Aerospace Fuel Tank Center of Mass and Slosh Baffle (Aerospace & Mechanical Engineering) Solution

An aerospace structural dynamics engineer at the UNLV Howard R. Hughes College of Engineering is designing a lightweight titanium fuel cell integrated into the wingtip structure of an autonomous reconnaissance aircraft.

The interior reservoir of the fuel cell is bounded by the three coordinate planes $x = 0$, $y = 0$, $z = 0$ and the upper inclined wing-skin aerodynamic fairing plane:

with mass density $\rho(x, y, z) = 120 z\text{ kg/m}^3$.

Step-by-Step Solution:

1. Extract Type I Regional Bounds:
   State the Type I Spatial Region Definition (defined in Triple Integrals over a General Region):

   $$E = \{(x, y, z) \mid 0 \le x \le 1, \, 0 \le y \le 1 - x, \, 0 \le z \le 1 - x - y\}$$

2. Evaluate Enclosed Volume:
   State the Volume via Triple Integration Formula:

   $$V(E) = \iiint_{E} 1 \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} 1 \, dz \, dy \, dx$$

   Inner $z$-integral:

   $$\int_{0}^{1-x-y} 1 \, dz = 1 - x - y$$

   Middle $y$-integral:

   $$\int_{0}^{1-x} (1 - x - y) \, dy = \left[ (1 - x)y - \frac{y^2}{2} \right]_{0}^{1-x} = (1 - x)^2 - \frac{(1 - x)^2}{2} = \frac{1}{2}(1 - x)^2$$

   Outer $x$-integral:

   $$V(E) = \int_{0}^{1} \frac{1}{2}(1 - x)^2 \, dx = \left[ -\frac{1}{6}(1 - x)^3 \right]_{0}^{1} = 0 - \left(-\frac{1}{6}(1)^3\right) = \frac{1}{6}\text{ m}^3$$

3. Evaluate Total Mass:
   State the Total Mass Formula with Variable Density:

   $$M = \iiint_{E} \rho(x, y, z) \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} 120 z \, dz \, dy \, dx$$

   Inner $z$-integral:

   $$\int_{0}^{1-x-y} 120 z \, dz = \left[ 60 z^2 \right]_{0}^{1-x-y} = 60(1 - x - y)^2$$

   Middle $y$-integral:

   $$\int_{0}^{1-x} 60(1 - x - y)^2 \, dy = \left[ -20(1 - x - y)^3 \right]_{0}^{1-x} = 0 - \left(-20(1 - x)^3\right) = 20(1 - x)^3$$

   Outer $x$-integral:

   $$M = \int_{0}^{1} 20(1 - x)^3 \, dx = \left[ -5(1 - x)^4 \right]_{0}^{1} = 0 - (-5(1)^4) = 5\text{ kg}$$

4. Compute First Moment and Center of Mass Coordinate $\bar{z}$:
   State the First Moment of Mass Formula about the $xy$-Plane:

   $$M_{xy} = \iiint_{E} z \rho(x, y, z) \, dV = \int_{0}^{1} \int_{0}^{1-x} \int_{0}^{1-x-y} 120 z^2 \, dz \, dy \, dx$$

   Inner $z$-integral:

   $$\int_{0}^{1-x-y} 120 z^2 \, dz = \left[ 40 z^3 \right]_{0}^{1-x-y} = 40(1 - x - y)^3$$

   Middle $y$-integral:

   $$\int_{0}^{1-x} 40(1 - x - y)^3 \, dy = \left[ -10(1 - x - y)^4 \right]_{0}^{1-x} = 10(1 - x)^4$$

   Outer $x$-integral:

   $$M_{xy} = \int_{0}^{1} 10(1 - x)^4 \, dx = \left[ -2(1 - x)^5 \right]_{0}^{1} = 2\text{ kg}\cdot\text{m}$$

   State the Center of Mass Coordinate Formula:

   $$\bar{z} = \frac{M_{xy}}{M} = \frac{2\text{ kg}\cdot\text{m}}{5\text{ kg}} = 0.4\text{ m}$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The cell encloses a total volume of $\frac{1}{6}\text{ m}^3 \approx 0.167\text{ m}^3$ containing $5\text{ kg}$ of fuel. Because the density $\rho(z) = 120z$ increases toward the upper reservoir ceiling, the vertical center of mass $\bar{z} = 0.4\text{ m}$ is located higher than that of a uniform-density tetrahedron (which would have its centroid at $\bar{z}_{\text{uniform}} = \frac{1}{4} = 0.25\text{ m}$).
• Engineering Application: In unmanned combat air vehicle (UAV) design, wingtip mass distribution strongly affects the wing's torsional natural frequency and roll moment of inertia $I_{xx}$. Locating the center of gravity at $\bar{z} = 0.4\text{ m}$ rather than $0.25\text{ m}$ alters the vertical moment arm to the wing chord neutral axis. Autopilot flight control laws must incorporate this vertical offset into aileron feedforward gains to prevent roll-yaw dynamic coupling during turbulent wind gusts.