Mathamatical model for evaporator performance?

redleader · 04-30-2003, 04:44 PM

It occured to me that its trivally easy to calculate the rate at which an ideal evaporator will use up its coolant, however I've never seen a model of the temperature the coolant will reach at equallibrium.

The way I figure its probably a differential equation that relates vapor pressure as a function of temperature to power, flow, surface area, air pressure etc. However I haven't been able to find anything on google.

Does anyone have the equation?

Volenti · 04-30-2003, 10:07 PM

You tried looking at sites/companies that sell commercial evaporative cooling/towers? I'll have a look as well.

Volenti · 05-01-2003, 08:26 AM

I've found this I havn't looked through it completly, but it's chockers full of formula

redleader · 05-04-2003, 03:17 AM

Interesting. I need to spend some time with it this summer and see how well it works.

IYIENACE · 03-17-2004, 09:54 PM

http://www.cti.org/toolkit.shtml

Ares · 03-20-2004, 12:53 PM

quick thoughts would be get the coeffiencients of water to convert a water to gas.

4.16joules/something... been too long. I can get the numbers on monday when I have my physics book handy.

you need to know some other info. bring the system to equilibrium with the PC on. now you have a steady temp going in and out of the PC. take those temps.

then you figure out how much heat your removing from the system using gallons per minute. Im sketchy on this, but its doable. knowing the difference in temp in a minute you have say 100gallons of 28degree water going to the PC, and 100gallons of 29degree water coming back. using thermodynamic formulas that I have to get on monday, you can figure out how much heat was put out in a minute to heat 100gallons of water 1 degree.

make sense?

alright, so thats how much heat the evaporater is taking out of the system. to find efficiency and what not, theres another coefficient, that says it takes this much heat to evaporate this much water. so knowing how much heat your removing, you can figure out how much water had to evaporate to remove that much heat.

in a complete guess, Im betting it would take SOOOOO little. to the point where Id be amazed if you were really evaporating that much water in a month. maybe a cup or so.ignoring inefficiencies, toss a cup of water in a pot and get it boiling on the stove, tell me how long it takes to COMPLETELY boil away a cup of water, and think of how hot that flame is. takes alot of energy to evaporate water. ALOT. like 100x as much as it takes to raise water 1degree.

pdf27 · 03-20-2004, 02:26 PM

What you want are steam tables. These define the relationship between vapour pressure of steam and temperature.

http://www.connel.com/freeware/steam.shtml

Energy dumped to atmosphere will be related to air flow rate, temperature of inlet air and inlet air relative humidity. With a perfect evaporator (which you won't get in reality), the total heat loss will be the latent heat of evaporation multiplied by the density of water vapour at the inlet temperature, multiplied by (1-inlet humidity fraction).

The problem here is that the temperature limit of an evaporator isn't set by the ideal situation. For instance, if you double the surface area used for evaporation if all else is held constant the mass of water evaporating will be doubled too. This will then increase the heat dumped to atmosphere, and drop the temperature.

You do then end up with a set of differential equations which define the relationship between vapour pressure, water temperature, droplet size, etc. Unfortunately, this doesn't give you the equilibrium temperature. The shape of the bong will also affect things - as the vapour pressure will not be constant within the bong, changing the rate at which water evaporates in different parts.
This means you end up with two coupled sets of differential equations - which you are very unlikely to be able to solve analytically. You'll have to derive both sets of equations (probably not that difficult), and then set up a solver to find the numerical solution. That is probably what the cooling tower programs you found do.
Just to complicate things, you'll also have heat transfer between the air and the liquid droplets, water/air and bong, etc.

lolito_fr · 03-20-2004, 02:40 PM

Quote:

theres another coefficient, that says it takes this much heat to evaporate this much water

Yes, roughly 680W to evaporate 1kg of water in 1 hour
thus a 100W heatload will be drinking 1kg (or liter) of water every 7hours or so (assuming the water is at ambient temperature - so no heat loss by convection)
hardly negligible

04-30-2003, 04:44 PM	#1
redleader Thermophile Join Date: Jun 2001 Location: The deserts of Tucson, Az Posts: 1,264	Mathamatical model for evaporator performance? It occured to me that its trivally easy to calculate the rate at which an ideal evaporator will use up its coolant, however I've never seen a model of the temperature the coolant will reach at equallibrium. The way I figure its probably a differential equation that relates vapor pressure as a function of temperature to power, flow, surface area, air pressure etc. However I haven't been able to find anything on google. Does anyone have the equation?

04-30-2003, 10:07 PM	#2
Volenti Cooling Savant Join Date: Jan 2002 Location: in a nice cool spot Posts: 427	You tried looking at sites/companies that sell commercial evaporative cooling/towers? I'll have a look as well. __________________ feel free to icq/msn me, I'm always willing to toss around ideas.

05-01-2003, 08:26 AM	#3
Volenti Cooling Savant Join Date: Jan 2002 Location: in a nice cool spot Posts: 427	I've found this I havn't looked through it completly, but it's chockers full of formula __________________ feel free to icq/msn me, I'm always willing to toss around ideas.

03-20-2004, 02:26 PM	#7
pdf27 Cooling Savant Join Date: Jan 2004 Location: Horsham, UK Posts: 140	What you want are steam tables. These define the relationship between vapour pressure of steam and temperature. http://www.connel.com/freeware/steam.shtml Energy dumped to atmosphere will be related to air flow rate, temperature of inlet air and inlet air relative humidity. With a perfect evaporator (which you won't get in reality), the total heat loss will be the latent heat of evaporation multiplied by the density of water vapour at the inlet temperature, multiplied by (1-inlet humidity fraction). The problem here is that the temperature limit of an evaporator isn't set by the ideal situation. For instance, if you double the surface area used for evaporation if all else is held constant the mass of water evaporating will be doubled too. This will then increase the heat dumped to atmosphere, and drop the temperature. You do then end up with a set of differential equations which define the relationship between vapour pressure, water temperature, droplet size, etc. Unfortunately, this doesn't give you the equilibrium temperature. The shape of the bong will also affect things - as the vapour pressure will not be constant within the bong, changing the rate at which water evaporates in different parts. This means you end up with two coupled sets of differential equations - which you are very unlikely to be able to solve analytically. You'll have to derive both sets of equations (probably not that difficult), and then set up a solver to find the numerical solution. That is probably what the cooling tower programs you found do. Just to complicate things, you'll also have heat transfer between the air and the liquid droplets, water/air and bong, etc. __________________ Member of the paramilitary wing of CAMRA

05-04-2003, 03:17 AM	#4
redleader Thermophile Join Date: Jun 2001 Location: The deserts of Tucson, Az Posts: 1,264	Interesting. I need to spend some time with it this summer and see how well it works.

03-17-2004, 09:54 PM	#5
IYIENACE Cooling Neophyte Join Date: Mar 2004 Location: Dallas Texas Posts: 14	http://www.cti.org/toolkit.shtml

03-20-2004, 12:53 PM	#6
Ares Cooling Neophyte Join Date: Mar 2004 Location: atlanta Posts: 28	quick thoughts would be get the coeffiencients of water to convert a water to gas. 4.16joules/something... been too long. I can get the numbers on monday when I have my physics book handy. you need to know some other info. bring the system to equilibrium with the PC on. now you have a steady temp going in and out of the PC. take those temps. then you figure out how much heat your removing from the system using gallons per minute. Im sketchy on this, but its doable. knowing the difference in temp in a minute you have say 100gallons of 28degree water going to the PC, and 100gallons of 29degree water coming back. using thermodynamic formulas that I have to get on monday, you can figure out how much heat was put out in a minute to heat 100gallons of water 1 degree. make sense? alright, so thats how much heat the evaporater is taking out of the system. to find efficiency and what not, theres another coefficient, that says it takes this much heat to evaporate this much water. so knowing how much heat your removing, you can figure out how much water had to evaporate to remove that much heat. in a complete guess, Im betting it would take SOOOOO little. to the point where Id be amazed if you were really evaporating that much water in a month. maybe a cup or so.ignoring inefficiencies, toss a cup of water in a pot and get it boiling on the stove, tell me how long it takes to COMPLETELY boil away a cup of water, and think of how hot that flame is. takes alot of energy to evaporate water. ALOT. like 100x as much as it takes to raise water 1degree.

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