Heat Transfer Calculator
Fast thermal analysis • 2026 standards
Heat Transfer Formulas:
Show the calculatorConduction: Fourier's Law: \( q = -kA \frac{dT}{dx} \)
Convection: Newton's Law: \( q = hA(T_s - T_∞) \)
Radiation: Stefan-Boltzmann Law: \( q = εσA(T_s^4 - T_∞^4) \)
Where:
- \( q \) = heat transfer rate (W)
- \( k \) = thermal conductivity (W/m·K)
- \( A \) = surface area (m²)
- \( h \) = convective heat transfer coefficient (W/m²·K)
- \( ε \) = emissivity (dimensionless)
- \( σ \) = Stefan-Boltzmann constant (5.67×10⁻⁸ W/m²·K⁴)
- \( T_s \) = surface temperature (K)
- \( T_∞ \) = ambient temperature (K)
These fundamental equations describe the three mechanisms of heat transfer: conduction (through materials), convection (between surfaces and fluids), and radiation (electromagnetic waves).
Example: For a steel plate with k=50 W/m·K, area=1 m², thickness=0.1 m, with T₁=100°C and T₂=20°C:
\( q = kA \frac{ΔT}{L} = 50 \times 1 \times \frac{80}{0.1} = 40,000 \) W
Thus, the heat transfer rate is 40,000 W.
Thermal Parameters
Advanced Options
Results
| Parameter | Value | Unit |
|---|---|---|
| Temperature Difference | 80.00 | K |
| Surface Area | 1.00 | m² |
| Thickness | 0.10 | m |
| Thermal Conductivity | 50.00 | W/m·K |
| Parameter | Value | Unit |
|---|---|---|
| Thermal Resistance | 0.002 | K/W |
| Heat Capacity | 0.00 | J/K |
| Energy Transfer | 0.00 | kWh |
| Biot Number | 0.00 | - |
Comprehensive Heat Transfer Guide
Heat transfer is the process of thermal energy moving from a region of higher temperature to a region of lower temperature. It occurs through three fundamental mechanisms: conduction (within materials), convection (between surfaces and fluids), and radiation (through electromagnetic waves). Understanding heat transfer is essential in engineering applications ranging from HVAC systems to electronics cooling.
Where each mechanism follows different physical laws:
- Conduction: Fourier's Law: \( q = -kA \frac{dT}{dx} \)
- Convection: Newton's Law: \( q = hA(T_s - T_∞) \)
- Radiation: Stefan-Boltzmann Law: \( q = εσA(T_s^4 - T_∞^4) \)
Heat transfer calculations are essential for various engineering applications:
- Building Design: Insulation requirements and HVAC system sizing
- Electronics: Component cooling and thermal management
- Power Generation: Heat exchanger design and efficiency optimization
- Manufacturing: Furnace design and material processing
- Aerospace: Thermal protection systems and environmental control
- Thermal Resistance: Combine resistances in series and parallel like electrical circuits
- Heat Exchangers: Consider log mean temperature difference (LMTD)
- Phase Change: Account for latent heat during melting/freezing
- Transient Effects: Consider time-dependent temperature changes
- Thermal Stress: Account for differential expansion in design
Heat Transfer Fundamentals
Process of thermal energy transfer from high to low temperature regions through three mechanisms.
\(q = q_{cond} + q_{conv} + q_{rad}\)
Where each component follows specific physical laws and material properties.
- Heat flows from hot to cold objects
- Conduction increases with temperature gradient
- Radiation depends on fourth power of temperature
Engineering Applications
Opposition to heat flow, analogous to electrical resistance in circuits.
- Calculate individual resistances
- Combine in series or parallel
- Apply Ohm's law equivalent
- Verify boundary conditions
- Steady-state vs transient analysis
- Material property variations
- Geometric complexities
- Boundary condition effects
Heat Transfer Learning Quiz
Which of the following heat transfer mechanisms can occur in a vacuum?
The answer is C) Radiation only. Radiation heat transfer occurs through electromagnetic waves and does not require a medium, so it can occur in a vacuum. Conduction requires direct contact between particles, and convection requires a fluid medium to carry the heat. This is why the sun can transfer heat across the vacuum of space to Earth through radiation.
Understanding the differences between heat transfer mechanisms is fundamental to thermal engineering. Conduction occurs through molecular vibrations in solids, convection involves the bulk movement of fluids, and radiation travels as electromagnetic waves. This knowledge is crucial for designing thermal systems in space applications, vacuum furnaces, and insulation systems.
Conduction: Heat transfer through direct contact between particles in a material
Convection: Heat transfer through the movement of fluids (liquids or gases)
Radiation: Heat transfer through electromagnetic waves
• Conduction needs a solid medium
• Convection needs a fluid medium
• Radiation works in vacuum
• Remember: CONduction needs CONtact, CONvection needs a CONducting fluid
• Radiation is the only mechanism that works in space
• All three can occur simultaneously in real systems
• Thinking convection can occur in a vacuum
• Assuming conduction can occur without matter
• Forgetting that radiation is temperature dependent
A steel pipe (k=50 W/m·K) with outer diameter 0.1 m and length 10 m carries steam at 200°C. The pipe is insulated with 0.05 m thick insulation (k=0.04 W/m·K). The ambient air temperature is 25°C with a convection coefficient of 10 W/m²·K. Calculate the heat loss per meter of pipe length.
This is a combined heat transfer problem involving conduction through the pipe wall and insulation, plus convection at the outer surface. We'll use the thermal resistance approach:
For cylindrical coordinates, thermal resistance is: R = ln(r₂/r₁) / (2πkL)
Given: r₁=0.05 m (inner radius), r₂=0.05 m (outer pipe), r₃=0.1 m (outer insulation)
R_pipe = ln(0.05/0.05) / (2π × 50 × 1) = 0 (negligible)
R_insulation = ln(0.1/0.05) / (2π × 0.04 × 1) = 0.693 / 0.251 = 2.76 K/W
R_convection = 1 / (h × A) = 1 / (10 × 2π × 0.1 × 1) = 1 / 6.28 = 0.159 K/W
Total resistance: R_total = 0 + 2.76 + 0.159 = 2.92 K/W
Heat loss per meter: q = ΔT / R_total = (200-25) / 2.92 = 60.0 W/m
This problem demonstrates the importance of thermal resistance networks in solving complex heat transfer problems. By treating each layer as a resistor in series, we can simplify the analysis. The key insight is recognizing that the steel pipe's resistance is negligible compared to the insulation, which is often the case in practical applications. The logarithmic relationship in cylindrical coordinates is crucial for accurate calculations.
Thermal Resistance: Opposition to heat flow, R = ΔT/q
Cylindrical Coordinates: Geometry where R = ln(r₂/r₁)/(2πkL)
Insulation: Material with low thermal conductivity to minimize heat transfer
• Use appropriate geometric formula for thermal resistance
• Combine resistances in series like electrical circuits
• Insulation effectiveness depends on thickness and conductivity
• For cylinders: R = ln(r₂/r₁)/(2πkL)
• Always check if metal resistances are negligible
• Insulation performance improves with thickness
• Using flat wall formulas for cylindrical geometry
• Forgetting to account for surface area in convection
• Adding resistances instead of combining in series
FAQ
Q: How do you determine which heat transfer mechanism dominates in a given situation?
A: The dominant heat transfer mechanism depends on the specific conditions. Generally:
Conduction: Dominates in solid materials with high temperature gradients. Calculate using thermal conductivity (k) and thickness (L). The thermal resistance is R_cond = L/(kA).
Convection: Dominates at fluid-solid interfaces. The heat transfer coefficient (h) ranges from 5-25 W/m²·K for natural convection to 100-10,000 W/m²·K for forced convection. Resistance is R_conv = 1/(hA).
Radiation: Becomes significant at high temperatures (T⁴ relationship). At room temperature, radiation is typically negligible, but becomes dominant above ~500°C. The net radiation heat transfer is q_rad = εσA(T_s⁴ - T_∞⁴).
To determine dominance, calculate the thermal resistances for each mechanism and compare them. The smallest resistance (highest conductance) corresponds to the dominant mechanism.
Q: What is the significance of the Biot number in heat transfer analysis?
A: The Biot number (Bi) is a dimensionless parameter that compares internal thermal resistance to external thermal resistance: Bi = hL_c/k
Where h is the convective heat transfer coefficient, L_c is the characteristic length (volume/surface area), and k is the thermal conductivity.
The significance is:
- Bi < 0.1: Lumped capacitance method applies (uniform temperature within object)
- 0.1 ≤ Bi ≤ 10: Both internal and external resistances are important
- Bi > 10: External resistance is negligible compared to internal resistance
For example, in a steel sphere (k=50 W/m·K) with h=10 W/m²·K and diameter 0.1 m: L_c = V/A = (4πr³/3)/(4πr²) = r/3 = 0.0167 m
Bi = (10 × 0.0167) / 50 = 0.0033
Since Bi < 0.1, we can assume uniform temperature distribution within the sphere during heating/cooling.