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Exercise set

Heat Sink and Forced Convection Sizing Exercises

Worked mechanical engineering exercises for heat sink and forced convection sizing, covering thermal resistance targets, airflow, convection coefficients, fin area, coolant flow, pressure drop, fan power, derating, and validation.

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Mechanical Engineering
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v1.0.0 ยท reviewed

These exercises practise heat sink and forced-convection sizing for electronics, power modules, compact machines, battery support equipment, and cooled enclosures. The goal is to connect power dissipation, temperature limits, airflow or coolant flow, thermal resistance, pressure drop, and validation evidence.

Assume simplified steady operation unless an exercise states otherwise. Real designs also require transient thermal impedance, nonuniform heat generation, contact resistance, enclosure recirculation, fan curves, altitude, fouling, manufacturing tolerance, acoustic limits, and safety review.

How to Use These Exercises

For each exercise, define:

  1. heat source and duty cycle;
  2. limiting temperature node;
  3. heat path from source to ambient or coolant;
  4. available temperature rise;
  5. airflow or coolant-flow boundary;
  6. validation measurement that would prove the result.

The most common mistake is selecting a heat sink from a catalog rating without matching the actual airflow, board orientation, enclosure restriction, ambient temperature, interface material, and neighboring heat sources.

Exercise 1: Required Sink-to-Ambient Thermal Resistance

A power module dissipates 180\ \text{W}. Maximum junction temperature is 125^\circ\text{C}. Design ambient temperature is 45^\circ\text{C}. Junction-to-case thermal resistance is 0.12^\circ\text{C/W} and case-to-sink interface resistance is 0.08^\circ\text{C/W}.

Find the maximum allowable sink-to-ambient thermal resistance.

Solution

Available temperature rise:

\Delta T_{allow}=125-45=80^\circ\text{C}

Allowable total resistance:

\displaystyle R_{\theta,total}=\frac{\Delta T_{allow}}{P}=\frac{80}{180}=0.444^\circ\text{C/W}

Known internal and interface resistance:

R_{\theta,known}=0.12+0.08=0.20^\circ\text{C/W}

Maximum sink-to-ambient resistance:

R_{\theta,SA}=0.444-0.20=0.244^\circ\text{C/W}

The heat sink and airflow path must provide no more than about 0.24^\circ\text{C/W}.

Engineering Comment

This is a demanding thermal requirement. The selected heat sink rating must be valid for the installed airflow and orientation. If a catalog value was measured in open air with ideal ducting, it may not apply inside an enclosure.

Exercise 2: Junction Temperature with a Candidate Heat Sink

Using the same module, a candidate forced-air heat sink has R_{\theta,SA}=0.30^\circ\text{C/W} under the expected fan flow. Estimate junction temperature.

Solution

Total resistance:

R_{\theta,total}=0.12+0.08+0.30=0.50^\circ\text{C/W}

Junction temperature:

T_j=T_a+P R_{\theta,total}
T_j=45+180(0.50)=135^\circ\text{C}

The candidate heat sink exceeds the 125^\circ\text{C} junction limit by 10^\circ\text{C}.

Engineering Comment

The design fails even though the heat sink may look large. Corrective options include more airflow, larger heat sink area, lower interface resistance, lower power dissipation, liquid cooling, lower ambient, or operating derating.

Exercise 3: Required Convection Area

A heat sink must reject 240\ \text{W} to forced air. The average allowable heat-sink surface temperature is 78^\circ\text{C} and inlet air is 38^\circ\text{C}. The estimated average convection coefficient is 55\ \text{W/(m}^2\text{K)}.

Estimate required effective convection area.

Solution

Use:

\dot{Q}=hA(T_s-T_\infty)

Rearrange:

\displaystyle A=\frac{\dot{Q}}{h(T_s-T_\infty)}
\displaystyle A=\frac{240}{55(78-38)}=0.109\ \text{m}^2

The heat sink needs about 0.11\ \text{m}^2 of effective convecting area.

Engineering Comment

Effective area is not just geometric fin area. Fin efficiency, flow bypass, blocked channels, dust, surface finish, and nonuniform base temperature can reduce useful area. The calculation is a sizing screen, not a final heat sink design.

Exercise 4: Airflow from Heat Balance

An enclosure must remove 1.8\ \text{kW} of heat with forced air. The design permits air to rise by 12^\circ\text{C} from inlet to outlet. Use air density 1.15\ \text{kg/m}^3 and heat capacity 1.0\ \text{kJ/(kg K)}.

Find required volumetric airflow.

Solution

Mass flow:

\displaystyle \dot{m}=\frac{\dot{Q}}{c_p\Delta T}
\displaystyle \dot{m}=\frac{1.8\ \text{kJ/s}}{1.0(12)}=0.150\ \text{kg/s}

Volumetric flow:

\displaystyle \dot{V}=\frac{\dot{m}}{\rho}=\frac{0.150}{1.15}=0.130\ \text{m}^3/\text{s}

The required airflow is about 0.13\ \text{m}^3/\text{s}.

Engineering Comment

This is bulk heat removal. It does not prove local component cooling. Air may bypass hot components, recirculate, short-circuit between inlet and outlet, or stagnate behind boards and cable bundles.

Exercise 5: Flow Regime in a Cooling Channel

Air flows through a rectangular cooling channel with hydraulic diameter D_h=8\ \text{mm}. Mean velocity is 6.0\ \text{m/s}. Use air density 1.15\ \text{kg/m}^3 and dynamic viscosity 1.9\times 10^{-5}\ \text{Pa s}.

Estimate Reynolds number and classify the flow.

Solution

Reynolds number:

\displaystyle Re=\frac{\rho v D_h}{\mu}
\displaystyle Re=\frac{1.15(6.0)(0.008)}{1.9\times 10^{-5}}=2905

The flow is near the transitional range for internal flow.

Engineering Comment

Heat-transfer correlations are sensitive near transition. A design based on a fully turbulent correlation may overpredict cooling. Geometry, roughness, entrance length, fan pulsation, and channel blockage should be reviewed.

Exercise 6: Liquid Coolant Flow for a Cold Plate

A cold plate removes 6.0\ \text{kW} from a power-electronics assembly. The coolant heat capacity is 3.7\ \text{kJ/(kg K)}. The allowable coolant temperature rise is 5^\circ\text{C}.

Find required coolant mass flow.

Solution

Use:

\dot{Q}=\dot{m}c_p\Delta T

Rearrange:

\displaystyle \dot{m}=\frac{\dot{Q}}{c_p\Delta T}
\displaystyle \dot{m}=\frac{6.0}{3.7(5)}=0.324\ \text{kg/s}

The required coolant mass flow is about 0.32\ \text{kg/s}.

Engineering Comment

The flow must be distributed through the cold plate, not merely supplied at the pump outlet. Manifold imbalance, trapped air, blocked microchannels, coolant degradation, or a partially closed valve can create local hot spots even if total flow looks sufficient.

Exercise 7: Pumping or Fan Power

A cooling duct requires volumetric airflow of 0.13\ \text{m}^3/\text{s}. The measured pressure rise across the fan is 420\ \text{Pa} and fan efficiency is 42\%.

Estimate fan shaft or electrical input power using the pressure-flow relation.

Solution

Fan power:

\displaystyle P_{fan}=\frac{\Delta p\dot{V}}{\eta}
\displaystyle P_{fan}=\frac{420(0.13)}{0.42}=130\ \text{W}

The fan input power is about 130\ \text{W} before any motor-drive or control losses not included in the efficiency.

Engineering Comment

Cooling power is not free. Fan or pump energy may heat the enclosure, reduce net efficiency, add acoustic noise, and create reliability concerns. Pressure drop should be measured with filters, grills, ducts, guards, and installed hardware in place.

Exercise 8: Ambient Derating

A heat-sinked device dissipates 120\ \text{W}. Its total thermal resistance to ambient is 0.55^\circ\text{C/W}. Maximum junction temperature is 125^\circ\text{C}.

Find maximum allowable ambient temperature.

Solution

Temperature rise:

\Delta T=P R_\theta=120(0.55)=66^\circ\text{C}

Maximum ambient:

T_{a,max}=T_{j,max}-\Delta T
T_{a,max}=125-66=59^\circ\text{C}

The device can dissipate 120\ \text{W} only if ambient stays at or below about 59^\circ\text{C} under the assumed cooling condition.

Engineering Comment

This is not the room ambient unless the heat sink actually sees room air. Inside an enclosure, local inlet air may be much hotter because of recirculation, neighboring components, solar load, or upstream heat sources.

Exercise 9: Validation from Measured Sink Temperature

A device dissipates 90\ \text{W}. During a thermal test, ambient air near the heat sink inlet is 40^\circ\text{C} and heat sink base temperature is 73^\circ\text{C}. The model predicted sink-to-ambient resistance of 0.30^\circ\text{C/W}.

Estimate measured sink-to-ambient resistance and compare with the model.

Solution

Measured resistance:

\displaystyle R_{\theta,SA,meas}=\frac{73-40}{90}=0.367^\circ\text{C/W}

Model error:

\displaystyle e_R=100\frac{0.367-0.30}{0.30}=22.3\%

The measured thermal resistance is about 22\% higher than predicted.

Engineering Comment

The difference is large enough to investigate. Possible causes include lower actual airflow, blocked fins, poor fan curve match, recirculation, incorrect power estimate, sensor location error, or a heat sink rating taken from a different test condition.

Release Checklist

Before accepting a forced-convection thermal design, confirm:

  1. component power is based on worst credible duty cycle;
  2. heat sink resistance matches installed airflow;
  3. interface material and mounting pressure are controlled;
  4. bulk airflow removes total heat;
  5. local airflow reaches hot components;
  6. pressure drop is measured in installed condition;
  7. fan or pump failure mode is defined;
  8. temperature is validated at the controlling node;
  9. margin remains under high ambient and fouled or degraded operation;
  10. sensor uncertainty does not erase the stated margin.

Heat sink and forced-convection sizing is successful when the calculated heat path, installed airflow, pressure drop, component temperature, and validation data agree inside the required operating envelope.

REF

See also

  1. Thermal Management, Heat Transfer, and Cooling Systems Mechanical Engineering
  2. Thermal Resistance Network Formula Sheet Mechanical Engineering
  3. Electronics Thermal Management Project Mechanical Engineering
  4. Thermal Runaway and Cooling Failure Case Study Mechanical Engineering
  5. Thermal Management and Heat Transfer Exercises Mechanical Engineering
  6. Thermal Energy Systems Formula Sheet Energy Engineering
  7. How Liquid Cooling Works in Data Centers Energy Engineering
  8. Data Center Power and Cooling Engineering Energy Engineering
  9. Power Electronics and Converter Systems Electrical Engineering
  10. Power and Sensor Interface Electronics Electronic Engineering
  11. Power and Sensor Interface Electronics Exercises Electronic Engineering
  12. Power and Sensor Interface Formula Sheet Electronic Engineering
  13. PCB Design, Power Integrity, and Signal Integrity Electronic Engineering
  14. PCB Design, Power Integrity, and Signal Integrity Exercises Electronic Engineering
  15. Electrical Machines and Motor Drive Systems Electrical Engineering
  16. Liquid Cooling Energy Engineering method
  17. Heat Duty Chemical Engineering quantity
  18. Heat Exchanger Energy Engineering device
  19. Heat Flux Energy Engineering quantity
  20. Junction Temperature Electronic Engineering quantity
  21. Thermal Stress Mechanical Engineering phenomenon
  22. Power Electrical Engineering quantity
  23. Flow Rate Mechanical Engineering quantity
  24. Density Materials Engineering quantity
  25. Viscosity Mechanical Engineering quantity
  26. Reynolds Number Mechanical Engineering quantity
  27. Nusselt Number Energy Engineering quantity
  28. Grashof Number Energy Engineering quantity
  29. Laminar Flow Mechanical Engineering phenomenon
  30. Turbulence Mechanical Engineering phenomenon
  31. Hydraulics Mechanical Engineering concept
  32. Thermocouple Electronic Engineering device
  33. Transducer Electronic Engineering device
  34. Reliability Industrial and Management Engineering metric
  35. Failure Mode Industrial and Management Engineering concept
  36. Validation Industrial and Management Engineering method
  37. Uncertainty Analysis Mathematical Engineering method
  38. Error Budget Electronic Engineering method
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