“…= 3.41 + (3.55 Γ π where V is the indoor air velocity at a height of 0.1 m above the ice, and h Conv [W/(m 2 ΛK)] is the heat transfer coefficient at the same height above the ice, calculated with V assumed to be 0.15 m/s. Then [19,20], (19) π πΆπππ£. = β πΆπππ£ Γ (π ππ -π π ), where q conv [W/m 2 ] is the convection heat transfer between the ice surface and indoor air, T S is the ice surface temperature [ΛK], and is the indoor air temperature at height 0.1 m. Furthermore,…”