Calculate heat transfer through walls, windows, or insulation materials.
Custom (Enter k-value)
Fiberglass Insulation
Water
Glass (Window)
Concrete
Brick
Pine Wood
Air (Still)
Steel
Aluminum
Temp Difference (ΔT):0 °C
Heat Loss Rate (Watts):0 W
Heat Loss Rate (kW):0 kW
Heat Loss (BTU/hr):0 BTU/hr
Energy Loss per 24h:0 kWh
function updateConductivity() {
var selector = document.getElementById("materialSelect");
var kInput = document.getElementById("thermalConductivity");
var selectedValue = selector.value;
if (selectedValue !== "custom") {
kInput.value = selectedValue;
} else {
kInput.value = "";
kInput.focus();
}
}
function calculateHeatLoss() {
// Get input values
var k = parseFloat(document.getElementById("thermalConductivity").value);
var area = parseFloat(document.getElementById("surfaceArea").value);
var thicknessCm = parseFloat(document.getElementById("thickness").value);
var tempIn = parseFloat(document.getElementById("tempInside").value);
var tempOut = parseFloat(document.getElementById("tempOutside").value);
// Validation
if (isNaN(k) || isNaN(area) || isNaN(thicknessCm) || isNaN(tempIn) || isNaN(tempOut)) {
alert("Please fill in all fields with valid numbers.");
return;
}
if (thicknessCm <= 0) {
alert("Thickness must be greater than zero.");
return;
}
if (area <= 0) {
alert("Surface area must be greater than zero.");
return;
}
// Calculations
// Formula: Q = (k * A * deltaT) / d
// d needs to be in meters
var thicknessM = thicknessCm / 100;
// Absolute difference in temperature
var deltaT = Math.abs(tempIn – tempOut);
// Heat loss in Watts (Joules per second)
var heatLossWatts = (k * area * deltaT) / thicknessM;
// Conversions
var heatLossKW = heatLossWatts / 1000;
var heatLossBTU = heatLossWatts * 3.412142; // 1 Watt = 3.412 BTU/hr
var energyPer24h = heatLossKW * 24; // kWh lost in a day
// Display Results
document.getElementById("resultDeltaT").innerHTML = deltaT.toFixed(1) + " °C";
document.getElementById("resultWatts").innerHTML = heatLossWatts.toFixed(2) + " W";
document.getElementById("resultKW").innerHTML = heatLossKW.toFixed(4) + " kW";
document.getElementById("resultBTU").innerHTML = heatLossBTU.toFixed(1) + " BTU/hr";
document.getElementById("resultKWh").innerHTML = energyPer24h.toFixed(2) + " kWh";
document.getElementById("hlResults").style.display = "block";
}
Understanding How to Calculate Rate of Heat Loss
Calculating the rate of heat loss is a fundamental concept in thermodynamics, building engineering, and energy efficiency analysis. Whether you are trying to determine the insulation requirements for a home, sizing a heater for a specific room, or analyzing thermal leakage in industrial pipes, knowing the rate at which heat escapes is crucial.
The Physics of Heat Conduction
This calculator uses Fourier's Law of Thermal Conduction to estimate the rate of heat transfer through a solid medium (like a wall, window, or insulation layer). The rate of heat loss depends on four primary factors:
Thermal Conductivity ($k$): A measure of how easily a material transmits heat. Metals like aluminum have high conductivity (bad insulators), while fiberglass and foam have low conductivity (good insulators).
Surface Area ($A$): The total area through which the heat is escaping. Larger walls allow more heat to escape.
Temperature Difference ($\Delta T$): The driving force of heat transfer. The greater the difference between the inside and outside temperatures, the faster heat will flow.
Thickness ($d$): The distance the heat must travel. Thicker materials slow down the rate of heat loss.
The Formula
The mathematical formula used to calculate the rate of heat loss via conduction is:
Q = (k × A × ΔT) / d
Where:
Q = Rate of heat loss in Watts (W)
k = Thermal conductivity in Watts per meter-Kelvin (W/m·K)
A = Surface area in square meters (m²)
ΔT = Temperature difference in degrees Celsius or Kelvin (°C)
d = Thickness of the material in meters (m)
Real-World Example Calculation
Let's assume you want to calculate the heat loss through a single-pane glass window on a cold winter day.
Material: Glass ($k \approx 0.8$ W/m·K)
Area: 2 square meters ($2 m^2$)
Thickness: 5 millimeters ($0.005 m$)
Inside Temp: 20°C
Outside Temp: 0°C
First, calculate the temperature difference ($\Delta T = 20 – 0 = 20$). Then apply the formula:
Q = (0.8 × 2 × 20) / 0.005
Q = 32 / 0.005 = 6400 Watts
This high number demonstrates why single-pane windows are very inefficient and why double glazing (which utilizes an air gap with a much lower $k$-value) is standard in modern construction.
Why This Matters for Energy Bills
The result of this calculation ($Q$) tells you how much energy is leaving your system every second. If your room loses heat at a rate of 2000 Watts (2 kW), your heating system must continuously supply 2000 Watts of power just to maintain the current temperature. By improving insulation (lowering $k$) or increasing wall thickness, you reduce $Q$, allowing you to use a smaller heater and consume less electricity or gas.