Heat Conduction Calculator: Multi-Layer Walls & Cylinders
Calculate steady-state heat transfer through plane walls and cylindrical pipes. Build thermal resistance networks with multi-layer walls and convection, and compute heat rate, heat flux, and U-values.
Last Updated: February 2026. R-values and U-factors referenced from ASHRAE Fundamentals 2021 and ISO 10456.
A thermal resistance calculator helps you select insulation thickness, verify U-values for code compliance, and compare wall assemblies. An architect specified R-19 insulation for 2×6 walls, assuming that matched the cavity depth. But the actual R-value after accounting for studs, drywall, sheathing, and air films came to R-14—30% worse than expected. Thermal bridging through framing members short-circuits insulation, and ignoring it means buildings fail energy code inspections.
This page covers steady-state conduction through plane and cylindrical walls, series resistance networks, and selection criteria for insulation materials. Whether you're sizing pipe insulation, designing building envelopes, or analyzing heat exchangers, the resistance-addition principle is the same—but material selection and code requirements differ by application.
Selection Guide: Insulation Materials by Application
| Material | k (W/m·K) | R per inch (ft²·°F·h/BTU) | Typical Application |
|---|---|---|---|
| Fiberglass batt | 0.040 | R-3.2 | Wall cavities, attics |
| Mineral wool | 0.035 | R-3.7 | Fire-rated assemblies, high-temp |
| XPS foam | 0.029 | R-5.0 | Below-grade, exterior sheathing |
| Polyiso foam | 0.023 | R-6.5 | Roofing, continuous insulation |
| Spray foam (closed) | 0.024 | R-6.0 | Air sealing + insulation |
| Calcium silicate | 0.055 | R-2.6 | High-temp pipe insulation |
Higher R per inch means less thickness needed—but cost, moisture resistance, fire rating, and installation constraints often drive selection more than thermal performance alone.
Multi-Layer Walls: Series Thermal Resistances
Heat flows through building walls in series: inside air film → drywall → insulation → sheathing → siding → outside air film. Each layer adds its own thermal resistance, and they sum directly: R_total = R₁ + R₂ + R₃ + ... For conduction through a uniform layer: R = L/(k·A), where L is thickness, k is thermal conductivity, and A is area.
Example: 2×6 wall assembly
| Inside air film | R-0.68 |
| ½" drywall | R-0.45 |
| 5.5" fiberglass | R-19 |
| ½" OSB sheathing | R-0.62 |
| Outside air film | R-0.17 |
| Total (clear cavity) | R-20.9 |
But this ignores framing. Studs at 16" o.c. occupy about 25% of the wall area, and wood (R-1.25/inch) has much lower R-value than insulation. The area-weighted average drops to R-14–15. This is why continuous insulation on the exterior is often required to meet code.
U-Value Selection for Building Envelopes
Building codes specify maximum U-factors (U = 1/R_total in metric, or 1/(R_total × A) for total heat flow). Lower U means better insulation. ASHRAE 90.1 and IECC set requirements by climate zone:
| Climate Zone | Wall U-factor max | Roof U-factor max | Approx. R-value |
|---|---|---|---|
| Zone 4 | 0.084 | 0.032 | R-12 wall, R-31 roof |
| Zone 5 | 0.064 | 0.032 | R-16 wall, R-31 roof |
| Zone 6 | 0.064 | 0.028 | R-16 wall, R-36 roof |
| Zone 7-8 | 0.057 | 0.028 | R-18 wall, R-36 roof |
Selection rule: Code U-factors are assembly values including framing. To hit U-0.064, you often need R-20 cavity insulation plus R-5 continuous exterior insulation—not just R-16 in the stud bays.
Cylindrical Conduction: Pipe Insulation Thickness
For pipes, heat flows radially through surfaces of increasing area. The resistance formula changes: R_cyl = ln(r₂/r₁)/(2πkL), where r₁ and r₂ are inner and outer radii, k is conductivity, and L is pipe length. The logarithm arises from integrating Fourier's law in cylindrical coordinates.
Example: 2" NPS pipe (r₁ = 30.5 mm), 2" insulation (r₂ = 81.5 mm), k = 0.04 W/m·K
R = ln(81.5/30.5)/(2π × 0.04 × 1) = 0.98/(0.251) = 3.9 K·m/W per meter length
Pipe insulation is specified by nominal pipe size and insulation thickness. ASTM C585 covers dimensional standards. For steam and hot water, thickness selection balances heat loss cost against insulation cost—economic thickness analysis.
Critical Radius of Insulation Concept
Counter-intuitively, adding insulation to a small pipe can increase heat loss. The critical radius r_cr = k/h (k = insulation conductivity, h = external convection coefficient) is the outer radius where heat loss is maximum. Below this radius, adding insulation increases the surface area faster than it adds thermal resistance.
For typical insulation (k = 0.04 W/m·K) with natural convection in air (h = 10 W/m²·K): r_cr = 0.04/10 = 4 mm. For pipes larger than 8 mm diameter, adding insulation always reduces heat loss. The critical radius concept matters for electrical wires and very small tubes—rarely for typical pipe insulation.
Practical impact: For most pipe insulation applications (NPS ½" and larger), the critical radius is exceeded by the bare pipe itself. Don't worry about it unless you're insulating instrument tubing or electrical conductors.
Worked Example: Heat Loss Through a Composite Wall
Problem: A wall consists of 100 mm brick (k = 0.7 W/m·K), 50 mm air gap (R = 0.18 m²·K/W), and 13 mm plasterboard (k = 0.16 W/m·K). Inside T = 20°C with h_i = 8 W/m²·K; outside T = 0°C with h_o = 25 W/m²·K. Find U-value and heat flux.
Step 1: Calculate individual resistances (per m²)
R_i = 1/h_i = 1/8 = 0.125 m²·K/W
R_plaster = L/k = 0.013/0.16 = 0.081 m²·K/W
R_air = 0.18 m²·K/W (given)
R_brick = L/k = 0.1/0.7 = 0.143 m²·K/W
R_o = 1/h_o = 1/25 = 0.04 m²·K/W
Step 2: Sum resistances
R_total = 0.125 + 0.081 + 0.18 + 0.143 + 0.04 = 0.569 m²·K/W
Step 3: Calculate U-value
U = 1/R_total = 1/0.569 = 1.76 W/m²·K
Step 4: Calculate heat flux
q″ = U × ΔT = 1.76 × 20 = 35.2 W/m²
This U = 1.76 W/m²·K is typical of uninsulated masonry walls—roughly 10× worse than code requirements. Adding 100 mm of mineral wool (k = 0.035) would add R = 2.86 m²·K/W, dropping U to 0.29 W/m²·K.
Material Selection: k Values by Application
Thermal conductivity spans orders of magnitude. Selection depends on temperature range, moisture exposure, fire requirements, and cost:
Low-Temperature (<100°C)
- • Fiberglass: 0.040 W/m·K, low cost
- • XPS/EPS foam: 0.029–0.038, moisture resistant
- • Polyiso: 0.023, highest R/inch
- • Spray foam: 0.024–0.038, air sealing
High-Temperature (>150°C)
- • Mineral wool: 0.035, rated to 650°C
- • Calcium silicate: 0.055, to 1000°C
- • Ceramic fiber: 0.08, to 1260°C
- • Microporous: 0.020, extreme high-temp
For building insulation, foam boards have the best R/inch but may require thermal barriers per fire code. For pipe insulation, material must handle operating temperature and be formable around pipes. Always check maximum service temperature before selecting.
Steady-State Assumption Limits
All resistance calculations here assume steady state—temperatures aren't changing with time. This is valid when boundary conditions are constant long enough for the temperature profile to stabilize. The characteristic time τ = L²/(α) (where α = k/(ρ·c_p) is thermal diffusivity) indicates how quickly a material responds.
| Material | α (m²/s) | τ for 100mm | Steady-state valid? |
|---|---|---|---|
| Aluminum | 8.4 × 10⁻⁵ | ~2 minutes | Yes, after short transient |
| Concrete | 5.2 × 10⁻⁷ | ~5 hours | Depends on load cycle |
| Fiberglass | 3.3 × 10⁻⁷ | ~8 hours | Marginal for daily cycles |
For daily heating/cooling cycles, massive walls (concrete, masonry) may never reach steady state—transient analysis is needed. For process equipment with constant operating temperature, steady state is usually valid after startup.
ASHRAE Fundamentals and ISO 10456
Two authoritative sources for thermal property data:
ASHRAE Handbook—Fundamentals
Chapter 26 covers heat transfer principles; Chapter 27 provides thermal properties of building materials including R-values, conductivities, and surface film coefficients. Updated every 4 years. The US industry standard for building thermal calculations.
ISO 10456:2007
"Building materials and products—Hygrothermal properties." Provides design values for thermal conductivity at reference conditions (23°C, dry) and methods for adjusting to other conditions. International standard used in Europe and many other countries.
ISO 6946:2017
"Building components and building elements—Thermal resistance and thermal transmittance—Calculation methods." Specifies how to calculate R-values for building assemblies including air layers, surface resistances, and thermal bridging corrections.
Practical note: Published k values are "declared" or "design" values that include safety margins. Test values on specific samples may differ. For code compliance, use values from approved sources (ASHRAE, manufacturer's ICC-ES reports, or tested per ASTM C518/C177).
Limitations and Assumptions
One-Dimensional Heat Flow
Calculations assume heat flows perpendicular to surfaces. Thermal bridging through studs, corners, and penetrations requires 2D/3D analysis or correction factors.
Constant Material Properties
k is assumed constant. In reality, conductivity varies with temperature (especially for insulation) and moisture content. Use temperature-corrected values for high-temp applications.
No Radiation
Radiant heat transfer is not modeled. Above ~200°C or across large air gaps, radiation becomes significant and must be added separately.
Perfect Contact Between Layers
Contact resistance at interfaces is ignored. For metal-to-metal joints or poor construction, add thermal contact resistance (typically 10⁻⁴ to 10⁻² m²·K/W).
Sources and References
- ASHRAE Handbook—Fundamentals – Thermal properties and R-values for building materials
- ISO 10456:2007 – Hygrothermal properties of building materials
- ISO 6946:2017 – Thermal resistance calculation methods
- Incropera, DeWitt, et al. Fundamentals of Heat and Mass Transfer, 8th ed. (2017)
- Engineering Toolbox – Thermal conductivity reference tables
All R-values and conductivities referenced to 24°C mean temperature unless otherwise noted. Foam insulation R-values are aged values per LTTR method.
Troubleshooting Thermal Resistance and U-Value Calculations
Real questions from engineers and builders stuck on R-value discrepancies, thermal bridging corrections, pipe vs. flat-wall formulas, and why calculated insulation performance doesn't match measured results.
I calculated R-19 for my wall but the energy auditor measured R-14—where did I lose 5 R-value?
Thermal bridging through studs. Wood studs (k ≈ 0.12 W/m·K) conduct heat much faster than fiberglass (k ≈ 0.04). In a typical 2×6 wall at 16" O.C., studs occupy about 25% of the area. The parallel path calculation gives R_effective = 1/(0.75/R_cavity + 0.25/R_stud). With R-19 batts between R-4 studs, you get R_eff ≈ 13–14. Steel studs are worse. This is why continuous exterior insulation exists—it covers the bridges.
My pipe insulation thickness came out to 2 inches, but the supplier only has 1.5 or 2.5—which should I pick?
Usually go thicker (2.5"), but check economics first. More insulation means higher material cost but lower heat loss and operating cost. Calculate the payback: if the extra 0.5" saves $X/year in energy and costs $Y to install, payback = Y/X. For high-temperature steam lines, err thicker—personnel safety matters too. For chilled water, thicker prevents condensation on the jacket surface.
I added 2 inches of insulation to our hot water tank but barely noticed a difference—did I install it wrong?
Check the top and bottom. Heat rises, so the top loses more than the sides. If you only wrapped the cylindrical surface and left the top bare, you might have insulated only 60% of the loss area. Also check for gaps at seams—a 1/4" gap can act like a thermal short circuit. Finally, compare to baseline: a well-insulated modern tank may already have R-12 built in, so adding R-8 externally only improves by 40%.
The critical radius formula gives 0.5 inches, but my pipe is 2 inches OD—does critical radius even matter here?
No, it doesn't matter for your case. Critical radius r_cr = k/h applies to small-diameter pipes where adding insulation initially increases surface area faster than thermal resistance. At 2" OD, you're well past r_cr ≈ 0.5" for typical insulation (k = 0.04, h = 10). Every inch of insulation you add will reduce heat loss. Critical radius is mainly a concern for thin wires and small tubing—anything under about 1" diameter.
My U-value calculation gives 0.28 W/m²·K but code requires 0.25—can I just round down?
Absolutely not. Code inspectors won't accept 0.28 when the limit is 0.25. You need to either add more insulation (maybe go from R-20 to R-23), reduce thermal bridging (add continuous insulation), or improve window specs if they're in the same assembly. The 0.03 difference seems small but represents 12% more heat loss than code allows. Recalculate with specific upgrades until you genuinely hit ≤ 0.25.
I'm designing a freezer wall and keep getting condensation between layers—how do I figure out where the dew point falls?
Calculate interface temperatures through the wall. At each layer boundary, T_interface = T_hot − (R_cumulative/R_total) × ΔT. Then compare each T_interface to the dew point of the warm-side air. Where T_interface < T_dew, moisture condenses. The fix: put a vapor barrier on the warm side, or rearrange layers so the temperature doesn't cross the dew point inside the wall. In freezer walls, the vapor barrier goes on the warm (outside) face.
My textbook uses R-value in ft²·°F·h/BTU but my software wants SI units—how do I convert without messing up?
R_SI = R_IP × 0.1761. So R-19 (IP) = 3.35 m²·K/W. Going the other way, R_IP = R_SI × 5.678. For U-values: U_SI = U_IP × 5.678 (since U = 1/R and the conversion factor inverts). Keep units consistent throughout any calculation—mixing IP and SI in the same formula is the most common error. When in doubt, convert everything to SI at the start and convert back at the end.
I specified mineral wool insulation but the contractor wants to substitute fiberglass—will my R-values change?
Slightly, but probably acceptable. Mineral wool (rock wool) typically has k ≈ 0.035–0.040 W/m·K; fiberglass batts are k ≈ 0.040–0.045. For the same thickness, mineral wool gives 5–15% higher R-value. If your design had margin, fiberglass may still work. If you were tight on code compliance, insist on mineral wool or increase thickness. Also consider: mineral wool is denser, handles moisture better, and has higher fire resistance.
The wall U-value I calculated doesn't match the ASHRAE table for the same construction—what's different?
ASHRAE tables include standard surface films (R_si ≈ 0.12, R_so ≈ 0.03 m²·K/W for still air and 24 km/h wind), air gaps, and thermal bridging corrections. If you only summed layer resistances without surface films, you're missing about R-0.15 total. If you ignored framing, you're overestimating R by 15–25%. ASHRAE values represent whole-assembly performance including these effects. Match your assumptions to theirs for comparison.
I'm getting different answers for the same pipe insulation problem using flat-wall vs. cylindrical formulas—which is right?
Cylindrical is right for pipes. The flat-wall formula R = L/(k×A) assumes constant cross-sectional area, but a cylindrical shell has area that increases with radius. The correct formula is R = ln(r₂/r₁)/(2πkL). For thin insulation on large pipes (r₂/r₁ < 1.5), flat-wall is close enough. For thick insulation on small pipes (r₂/r₁ > 2), the error exceeds 20%. Always use cylindrical for pipes under 6" diameter with more than 1" insulation.