How to Calculate Insulation Thickness: Pipe Formula & Example

Direct answer

To calculate insulation thickness, do not start from a catalogue thickness. Build the thermal resistance network for the real geometry, set the governing design constraint, then solve for thickness.

For a pipe, heat flows through the internal film resistance, one or more cylindrical insulation layers, and the external convection-plus-radiation resistance connected in series. The per-metre heat flow is:

q = deltaT / R_total = deltaT / [1 / (2π r1 hi) + ln(r2/r1) / (2π k) + 1 / (2π r2 ho)]

Solve for the outer radius r2; the required thickness is t = r2 − r1. Use the heat-loss method when the governing limit is allowable energy loss, the surface-temperature method when the limit is personnel protection or process temperature, and the economic-thickness method when the objective is the lowest life-cycle cost.

For the worked example—DN100 pipe carrying 400 °C fluid in 20 °C ambient air, with a target outer surface temperature of 55 °C—the calculated continuous thicknesses are approximately 20.4 mm for rock wool and 29.1 mm for calcium silicate. These are calculation outputs for stated assumptions, not specifications. Round up to a commercially available size and verify against the current manufacturer data sheet, project code, personnel-protection requirements, and local regulations.

Which thickness should you calculate first: heat loss, safety, condensation, or cost?

The governing requirement—not the calculator—selects the thickness. Compare every applicable method, then adopt the thickest result that meets project, safety, and maintainability requirements.

“Insulation thickness” has at least four different engineering meanings:

1. Heat-loss-limited thickness: the layer must keep energy loss, fuel consumption, or fluid cooling within an allowable value.
2. Surface-temperature-limited thickness: the outer surface must remain below a specified touch-temperature, burn-protection, or process requirement.
3. Condensation-limited thickness: the cold outer surface must stay above the design dew point, with a continuous vapor-retarder system.
4. Economic thickness: the layer minimizes the combination of installed insulation cost and ongoing energy cost over a defined study period.

These methods can produce different answers. A chilled-water line may be inexpensive at 25 mm but still sweat; a high-temperature line may pass an energy target yet remain too hot for safe maintenance access. ASTM C680 provides calculation methodologies for estimating heat gain or loss and surface temperatures of insulated flat, cylindrical, and spherical systems under one-dimensional, steady- or quasi-steady-state conditions. ISO 12241:2022 gives rules for calculating heat-transfer-related properties of building equipment and industrial installations, predominantly under steady-state conditions, and includes a simplified approach for thermal bridges.

What governs an insulation thickness calculation?

ISO 12241 is a calculation standard, not a universal thickness table. It sets out how to calculate steady-state heat transfer, surface conditions, and simplified thermal-bridge effects; it does not replace national energy codes, occupational safety rules, or project specifications. ISO 12241:2022 is the third edition, published in 2022, and supersedes ISO 12241:2008.

Pipe insulation thickness formula and variable definitions

A cylindrical calculation is mandatory when the insulation thickness becomes significant relative to the pipe radius. The area term makes radius and thickness coupled, so the formula must be solved iteratively or algebraically for r2.

Standard pipe formula

For a single insulation layer, steady radial heat transfer per unit length is:

q = (Tf − Tamb) / [1 / (2π r1 hi) + ln(r2/r1) / (2π k) + 1 / (2π r2 ho)]

where:

• q = heat loss or gain, W/m
• Tf = bulk fluid temperature, K or °C
• Tamb = ambient air temperature, K or °C
• r1 = outside radius of the pipe or innermost layer, m
• r2 = outside radius of the insulation, m
• t = insulation thickness = r2 − r1, m
• k = design thermal conductivity of the insulation layer, W/(m·K)
• hi = combined internal convection coefficient, W/(m²·K)
• ho = combined external convection and radiation coefficient, W/(m²·K)

The corresponding total thermal resistance per metre is:

R_total = 1 / (2π r1 hi) + ln(r2/r1) / (2π k) + 1 / (2π r2 ho)

This expression is the series sum of three resistances:

1. internal film resistance;
2. logarithmic cylindrical conduction resistance;
3. external surface resistance, whose area grows with r2.

Surface coefficients are often written as a combined value:

ho = h_c + h_r

where forced or natural convection is h_c, and linearised radiation is:

h_r = epsilon sigma (Ts^4 − Tamb⁴) / (Ts − Tamb)

In this radiation term, epsilon is surface emissivity, sigma is the Stefan–Boltzmann constant 5.670374419×10⁻⁸ W/(m²·K⁴), and absolute temperatures must be used. This form is especially useful because it lets you enter a measured or allowed surface temperature directly.

Do not substitute the pipe nominal size for the actual outside diameter. DN100 commonly indicates a nominal size, but the actual outside diameter can differ by standard and manufacturing route. The insulation is sized to the pipe’s outside surface, so use the measured or specified OD in calculations and procurement documents.

Cylindrical formula in terms of diameters

If you prefer diameters, define:

• D₁ = 2r1 = pipe outside diameter, m
• D₂ = 2r2 = insulation outer diameter, m
• t = (D₂ − D₁) / 2, m

Then:

q = (Tf − Tamb) / [2 / (D₁ hi) + ln(D₂/D₁) / k + 2 / (D₂ ho)]

and the required thickness is still:

t = (D₂ − D₁) / 2

This is the same thermal model written with diameters. It is not a flat-plate approximation.

Multi-layer pipe formula

For multiple insulation layers, add one conduction term for each layer:

R_total = 1 / (2π r1 hi) + sum [ln(r_j+1/r_j) / (2π k_j)] + 1 / (2π r_last ho)

Thermal contact resistance, metal bands, weather barriers, air gaps, and high-emissivity liners may be added when their effects are quantified. ISO 12241 includes a simplified treatment of thermal bridges, but complex supports, penetrations, and multilayer hygrothermal interactions should be modeled individually.

Flat-surface simplification: when you can use it

Use the flat-plate formula only when the pipe radius is large relative to the thickness. For most industrial piping, use the logarithmic cylindrical expression.

For a plane wall or large flat surface, thermal resistance is:

R = 1/hi + t/k + 1/ho

and the heat flux is:

q_flux = (T_hot − T_cold) / R

where q_flux is in W/m² and t is the flat insulation thickness in metres.

ASTM C680 addresses rectangular, cylindrical, and spherical coordinate systems, making it suitable for boilers, furnaces, refrigerated chambers, piping circuits, and stored fluids such as LNG.

A common practical rule is to compare the insulation thickness with the inside radius. The flat approximation becomes progressively worse as t/r1 increases. If the selected thickness is a substantial fraction of the pipe radius, or if you are solving for thickness rather than merely checking heat flux, use the cylindrical formula. The flat formula may still be useful for early screening; record it as a screening value, not a final design value.

Heat-loss method: calculate thickness from allowable heat loss

When energy conservation, process cooling, or carbon limits govern the design, start from the allowable heat flow and solve the resistance network for the required radius.

Direct algebraic solution

If you can neglect the internal film resistance and use a known combined external coefficient ho, the allowable heat flow per metre is q_allow. The required outer radius satisfies:

ln(r2/r1) / (2π k) + 1 / (2π r2 ho) = (Tf − Tamb) / q_allow

This cannot generally be rearranged into a clean closed form for r2, so engineers solve it iteratively:

1. Estimate r2.
2. Calculate q.
3. Adjust r2 until q equals or falls below q_allow.
4. Calculate the resulting surface temperature as a check.
5. If the surface temperature changes k or h_r materially, repeat the calculation.

ASTM C680 explicitly notes that thermal conductivity depends on temperature and that accuracy depends strongly on input data. ISO 23993:2008 remains confirmed and gives methods to calculate design thermal conductivity from declared thermal conductivity for operating temperatures from −200 °C to +800 °C.

Why the iterative check matters

k should be selected at the appropriate mean insulation temperature, not at the fluid temperature or ambient temperature alone. For high-temperature service, a temperature correction can change the apparent performance materially. ISO 23993 exists specifically to derive a design conductivity from declared values for use in thermal-performance calculations.

The worked example below keeps k constant at an illustrative mean-temperature value. A real calculation should:

• select the grade and density;
• obtain the declared conductivity curve;
• estimate mean temperatures for each layer;
• calculate surface temperature;
• revise conductivity and repeat.

Surface-temperature or burn-protection method: calculate thickness from allowable surface temperature

When a maximum outer-surface temperature governs the design, use the external film and radiation resistance to find the allowable heat flux, then solve the cylindrical resistance network for thickness.

Direct solution used in the worked example

Let the governing surface temperature be Ts. The heat lost from the outer surface is:

q = ho × 2π r2 × (Ts − Tamb)

The total resistance required is therefore:

R_total = (Tf − Tamb) / q

and the insulation resistance required is:

R_ins = R_total − 1 / (2π r1 hi) − 1 / (2π r2 ho)

Then solve:

ln(r2/r1) / (2π k) = R_ins

Because ho may include radiation, which depends on Ts, and because commercial thicknesses change r2, this is normally solved iteratively. For a specified Ts, a direct iterative formulation is:

r2 = r1 exp[2π k (R_total − 1 / (2π r1 hi) − 1 / (2π r2 ho))]

Use a spreadsheet or solver to update r2 until the equation closes. Then set:

t = r2 − r1

Safety limit is a project requirement, not a universal constant

A surface-temperature calculation answers a thermal question, not a human-burn question by itself. Skin contact time, contact area, gloving, maintenance activity, and applicable ergonomics or safety standards also matter. For personnel protection, ISO 13732-1 covers methods for assessing human responses to contact with hot surfaces; its applicable edition should be confirmed for the project. ASTM C680 is referenced for use with ASTM C1055 and ASTM C1057 for burn-hazard evaluation of heated surfaces.

Therefore, record the governing surface-temperature limit in the calculation as a project input. Do not present an engineering thermal model as a substitute for a safety study.

Economic-thickness method: calculate the lowest life-cycle cost

Economic thickness is the layer that gives the lowest total cost—not the thinnest layer and not simply the thickest affordable layer. ISO 12241 provides the thermal calculation basis; cost optimization must add project-specific economics.

Minimum-total-cost method

For a given diameter and material, calculate heat loss over a thickness range using the cylindrical thermal model. Then build the annual cost:

C_total(t) = C_capital(t) + C_maintenance(t) + C_energy(t) + C_other(t)

A common annualised form is:

C_capital(t) = C_installed(t) × CRF

where the capital recovery factor is:

CRF = i(1 + i)ⁿ / [(1 + i)ⁿ − 1]

In this expression:

• i = real discount or interest rate, per year
• n = study period, years
• C_installed(t) = installed insulation cost for thickness t

Annual energy cost for a hot pipe is:

C_energy(t) = q(t) × L × H × P / η

where:

• L = pipe length, m
• H = annual operating hours, h/yr
• P = energy unit cost, USD/MWh or local currency per energy unit
• η = system generation, boiler, or plant efficiency
• q(t) = heat loss at thickness t, W/m

Maintenance, replacement, downtime, carbon costs, fuel escalation, tax effects, and residual value may be added if the project accounting model requires them.

The economic thickness is the t that minimizes C_total(t) after the cost curve is evaluated over available product thicknesses.

What ISO 12241 provides and what it does not provide

ISO 12241:2022 is the current calculation-rules document for thermal performance of building equipment and industrial installations. It supports the thermal side of economic-thickness work: heat flow, temperatures, surface checks, and simplified thermal bridges. It does not prescribe a universal fuel price, escalation rate, discount rate, maintenance cost, or payback rule.

Economic thickness therefore changes when any of the following change:

• insulation material and installed cost;
• pipe diameter;
• fluid and ambient temperatures;
• operating hours;
• fuel or energy price;
• boiler or plant efficiency;
• financing and discount rate;
• study period;
• maintenance and replacement assumptions.

A project with 2,000 annual operating hours and low fuel cost can have a much thinner economic thickness than a similar project operating 8,000 hours at high fuel cost. The thermal model may be the same; the cost answer will not be.

Cost-sensitivity warning from the worked example

For the DN100 example discussed below, an illustrative installed cost of USD 200/m³, 8,000 h/yr, USD 0.12/kWh thermal, 85% generation efficiency, 7% discount, and a 10-year study period produces a curve that is still falling at 160 mm. That output is a warning, not a recommendation: the input price is too low relative to the avoided-energy value, so the modeled optimum is pushed to the upper thickness bound.

In an actual specification:

1. use current installed cost quotations;
2. include jacket, bands, vapor retarder, labour, fittings, and access;
3. compare at least three to five discrete commercial thicknesses;
4. apply actual fuel cost and operating hours;
5. check whether maintenance, replacement, or fuel escalation changes the minimum;
6. enforce safety, condensation, code, and mechanical constraints before selecting the cost minimum.

Critical radius: can insulation ever be too thick?

“Can insulation be too thick?” has three different engineering answers: critical radius, condensation, and economic limit. For a conventional hot DN100 pipe, simply adding more thickness does not initially increase heat loss.

Critical radius for a cylinder

For a bare cylinder with negligible internal resistance and constant k, the heat flow per metre is:

q = (Tf − Tamb) / [ln(r2/r1) / (2π k) + 1 / (2π r2 ho)]

Differentiating with respect to r2 gives a maximum heat flow when:

r_cr = k / ho

This is the critical radius. If the bare outer radius is less than r_cr, adding a small amount of insulation can initially increase heat loss because the growing surface area initially outweighs the added conduction resistance. Once r2 exceeds r_cr, further thickness reduces heat loss.

These values are illustrative. A larger ho—for example, from wind or forced convection—reduces the critical radius. The important design conclusion is that r_cr is usually small relative to industrial piping.

Why DN100 is not a critical-radius concern in the example

The DN100 example uses an actual pipe outside radius of 50 mm. Both illustrative critical radii are below 6 mm, so even the first millimetre of insulation increases total resistance and immediately reduces heat loss. The figure illustrates that result; it does not prove the same outcome for every small tube, wire, or cryogenic line.

For small-diameter electric wires, thin process tubes, and some cryogenic tubing, critical radius can matter. In those cases:

• calculate k/ho;
• compare it with the actual bare outer radius;
• use a heat-transfer program rather than assuming a chart thickness;
• recheck the surface temperature;
• consider jacketing, spacing, and maintenance access as separate constraints.

The other “too thick” limits

Condensation can impose a lower thermal-transmittance limit, not a thickness limit. On a cold line, increasing insulation generally helps keep the outer surface above dew point. However, insulation alone does not stop water vapor. An interrupted vapor retarder, permeable jacket seam, or warm-side breach can allow moist air into the system.

Economic thickness sets a cost limit. Beyond the cost-minimising layer, additional material and labour cost more than the additional energy saved.

Mechanical and constructability limits also matter. Very thick systems may be difficult to support, inspect, repair, route through structures, or protect from impact. Thickness that is thermally correct but mechanically impractical is not a complete design.

Worked DN100 example: rock wool vs calcium silicate

The worked example shows the surface-temperature method with two materials. It does not imply that either material is suitable for every 400 °C application or that the calculated value is a commercial standard size.

Stated assumptions

These values are deliberately simple so the calculation can be followed. They are not site measurements, manufacturer guarantees, or code values.

Step 1: Calculate the governing heat flux

With Ts = 55 °C:

deltaTsurface = 55 − 20 = 35 K

For a 1 m length:

A_surface = 2π r2

The heat loss must satisfy:

q = ho × 2π r2 × 35

Because radiation depends on surface temperature, use:

h_r = epsilonsigma(Ts^4 − Tamb⁴)/(Ts − Tamb)

At epsilon = 0.85:

h_r ≈ 5.798 W/(m²·K)

ho = h_c + h_r = 11 + 5.798 = 16.798 W/(m²·K)

So the allowable heat flux per unit area is:

q_flux = 16.798 × 35 ≈ 588 W/m²

and the allowable heat flow per metre is:

q = 588 × 2π r2

The exact value depends on r2, which is the unknown being solved.

Step 2: Calculate total thermal resistance

deltaT_total = 400 − 20 = 380 K

R_total = 380 / q = 380 / (16.798 × 2π r2) = 3.605 / r2 K·m/W

Step 3: Subtract internal and external resistances

Internal resistance:

R_internal = 1 / (2π × 0.05 × 2500) ≈ 0.001273 K·m/W

External resistance:

R_external = 1 / (16.798 × 2π r2) = 0.009486 / r2 K·m/W

Required insulation resistance:

R_ins = 3.605/r2 − 0.001273 − 0.009486/r2 = 3.5955/r2 − 0.001273

Step 4: Solve for rock wool

Rock wool conductivity is k = 0.041 W/(m·K). Set:

ln(r2/0.05) / (2π × 0.041) = 3.5955/r2 − 0.001273

Solving iteratively gives:

• r2 ≈ 70.36 mm
• t = r2 − r1 ≈ 20.36 mm
• q ≈ 259.9 W/m
• Tsurface ≈ 55.0 °C

Step 5: Solve for calcium silicate

Calcium silicate conductivity is k = 0.062 W/(m·K). Set:

ln(r2/0.05) / (2π × 0.062) = 3.5955/r2 − 0.001273

Solving iteratively gives:

• r2 ≈ 79.14 mm
• t = r2 − r1 ≈ 29.14 mm
• q ≈ 292.4 W/m
• Tsurface ≈ 55.0 °C

Worked-example result

The calculated thickness is not automatically the ordered thickness. In practice, round up to the next available product thickness, then recheck. If 25 mm rock wool gives a surface temperature below 55 °C, it may satisfy this thermal target; it must still satisfy material temperature compatibility, mechanical protection, corrosion-under-insulation provisions, code requirements, and site safety rules.

The example also demonstrates why a single “insulation thickness for 400 °C” value is invalid. Two materials with different k values produce two different thicknesses. A different surface-temperature limit, wind condition, emissivity, or jacket type would change both results again.

Cold-surface and anti-condensation thickness

Anti-condensation thickness requires a dew-point analysis, not only a thermal-resistance value. Keeping the surface above dew point is necessary, but continuous vapor control is what prevents internal moisture accumulation.

For a cylindrical cold line, the same steady thermal model applies, with heat gain reversed:

q = (Tamb − T_pipe) / [1 / (2π r1 hi) + ln(r2/r1) / (2π k) + 1 / (2π r2 ho)]

The outer-surface temperature is:

Tsurface = Tamb − q / (2π r2 ho)

To avoid surface condensation under the design condition:

Tsurface > T_dp

or, written as a required total thermal resistance:

R_total > (Tamb − T_pipe) / [2π r2 ho (T_dp − Tamb)]

This form shows why a colder pipe, a higher dew point, or a lower external surface coefficient drives a larger thermal-resistance requirement. In practice, calculate a safety margin above dew point according to the project specification rather than relying on equality.

Dew point depends on climate and operating mode

Design dew point should be selected from local weather data, microclimate, indoor humidity control, and the expected coincidence of temperature and humidity. Design guidance for cold mechanical insulation commonly evaluates conditions such as extreme dew points with coincident dry-bulb temperatures and may use 80–90% relative humidity as an illustrative range for screening. Those values are not universal project design values.

Outdoor systems can encounter saturated or near-saturated air at times when no reasonable thickness can economically prevent every surface-dew event. In that case, design for an acceptable frequency, increase air movement or sheltering, specify a high-emissivity jacket if thermally appropriate, and protect the substrate through a continuous weather and vapor-retarder system.

Surface condensation and interstitial condensation are separate risks

Surface condensation control is only a surface-temperature check. Moisture can still diffuse through or around the insulation and condense inside the system if the vapor-retarder system is incomplete.

For cold service, specify:

• a continuous vapor retarder on the warm side of the insulation;
• sealed laps, joints, penetrations, and terminations;
• compatible mastics, adhesives, and jacketing;
• closures at valves, supports, and hangers;
• protection from rainwater and washwater;
• a drainage path where the design permits it;
• inspection and repair procedures.

Closed-cell foam, elastomeric, or other moisture-resistant insulations may reduce liquid-water uptake, but permeability and seam continuity still matter. A high-performance blanket placed behind an interrupted metal jacket is not automatically a complete vapor-control system.

Material temperature ranges and selection boundaries

Material class and temperature capability should be checked before conductivity is entered into the formula. A low conductivity value cannot compensate for a material operating beyond its recommended limit.

The table uses the requested classification-temperature boundaries as selection guidance, not manufacturer guarantees.

One rock-wool pipe-section data sheet declares a maximum service temperature of 750 °C and a recommended service temperature of 650 °C, illustrating why “≤650 °C” should be treated as a recommended-service boundary rather than a generic property of all rock wool. A published ceramic-fiber data sheet lists blanket temperature grades up to 1430 °C. Those examples support the use of grade-specific classification boundaries but do not establish a universal property for every product in a material family. For material-specific data sheets, temperature limits, and product forms, see the rock wool, glass wool, calcium silicate, ceramic fiber, and aerogel hub pages.

The upper limit is not automatically the recommended operating temperature

A classification or maximum-service temperature describes a material capability under the manufacturer’s defined test or product condition. It is not a recommendation to operate every product continuously at that temperature.

For each grade, obtain:

• continuous service temperature;
• linear shrinkage;
• chemical compatibility;
• thermal conductivity at the expected mean temperature;
• compressive resistance;
• fire reaction and regulatory status;
• moisture behaviour;
• fiber-safety and PPE requirements.

Then calculate thickness with the declared design conductivity, not the boundary temperature.

Illustrative hot-service thickness starting points

These ranges are deliberately wide. Ambient temperature, wind, pipe diameter, surface-temperature target, energy value, insulation quality, and code requirements can change the final value by much more than one standard product step.

Selection matrix by application

Use the matrix to choose candidate materials, then perform a thermal calculation. It is not a substitute for heat-transfer analysis or project specifications.

The material decision also interacts with installation. A technically thinner product that cannot be sealed, supported, or inspected reliably may be a poor project choice. Conversely, a thicker conventional material may be preferred where maintenance access, repairability, or procurement simplicity is more valuable than space savings. Related reading: for a broader material map see Types of Thermal Insulation Materials; for pipe-specific selection see Industrial Pipe Insulation Types; for cost optimisation see Power-Plant Economic Thickness; for calcium silicate pipe data see Calcium Silicate Pipe Insulation; and for application notes see Piping Applications and Power Generation.

Thermal bridges, jacketing, and quality factors

Thermal bridges and support details can defeat a nominally correct insulation thickness. Calculate their added loss separately and show the correction on the drawings.

ISO 12241 includes a simplified approach for thermal bridges. Depending on the project and standard adopted, additional loss may be represented by:

• an equivalent added thermal conductivity for the insulation system;
• an additional linear thermal transmittance for pipes;
• an equivalent length added to pipe runs;
• a separately calculated fitting or support loss.

Examples include:

• hangers and saddles;
• bands and wires;
• jacket seams and fasteners;
• nozzles, flanges, and valves;
• penetrations;
• support shoes;
• anchors for hot-face refractory;
• voids, compression, and gaps.

ISO 6946:2017 is the current ISO method for thermal resistance and thermal transmittance of building components and elements. It can provide an approximate treatment for inhomogeneous layers, including metal fasteners, through a correction term, but explicitly excludes other cases where insulation is bridged by metal. For industrial pipe supports and complex penetrations, ISO 6946 alone is usually insufficient.

What to specify for jacketing

Jacketing is part of the thermal and durability system. For each project, specify:

• jacket material, thickness, and finish;
• surface emissivity;
• seam orientation and lap width;
• sealant and fastener compatibility;
• weather-barrier continuity;
• UV, chemical, and washdown resistance;
• expansion provisions;
• secure fixing at valves and penetrations.

Document assumptions that affect every thickness

A rigorous calculation should state:

1. inside and outside diameters;
2. fluid and ambient temperatures;
3. wind speed and orientation;
4. indoor or outdoor location;
5. relative humidity and design dew point for cold systems;
6. surface-temperature limit;
7. jacket emissivity and color;
8. conductivity source, temperature correction, and mean temperature;
9. material temperature compatibility;
10. operating hours and energy cost;
11. installed-cost basis;
12. study period and discount rate;
13. thermal-bridge allowances;
14. codes and project specifications.

Step-by-step engineering checklist

Follow the same sequence for every project: define the constraint, model the system, solve iteratively, round up conservatively, then verify surface, safety, and constructability conditions.

1. Define the real objective: heat loss, surface temperature, condensation control, or cost.
2. Collect actual pipe diameter, wall construction, fluid temperature, and environment.
3. Select candidate materials and confirm current data-sheet temperature limits.
4. Obtain declared conductivity and apply design-temperature correction where required.
5. Estimate internal film coefficient and external convection plus radiation.
6. Build the series-resistance model for pipe, multilayer, or flat geometry.
7. Solve for thickness; iterate if conductivity or radiation coefficient changes materially.
8. Calculate final surface temperature, heat flux, and condensation margin.
9. Add thermal bridges for supports, hangers, valves, and penetrations.
10. Compare available commercial thicknesses and select a conservative available size.
11. Recheck the governing requirement with the rounded-up thickness.
12. Verify fire, personnel protection, corrosion, environmental, and local-code requirements.
13. Specify jacketing, sealants, vapor retarders, and installation tolerances.
14. Record assumptions, references, and revision date in the calculation file.