3 ms·

That’s why it absorbs so much energy when you wear it. If it didn’t take that much energy, it wouldn’t be that useful. But heat pumps are actually extraordinar

by ericd 5y ago

That’s why it absorbs so much energy when you wear it. If it didn’t take that much energy, it wouldn’t be that useful. But heat pumps are actually extraordinarily efficient, so ice production isn’t bad compared to the cooling it would give the wearer. Which is one of the reasons many buildings build up big blocks of ice at night and then use them to cool themselves throughout the day.

LorenPechtel 5y ago
Building up big blocks of ice at night is inherently less efficient than cooling only as much as needed. The reason it's done is to take advantage of night time power being cheaper. On the scale those buildings operate at it's worth it.
- Gravityloss 5y ago
  From a naive perspective, energy is energy, so melting 1 kg of ice eats 300 kJ of energy, and heating 8 kg of water from 20 C to 30C also takes 300 kJ. But if the ambient temperature is 30 C, how much did it take to make that 1 kg of ice or that 8 kg of 20C water? With a heat pump, like your refrigerator or freezer, producing cool water is much easier with high efficiency multipliers, while producing ice is a lot harder. If the outside temperature is 30 C, the the ice machine radiator must be at a higher temperature, say, at 40 C, to radiate heat, and the cold end must be below freezing to actually freeze the water. So, the radiator might be at 40C and the cooler at -10 C. So, the compressor must transport the 300 kJ of heat energy "against the grain" from cool to hot, a temperature difference of 50 C. With the refrigerator, the hot end might be at 40 C and the cool end at 10 C, meaning the difference is only 30 C. COP or Coefficient power is the amount of energy (J) the heat engine uses to transfer 1 Joule of energy. Some formulas: eff = 1 - T_cold / T_hot Refrigerator: T_cold = 10 C = 280 K, T_hot = 40 C = 310 K eff = 1 - 280/310 = 0.0968 COP = 1/eff = 10 Freezer: T_cold = -10 C = 260 K, T_hot = 40 C = 310 K eff = 1- 260/310 = 0.161 COP = 1/eff = 6.2
- ericd 5y ago
  You're right that it's mainly a power time shifting scheme. But there's no fundamental reason that I can see that it should be much less efficient than on-demand cooling - you can always insulate the ice storage well, and air tends to be cooler at night, so your heat pumps should be somewhat more efficient at radiating heat into the air than during the day.

by ericd 5y ago

LorenPechtel 5y ago

Building up big blocks of ice at night is inherently less efficient than cooling only as much as needed. The reason it's done is to take advantage of night time power being cheaper. On the scale those buildings operate at it's worth it.

Gravityloss 5y ago
From a naive perspective, energy is energy, so melting 1 kg of ice eats 300 kJ of energy, and heating 8 kg of water from 20 C to 30C also takes 300 kJ. But if the ambient temperature is 30 C, how much did it take to make that 1 kg of ice or that 8 kg of 20C water? With a heat pump, like your refrigerator or freezer, producing cool water is much easier with high efficiency multipliers, while producing ice is a lot harder. If the outside temperature is 30 C, the the ice machine radiator must be at a higher temperature, say, at 40 C, to radiate heat, and the cold end must be below freezing to actually freeze the water. So, the radiator might be at 40C and the cooler at -10 C. So, the compressor must transport the 300 kJ of heat energy "against the grain" from cool to hot, a temperature difference of 50 C. With the refrigerator, the hot end might be at 40 C and the cool end at 10 C, meaning the difference is only 30 C. COP or Coefficient power is the amount of energy (J) the heat engine uses to transfer 1 Joule of energy. Some formulas: eff = 1 - T_cold / T_hot Refrigerator: T_cold = 10 C = 280 K, T_hot = 40 C = 310 K eff = 1 - 280/310 = 0.0968 COP = 1/eff = 10 Freezer: T_cold = -10 C = 260 K, T_hot = 40 C = 310 K eff = 1- 260/310 = 0.161 COP = 1/eff = 6.2
ericd 5y ago
You're right that it's mainly a power time shifting scheme. But there's no fundamental reason that I can see that it should be much less efficient than on-demand cooling - you can always insulate the ice storage well, and air tends to be cooler at night, so your heat pumps should be somewhat more efficient at radiating heat into the air than during the day.