A classroom has 15 m² of glazing. The existing glazing has a U-value of 5.0 W/m²K, while a proposed double-glazing system has a U-value of 2.0 W/m²K.

The indoor design temperature is 22°C, and the outdoor winter design temperature is 6°C.

Assume that all other sources of heat loss remain unchanged and can be ignored for this calculation.

Calculate the reduction in heating load (in kW) resulting from the glazing upgrade.

Question

A classroom has 15 m² of glazing. The existing glazing has a U-value of 5.0 W/m²K, while a proposed double-glazing system has a U-value of 2.0 W/m²K.

The indoor design temperature is 22°C, and the outdoor winter design temperature is 6°C.

Assume that all other sources of heat loss remain unchanged and can be ignored for this calculation.

Calculate the reduction in heating load (in kW) resulting from the glazing upgrade.

BlackTom AI · Accepted Answer

Heat transfer through glazing is given by Q = U × A × ΔT, so the reduction equals the change in U-value times area and temperature difference.

1. Relevant knowledge points and concepts
- U-value (W/m²K) quantifies heat transfer per unit area per degree temperature difference; lower U means better insulation.
- Steady-state conductive heat loss through a glazing element is Q = U × A × ΔT (Watts).
- Reduction in heating load from upgrading glazing = (U_old − U_new) × A × ΔT.
- Temperatures in °C can be used for ΔT since differences are the same in Kelvin.

2. Complete reasoning process step by step
- Determine the temperature difference: ΔT = T_indoor − T_outdoor = 22°C − 6°C.
- Compute the change in U-value: ΔU = U_existing − U_proposed = 5.0 − 2.0.
- Multiply ΔU by glazing area A and ΔT to get the saved heat loss in watts.
- Convert watts to kilowatts for the final answer.

3. All intermediate calculation steps
- ΔT = 22 − 6 = 16 K
- ΔU = 5.0 − 2.0 = 3.0 W/m²K
- Saved heat loss Q = ΔU × A × ΔT = 3.0 W/m²K × 15 m² × 16 K
- Q = 3.0 × 15 × 16 = 720 W
- Convert to kW: 720 W ÷ 1000 = 0.72 kW

4. Why the correct answer is correct
Because the upgrade reduces the overall heat-transfer coefficient by 3.0 W/m²K across 15 m², and with a 16 K indoor–outdoor difference, the steady-state saved heat is exactly 720 W (0.72 kW). The relation Q = U·A·ΔT is the appropriate linear model for conductive/convective glazing heat losses under the stated assumptions.

5. Why other answers (if any) would be incorrect
Common mistakes would be using the absolute U-value rather than the difference, using the wrong ΔT, omitting the area, or failing to convert to kW. Any answer not equal to 0.72 kW likely results from one of these errors.

6. Conclusion confirming the final answer
The calculated reduction in heating load is 0.72 kW, which matches the provided correct answer.

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