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I was thinking of letting it evaporate too. It seems this is the less costly option. No need to truck it anywhere, just dump it in big artificial pools. Once in

by mrb 13y ago

I was thinking of letting it evaporate too. It seems this is the less costly option. No need to truck it anywhere, just dump it in big artificial pools. Once in a while, collect the solid waste from the pools (lactose, mineral, proteins). They say this waste is dangerous for the environment, but maybe not when it is in solid form?

icegreentea 13y ago
Here are the following problems. Firstly, the sheer scale. One of the major producers makes 150 million gallons a year. Or half a million gallon a day, or roughly an olympic sized swimming pool. Or... lets go all out. Half a million gallons is roughly 2 million liters, or 2 million kilograms of water. Assume air temperature of 20 degrees. To boil off the water, its going to take 4 kilojoules per kilogram per degree celcius.... so thats 640 GJ already. Ignore heat of vaporization. Incident solar radiation at like ... 1000W per m^2. 12 hours -> 43200 seconds per day, so 1 m^2 receives 44MJ is a day... which means you need about 15000m^2 of surface area to evaporate off a day's worth of production in a day. Optimistically.
- rdl 13y ago
  Which really doesn't seem like that big a deal for a farm or large factory -- it's a 100m x 150m pad. Presumably you could keep it sanitary by using plastic sheeting over a concrete pad.
- takluyver 13y ago
  Actually, I think the heat of vaporisation is more than the energy to heat the water to boiling: 2.26 MJ/kg, which takes us to ~4TJ/day. The 1000W solar radiation is also for the sun directly overhead without clouds, which obviously can't happen for 12 hours a day. I've found a source saying Bangkok gets something like 20MJ/m^2/day, and higher latitudes presumably have less. With those figures we're at 200000m^2 of evaporation area, and it's still optimistic (what about rain?).
- nijk 13y ago
  That is a completely reasonable size for an industrial pond.
- ars 13y ago
  It's not necessary to heat water in order to evaporate it. The energy calculations are completely off. The main thing you need is low humidity, and/or lots of wind.

by mrb 13y ago

icegreentea 13y ago

Here are the following problems. Firstly, the sheer scale. One of the major producers makes 150 million gallons a year. Or half a million gallon a day, or roughly an olympic sized swimming pool. Or... lets go all out. Half a million gallons is roughly 2 million liters, or 2 million kilograms of water. Assume air temperature of 20 degrees. To boil off the water, its going to take 4 kilojoules per kilogram per degree celcius.... so thats 640 GJ already. Ignore heat of vaporization. Incident solar radiation at like ... 1000W per m^2. 12 hours -> 43200 seconds per day, so 1 m^2 receives 44MJ is a day... which means you need about 15000m^2 of surface area to evaporate off a day's worth of production in a day. Optimistically.

rdl 13y ago
Which really doesn't seem like that big a deal for a farm or large factory -- it's a 100m x 150m pad. Presumably you could keep it sanitary by using plastic sheeting over a concrete pad.
takluyver 13y ago
Actually, I think the heat of vaporisation is more than the energy to heat the water to boiling: 2.26 MJ/kg, which takes us to ~4TJ/day. The 1000W solar radiation is also for the sun directly overhead without clouds, which obviously can't happen for 12 hours a day. I've found a source saying Bangkok gets something like 20MJ/m^2/day, and higher latitudes presumably have less. With those figures we're at 200000m^2 of evaporation area, and it's still optimistic (what about rain?).
nijk 13y ago
That is a completely reasonable size for an industrial pond.
ars 13y ago
It's not necessary to heat water in order to evaporate it. The energy calculations are completely off. The main thing you need is low humidity, and/or lots of wind.