Question about the Propane powered RV refrigerator

[pierhogunn]

is it possible to replace the heat source in an RV, or a larger scale propane powered refrigerator with the heat from the focal point of a parabolic reflector that is 7' wide and get it to work?

the only reason that I ask, is that I figure I could use the projects old satellite dish to be the heat source, and unlike the normal intended momentary use for an RV refrigerator, just run this thing continuously for the 6 - 8 hours of full sun that we can get during the day here in Charlotte ( barring cloudy days)

The idea is that if this thing is running hard for 6 - 8 hours I could reduce the temperaure of a volume of water by an appreciable amount, hopefully something useful to cool the small volume of air in the www.ecobox.me home concept.

shoot, if there is enough heat to spare, I imagine that you could run 2 or 3 of these refrigeration units...

If I could find some more documentation on it, I would rather use a lithium bromide based absorption system, since it would use lower temps in the process, and you can use evacuated solar tubes which collect long-wave infrared even on cloudy days...

I just wish I knew what the math was to figure this out, and if I knew what that was, I wish I could actually do the math... ( I don't do math, I have nothing against it, I just stink at it)

[master53yoda]

here is a piece of info that you can think about the obtainable capacity of a properly oriented collector averaged over 8 hours of operation (full sun ) is between 400 and 800 BTUH per square yard. calc out the area of the collector and that will give you what you need. check the wattage of the RV heater and multiply the wattage * 3.4 and it will give you BTUH. An RV frig would takes about 4 to 6 hours to start operating . so that would only give 6 to 8 HRs of operation per day because they would have to start the cycle over each day. Not much usable heat.

[LCHIEN]

the energy from the sun hitting the earth has a flux density of 680 watts per square meter.
Figure the area of the collector in square meters and multiply times 680, this will give the maximum energy you can collect.
Now you start taking deductions. Only about 30% of solar energy reaches the surface, the rest is reflected by the atmosphere or clounds. You have to take further deductions if the sun is not directly overhead, any angle increases the amount of atmosphere you have to traverse which refleccts more enery and more cloud cover. If you are near the poles of the earth probably only 10% of the energy reaches the surface. (related by the trigonometric Sine of the azimuth)
Then you have to allow for nighttime and cloudy days, nighttime averages 50% loss but again if you are near the artic circle, the geometry says you can get 0% in the winter and 100% in the summer. Cloudy days can be 0% in sunny parts of the world and 90% loss in perpetually cloudy locales.

So by now your 680 watts per square meter has dropped to probably about 50-100 watts per square meter for most of the middle North America.

off the top of my head a 7-foot dish has a surface area of a bit over 3 square meters (using pi x r-squared and 9 sq ft/sq meter).
You'll get 150 watts for 12 hours a day or 1.8 KWH (about 20 cents worth at todays prices).


Then you have to take into account transfer efficiencies.
Solar to voltaic cells runs 18% efficient. reflectors probably run 80% efficient.
Heat exchanger transfer pipes, maybe 50% if they're good. You'll have losses due to insulation and temperature differences.

All the percentages multiply to give you a very small part of the original incoming power.

When you figure out what you have, see if its enough to power what you need.
Convert watts to joules and calories. remember it takes 1 calorie to raise 1 cubic centimeter of water 1 degree C (or 1.8 degrees F).
Good luck

[mpc]

RV refrigerators work on the ammonia cycle... kinda odd compared to your home refrigerator that uses a compressor to drive the cycle. Your car air conditioner works like the home refrigerator too.

This Wikipedia entry gives an overview of the ammonia cycle type of refrigeration:
refrigeration cycles

This will give you an idea how the RV 'fridge should work and help you create ideas for your alternate heat source idea.

A friend has solar water heating panels on his roof... they are pretty darn large, especially compared to the typical boiler element size of a gas water heater. And this is in sunshine rich SoCal. That gives me an idea of how solar collection energy rates compare to flames. Your old satellite dish might not be big enough to do much except prove your concept is sound or not... but it probably won't be large enough to sustain a lot of RV style refrigeration. Good luck though, I hope it does work for you if you try it.

mpc

[master53yoda]

Lchien has the hard data, I was not able to find that actual watt calc and could only remember the 400 or so that the panels picked up extra when the sun was on them.( the system I had worked with were refrigerated panels. the thing I think that is being missed here is that temperature (Heat density) and the amount of heat (BTUs or Joules ) are not necessarily related. The reflector would concentrate the defused energy into a very small location making it very hot but that does not equate to any large amount of energy. An example is that 212 degree steam has approxiamatly 3 times the amount of energy as 212 degree water. In refrigeration you can have a liguid at 100 degrees with only 15 btu per lb and a vapor at 30 degrees holding 85 btus per lb. The total energy term is enthalpy.

Temerature and the usable amount of energy is not necessarily what you would think it is.

[Kristofor]

the energy from the sun hitting the earth has a flux density of 680 watts per square meter.

The numbers I've seen are about twice that for solar power density (~1.4kW/m^2). I'm thinking the number you used averaged in 12hrs of night already...

Of the ~30% that's reflected the majority (~2/3) is being reflected off of clouds not the rest of the atmosphere or the ground, water, ice, etc... Since the atmosphere is so thin (relatively) even at a high angle of incidence for inhabited areas it's not sapping more than ~10%. The ratio of cloud cover/sunny days at your location will be a more important consideration.

Still, during the day it would not be unreasonable to hit 800-1000W/m^2 in much of the USA. Transfer of solar heat to a fluid is often much more efficient than photovoltaic cells, but still well below 100%. And once you're into the physical systems you have all of those inefficiencies too (not unique to solar though).

But, after doing all of that evaluation you need to ask the question "what am I trying to accomplish". If it's to show some interesting systems for a hobbyist to play with that should be an achievable goal. But if it is to provide a cost effective, reliable solution I don't think we can get there from here with this type of homebrew kludge...

[pierhogunn]

Kristofor, I believe you just answered the million dollar quesiton, is this feasable...

probably not is what I am hearing.... thank goodness this is for a college, and is considered a classroom.

so, now to locate an RV refrigerator and start playing with the mylar film.

thanks for the math guys, I'll try later today to condense that and start playing

even if this only works for 6 - 8 hours a day, I am hoping to over time, maybe 2 weeks or so drop the temperature of a 500 gallon tank that has been insulated with about 7 or 8" of spray-foam

so, I might be able to capture about 1kwH out of this thing?

I got to get a bigger dish, or make a 16' wide heliostat

[billwmeyer]

I know nothing about this topic, but I remember reading an article one day about using mirrors to concentrate more sunlight into an area. I don't know if it is correct or feasable, but I thought I would throw it out anyway since you are in a classroom environment anyway.

Bill

[LCHIEN]

i was just doing a "napkin" analysis - e.g. what we do when we have an idea over lunch and then do some order of magnitude calculations without using a calculator just to see whats feasible.

A mirror bill suggested is what pier was suggesting with a reflectorised former satellite dish.

In reality the parabolic reflector simply takes X number of watts from a large area and concentates it to X watts in a smaller area (less any losses from being a less than perfect reflector). This is only useful if you have a small collector. If you can make your collector the size of the dish then there's no benefit to the dish reflector, seems to me.
The focal point may be hotter but if flowing fluid through a small collector vs. flowing the same rate through a larger collector the heat transferred to the fluid will be exactly the same.

[pierhogunn]

anyone here got any plans for a stirling engine they wouldn't mind sharing?

I figure, for just testing / playing I could stick a small stirling engine's hot side at the focal point, and have it turn a very small alternator to maybe get 250 - 500 W of 12V juice out of it

[pierhogunn]

okay maybe someone can explain this to me

okay, what exactly does 5.04 KwH/sq.meter/day mean

does it mean that I get 5.04Kw of power over the span of one hour for each square meter of surface I have to collect it averaged out over the space of a 24 hour day?

so (((5.04 KW/H )/1)/24))

I am confused, math is not my strongest suit

does this mean that I can run 50 100W light bulbs at the same time for 24 hours straight on the energy collected by a single square meter of my super efficient ( and lossless) solar power system?

[Kristofor]

A kW is a unit of power (Current X Voltage) = 1 Amp x 1Volt x 1000 (feel free to spread that 1000 between volts and amps so long as the product equals 1000)

A kWh is a unit of energy equal to a 1 kW flowing for 1 hour.

So your 50x 100W light bulbs would use (50 x 100W = ) 5000W, and would run for 1 hour on 5kWh of energy.

You could run 25 for 2 hours, or 5 for 10 hours, etc.

The per m^2 per day means that you generate ~5kWh for every square meter of collector that you have per day. So to run all of the lightbulbs for two hours you either need twice the collector area for a single day, or two days to gather the energy.

[Russianwolf]

I'm not an engineer, but I know for a fact that using a 6 inch first surface mirror (from my reflecting telescope) which has a focal length of 4 foot, I can start a fire in paper or other thin kindling in about 2 second on a clear day. Getting a solid piece of wood started can be done with a steady hand and a couple minutes.