Ice Rink Thermodynamics

I have no answer for you... But, thermodinamic should read Thermodynamic. Sorry to nitpick.

There may be evaporation/sublimation from light or heat impinging on the ice, but this may be more than cancelled by water vapor from the air condensing onto the cold ice surface.

Evaporation is vaporization of a liquid.

So, if the refrigeration equipment is operating properly, there will be no evaporation.

What do suppose the wind velocity to be in an enclosed building, in m/s?

[edit] As I read Tornado's comment again, I think he is probably correct in his opinion. There may be a little bit of loss due to influences unknown.

Will the ice surface 'radiate' cooling on the ceiling, people around ? How that can be calculate ?

I think that the air velocity in an enclosed builduing it is around 0.2 m/s

If I understand you question correctly.

RE: ...radiate cooling; well, you can't really radiate cold, you can only move heat.

RE: ..on the ceiling...; cold air is denser that warm air. Which will be inclined to go (in this case stay) towards the ceiling, warm air or cold? So, is this convection or not?

"I think that the air velocity in an enclosed builduing it is around 0.2 m/s" Okay.

- if I put 0.3 (litres/sm) of hot water (50 C) on ice rink (-5 C), how long will take to the water to be 0 grade C ? How much evaporation moisture will be in the air (5 C) and how long will be the time for evaporation ?

I've try this but i'm not sure:
Heat load= 1000*0.3(liters/cm)*(4.2*(50-0)+333+2*(0-(-5))
Heat load=165.9 kJ
- evaporation heat of water =2270(kJ/kg)
so the moisture will be: 165.9/2270=0.07 kg

How much evaporation moisture will be in the air (5 C) and how long will be the time for evaporation ?

maybe the moisture it is: (1000*0.3(liters/cm)*(4.2*(50-0))/2270

It depends on whether the water container is shallow or deep.

the water container is shallow

Actually, the ice will sublimate only if the absolute humidity (AH) gradient from the ice surface upwards is negative. Inherently, the relative humidity (RH) directly at the ice surface will be 100%. Moving up and away from the surface, the RH will most likely decrease as the temperature increases. In the case of a positive AH gradient, humidity will be transported towards the ice where it will form frost and the heat of sublimation will be released, increasing the cooling charge on the cooling unit. If the AH gradient is negative, the ice surface will sublimate and the humidity will be transported into the air thus absorbing the heat of sublimation and decreasing the cooling charge on the cooling unit.

Yes, and to put some figures on it - if the ice is at -5°C, the saturated vapour pressure is ~ 4mb. If the air temperature is say 10°C, saturated VP ~ 12mb. So there is equilibrium when the RH is ~33%. Actual VP = saturated vapour pressure x RH = 4mb. The actual VP falls as the temperature falls from 10 to -5°C, but only slightly.

Codey