Microprocessor Heat Sink Volumetric Thermal Dissipation (Computer & Mechanical Engineering) Solution

A computer systems thermal architect at the UNLV Howard R. Hughes College of Engineering is designing a high-conductivity copper heat-spreader block mounted directly above an enterprise artificial intelligence (AI) accelerator microprocessor.

The heat-spreader block occupies a 3D rectangular box domain in centimeters:

Step-by-Step Solution:

1. Calculate Geometric Volume:
   State the Volume of a Rectangular Box Formula (defined in Triple Integrals over a Rectangular Region):

   $$V(B) = (b - a)(d - c)(s - r)$$
   Substitute the domain limits:
   $$V(B) = (4 - 0)(4 - 0)(2 - 0) = 4 \times 4 \times 2 = 32\text{ cm}^3$$

2. Set Up the Triple Integral for Total Heat Dissipation:
   State the Total Quantity Accumulation Formula via Triple Integration:

   $$P_{\text{total}} = \iiint_{B} q(x, y, z) \, dV$$
   Using the differential volume element $dV = dz \, dy \, dx$:
   $$P_{\text{total}} = \int_{0}^{4} \int_{0}^{4} \int_{0}^{2} 3x^2 y z \, dz \, dy \, dx$$

3. Evaluate the Triple Integral:
   Because the integrand is algebraically separable and the boundaries are constants, apply Separability for Triple Integrals over Rectangular Boxes:

   $$P_{\text{total}} = 3 \left( \int_{0}^{4} x^2 \, dx \right) \left( \int_{0}^{4} y \, dy \right) \left( \int_{0}^{2} z \, dz \right)$$

   Evaluate the $x$-integral:

   $$\int_{0}^{4} x^2 \, dx = \left[ \frac{x^3}{3} \right]_{0}^{4} = \frac{64}{3} - 0 = \frac{64}{3}$$

   Evaluate the $y$-integral:

   $$\int_{0}^{4} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{4} = \frac{16}{2} - 0 = 8$$

   Evaluate the $z$-integral:

   $$\int_{0}^{2} z \, dz = \left[ \frac{z^2}{2} \right]_{0}^{2} = \frac{4}{2} - 0 = 2$$

   Multiply the evaluated factors:

   $$P_{\text{total}} = 3 \times \left(\frac{64}{3}\right) \times 8 \times 2 = 64 \times 16 = 1024\text{ W}$$

4. Compute Average Volumetric Generation Rate:
   State the Average Value Formula for a 3D Function:

   $$\overline{q} = \frac{1}{V(B)} \iiint_{B} q(x, y, z) \, dV = \frac{P_{\text{total}}}{V(B)}$$
   Substitute the calculated values:
   $$\overline{q} = \frac{1024\text{ W}}{32\text{ cm}^3} = 32\text{ W/cm}^3$$

Meaning of the Answer & Real-Life Application:

• Quantitative Interpretation: The total heat generation rate across the copper spreader block is $1024\text{ W}$ (approximately $1.024\text{ kW}$), with an average volumetric energy density of $32\text{ W/cm}^3$.
• Engineering Application: In data-center thermal management, an instantaneous heat load of over $1\text{ kW}$ concentrated on a $4\text{ cm} \times 4\text{ cm}$ footprint represents extreme thermal flux ($\approx 64\text{ W/cm}^2$). Knowing the exact total thermal load allows cooling system engineers to size the coolant volumetric flow rate $\dot{V} = \frac{P_{\text{total}}}{\rho c_p \Delta T}$ in a pumped liquid cold plate to ensure the junction temperature $T_j$ never exceeds the $85^\circ\text{C}$ silicon degradation threshold.