# How effectively will this evaporation setup cool down the water?

I devised a setup, and I show it in the diagram below. Suppose I have these specifications:

1. Water in the bottle is 5 liters and it's initial temperature is 28°C.
2. The bottle capacity is 6 liters.
3. The tank water is at an initial temperature of 35°C.
4. The air inlet fan AND the vapor exhaust have a CFM of 1000 (one is in negative I guess). The circular vent cross-sectional radius is 1ft.
5. The cloth surface area inside the enclosed space is $$1m^2$$ (I don't know if this is realistic considering the bottle size. If not, just use a suitable one in calculations.)
6. Ambient temperature outside enclosed space is 35°C.
7. Enclosed space volume is $$5.8m^3$$
8. (Don't ask about humidity inside enclosed space, as it is supposed to be constant since one fan will cause vapor, the other takes it out. For calculations, use a random, suitable, humidity %).

To bring the water temperature down to 3°C, how long will I have to run the inlet and exhaust fans? Please don't close this outright. It took a lot of energy to make that diagram, so if anything is missing, just comment and I'll update.

• Do you have a time constraint? Apr 29 '20 at 6:26
• FWIW if the worst happens closed questions can be reopened if update adequately. | Is this an assignment of a real world problem? Either is OK but we approach each differently. || Your CFM is WRONG. CFM = cubic feet/minute. 5000 cfm is about 3000 litres or 3 cubic metres/second. If the whole side was inlet on one side and exhaust on the other you'd get about 1 m/s flow. Typical vent sizes would blow a hurricane .| It does not look very air use efficient as drawn - where did you get that flow idea from ? Apr 29 '20 at 8:34
• @SolarMike Time is what I want to know. It is what will vary. Apr 29 '20 at 9:45
• @RussellMcMahon It is not really "real world" because I am trying to come up with something useful to replace a fridge (a big task). I am not really applying it physically right now | Why is the CFM wrong? Maybe you misunderstood me. I'm saying that the inlet is 5000 cfm, the exhaust is 5000 cfm in the opposite direction. | What can I do to increase the efficiency? If you're saying remove the exhaust, then that's impossible as the vapor needs to get out or the humidity will prevent more water to evaporate. Apr 29 '20 at 9:51
• I'm saying that if CFM = cubic feet per minute then 5000 cfm is a huge rate of flow. Where did you get your figures from ? (size, cfm etc) Apr 29 '20 at 10:27

Your proposed device is an adaptation of a Coolgardie Safe, devised in the 1890s. To be effective, a continual supply of moving air is required to keep the items with the devices cool.

Your proposed device has the water tank on the bottom and relies on capillary action for the cloth to absorb the water. A Coolgardie Safe has the water tank on top and relies on gravity to wet the cloth. Also your device has the item inside draped by the cloth, whereas the Coolgardie Safe has the cloth attached to the frame of the device, creating a large cooling zone.

Depending on atmospheric humidity levels and levels of airflow, reducing the temperature by 3 $$\sf^\circ$$C, could take between one hour to and 1.75 hours.

Reducing the temperature to 3 $$\sf^\circ$$C won't happen because of dew point consideration.

Your proposed device and the Coolgardie Safe operate on the same principles as evaporate air conditioners/coolers. They do not have the capability to achieve such low temperatures.

• @SolarMike: You're right, I misread the question. I'll change my answer
– Fred
Apr 29 '20 at 20:29
• I like the example though - I thought of the zeer pot and clay wine coolers.. plus 1... Apr 29 '20 at 20:32
• @SolarMike: Thanks for the heads up about zeer pots. I've learned something today.
– Fred
Apr 29 '20 at 20:43
• Can you give me a simple example of how the dew point is going to prevent the temperature reaching 3°C? Something like "when the temperature reaches [temperature], the dew point is [temperature], and beyond this it's not possible". Thanks. May 2 '20 at 6:27

Think about it, evaporation, dew point?

Under an atmospheric pressure condition, the dew point that I guess is going to be very close to the ambient temperature is the definition of the lowest temperature beyond that no vapors.

If you could vacuum the room or the container then you could reduce the dew point.

• What if I removed the air inlet? That way the exhaust on its own, being powerful enough, would effectively create a constant almost-vacuum enclosed space. Will that help? May 2 '20 at 6:30
• By vacuum i mean industrial vacuum. Your contribution can not produce more than 1 -2 % vacuum. I am talking about 75% vacuum. May 2 '20 at 17:15
• That's a shame. So how low CAN the temperature go? Considering my model, that is. May 3 '20 at 17:05