PIR insulation
One hall, two temperature zones: comparing heat loss through panel walls
At a glance
Using the same sandwich panel across a facade does not mean every room loses heat through its wall at the same rate. Wall area and the temperature difference across it matter too. A heated workshop beside a cooler store illustrates how to use the panel U-value to compare two wall areas. We explain why the declared conductivity of the foam does not belong in that calculation and why a result in watts is not an annual energy figure. The example covers the panel fields only, excluding doors, windows and junctions.
Split the elevation by the rooms behind it
Mark the workshop and store on the floor plan and trace each room’s external wall. If the rooms share an internal partition, do not add it to the external areas. Heat exchange between the rooms is a separate part of the heat balance.
Choose the right thermal property
For panel fields, use the relevant U-value for the selected panel or the value adopted in the calculation for that wall area. Core conductivity is a material property. Do not multiply it by wall area in place of U, even if both figures come from one sheet.
In the example, assume U = 0.25 W/(m²·K) for both wall areas and an outdoor temperature of 0°C. These are illustrative conditions for one comparison, not a product declaration or the heating design temperature for any location.
Workshop: smaller wall, higher temperature
After openings are deducted, the workshop has 40 m² of opaque panel wall. At 20°C inside, the difference is 20 K. Under steady conditions, multiply U by area and temperature difference: 0.25 × 40 × 20 = 200 W.
The 200 W figure does not size the workshop heater. Roof, floor, windows and air exchange are outside this calculation. The result compares one identified part of the envelope under the temperatures stated.
Store: larger wall, smaller temperature difference
For the store, take 60 m² of opaque wall and 10°C inside. With the same U-value and weather, the result is 0.25 × 60 × 10 = 150 W. Here the larger facade area loses heat more slowly than the wall of the warmer workshop.
The two fields together account for 350 W. Applying the workshop temperature to all 100 m² by mistake would give 500 W. The 150 W discrepancy comes from assigning the wrong indoor temperature to the store, not from poor insulation.
What would a different panel change?
If an alternative gives U = 0.20 W/(m²·K) for both fields, the workshop contributes 160 W and the store 120 W. The combined difference at this operating point is 70 W. Keep the areas and temperatures fixed to isolate the effect of changing U.
A substitute panel may have a different joint or fixing arrangement. Treat 350 W and 280 W as results for the wall fields. The specific junctions and remaining elements still need assessment for both design options.
Report the result without promising annual savings
Attach a sketch identifying the rooms, net areas excluding openings and temperatures used. List the exclusions as well. Someone else can then reproduce the 350 W result or see what changes if the internal partition moves.
Converting watts into energy requires the duration of each set of conditions. A workshop may only be heated during a shift, and the outdoor temperature varies through the year. Multiplying the example’s 70 W difference by every hour of the year would not predict the heating bill.
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