Heat Transfer · Chapter 15
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Heat Transfer · Chapter Fifteen

Cooling of Electronic Equipment

Every current that flows through a resistance generates heat. As electronics shrink, the heat per unit volume rivals that of a nuclear reactor — and failure rates climb steeply with temperature. Cooling is now a first-order design constraint.

This chapter applies the whole toolkit of heat transfer to electronics: estimating the cooling load, the thermal environment, and the ladder of cooling methods — conduction, natural and forced air, liquid, and immersion — chosen by escalating heat flux.

Learning objectives

  • Translate electrical operating states into a defensible worst-case cooling load.
  • Construct junction-to-ambient thermal-resistance networks with correct reference nodes.
  • Select passive air, forced air, liquid, or immersion cooling from heat flux and environment.
  • Estimate airflow or coolant flow using energy balances and state-dependent properties.
  • Audit junction limits, thermal margins, fan/pump assumptions, and failure scenarios.
15–1

Introduction and History

Thermal control determines the reliability of electronic systems. Component failure rate rises roughly exponentially with temperature, so keeping junctions cool directly extends service life.

Key Idea A widely cited rule of thumb holds that failure rate doubles for roughly every 10 °C rise in operating temperature. Thermal design is therefore reliability design — not an afterthought.

Temperature-acceleration rules are screening heuristics, not universal lifetime laws. Reliability predictions require component-specific failure mechanisms, duty cycles, cycling fatigue, and qualified data. Thermal design therefore controls both peak junction temperature and spatial/temporal gradients.

15–2

Manufacturing of Electronic Equipment

Heat is generated at the chip and must travel through a stack of packaging layers to reach the coolant. Each layer adds thermal resistance along the path.

The Thermal Path

Heat flows from the semiconductor junction through the case, often a heat sink, and finally to the ambient. Manufacturers quote junction-to-case and case-to-ambient resistances so designers can chain them like any resistance network.

Physical meaning

Thermal resistance is a temperature rise per watt, K/W. Nodes such as junction, case, sink, local air, and room air must be distinguished: using room temperature with a resistance referenced to local inlet air can hide recirculation heating.

15–3

Cooling Load of Electronic Equipment

For equipment inside a control volume, electrical input ultimately becomes heat unless useful energy leaves in another form (optical, mechanical, RF, or stored energy). Build the load from credible simultaneous operating states, conversion losses, and transient peaks rather than simply adding every nameplate maximum.

Cooling load equals dissipated power
\[ \dot{Q}_{\text{cooling}} = \dot{W}_{\text{electrical}} = V\,I \]
15–4

Thermal Environment

The environment — ambient temperature, altitude, airflow, and any enclosure — sets the heat-rejection conditions. Military, aerospace, and consumer environments impose very different limits.

Junction Temperature Limit

Each device has a maximum allowable junction temperature (often around 85–125 °C for silicon). The cooling system must keep the junction below this limit under worst-case ambient and load, with margin for reliability.

Ambient specification must include inlet temperature, altitude, contamination, solar load, enclosure class, and loss of airflow. Air density and fan operating point change with altitude and filter loading; radiation uses absolute temperatures and actual surface emissivity.

15–5

Electronics Cooling in Different Applications

The right method depends on the heat flux to be removed. As power density rises, designs climb a ladder from passive conduction to aggressive liquid immersion.

increasing heat flux → conduction natural air forced air liquid immersion
Figure 15–1The cooling-method ladder: as heat flux rises, designs progress from passive conduction through air cooling to liquid and immersion cooling.
15–6

Conduction Cooling

In conduction cooling, heat spreads through solid paths — leads, boards, frames, and heat sinks — to a cooler structure. The design problem is minimizing the thermal resistance from junction to ambient.

Junction temperature from a resistance chain
\[ T_{\text{junction}} = T_{\text{ambient}} + \dot{Q}\,(R_{\text{j-c}} + R_{\text{c-s}} + R_{\text{s-a}}) \]
\(R_{\text{j-c}}\), \(R_{\text{c-s}}\), \(R_{\text{s-a}}\) are junction-to-case, case-to-sink, and sink-to-ambient resistances in series.
EXAMPLE 1Thermal Margin for a Motor-Drive Module
Problem. A motor-drive module dissipates 42 W into a heat sink. Manufacturer and interface data give \(R_{j-c}=0.38\), \(R_{c-s}=0.12\), and \(R_{s-a}=0.72\) K/W. At 46 °C inlet air, find junction temperature and margin to a 125 °C limit.

Strategy

One dominant junction-to-air path with quoted series resistances is a steady thermal-circuit problem.

Givens

  • \(\dot Q=42\) W; resistances \(0.38,0.12,0.72\) K/W.
  • \(T_a=46^\circ C,\ T_{j,max}=125^\circ C\).

Physical/model interpretation

Steady heat flow, all module power follows the stated path, and sink resistance is referenced to inlet air. Temperature rises in the direction of heat-flow resistance from air to junction.

Numbered derivation steps
1. Combine series elements. \(R_{j-a}=0.38+0.12+0.72=1.22\ {\rm K/W}\).
2. Find rise. \(\Delta T_{j-a}=\dot Q R_{j-a}=(42)(1.22)=51.24\) K.
3. Find junction temperature. \(T_j=46+51.24=97.2^\circ C\).
4. Find thermal margin. \(M=125-97.2=27.8\) K.

Unit/sign/limit checks

Units: Multiplying the 42 W heat rate by the total resistance 1.22 K/W cancels watts and gives a 51.24 K temperature rise, so adding it to 46 °C is dimensionally valid.

Signs: A dissipating junction must be hotter than its sink and inlet air; the positive rise confirms this.

Limits: At zero power \(T_j\to T_a\); the computed value is below the stated limit.

Final answer: \(\boxed{T_j=97.2^\circ C,\ {\rm margin}=27.8\ {\rm K}}\).

Engineering insight

Margin must absorb interface aging, sink tolerances, blocked fins, and power uncertainty; a positive nominal margin alone does not qualify the product.

15–7

Air Cooling: Natural Convection and Radiation

For low-power devices, buoyant air and radiation alone can carry the heat away — silent, reliable, and free of moving parts. Both mechanisms act together on exposed surfaces.

Combined natural convection and radiation
\[ \dot{Q} = h_{\text{conv}}A_s(T_s - T_\infty) + \varepsilon\sigma A_s(T_s^4 - T_{\text{surr}}^4) \]
Course Note

In natural-convection electronics, radiation is not negligible — it often carries 25–40% of the load. That is why painting a heat sink matte black (high \(\varepsilon\)) can meaningfully help a passive design, while doing nothing for a fan-cooled one where convection dominates.

Natural-convection coefficients depend on orientation and properties at the film temperature \(T_f=(T_s+T_\infty)/2\). Verify the Rayleigh-number range of the selected correlation. Radiation exchanges with surrounding surfaces, not necessarily with the air, and all radiative temperatures must be absolute.

15–8

Air Cooling: Forced Convection

Adding a fan raises the convection coefficient several-fold, handling much higher heat fluxes. The fan's airflow and the device's allowable temperature rise set the achievable cooling.

Fan flow rate from the air-side energy balance
\[ \dot{Q} = \dot{m}\,c_p\,(T_{\text{out}} - T_{\text{in}}) = \rho\,\dot{V}\,c_p\,\Delta T_{\text{air}} \]
EXAMPLE 2Airflow for a Telecom Shelf
Problem. A ventilated telecom shelf releases 620 W to air entering at 32 °C. Limit bulk exhaust rise to 12 K. At the 38 °C mean air temperature, use \(\rho=1.13\) kg/m³ and \(c_p=1007\) J/kg·K. Find ideal volume flow and a selection flow with 25% allowance.

Strategy

The target is bulk airflow rather than a surface coefficient, so use a steady air-side energy balance at the mean-property temperature.

Givens

  • \(\dot Q=620\) W, \(\Delta T_b=12\) K.
  • \(\rho=1.13\) kg/m³, \(c_p=1007\) J/kgK; allowance 25%.

Physical/model interpretation

Steady, well-mixed outlet; all shelf heat enters the airflow. The allowance represents bypass/leakage but not fan pressure-drop performance.

Numbered derivation steps
1. Write balance. \(\dot Q=\rho\dot V c_p\Delta T_b\).
2. Solve ideal flow. \(\dot V=620/[(1.13)(1007)(12)]=0.0454\ {\rm m^3/s}\).
3. Convert for selection. \(0.0454\ {\rm m^3/s}\times2118.9=96.2\) CFM.
4. Apply allowance. \(\dot V_{sel}=1.25(0.0454)=0.0568\ {\rm m^3/s}=120\) CFM.

Unit/sign/limit checks

Units: In \(\dot V=\dot Q/(\rho c_p\Delta T)\), W divided by \(({\rm kg/m^3})({\rm J/kgK})({\rm K})\) reduces to m³/s, confirming the result is dimensionally a volume flow rate.

Signs: Positive heat input produces \(T_{out}>T_{in}\).

Limits: Required flow grows without bound as allowed rise tends to zero and halves if allowed rise doubles.

Final answer: \(\boxed{\dot V_{ideal}=0.0454\ {\rm m^3/s};\ \dot V_{sel}=0.0568\ {\rm m^3/s}\ (120\ CFM)}\).

Engineering insight

Select a fan at 120 CFM on the system pressure-drop curve, not at free delivery; verify hot spots and recirculation separately.

Problem-solving strategy

  1. Define the thermal limit, worst credible power state, inlet condition, and failure case.
  2. Sketch heat-flow paths and nodes; distinguish series and parallel resistances.
  3. Choose the least intensive cooling method that can meet heat flux and environmental constraints.
  4. Evaluate air/coolant properties at a justified mean or film temperature and check correlation ranges.
  5. Audit junction margin, energy closure, pressure drop, redundancy, fouling, and manufacturability.
15–9

Liquid Cooling

When air cannot keep up, liquids — with far higher density and specific heat — carry heat to a remote radiator. Cold plates conduct heat from components into a circulating coolant.

Why Liquids Win at High Flux

A liquid's volumetric heat capacity is hundreds of times that of air, so a small flow removes large power with a modest temperature rise. The trade-offs are pumps, plumbing, and the risk of leaks near electronics.

EXAMPLE 3Coolant Flow and Pump Heat for an AI Cold Plate
Problem. A cold plate removes 2.40 kW from an accelerator. A 30% propylene-glycol coolant may warm 6.0 K; at its mean temperature take \(c_p=3.75\) kJ/kg·K and \(\rho=1030\) kg/m³. The loop drop is 48 kPa and pump efficiency is 55%. Find mass flow, volume flow, and pump heat delivered to the fluid.

Strategy

Use the coolant sensible-energy balance, then hydraulic power divided by efficiency; the difference between input and useful hydraulic power becomes pump heat.

Givens

  • \(\dot Q=2400\) W, \(\Delta T=6.0\) K, \(c_p=3750\) J/kgK, \(\rho=1030\) kg/m³.
  • \(\Delta p=48{,}000\) Pa, \(\eta_p=0.55\).

Physical/model interpretation

Steady single-phase incompressible loop; properties are supplied at mean temperature. Pump inefficiency is assumed deposited in the coolant.

Numbered derivation steps
1. Mass flow. \(\dot m=\dot Q/(c_p\Delta T)=2400/[(3750)(6.0)]=0.1067\ {\rm kg/s}\).
2. Volume flow. \(\dot V=\dot m/\rho=0.1067/1030=1.036\times10^{-4}\ {\rm m^3/s}=6.21\ {\rm L/min}\).
3. Hydraulic power. \(\dot W_{hyd}=\Delta p\dot V=(48{,}000)(1.036\times10^{-4})=4.97\) W.
4. Pump loss to fluid. \(\dot W_{in}=4.97/0.55=9.04\) W; \(\dot Q_{pump}=9.04-4.97=4.07\) W.

Unit/sign/limit checks

Units: kg/s divided by kg/m³ gives m³/s; Pa m³/s = W.

Signs: Both accelerator and pump losses add heat to the coolant.

Limits: At 100% pump efficiency the calculated pump heat tends to zero; 4.07 W is small versus 2.40 kW.

Final answer: \(\boxed{\dot m=0.1067\ {\rm kg/s},\ \dot V=6.21\ {\rm L/min},\ \dot Q_{pump}=4.07\ {\rm W}}\).

Engineering insight

Final design still requires cold-plate thermal resistance, channel pressure-drop validation, corrosion compatibility, leak detection, pump curves, and redundancy review.

15–10

Immersion Cooling

At the extreme, components are submerged directly in a dielectric liquid. Heat can be removed by single-phase circulation or, most aggressively, by boiling the coolant right at the chip surface.

Key Idea Immersion cooling exploits the latent heat of a dielectric fluid: boiling at the component holds its surface near the coolant's saturation temperature while absorbing enormous heat flux. It sits at the top of the cooling ladder for the densest, highest-power electronics.

Two-phase immersion is constrained by critical heat flux, saturation pressure, fluid compatibility, vapor containment, and condenser capacity. Correlations require saturation-state properties and valid surface/fluid combinations; crossing critical heat flux can cause a destructive temperature excursion.

Chapter recap

  • Start from the true heat load and the actual thermal environment.
  • Resistance networks expose bottlenecks and convert device power into node temperatures.
  • Escalate from passive to forced air, liquid, or immersion only as heat flux and constraints require.
  • Before moving on: be able to define nodes, calculate junction margin and flow, select properties, and identify the validation still required.
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