2 computer fans are installed in the hood blowing air into the hood. The hot air is exhausted out in the back. An oscillating fan is on the hood to distribute the air around the room.
Here is my calculation:
90 gallon of water is 368 kg
2L of water is 2.16 kg
Assumptions:
1) Assuming tank tmp gets from 80 - 83 in 2 hours (difference is 2 degrees Celcius).
2) Assuming iced water is at 0 deg F
Using the famous thermodyanmic equation: Q = c mass (Delta temp in C), c = 4.186 kJ/kg for water.
Using assumption 1:
Q (80 - 83F; 2 deg C change) = 4.186 kJ/kg * 368 kg * 2= 3082 kJ
So, in 2 hours, the water will gain 3082 kJ of heat. That's 1541 kJ per hour.
Using assumption 2:
Q (-17C - 0C; 0 - 32F): 4.186 kJ/kg * 2.16 kg * 17 = 154 kJ
Q (Phase Change): 334 kJ/kg * 2.16kg = 721kJ
Q (0 - 27C; 32-80F): 4.186 kJ/kg * 2.16 kg * 27 = 244 kJ
Total = 154 + 721 + 244 = 1119 kJ
Summary:
In 1 hour, the heat build-up in my tank is 1541 kJ. Each 2L of ice at 0F can reduce heat by 1119 kJ. So, the net gain in heat is 442 kJ (1541 - 1119). If you put in 2 2L bottles (~ 1 gallon) per hour, you can reduce the temperature.