Specification 44 Calculation of heating load limit, cooling load limit and thermal energy load limit

NCC 2022 Volume Two - Building Code of Australia Class 1 and 10 buildings

This Specification contains the method of calculating the heating load limit, cooling load limit and thermal energy load limit for compliance with J1P2 and H6P1.

The heating load limit of a space, measured in MJ/m².annum, is equal to the greater of—

Table S44C2 Area adjustment factors for the heating load limit

Total area of habitable rooms (A_H)	Area adjustment factor (F_H)
≤ 50 m²	1.37
> 50 m² to ≤ 350 m²	$(5.11 \times 1 0^{- 6}) A_{H}^{2} - (3.82 \times 10^{- 3}) A_{H} + 1.55$
> 350 m²	0.84

(1) The cooling load limit of a space, measured in MJ/m².annum, is calculated in accordance with the following formula:

C L L = (5.4 + 0.00617 \times (C D H + 1.85 D G H)) \times F_{C}

(2) In the formula at (1)—

$C L L$
= the cooling load limit (MJ/m².annum); and
$C D H$
= the total annual cooling degree hours of the building location; and
$D G H$
= the total annual dehumidification gram hours of the building location; and
$F_{C}$
= the area adjustment factor for the cooling load limit, determined in accordance with Table S44C3.

Table S44C3 Area adjustment factors for the cooling load limit

Total area of the habitable rooms (A_H)	Area adjustment factor (F_C)
≤ 50 m²	1.34
> 50 m² and ≤ 200 m²	$(1.29 \times 10^{- 5}) A_{H}^{2} - (5.55 \times 10^{- 3}) A_{H} + 1.58$
> 200 m² and ≤ 1000 m²	$(3.76 \times 10^{- 7}) A_{H}^{2} - (7.82 \times 10^{- 4}) A_{H} + 1.12$
> 1000 m²	0.71

(1) The thermal energy load limit of a space, measured in MJ/m².annum, is calculated in accordance with the following formula:

T L L = \frac{19.3 H L L + 22.6 C L L - 8.4}{T_{r} + 10.74} - 15

(2) In the formula at (1)—