the tank ends.

$$\begin{gather*} t_{k}=\frac{K p_{t} a}{S_{w} e_{w}} \tag{8-17}\\ t_{\mathrm{cr}}=\frac{p_{t} R}{2 S_{w} e_{w}} \tag{8-18}\\ t_{\mathrm{e}}=\frac{\left(t_{k}+t_{\mathrm{cr}}\right)}{2}=\frac{p_{t} a\left(K+\frac{k}{2}\right)}{2 S_{w}} \tag{8-19}\\ t_{\mathrm{s}}=\frac{p_{t} a\left(K+\frac{1}{2}\right)}{2 S_{w}} \tag{8-20}\\ \frac{t_{e}}{t_{s}}=\frac{\left(K+\frac{k}{2}\right)}{\left(K+\frac{1}{2}\right)} \tag{8-21}\\ \frac{t_{\mathrm{e}}}{t_{\mathrm{c}}}=\frac{\left(K+\frac{k}{2}\right)}{2} \tag{8-22} \end{gather*}$$

where $k=$ tank end ellipse ratio $=a / b ; k=1$ for a spherical end $R=$ tank end crown radius, in $=k a ; R=a$ for a spherical end $K=$ stress factor, a function of the ellipse ratio $k$. Figure 8.7 presents a $K$ versus $k$ curve for combined membrane, discontinuity, and local bending stresses $t_{k}=$ wall thickness at the knuckle, in $t_{\mathrm{cr}}=$ wall thickness at the crown, in $t_{\mathrm{e}}=$ equivalent wall thickness of an ellipsoidal tank end, in $t_{\mathrm{s}}=$ equivalent wall thickness of a spherical tank end, in $t_{c}=$ wall thickness of a cylindrical tank section, in (3) Wall surface area:

Ellipsoidal tank end surface area, in ${ }^{2}$ :

$$\begin{equation*} A_{e}=a^{2}+\frac{\pi b^{2} \ln \left[\frac{(1+e)}{(1-e)}\right]}{2 \mathrm{e}} \tag{8-23} \end{equation*}$$

Spherical tank end surface area, in ${ }^{2}$ :

$$\begin{equation*} A_{S}=2 \pi \mathrm{a}^{2} \tag{8-24} \end{equation*}$$

Figure 8-7.-Ellipse ratio $k$ versus knuckle stress factor $K$, compression stress $-K$, and parameter $E^{\prime}$. (From ARS paper, "Design Criteria and Analyses for Thin-Walled Pressurized Vessels and Interstage Structures," by T. J. Hart.)

where

$$e=\text { eccentricity }=\frac{\sqrt{a^{2}-b^{2}}}{a}=\sqrt{1-\frac{1}{k^{2}}}$$

(4) Weight:

Ellipsoidal tank end weight, lb:

$$\begin{equation*} W_{e}=\frac{\pi a^{2} t_{e} E^{\prime} \rho}{2 k} \tag{8-25} \end{equation*}$$

Spherical tank end weight, lb:

$$\begin{equation*} W_{S}=2 \pi a^{2} t_{S} \rho \tag{8-26} \end{equation*}$$

where

$$E^{\prime}=\text { design factor }=2 k+\frac{1}{\sqrt{k^{2}-1}} \ln \frac{k+\sqrt{k^{2}-1}}{k-\sqrt{k^{2}-1}}$$

(see fig. 8-7) (5) Critical pressure due to external loading, psi: For an ellipsoidal tank end, it may be approximated as

$$\begin{equation*} p_{\text {cre }}=\frac{C_{b} 2 E t_{e}^{2}}{R^{2}} \tag{8-27} \end{equation*}$$

For a spherical tank end

$$\begin{equation*} p_{\mathrm{crs}}=\frac{0.342 E t_{\mathrm{s}}^{2}}{\mathrm{a}^{2}}(\text { approximately }) \tag{8-28} \end{equation*}$$

where $C_{b}=$ buckling coefficient, a function of $R / t_{e}$. ranging from 0.05 to 0.10

Cylindrical Tank Section

(1) Volume, in ${ }^{3}$ :

$$\begin{equation*} V_{c}=\pi a^{2} l_{c} \tag{8-29} \end{equation*}$$

where a = radius, in $I_{c}=$ length, in (2) Wall thickness, in, required to withstand membrane stresses due to internal tank pressure:

$$\begin{equation*} t_{c}=\frac{p_{t} \mathrm{a}}{S_{W} e_{W}} \tag{8-30} \end{equation*}$$

(3) Discontinuity stresses.-The discontinuity at the juncture between the cylindrical tank section and the tank ends will cause bending and shear loads along the cylindrical circumference at the juncture, and the adjacent areas. These discontinuity stresses are superimposed upon the membrane stresses and comprise: (a) axial bending stress; (b) hoop bending stress; (c) additional hoop stress due to the shear load; (d) shear stress. Discontinuity stresses fade out rapidly, so that they become negligibly small a short distance from the juncture. Detail analyses of discontinuity stresses can be found in standard textbooks. In general, buildup of wall thickness of less than $0.5 t_{c}$, near the juncture, will suffice for most designs, with only small weight penalty. (4) Wall surface area, in $^{2}$ :

$$\begin{equation*} A_{c}=2 \pi a l_{c} \tag{8-31} \end{equation*}$$

(5) Weight, lb:

$$\begin{equation*} W_{c}=2 \pi a l_{c} t_{c} \rho \tag{S-32} \end{equation*}$$

(6) External loading critical pressure, psi :

For a short tank (i.e., $I_{c}<4.9 a \sqrt{a / t_{c}}$ ):

$$\begin{equation*} p_{\text {crc }}=0.807 \frac{E t_{c}{ }^{2}}{l_{c} a} \sqrt[4]{\left(\frac{1}{1-v^{2}}\right)^{3} \frac{t_{c}{ }^{2}}{a^{2}}} \tag{8-33} \end{equation*}$$

For long tanks (i.e., $l_{c} \geqq 4.9 a \sqrt{a / t_{c}}$ ):

$$\begin{equation*} p_{\text {crc }}=\frac{E t_{c}^{3}}{4\left(1-v^{2}\right) a^{3}} \tag{8-34} \end{equation*}$$

where $E=$ modulus of elasticity.

Sample Calculation (8-3)

The following design data are specified for the A-4 stage propulsion system, which employs a cylindrical propellant tank section with ellipsoidal ends (preliminary layout shown in fig. 3-10).

Required design volume of the oxidizer tank,

$$V_{\mathrm{to}}=120 \mathrm{ft}^{3}$$

Maximum oxidizer tank working pressure,

$$p_{\mathrm{to}}=180 \mathrm{psia}$$

Required design volume of the fuel tank,

$$V_{\mathrm{tf}}=143.5 \mathrm{ft}^{3}$$

Maximum fuel tank working pressure, $p_{\text {tf }}=170$ psia Internal radius of the cylindrical tank section,

$$a=41 \mathrm{in}$$

Tank construction material, aluminum alloy

$$\text { 6066-T6: } F_{y}=50000 \mathrm{psi} ; F_{u}=57000 \mathrm{psi} ;$$ $$\rho=0.101 \mathrm{lb} / \mathrm{in}^{3} ; \quad E=10.4 \times 10^{6} \mathrm{psi} ; \quad v=0.36$$

Weld efficiency, $e_{W}=100$ percent. Determine the following: (a) Required internal tank dimensions (b) Required thickness of the tank walls at various sections, considering internal pressure loads, discontinuity, and local bending stresses (c) Approximate weight of the tankage (d) Critical external loading pressures, using a buckling coefficient, $C_{b}=0.10$ for the tank ends

Solution

(a) Since the oxidizer tank consists of two ellipsoidal ends without a cylindrical section, equation (8-15) may be applied:

$$V_{\text {to }}=\frac{2 \times 2 \pi \mathrm{a}^{2} b}{3}$$

We rearrange to obtain the minor elliptical half-diameter of the tank ends:

$$b=\frac{3 \times V_{\mathrm{to}}}{2 \times 2 \pi \mathrm{a}^{2}}=\frac{3 \times 120 \times 1728}{4 \times \pi \times(41)^{2}}=29.4 \mathrm{in}$$

The tank end ellipse ratio:

$$k=\frac{a}{b}=\frac{41}{29.4}=1.395$$

Since we use an ellipsoidal end of the same proportion at the fuel tank top, the fuel tank volume may be treated as the volume of a cylindrical tank section with the length $l_{c}$. From equation (8-29), the volume of the fuel tank:

$$\begin{aligned} V_{\mathrm{tf}} & =\pi \mathrm{a}^{2} l_{\mathrm{c}} \\ l_{\mathrm{c}} & =\frac{V_{\mathrm{tf}}}{\pi \mathrm{a}^{2}}=\frac{143.5 \times 1728}{\pi \times(41)^{2}}=46.9 \mathrm{in} \end{aligned}$$

To summarize the internal dimensions of the tankage:

$$\mathrm{a}=41 \mathrm{in}, \quad b=29.4 \mathrm{in}, \quad k=1.395, l_{\mathrm{c}}=46.9 \mathrm{in}$$

(b) We assume that certain missions of the A-4 vehicle require it to be man rated. From equations (8-6) and (8-7), the maximum allowable working stresses are derived:

$$\begin{aligned} & S_{w}=\frac{F_{y}}{1.33}=\frac{50000}{1.33}=37600 \mathrm{psi} \\ & S_{w}=\frac{F_{u}}{1.65}=\frac{57000}{1.65}=34560 \mathrm{psi} \end{aligned}$$

We use the lower value of 34560 psi . From figure 8-6, the tank end stress factor $K$ of the combined stresses is 0.80 for an ellipse ratio $k$ of 1.395 . From equation (8-17), the re- quired wall thickness at the knucklo of the oxidizer tank end

$$t_{\mathrm{ko}}=\frac{K p_{t} \mathrm{a}}{S_{w}}=\frac{0.80 \times 180 \times 41}{34560}=0.171 \mathrm{in}$$

From equation (8-18), the required wall thickness at the crown of the oxidizer tank end

$$t_{\mathrm{cro}}=\frac{p_{t} R}{2 S_{w}}=\frac{180 \times 1.395 \times 41}{2 \times 34560}=0.149 \mathrm{in}$$

From equation (8-19), the equivalent wall thickness of the oxidizer tank end

$$t_{\mathrm{eo}}=\frac{\left(t_{\mathrm{kO}}+t_{\mathrm{crO}}\right)}{2}=\frac{(0.171+0149)}{2}=0.16 \mathrm{in}$$

NOTE: In some designs, weight has been saved by taking advantage of the fact that the bulkhead common to both tanks is subject to a relatively small differential pressure in operation. Such systems, however, require more elaborate pressurization and loading systems including interlocks. In case of malfunction, the coumon bulkhead may suffer serious damage.

The required wall thickness at the knuckle of the fuel tank end:

$$t_{\mathrm{kf}}=\frac{t_{\mathrm{ko}} p_{\mathrm{tf}}}{p_{\mathrm{to}}}=\frac{0.171 \times 170}{180}=0.162 \mathrm{in}$$

The required wall thickness at the crown of the fuel tank end:

$$t_{\mathrm{crf}}=\frac{t_{\mathrm{cro}} \mathrm{p}_{\mathrm{tf}}}{\mathrm{p}_{\mathrm{to}}}=\frac{0.149 \times 170}{180}=0.141 \mathrm{in}$$

The equivalent wall thickness of the fuel tank end:

$$t_{\mathrm{ef}}=\frac{\left(t_{\mathrm{kf}}+t_{\mathrm{crf}}\right)}{2}=\frac{(0.162+0.141)}{2}=0.152$$

From equation (8-30), the required wall thickness of the cylindrical tank section:

$$t_{c}=\frac{p_{\mathrm{tfa}}}{S_{W}}=\frac{170 \times 41}{34560}=0.202 \mathrm{in}$$

Provide a buildup of $0.4 t_{c}$ on the cylindrical tank section wall near the juncture to allow for discontinuity stresses: $t_{\mathrm{cj}}=t_{c}+0.4 t=0.202+0.4 \times 0.202=0.283 \mathrm{in}$ To summarize: $t_{\mathrm{ko}}=0.171 \mathrm{in}, t_{\mathrm{cro}}=0.149 \mathrm{in}, t_{\mathrm{eo}}=0.16 \mathrm{in} t_{\mathrm{kf}}=0.162 \mathrm{in}, t_{\mathrm{crf}}=0.141, t_{\mathrm{ef}}=0.152 \mathrm{in} t_{c}=0.202 \mathrm{in}, t_{\mathrm{cj}}=0.283 \mathrm{in}$ (c) From equation (8-25), the weight of the oxidizer tank end:

$$W_{\mathrm{eo}}=\frac{\pi \mathrm{a}^{2} t_{\mathrm{eo}} E^{\prime} \rho}{2 k}$$

From figure 8-7, $E^{\prime}$ is 4.56 , for $k=1.395$ :

$$W_{\mathrm{eo}}=\frac{\pi(41)^{2} \times 0.16 \times 4.56 \times 0.101}{2 \times 1.395}=139.5 \mathrm{lb}$$

The weight of the fuel tank end:

$$W_{\mathrm{ef}}=\frac{W_{\mathrm{eo}} t_{\mathrm{ef}}}{t_{\mathrm{eo}}}=\frac{139.5 \times 0.152}{0.16}=132.8 \mathrm{lb}$$

From equation (8-32), the weight of the cylindrical tank section: $W_{c}=2 \pi a l_{c} t_{c} \rho=2 \times 41 \times 46.9 \times 0.202 \times 0.101=246.4 \mathrm{lb}$ Add 4 percent of overall tankage weight for local wall thickness buildups, to allow for welded joints, discontinuity stresses, etc., and for tolerances during fabrication.

Approximate overall weight of the tankage (less accessories):

$$\begin{aligned} W_{t} & =1.04\left(2 \times W_{\mathrm{eO}}+W_{\mathrm{ef}}+W_{C}\right) \\ & =1.04(2 \times 139.5+132.8+246.4)=685 \mathrm{lb} \end{aligned}$$

(d) From equation (8-27), the critical external loading pressure for the oxidizer ends

$$\begin{aligned} p_{\text {creo }} & =\frac{C_{b} 2 E t_{\mathrm{eo}}{ }^{2}}{R^{2}}=\frac{0.10 \times 2 \times 10.4 \times 10^{6} \times(0.16)^{2}}{(1.395 \times 41)^{2}} \\ & =16.3 \mathrm{psi} \end{aligned}$$

The critical external loading pressure for the fuel tank end:

$$\begin{aligned} p_{\text {cref }} & =\frac{C_{b} 2 E t_{\mathrm{ef}}{ }^{2}}{R^{2}}=\frac{0.10 \times 2 \times 10.4 \times 10^{6} \times(0.152)^{2}}{(1.395 \times 41)^{2}} \\ & =14.7 \mathrm{psi} \end{aligned}$$

We find that $l_{c}<4.9 a \sqrt{a / t_{c}}$. Thus equation (8-33) applies, yielding an external loading critical pressure for the cylindrical tank section

$$\begin{aligned} p_{\mathrm{crc}} & =\frac{0.807 \times 10.4 \times 10^{6} \times(0.202)^{2}}{46.9 \times 41} \\ & \quad \times \sqrt[4]{\left(\frac{1}{1-(0.36)^{2}}\right)^{3}\left(\frac{0.202}{41}\right)^{2}} \\ & =13.9 \mathrm{psi} \end{aligned}$$

Axial Compressive Loading on the Cylindrical Tank Section

In integrated propellant tank designs (figs. 8-2 and 8-3), the cylindrical tank section must withstand large axial compressive loads during vehicle handling and operation. If the tank is not pressurized, i.e., if tank pressure = anibient pressure, the critical axial compressive stress for an unstiffened cylindrical tank may be calculated as

$$\begin{equation*} S_{c}=\left[9\left(\frac{t_{c}}{a}\right)^{1.6} \div 0.16\left(\frac{t_{c}}{l_{c}}\right)^{1.3}\right] E \tag{8-35} \end{equation*}$$

where $S_{c}=$ critical axial compressive stress, psi. This is the axial compressive stress that will cause the tank to buckle

One method of increasing the axial loadcarrying ability of a cylindrical tank section with minimum weight penalty is to pressurize the tank. This is known as pressure stabilization. Internal pressure will raise the critical buckling stress of a tank; or it may be used to counterbalance an axial compressive load $F_{\mathrm{a}}, \mathrm{lb}$, where

$$\begin{equation*} F_{\mathrm{a}}=\pi \mathrm{a}^{2} p_{\mathrm{t}} \tag{8-36} \end{equation*}$$

Pressurization will also reduce tank failures from very large bending loads. However, if the pressure is ever permitted to drop below a value necessary to carry the axial and bending loads, the tank will collapse and the vehicle will probably be damaged beyond repair.

An alternative method of increasing the external load-carrying ability of a cylindrical tank is to make it self-supporting. This involves stiffening the cylindrical skin by means of longitudinal and circumferential members, or honeycomb structures. The members may be either separate stiffeners welded to the tank wall, or may be made integral with the wall by machining or chemically milling a thicker sheet.

Sample Calculation (8-4)

For the A-4 stage tankage, calculate: (a) Critical axial compressive load of the cylindrical tank section with no internal pressure. (b) Required internal tank pressure to offset an axial compressive load of 100000 pounds with no compressive stress on the cylindrical tank section.

Solution

(a) From equation (8-35), the critical compressive stress of the cylindrical tank section:

$$\begin{aligned} S_{c} & =\left[9\left(\frac{0.202}{41}\right)^{1.6}+0.16\left(\frac{0.202}{46.9}\right)^{1.3}\right] \times 10.4 \times 10^{5} \\ & =20360 \mathrm{psi} \end{aligned}$$

The critical axial compressive load of the cylindrical tank section:

$$\begin{aligned} F_{c} & =S_{c} \times 2 \pi a t_{c}=20360 \times 2 \pi \times 41 \times 0.202 \\ & =1060000 \mathrm{lb} \end{aligned}$$

From the results it is obvious that the A-4 stage tankage is capable of withstanding a substantial axial compressive load without internal pressurization. (b) From equation (8-36), the required internal tank pressure:

$$p_{t}=\frac{F_{a}}{\pi a^{2}}=\frac{100000}{\pi \times(41)^{2}}=18.95 \mathrm{psi}$$

Water Hammer Effects Due to Impact

When a loaded propellant tank is subject to an impact force, a water-hammer effect occurs within the tank. This effect produces a surge of the tank internal pressure. For very short impacting times (less than $1.2 \times 10^{-3} \mathrm{sec}$ ), the following correlations are established for cylindrically shaped propellant tanks:

$$\begin{equation*} c=\frac{p_{\mathrm{s}}=\frac{c \dot{w}}{\pi a^{2} g}}{\sqrt{1+\left[\frac{2.5 E_{\mathrm{p}} a(1-0.8 v)}{E t_{c}}\right]}} \tag{8-37} \end{equation*}$$

where $p_{\mathbf{S}}=$ pressure surge due to the impact, psi $\dot{w}=$ equivalent flow rate of the propellant due to the impact, $\mathrm{lb} / \mathrm{sec}$ a = radius of the cylindrical tank, in $t_{c}=$ wall thickness of the cylindrical tank, in $g=$ gravitational constant, $32.2 \mathrm{ft} / \mathrm{sec}^{2}$ $c=$ acoustic velocity of the restrained propellant, in/sec $c^{\prime}=$ free acoustic velocity of the propellant, in/sec $E_{p}=$ compressive modulus of elasticity of the propellant, psi $E=$ modulus of elasticity of the tank construction material, psi $v=$ Poisson's ratio of the tank construction material In many prepackaged liquid applications, the propellant tankage is required to withstand certain impact loads, as specified by the height of the drop tests. The details of estimating tank pressure surges from a free-fall impact are illustrated by sample calculation (8-5).

Sample Calculation (8-5)

The following data are given for the cylindrical fuel tank of a prepackaged storable liquid propulsion system:

Fuel, $\mathrm{N}_{2} \mathrm{H}_{4}$ Fuel density, $\rho_{p}=63.17 \mathrm{lb} / \mathrm{ft}^{3}$ Compressive modulus of elasticity of the fuel, $E_{p}=6.06 \times 10^{5}$ Free acoustic velocity of the fuel, $c^{\prime}=80100$ in/sec Tank construction material, aluminum alloy 6066-T6 Modulus of elasticity of the tank construction material, $E=10.4 \times 10^{6}$

Poisson's ratio of the tank construction material, $v=0.36$ Radius of the cylindrical tank, $a=4$ in Length of the cylindrical tank, $l_{c}=50 \mathrm{in}$ Wall thickness of the cylindrical tank, $t_{c} =0.167 \mathrm{in}$ For a tank falling in direction of its longitudinal axis, estimate: (a) Tank pressure surge due to the impact after a 6-foot free drop (b) Tank pressure surge due to the impact after a 20 -foot free drop

Solution

(a) For a 6 -foot free drop the final velocity at impact:

$$V=\sqrt{2 g h}=\sqrt{2 \times 32.2 \times 6}=19.65 \mathrm{fps}$$

This yields an equivalent propellant flow rate due to impact of $\dot{W}=\rho_{p} \pi a^{2} V=63.17 \times \pi \times\left(\frac{4}{12}\right)^{2} \times 19.65=434.5 \mathrm{lb} / \mathrm{sec}$ From equation (8-38), the acoustic velocity of the restrained fuel

$$\begin{aligned} c & =\frac{80100}{\sqrt{1+\left[\frac{2.5 \times 6.06 \times 10^{5} \times 4(1-0.8 \times 0.36)}{10.4 \times 10^{6} \times 0.167}\right]}} \\ & =43100 \mathrm{in} / \mathrm{sec} \end{aligned}$$

Impact time delay in the tank: $\frac{l_{c}}{c}=\frac{50}{43100}=1.16 \times 10^{-3} \mathrm{sec}\left(\right.$ i.e., $\left.<1.2 \times 10^{-3} \mathrm{sec}\right)$

Using equation (8-37), we obtain the tank pressure surge due to impact after a 6 -foot free drop:

$$p_{S}=\frac{c \dot{W}}{\pi a^{2} g}=\frac{43100 \times 434.5}{\pi \times(4)^{2} \times 386}=964 \mathrm{psi}$$

(b) After a 20 -foot free drop, the final velocity at impact is

$$V=\sqrt{2 g h}=\sqrt{2 \times 32.2 \times 20}=35.9 \mathrm{fps}$$

The equivalent propellant flow rate due to impact: $\dot{w}=\rho_{p} \pi \mathrm{a}^{2} V=63.25 \times \pi \times\left(\frac{4}{12}\right)^{2} \times 35.9=792.5 \mathrm{lb} / \mathrm{sec}$ The tank pressure surge due to impact after a 20 -foot free drop:

$$p_{S}=\frac{c \dot{w}}{\pi a^{2} g}=\frac{43100 \times 792.5}{\pi \times(4)^{2} \times 386}=1765 \mathrm{psi}$$