Cryogenic Propellant Tank Bulkhead Surface Area Minimization (Mechanical & Aerospace Engineering) Solution

Problem Overview

An aerospace engineering team optimizes the dimensions of a closed cylindrical upper-stage liquid oxygen (LOX) tank holding fixed internal volume $V_0 = 54\pi\text{ m}^3$ . We minimize total surface area $S = 2\pi r^2 + 2\pi rh$ , prove global optimality via the Second Derivative Test, verify the $h = 2r$ aspect ratio theorem, and compute total structural skin mass.

Step-by-Step Mathematical Solution

Step 1: Formulate Governing Equations

The governing process follows the Applied Optimization Modeling Framework: (Defined in Applied Optimization Modeling Framework.)

Objective Function (Minimize Surface Area): A closed right circular cylinder has two flat circular bulkheads and one cylindrical barrel: $S(r, h) = 2\pi r^2 + 2\pi rh$ (Formula defined in Geometric Volume and Surface Area Formulas.)
Internal Propellant Volume Constraint: $V = \pi r^2 h = 54\pi\text{ m}^3 \implies r^2 h = 54$

Step 2: Variable Reduction (Substitution)

Solve the volume constraint for the barrel height $h$ :

h = \frac{54}{r^2}

Substitute into the surface area equation:

S(r) = 2\pi r^2 + 2\pi r\left(\frac{54}{r^2}\right) = 2\pi r^2 + \frac{108\pi}{r} = 2\pi r^2 + 108\pi r^{-1}

Step 3: Establish the Context-Driven Domain

(Governed by Context-Driven Domain in Geometric Optimization.)

Physical radius must be positive: $r > 0$ .
Feasible physical domain is the open interval $r \in (0, \infty)$ .

Step 4: Differentiate and Find Critical Numbers

Differentiate $S(r)$ with respect to $r$ using the Power Rule:

S'(r) = \frac{d}{dr}\left[2\pi r^2 + 108\pi r^{-1}\right] = 4\pi r - \frac{108\pi}{r^2}

Set

S'(r) = 0

to identify stationary points:

4\pi r - \frac{108\pi}{r^2} = 0 \implies 4\pi r = \frac{108\pi}{r^2}

Divide both sides by

\pi

and multiply by

r^2

4r^3 = 108 \implies r^3 = 27

r = \sqrt[3]{27} = 3\text{ meters}

Step 5: Verify Global Minimum via Second Derivative Test

Compute the second derivative:

S''(r) = \frac{d}{dr}\left[4\pi r - 108\pi r^{-2}\right] = 4\pi + \frac{216\pi}{r^3}

Evaluate at

r = 3\text{ m}

S''(3) = 4\pi + \frac{216\pi}{(3)^3} = 4\pi + \frac{216\pi}{27} = 4\pi + 8\pi = 12\pi > 0

Since

S''(3) = 12\pi > 0

and

r = 3\text{ m}

is the unique critical point on

(0, \infty)

, the Second Derivative Test guarantees that

r = 3\text{ m}

achieves the absolute minimum surface area.

Step 6: Verify Optimal Height and Aspect Ratio Theorem

Calculate the barrel height:

h = \frac{54}{(3)^2} = \frac{54}{9} = 6\text{ meters}

Evaluate the ratio of height to radius:

\frac{h}{r} = \frac{6\text{ m}}{3\text{ m}} = 2 \implies h = 2r = d

The optimal height is exactly equal to the diameter of the cylinder (

6\text{ meters}

), confirming the general optimality principle formalized in Optimal Aspect Ratios in Geometric Containers.

Step 7: Minimum Surface Area and Tank Skin Mass

Minimum Surface Area: $S(3) = 2\pi (3)^2 + \frac{108\pi}{3} = 18\pi + 36\pi = 54\pi\text{ m}^2 \approx 169.65\text{ m}^2$
Structural Dry Skin Mass: Given areal mass density $\rho_A = 15\text{ kg/m}^2$ : $m_{\text{skin}} = \rho_A \cdot S_{\min} = 15 \cdot 54\pi = 810\pi\text{ kg} \approx 2,544.7\text{ kg}$

Aerospace Design Note: In launch vehicle structures, a 6-meter diameter tank perfectly fits within modern heavy-lift rocket payload envelopes (such as 5.4m to 7m fairings), minimizing parasitic aerodynamic drag during ascent while minimizing dry structural mass.