Steps for Refrigeration Load Calculation for Ice Rink
Let's assume the following:
- Size of the ice rink = 10 m (l) X 5 m (b) X 0.5m (t)
- Density of ice (d)= 0.9 gm/cc = 900 kg/cubic metre
- Time duration for converting water to ice (T) = 5 hrs = 18000 sec.
- Ambient temperature (Tamb) = 20 degree C
- Final temperature of ice (Tice) = -4 degree C
- So, the weight of ice (M) = 22500 Kg.
- Specific heat of water (Swater) = 4186 J/kg -degree C
- Specific heat of ice (Sice) = 2093 J/kg -degree C
- Specific latent heat for fusion for ice (L) = 334000 J/Kg
- Thermal conductivity of ice (Kice) = 2.25 W/m-K
- Convection heat transfer coefficient for air (Uair) = 50 W/m2-K
Refrigeration load required to cool down the water from ambient temperature to zero degree centigrade: (R1) = M * Swater * (Tamb – 0) = 1883700 J.
Refrigeration load required to convert the zero degree water to zero degree ice: (R2) = M*L = 7515000 J.
Refrigeration load required to cool down the ice further to – 4 degree centigrade: (R3) = M*Sice*(0-Tice) = 188370 J.
Now the time required for the above three process is say 5 hrs. (18000 Sec.). So, the cumulative refrigeration load for the above three process KW will be R4 = (R1+R2+R3)/ (T*1000) = 532.615 KW.
In order to calculate the refrigeration load (R5) required to prevent the ice from loosing its cool, we first need to calculate the overall heat transfer coefficient (U) of the ice and air:
U = 1 / [(1/Uair) + (t / Kice)] = 4.128440367 W / M2 – K
R5 = [U * l * b * (Tamb – Tice)]/1000 = 4.954 KW
So the final refrigeration load (R)for the ice rink is = R4 + R5 = 537.569 KW