Heat Transfer Calculator
Heat transfer occurs through three fundamental mechanisms: conduction, convection, and radiation. Understanding these modes is essential for thermal management in engineering, building design, manufacturing, and everyday applications.
- Conduction: Direct molecular transfer through solids/liquids. Needs physical contact. Example: Spoon in hot soup.
- Convection: Heat carried by fluid motion (liquid/gas). Natural (buoyancy) or forced (fan/pump). Example: Room heater circulation.
- Radiation: Electromagnetic waves through vacuum/air. No medium needed. Example: Sun warming Earth.
Thermal resistance concept: Analogous to electrical resistance. Higher resistance = less heat flow. For conduction: R = d/(kA). For convection: R = 1/(hA). Total resistance in series: R_total = R₁ + R₂ + R₃.
Select heat transfer mode and enter required parameters:
- Conduction: Enter k, A, ΔT, d → Q = kAΔT/d
- Convection: Enter h, A, ΔT → Q = hAΔT
- Radiation: Enter ε, A, T₁, T₂ → Q = εσA(T₁⁴ - T₂⁴)
The calculator provides:
- Accurate calculations for all three heat transfer modes
- Material presets for common thermal conductivities and emissivities
- Unit conversions (SI & Imperial units)
- Thermal resistance calculation (R-value)
- Real-time results as you type
- Educational formulas and explanations
Reference values at room temperature (20°C):
| Material | Thermal Conductivity (k) | Density | Specific Heat | Common Use |
|---|---|---|---|---|
| Copper | 385 W/(m·K) | 8960 kg/m³ | 385 J/(kg·K) | Heat sinks, wiring |
| Aluminum | 205 W/(m·K) | 2700 kg/m³ | 900 J/(kg·K) | Radiators, cookware |
| Steel (carbon) | 50 W/(m·K) | 7850 kg/m³ | 500 J/(kg·K) | Structures, machinery |
| Glass | 0.8 W/(m·K) | 2500 kg/m³ | 840 J/(kg·K) | Windows, containers |
| Water | 0.6 W/(m·K) | 1000 kg/m³ | 4186 J/(kg·K) | Coolant, heating |
| Wood (oak) | 0.15 W/(m·K) | 750 kg/m³ | 2400 J/(kg·K) | Construction, furniture |
| Brick | 0.7 W/(m·K) | 1800 kg/m³ | 800 J/(kg·K) | Building walls |
| Fiberglass | 0.04 W/(m·K) | 50 kg/m³ | 840 J/(kg·K) | Wall insulation |
| Air (still) | 0.026 W/(m·K) | 1.2 kg/m³ | 1005 J/(kg·K) | Natural insulation |
| Aerogel | 0.016 W/(m·K) | 3 kg/m³ | 1000 J/(kg·K) | Advanced insulation |
Excellent insulators (k < 0.1): Aerogel, vacuum panels, polyurethane foam
Good insulators (0.1 < k < 0.5): Fiberglass, mineral wool, cork, wood
Moderate insulators (0.5 < k < 2): Brick, concrete, glass, water
Poor insulators (k > 2): Most metals (good conductors)
Below are answers to frequently asked questions about heat transfer calculations:
For multiple layers in series, calculate thermal resistance for each layer and sum:
- Calculate R for each layer: R = thickness/(k × area)
- Total resistance: R_total = R₁ + R₂ + R₃ + ...
- Heat transfer: Q = ΔT / R_total
Example: Wall with brick (0.1m, k=0.7), insulation (0.05m, k=0.04), plaster (0.02m, k=0.5). Area=10m², ΔT=20K.
R_brick = 0.1/(0.7×10)=0.0143 K/W
R_insulation = 0.05/(0.04×10)=0.125 K/W
R_plaster = 0.02/(0.5×10)=0.004 K/W
R_total = 0.1433 K/W
Q = 20/0.1433 = 139.6 W
R-value (thermal resistance) and k-value (conductivity) are related through thickness:
R (m²·K/W) = thickness (m) / k (W/(m·K))
R (ft²·°F·h/Btu) = thickness (ft) / k (Btu·in/(h·ft²·°F))
1 m²·K/W = 5.678 ft²·°F·h/Btu
Common R-values: R-13 insulation = 2.3 m²·K/W
Example: Fiberglass insulation k=0.04 W/(m·K), thickness=0.1m → R = 0.1/0.04 = 2.5 m²·K/W = R-14.2 in US units.
HVAC design involves calculating heating/cooling loads through building envelopes:
| Component | Heat Transfer Mode | Calculation | Design Consideration |
|---|---|---|---|
| Windows | Conduction + Radiation | Q = U × A × ΔT | Double/triple glazing, low-E coatings |
| Walls | Conduction | Q = A × ΔT / R | Insulation thickness, vapor barriers |
| Roof | Conduction + Radiation | Q = kAΔT/d + radiation | Radiant barriers, ventilation |
| Ventilation | Convection | Q = ṁ × c_p × ΔT | Air changes/hour, heat recovery |
| People/Equipment | Internal Gains | Q = sensible + latent heat | Occupancy, equipment power |
Total heating load: Sum of all heat losses + ventilation load - internal gains.
ASHRAE standards: Provide detailed calculation methods for different climates and building types.
Electronic components require efficient heat removal to prevent overheating:
- Conduction: Heat sinks, thermal paste, heat pipes
- Convection: Natural (fins) or forced (fans, liquid cooling)
- Radiation: Often negligible at electronics temperatures
- Phase change: Heat pipes, vapor chambers (very effective)
Thermal resistance network: From junction to ambient: R_jc (junction-case) + R_cs (case-sink) + R_sa (sink-ambient).
Example: CPU 100W, R_jc=0.5 K/W, R_cs=0.1 K/W, R_sa=0.4 K/W. ΔT = 100W × (0.5+0.1+0.4) = 100K temperature rise.