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Microprocessor Heat Sink Thermal Gradients (Mechanical & Computer Engineering) Solution

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Problem Overview

We analyze the temperature field T(x,y)=120−4x2−2y2T(x, y) = 120 - 4x^2 - 2y^2 at sensor point P(2,3)P(2, 3). We evaluate the gradient ∇T\nabla T, find the maximum rate of change, establish the direction of steepest cooling (−∇T-\nabla T), determine constant-temperature tangent vectors, and link results to Fourier heat conduction in micro-channel heat sink design.

Step-by-Step Mathematical Solution

1. Compute Temperature Gradient and Maximum Rate of Increase

State the Gradient Formula (defined in The Gradient Vector):

∇T(x,y)=⟨∂T∂x,∂T∂y⟩\nabla T(x, y) = \left\langle \frac{\partial T}{\partial x}, \frac{\partial T}{\partial y} \right\rangle

Compute the first partial derivatives:

Tx=∂∂x[120−4x2−2y2]=−8xT_x = \frac{\partial}{\partial x}[120 - 4x^2 - 2y^2] = -8x
Ty=∂∂y[120−4x2−2y2]=−4yT_y = \frac{\partial}{\partial y}[120 - 4x^2 - 2y^2] = -4y

Evaluate at sensor location P(2,3)P(2, 3):

Tx(2,3)=−8(2)=−16∘C/mmT_x(2, 3) = -8(2) = -16^{\circ}\text{C/mm}
Ty(2,3)=−4(3)=−12∘C/mmT_y(2, 3) = -4(3) = -12^{\circ}\text{C/mm}
∇T(2,3)=⟨−16,−12⟩∘C/mm\nabla T(2, 3) = \langle -16, -12 \rangle^{\circ}\text{C/mm}

Compute the maximum instantaneous rate of temperature increase:

∥∇T(2,3)∥=(−16)2+(−12)2=256+144=400=20∘C/mm\|\nabla T(2, 3)\| = \sqrt{(-16)^2 + (-12)^2} = \sqrt{256 + 144} = \sqrt{400} = 20^{\circ}\text{C/mm}

2. Direction of Steepest Cooling (Steepest Descent)

The direction of most rapid temperature drop is strictly opposite to the gradient vector:

v⃗cool=−∇T(2,3)=−⟨−16,−12⟩=⟨16,12⟩\vec{v}_{\text{cool}} = -\nabla T(2, 3) = -\langle -16, -12 \rangle = \langle 16, 12 \rangle

Normalize to obtain the unit vector direction u⃗cool\vec{u}_{\text{cool}}:

∥v⃗cool∥=20\|\vec{v}_{\text{cool}}\| = 20
u⃗cool=⟨16,12⟩20=⟨45,35⟩=⟨0.8,0.6⟩\vec{u}_{\text{cool}} = \frac{\langle 16, 12 \rangle}{20} = \left\langle \frac{4}{5}, \frac{3}{5} \right\rangle = \langle 0.8, 0.6 \rangle
Moving along unit heading ⟨0.8,0.6⟩\langle 0.8, 0.6 \rangle yields the steepest temperature drop at rate −∥∇T∥=−20∘C/mm-\|\nabla T\| = -20^{\circ}\text{C/mm}.

3. Directions of Zero Temperature Change (Isothermal Tangents)

Zero change in temperature occurs in directions orthogonal to ∇T=⟨−16,−12⟩\nabla T = \langle -16, -12 \rangle. For any 2D vector ⟨a,b⟩\langle a, b \rangle, orthogonal vectors are given by ⟨−b,a⟩\langle -b, a \rangle and ⟨b,−a⟩\langle b, -a \rangle:

w⃗1=⟨12,−16⟩,w⃗2=⟨−12,16⟩\vec{w}_1 = \langle 12, -16 \rangle, \qquad \vec{w}_2 = \langle -12, 16 \rangle

Normalize to unit vectors:

∥w⃗1∥=122+(−16)2=20\|\vec{w}_1\| = \sqrt{12^2 + (-16)^2} = 20
u⃗iso,1=⟨1220,−1620⟩=⟨35,−45⟩=⟨0.6,−0.8⟩\vec{u}_{\text{iso}, 1} = \left\langle \frac{12}{20}, -\frac{16}{20} \right\rangle = \left\langle \frac{3}{5}, -\frac{4}{5} \right\rangle = \langle 0.6, -0.8 \rangle
u⃗iso,2=−u⃗iso,1=⟨−35,45⟩=⟨−0.6,0.8⟩\vec{u}_{\text{iso}, 2} = -\vec{u}_{\text{iso}, 1} = \left\langle -\frac{3}{5}, \frac{4}{5} \right\rangle = \langle -0.6, 0.8 \rangle
Moving along either u⃗iso,1\vec{u}_{\text{iso}, 1} or u⃗iso,2\vec{u}_{\text{iso}, 2} yields Du⃗T=0∘C/mmD_{\vec{u}} T = 0^{\circ}\text{C/mm}.

4. Fourier's Law & Heat Sink Channel Alignment

  • Thermodynamic Basis: Conductive heat flow obeys Fourier's Law: q⃗=−k∇T\vec{q} = -k\nabla T. Because −k<0-k < 0, heat flux flows strictly in the direction of steepest cooling (−∇T=⟨16,12⟩-\nabla T = \langle 16, 12 \rangle).
  • Engineering Impact: Orienting liquid micro-channels or copper cooling fins parallel to u⃗cool=⟨0.8,0.6⟩\vec{u}_{\text{cool}} = \langle 0.8, 0.6 \rangle aligns the coolant path with the natural thermal conduction vector. This maximizes the convective heat transfer coefficient, prevents local hotspot accumulation, and minimizes thermal stresses that cause silicon fracture.