Pipe Insulation Thickness Calculator

ANA›Life Services Authority›National Calculator Authority›Pipe Insulation Thickness Calculator

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Pipe Insulation Thickness Calculator

Calculate the minimum insulation thickness required for a pipe to limit heat loss to a desired value, based on cylindrical heat conduction principles.

Pipe Outer Diameter (mm)

Pipe Length (m)

Pipe Surface Temperature (°C)

Ambient Temperature (°C)

Insulation Thermal Conductivity k (W/m·K)

External Surface Heat Transfer Coefficient h (W/m²·K)

Maximum Allowable Heat Loss (W/m)

Insulation Material

-- Custom (enter k above) -- Mineral Wool (k = 0.040 W/m·K) Glass Wool (k = 0.035 W/m·K) Polyurethane Foam (k = 0.025 W/m·K) Expanded Polystyrene (k = 0.033 W/m·K) Aerogel Blanket (k = 0.022 W/m·K) Calcium Silicate (k = 0.050 W/m·K) Cellular Glass (k = 0.060 W/m·K)

Calculate Insulation Thickness

Results will appear here.

function pipSetK() { var sel = document.getElementById('pip-insul-type'); var kField = document.getElementById('pip-k-insul'); if (sel.value !== '') { kField.value = sel.value; } }

function pipCalc() { var resultDiv = document.getElementById('pip-result');

// --- Read inputs --- var doPipe = parseFloat(document.getElementById('pip-pipe-od').value); var L = parseFloat(document.getElementById('pip-pipe-length').value); var Tp = parseFloat(document.getElementById('pip-temp-pipe').value); var Ta = parseFloat(document.getElementById('pip-temp-amb').value); var k = parseFloat(document.getElementById('pip-k-insul').value); var h = parseFloat(document.getElementById('pip-h-ext').value); var qMax = parseFloat(document.getElementById('pip-max-qloss').value);

// --- Validation --- var errors = []; if (isNaN(doPipe) || doPipe 0) { resultDiv.innerHTML = 'Input Errors:' + errors.map(function(e){ return ''; }).join('') + ''; return; }

var dT = Tp - Ta; if (Math.abs(dT) Pipe and ambient temperatures are equal — no heat loss occurs.'; return; }

// Convert pipe OD from mm to m var r1 = (doPipe / 1000) / 2; // inner radius of insulation = pipe outer radius

// --- Solve for r2 (outer radius of insulation) --- // Heat loss per unit length for a cylinder with convection on outer surface: // q = 2π·ΔT / [ ln(r2/r1)/k + 1/(h·r2) ] // // To find r2 such that q = qMax (W/m). // Rearranged: ln(r2/r1)/k + 1/(h·r2) = 2π·ΔT / qMax // // Solve numerically using bisection method.

var target = (2 * Math.PI * Math.abs(dT)) / qMax; // total thermal resistance needed (m·K/W per unit length)

// f(r2) = ln(r2/r1)/k + 1/(h*r2) - target = 0 function f(r2) { return Math.log(r2 / r1) / k + 1.0 / (h * r2) - target; }

// Check if insulation is even needed (bare pipe heat loss) var qBare = (2 * Math.PI * Math.abs(dT)) / (1.0 / (h * r1)); // qBare = 2π·ΔT·h·r1 (W/m)

var insThickness_mm; var r2_solution; var noInsulNeeded = false;

if (qBare r1 // Lower bound: r1 (no insulation) // Upper bound: search for a large enough r2 var rLow = r1 * 1.0001; var rHigh = r1;

// Expand upper bound until f(rHigh) > 0 // As r2 → ∞, ln(r2/r1)/k → ∞, so f → +∞ // At r2 = r1+ε, f ≈ 1/(h*r1) - target // We need f(rLow) 0

// Check direction if (f(rLow) > 0) { // Even tiny insulation exceeds target — shouldn't happen if qBare > qMax // This means target is very small; set r2 just above r1 r2_solution = rLow; } else { rHigh = r1 * 2; var iter = 0; while (f(rHigh) Could not converge — the required heat loss may be unachievable with this insulation material and geometry.'; return; }

// Bisection for (var i = 0; i r1) { criticalNote = '⚠ Critical Radius Note: The critical radius of insulation is ' + r_cr_mm.toFixed(1) + ' mm (outer radius). Adding insulation up to this radius will increase heat loss. ' + 'Insulation is only effective beyond this radius.

'; }

// Insulation outer diameter var d2_mm = r2_solution * 2 * 1000;

// Thermal resistance components var R_insul = noInsulNeeded ? 0 : Math.log(r2_solution / r1) / k; var R_conv = 1.0 / (h * r2_solution); var R_total = R_insul + R_conv;

// Annual energy loss (assuming continuous operation) var annualEnergy_kWh = qTotal_kW * 8760;

resultDiv.innerHTML = '### Results ' + (noInsulNeeded ? '✔ No insulation required. The bare pipe heat loss (' + qBare.toFixed(2) + ' W/m) is already within the allowable limit.

'
'') + criticalNote + '' + 'Required Insulation Thickness' + '' + insThickness_mm.toFixed(1) + ' mm' + 'Insulation Outer Diameter' + '' + d2_mm.toFixed(1) + ' mm' + 'Actual Heat Loss (per unit length)' + '' + qActual_per_m.toFixed(2) + ' W/m' + 'Total Heat Loss (full pipe length)' + '' + qTotal_W.toFixed(1) + ' W (' + qTotal_kW.toFixed(3) + ' kW)' + 'Estimated Annual Heat Loss' + '' + annualEnergy_kWh.toFixed(0) + ' kWh/year' + 'Thermal Resistance — Insulation (per m length)' + '' + R_insul.toFixed(4) + ' m·K/W' + 'Thermal Resistance — Convection (per m length)' + '' + R_conv.toFixed(4) + ' m·K/W' + 'Total Thermal Resistance (per m length)' + '' + R_total.toFixed(4) + ' m·K/W' + ''; }

#### Formulas Used

Heat loss per unit length through a cylindrical insulation layer with outer convection:

q = 2π·ΔT / [ ln(r₂/r₁)/k + 1/(h·r₂) ]

Where:

Critical radius of insulation (adding insulation below this radius increases heat loss):

rcr = k / h

The required r₂ is found by bisection, solving for q = qmax.

#### Assumptions & References

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