Heat Transfer Calculator | Conduction, Convection & Radiation

Heat Transfer Calculator | ProEngCalc

🔥 Thermodynamics

Heat Transfer Calculator

Calculate conductive, convective, and radiative heat transfer rates

Reference: Fourier’s Law | Newton’s Law of Cooling | Stefan-Boltzmann Law

Heat Transfer Mode

Heat Transfer Rate (Q)

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📐 Solution Breakdown

Variable Definitions

Symbol	Variable	Unit	Mode
Q	Heat Transfer Rate	W (J/s)	All
k	Thermal Conductivity	W/m·K	Conduction
A	Cross-Section / Surface Area	m²	All
ΔT	Temperature Difference	K or °C	All
Δx	Thickness (conduction path)	m	Conduction
h	Convection Coefficient	W/m²·K	Convection
ε	Emissivity	0–1 dimensionless	Radiation
σ	Stefan-Boltzmann Constant	5.67×10⁻⁸ W/m²·K⁴	Radiation

⚠ Assumptions & Limits

Conduction: Assumes steady-state, one-dimensional conduction through a uniform slab with constant thermal conductivity
Convection: Newton’s Law of Cooling applies for forced and natural convection. The convection coefficient h is empirically determined and highly dependent on flow conditions and geometry
Radiation: Assumes blackbody or graybody radiation with uniform emissivity. View factor assumed = 1 (surface completely surrounded by environment). For partial view factors, multiply result by F₁₂
Temperatures for radiation must be in Kelvin (absolute). This calculator handles the conversion automatically
Combined heat transfer modes must be summed separately for total heat flux

The Three Modes of Heat Transfer

Heat transfer is the movement of thermal energy from a region of higher temperature to lower temperature. There are three fundamental mechanisms: conduction (through solid matter), convection (by fluid motion), and radiation (via electromagnetic waves). Every thermal engineering problem involves one or more of these modes, and understanding which dominates in a given situation is the first step to an accurate analysis.

Conduction (Fourier)

Q = kA(ΔT/Δx)

Convection (Newton)

Q = hA(T_s − T_∞)

Radiation (S-B Law)

Q = εσA(T₁⁴ − T₂⁴)

Heat Flux

q = Q / A (W/m²)

Thermal Resistance

R_th = Δx / (kA)

Overall U-value

Q = UA(ΔT)

Typical Convection Coefficient Reference

Condition	h (W/m²·K)	Application
Free convection — air	2–25	Passive cooling, natural ventilation
Forced convection — air	25–250	Fans, HVAC ducts, heat sinks
Free convection — water	200–1,000	Immersion cooling, tanks
Forced convection — water	1,000–15,000	Heat exchangers, cooling jackets
Boiling water	2,500–35,000	Evaporators, steam generators
Condensing steam	5,000–100,000	Condensers, heat exchangers
Liquid metals	10,000–100,000	Nuclear reactor cooling

Worked Examples

Example 1 — Insulated Pipe Heat Loss (Conduction)

A steam pipe (inner wall 180°C) is covered with 75mm of mineral wool insulation (k = 0.04 W/m·K). Pipe outer diameter before insulation = 100mm. What is the heat loss per metre of pipe length?

Given

T_inner = 180°C | T_outer = 25°C | Δx = 0.075 m | k = 0.04 W/m·K

Area (per metre)

A = π × D_mid × L = π × 0.175 × 1 = 0.550 m² (using mean diameter)

Heat Loss

Q = kAΔT/Δx = 0.04 × 0.550 × (180−25) / 0.075

Solution

Q ≈ 45.5 W/m of pipe length

For accurate cylindrical pipe calculations use the log-mean area: Q = 2πkL(T₁−T₂)/ln(r₂/r₁). The flat-slab approximation used here overestimates by ~15% for thick insulation. Use this calculator for preliminary sizing then verify with cylindrical coordinates.

Example 2 — Electronic Component Cooling (Convection)

A power transistor dissipates 45W and has a case surface area of 0.008 m². A heat sink with fan provides h = 85 W/m²·K. What is the steady-state case temperature if ambient is 40°C?

Given

Q = 45 W | A = 0.008 m² | h = 85 W/m²·K | T_∞ = 40°C

Rearrange

T_s = T_∞ + Q/(hA) = 40 + 45/(85 × 0.008)

Solution

T_case = 40 + 66.2 = 106.2°C

106°C exceeds most semiconductor junction limits. Options: increase heat sink area, improve h with better fan (h→200 W/m²·K gives T_s = 68°C), or use a larger heat sink. Target T_case < 70°C for reliability — this design requires significant improvement.

Example 3 — Industrial Furnace Wall Radiation

A furnace wall (steel, ε = 0.7) at 800°C radiates heat to the surrounding environment at 25°C. Surface area = 4 m². Calculate radiative heat loss.

Convert to Kelvin

T₁ = 800 + 273.15 = 1073.15 K | T₂ = 25 + 273.15 = 298.15 K

Radiation

Q = 0.7 × 5.67×10⁻⁸ × 4 × (1073.15⁴ − 298.15⁴)

T⁴ values

T₁⁴ = 1.325×10¹² | T₂⁴ = 7.898×10⁹

Solution

Q = 0.7 × 5.67×10⁻⁸ × 4 × 1.317×10¹² = 209,300 W = 209 kW

Radiation dominates heat transfer at high temperatures because it scales as T⁴. At 800°C the T₁⁴ term is 167× larger than T₂⁴, so the environment temperature barely matters. Compare to convection at the same condition: Q_conv = 25×4×775 = 77,500 W — radiation is 2.7× larger than forced air convection.

Real World Applications

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Electronics Cooling

Thermal management of CPUs, power electronics, and LED drivers — combining conduction through heat spreaders with forced convection via heat sinks.

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Heat Exchanger Design

Shell-and-tube, plate, and air-cooled heat exchangers use convection on both sides with conduction through the tube wall.

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Building Insulation

U-value calculations for walls, roofs, and windows combining conduction through multiple layers with surface convection coefficients.

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Furnace & Kiln Design

High-temperature process equipment where radiation dominates heat transfer and refractory lining thickness is sized for conductive heat loss.

Common Mistakes Engineers Make

❌ Using Celsius Instead of Kelvin for Radiation

The Stefan-Boltzmann law requires absolute temperature (Kelvin) raised to the fourth power. Using Celsius gives completely wrong results. 100°C = 373.15 K — these are very different numbers when raised to the power of 4. Always convert to Kelvin: K = °C + 273.15.

❌ Assuming Emissivity = 1 for All Surfaces

Emissivity varies enormously: polished aluminum ε = 0.05, oxidized steel ε = 0.7–0.8, blackbody ε = 1.0. Assuming ε = 1 for a polished metal surface overestimates radiation heat transfer by 15–20×. Always use actual surface emissivity from published tables.

❌ Guessing the Convection Coefficient h

The convection coefficient h can range from 2 to 100,000 W/m²·K depending on fluid, flow velocity, geometry, and phase change. Using a generic value from a table without verifying the flow regime and geometry can produce errors of an order of magnitude. For important designs, calculate h from Nusselt number correlations (Dittus-Boelter, Churchill-Bernstein, etc.) appropriate for your specific geometry.

Frequently Asked Questions

What is thermal resistance and how do I use it for multi-layer problems?

Thermal resistance R_th = Δx/(kA) for conduction, 1/(hA) for convection, and 1/(εσA(T₁²+T₂²)(T₁+T₂)) for radiation (linearized). For layers in series (wall + insulation + surface film), total resistance = R₁ + R₂ + R₃ and Q = ΔT_total/R_total. This is exactly analogous to electrical resistance — Ohm’s law with temperature as voltage and heat flow as current.

When does radiation become significant compared to convection?

Radiation scales with T⁴ and convection scales with ΔT. At low temperatures (below ~300°C) and with forced convection, radiation is typically negligible. Above 500°C radiation usually dominates. At ambient temperatures with free convection, radiation can contribute 30–50% of total heat loss from an uninsulated surface — this is why radiation is included in building heat loss calculations even at room temperature.

What is the overall heat transfer coefficient (U-value) and how is it calculated?

The U-value combines all heat transfer resistances from hot fluid to cold fluid: 1/U = 1/h₁ + Δx₁/k₁ + Δx₂/k₂ + … + 1/h₂ (per unit area). Used in Q = UA×LMTD for heat exchanger design. For a building wall: 1/U = 1/h_inside + Σ(Δx/k) + 1/h_outside. Typical well-insulated wall: U ≈ 0.3 W/m²·K. Double-glazed window: U ≈ 1.4–2.0 W/m²·K.