Conductive heat loss calculator API

Compute conductive heat loss through a wall (Fourier's law Q = k·A·ΔT/d) and its thermal resistance (R = d/(k·A)). Answers 'heat loss through 0.1m insulation','thermal resistance of this wall','watts lost across a temperature difference'.

Price$0.01per request

MethodPOST

Route/v1/calc/eng-heat-loss

StatusLive

MIME typeapplication/json

Rate limit120/minute

CacheNo cache

calcengineeringthermalheat-lossfourierconductioninsulationhvac

API URLhttps://x402.hexl.dev/v1/calc/eng-heat-loss

Integration docs

Example request

{
  "conductivityWmK": 0.04,
  "areaM2": 10,
  "thicknessM": 0.1,
  "deltaTempK": 20
}

Example response

{
  "conductivityWmK": 0.04,
  "areaM2": 10,
  "thicknessM": 0.1,
  "deltaTempK": 20,
  "heatLossW": 80,
  "thermalResistanceKW": 0.25,
  "formula": "Q = k*A*ΔT / d, R = d/(k*A)"
}

Input schema

{
  "type": "object",
  "required": [
    "conductivityWmK",
    "areaM2",
    "thicknessM",
    "deltaTempK"
  ],
  "properties": {
    "conductivityWmK": {
      "type": "number",
      "description": "thermal conductivity k (W/m·K)",
      "examples": [
        0.04
      ]
    },
    "areaM2": {
      "type": "number",
      "description": "area (m²)",
      "examples": [
        10
      ]
    },
    "thicknessM": {
      "type": "number",
      "description": "wall thickness (m)",
      "examples": [
        0.1
      ]
    },
    "deltaTempK": {
      "type": "number",
      "description": "temperature difference (K)",
      "examples": [
        20
      ]
    }
  }
}

Output schema

{
  "type": "object",
  "additionalProperties": true
}