I know someone who can get me new auto parts at cheaper than junkyard prices. So, I thought about using a heater core/melting ice setup myself. I've seen the great reveiws for the Koolerair. I've put some thought into this and decided against it. My math below is why. I could be wrong. I always seem to miss some mundane detail.
Humans put off about 250 BTUs of heat per hour when sleeping.
(
http://www.answerbag.com/q_view/514275)
Ice requires 144 BTUs to melt one pound.
(
http://www.elitesoft.com/sci.hvac/iticehou.html)
So, to remove the heat generated by two people (and ignoring that it is hot outside) you would need 250*2/144=3.47222 pounds of ice per hour. Eight hours would be 27 pounds melted but still around 33 degrees.
The cold water (melted ice) warming to 72 degrees would consume another 40 BTUs per pound. (same link as above) 27*40/500=2.16 more hours until the 27 pounds water is 72 degrees.
But none of this accounts for air replacement. And instead of figuring out how to keep a closed system cool, instead, let's see what it would take to take outside air and cool it as it enters and blows on the occupants who are down low and hopefully, the warm air is rising and being exhausted out the top. For this, we need three variables. We need the outside temp, the desired temp of the air as it blows inside, and we need the volume of air exchanged.
Since it was over 80 degrees before 8am when I got in my truck for work this morning, lets use an outside temp of 90 degrees, a desired breeze of 75 degrees, and an exchange rate of 7*4*5=140 cubic feet exchanged each five minutes. (I make this up as I go along. I don't know if that is enough air exchange to be comfortable)
It takes 0.018 BTUs to raise (or give off to drop) one cubic foot of air one degree.
(
http://wiki.answers.com/Q/How_many_btu%27s_does_it_take_to_heat_up_1_cubic_foot_of_air_1_degree)
So we need to drop 15 degrees * 140 cubic feet * 12 times per hour * 0.018 BTUs = 453.6 BTUs per hour.
Divide 453.6/144=3.15 pounds of ice melted per hour. This is only .32 pounds less per hour than the first scenario.
So yes, you can be more comfortable using this than using just a fan, but to get AC quality cooling, you'll need a lot of ice. Other things to consider are the size of the heater core/radiator. Maybe it would be better to use several smaller heater cores in series instead of one bigger one. It won't change the math above, but it might (big might) change how much humidity is removed as the air passes over the fins. The slower the air moves, the more it should be chilled by the cores and more humidity should condense. But the faster the breeze moves across your skin, the greater the "wind chill" factor. And if you can dry the air more, than your body's perspiration evaporates quicker cooling you off quicker.
Now as some people said...if you are near an unlimited supply of cold water and just want to pump it through, it should work. I wouldn't want a big mud puddle left from using a campgrounds hose connection all night and I'm not sure they would want you to use up their water for that purpose, but it would work. I like the cold stream idea as well. But since I live in FL, unless you tap into a spring, the water will be too warm to do any cooling. Even the spring water is around 70 degrees here.
digress here...
How about sticking a water bed mattress into your teardrop. Fill it with tap water as long as it is cool. Say it is 5 wide by 6 long by .5 deep. That's 15 cubic feet of water. 15 * 62 pounds per cubic foot = 930 pounds of water. (I hope you have GOOD suspension) Oh, and a BTU is the energy required to raise one pound of water one degree. (at 59 degrees. At other temps it is close as long as it is not 32 or 212 degrees where the change of state requires lots more energy.) So, you have 930 BTUs absorbed for every degree the water can raise until it is too warm to be comfortable. Then just sleep on the cold mattress. Say it is 70 degree water and it would still be comfortable at 75. Then you have 930*5=4650 BTUs total and for two people (500 BTUs/hour) it could last one night at 9.3 hours. But when you consider that most of your heat will go up and out the vent, then the cool mattress will stay cool even longer. If it is too cold to sleep on, place a couple sheet between you and the mattress.
back on topic...
I guess we could ask people who have AC units what the smallest BTU rating is that would be considered comfortable in a teardrop. A PetKool removes 2500 BTUs/hour? That's the same BTUs removed as melting 13.6 pounds of ice per hour and raising the melted water to 72 degrees. Does anyone have a plug-in AC unit that doesn't cut it? If so, what is it's BTU rating? Is the teardrop insulated? What were the weather conditions when it didn't work? From what I have read in this forum, I think the PetKool has the lowest BTU rating and it still may be overkill for a teardrop except in the most severe heat.
Disclaimer. I did most of this math as I was typing this out. If you see a mistake, please, please, please correct me. I'm a computer guy, not an AC guy. So I tried to state my sources. If they are wrong, talk to the guys who wrote the Internet.