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

Formulas & Definitions

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