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

Formulas & Definitions

Problem Overview

An aerospace engineering team optimizes the dimensions of a closed cylindrical upper-stage liquid oxygen (LOX) tank holding fixed internal volume V0=54π m3V_0 = 54\pi\text{ m}^3. We minimize total surface area S=2πr2+2πrhS = 2\pi r^2 + 2\pi rh, prove global optimality via the Second Derivative Test, verify the h=2rh = 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.)

  1. Objective Function (Minimize Surface Area): A closed right circular cylinder has two flat circular bulkheads and one cylindrical barrel:
    S(r,h)=2πr2+2πrhS(r, h) = 2\pi r^2 + 2\pi rh
    (Formula defined in Geometric Volume and Surface Area Formulas.)
  2. Internal Propellant Volume Constraint:
    V=πr2h=54π m3  ⟹  r2h=54V = \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 hh:

h=54r2h = \frac{54}{r^2}
Substitute into the surface area equation:
S(r)=2πr2+2πr(54r2)=2πr2+108πr=2πr2+108πr−1S(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.)

Step 4: Differentiate and Find Critical Numbers

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

S′(r)=ddr[2πr2+108πr−1]=4πr−108πr2S'(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)=0S'(r) = 0 to identify stationary points:
4πr−108πr2=0  ⟹  4πr=108πr24\pi r - \frac{108\pi}{r^2} = 0 \implies 4\pi r = \frac{108\pi}{r^2}
Divide both sides by π\pi and multiply by r2r^2:
4r3=108  ⟹  r3=274r^3 = 108 \implies r^3 = 27
r=273=3 metersr = \sqrt[3]{27} = 3\text{ meters}

Step 5: Verify Global Minimum via Second Derivative Test

Compute the second derivative:

S′′(r)=ddr[4πr−108πr−2]=4π+216πr3S''(r) = \frac{d}{dr}\left[4\pi r - 108\pi r^{-2}\right] = 4\pi + \frac{216\pi}{r^3}
Evaluate at r=3 mr = 3\text{ m}:
S′′(3)=4π+216π(3)3=4π+216π27=4π+8π=12π>0S''(3) = 4\pi + \frac{216\pi}{(3)^3} = 4\pi + \frac{216\pi}{27} = 4\pi + 8\pi = 12\pi > 0
Since S′′(3)=12π>0S''(3) = 12\pi > 0 and r=3 mr = 3\text{ m} is the unique critical point on (0,∞)(0, \infty), the Second Derivative Test guarantees that r=3 mr = 3\text{ m} achieves the absolute minimum surface area.

Step 6: Verify Optimal Height and Aspect Ratio Theorem

Calculate the barrel height:

h=54(3)2=549=6 metersh = \frac{54}{(3)^2} = \frac{54}{9} = 6\text{ meters}
Evaluate the ratio of height to radius:
hr=6 m3 m=2  ⟹  h=2r=d\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 meters6\text{ meters}), confirming the general optimality principle formalized in Optimal Aspect Ratios in Geometric Containers.

Step 7: Minimum Surface Area and Tank Skin Mass

  1. Minimum Surface Area:
    S(3)=2π(3)2+108π3=18π+36π=54π m2≈169.65 m2S(3) = 2\pi (3)^2 + \frac{108\pi}{3} = 18\pi + 36\pi = 54\pi\text{ m}^2 \approx 169.65\text{ m}^2
  2. Structural Dry Skin Mass: Given areal mass density ρA=15 kg/m2\rho_A = 15\text{ kg/m}^2:
    mskin=ρA⋅Smin⁡=15⋅54π=810π kg≈2,544.7 kgm_{\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.