I don't have money for a nutrient chiller right now, and nutrient temperatures have been on the rise—reaching as high as 74F—I'd really prefer it to be around 68F—so what's a grower to do? Improvise! I used my freezer and some brain-stretching math to get to grips with lowering the nutrient solution using plain old ice.
So—inside my nutrient reservoir there's 30 gallons of nutrient solution. My Bluelab Guardian Connect monitor informed me that it was getting just a touch warm—around 23C or 73.4F. Now, an experienced grower buddy of mine said this was OKAY as long as I agitated (bubbled) the nutrient solution. But I wanted to see if I could lower the nutrient temperature with ice from my freezer, as I had plenty of room in there. Of course, ice will lower the temperature of your nutrient solution. But then the question arose: just how much ice was needed?
I asked some of my peers on the Just4Growers Facebook group. There are so many potential variables to take into account:
Volume of the nutrient solution.
Starting temperature of the nutrient solution.
Temperature of the ice.
Even the shape of the ice was mentioned!
But we'd only just begun...
From my own experiments, I'd found that 4.5 litres of ice at -17 °C was sufficient to reduce the temperature of my nutrient solution by around 4 °C.
However I wanted to get into the real science of it—without melting my brain in the process. I decided to forget about ice shape and the effect of the mineral nutrients (conductivity) and, with the help of my Facebook buddy Nigel Davenport
, I Googled
some info on energy conservation and the laws of thermodynamics
I learned that there were, in fact, three processes taking place:Process 1: Ice melting to a liquidProcess 2: Water gives energy to melt iceProcess 3: Two liquids of different temperatures mix and find equilibriumProcess 1 - ice melting:
Mass of ice: 4,500 grams
Initial temperature: -16 °C
Temperature change to turn into water: 16 °C
Specific heat of ice: 2060 J/kg C (I guess this is a constant that physicists just know - heheh)
Q1 = Ice mass x specific heat of ice x temperature change
Q1 = 4.5 x 2060 x 16 = 148,320 Joules
Process 2: the heat required to melt the ice is determined by the specific latent heat of fusion:
The specific latent heat of fusion is 334,000 J/kg. So:
Q2: 4.5 x 334,000 = 1,503,000 JoulesProcess 3: The ice now becomes the water and we need to increase the temperature of the water from 0 to the final temperature - T
. The heat required to increase the temperature of the water is determined by the specific heat of the water:
The specific heat of the water is 4186 J/kg C (another constant, I presume, based on the masses of the ice and water being equal). So:
Q3 = ice mass x specific heat of water x T
Q3 = 4.5 x 4186 x T
Q3 = 18,837 T
Total heat transfer from ice (QI) = 148,320 + 1,503,000 + 18,837 T = 1,651,320 + 18,837T
Now, for the final bit regarding the temperature decrease of the water:
The mass of our water is 30 gallons or 113.562 kg
We already know that the specific heat of water is 4186 J / kg C
The change in water temperature can be expressed as:
ΔT = 23 - T
QII = Mass of water x Specific heat of water (23 - T) = 113.562 x 4186 (23 - T) = 475,370 (23 - T)
So the energy conversion takes the form:
1,651,320 + 18,837 T = QII = 475,370 x (23 - T)
1,651,320+ 18,837 T = 10,933,510 - 475,370 T
18,837 T + 475,370 T = 10,933,510 - 1,651,320
494, 207 T = 9,282,190
T = 9,282,190 / 494, 207
T = 18.78 degrees celsius.
Yes, it appears around 4.5 litres will do just fine.
By the way, my plants are doing fine! Make sure you check my YouTube channel - Just4Growers
— for regular video updates.
All that made my brain ache. Perhaps I should ask Santa for a nutrient chiller for Christmas?