Insulation Thickness from Heat Loss Calculator: worked example
This example is generated from PROFENG's own default input set and the implemented relationship. It is not copied from a cited publication.
Relationship
L = kAΔT/Q̇
This calculator determines required thickness from thermal conductivity, area, temperature difference, allowable heat loss. Use it to screen energy, temperature and heat-transfer relationships with explicitly stated properties and boundary conditions.
Inputs
- Thermal conductivity (k) = 0.04 W/(m·K)
- Area (area) = 50 m²
- Temperature difference (deltaT) = 20 K
- Allowable heat loss (heat) = 400 W
Working
(0.04)*(50)*(20)/(400) = 0.1 m
Required thickness: 0.1 m
Assumptions and limits
- Properties entered by the user are treated as constant over the calculation unless the formula explicitly varies them.
- Temperatures and temperature differences must use the unit convention shown by the calculator.
- The model is lumped or one-dimensional where implied by the displayed relationship.
- Steady one-dimensional plane-wall conduction is assumed.
- Thermal conductivity is treated as constant.
- Surface-film, contact and radiation resistances are not represented.
Formula source and acknowledgement
Theodore L. Bergman, Adrienne S. Lavine, Frank P. Incropera, David P. DeWitt, Fundamentals of Heat and Mass Transfer, 8th, 2018, Wiley. Credit belongs to the cited author(s)/publisher; PROFENG presents its own implementation and does not imply endorsement.
Automated verification passed – 30 September 2026 – six numerical cases checked, with dimensional checks and an independent implementation where applicable.
This worked example is for engineering screening and education. Replace defaults with traceable project data and confirm the method, assumptions, standards and equipment information before design, procurement or operation.