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Free energy tricks with water As an exercise, lets build an ideal 2.5 horsepower vacuum water motor for a lawnmower... The easy part is figuring the water consumption. A small power lawnmower usually has a one quart fuel tank. Since we know water only yields one sixteenth as much energy per gallon, we will make the tank 16x bigger, or four gallons. Water is also heavier (denser) than gasoline, so where the gasoline powered mower takes (1/4 x 6) 1.5 pounds of fuel, the water powered mower will take (8 x 4) 32 pounds of fuel. That's more than some power mowers weigh altogether! One horsepower equals approximately 44 BTUs per minute. Good conversion efficiency will be around 20%, so we'll need (44 x 2.5 / 0.20) 550 BTUs per minute to run this mower. At 1,000 BTUs per pound, or 8,000 BTUs per gallon, our mower will run (32,000 / 550) 58 minutes, or about one hour on a four gallon tank of water. Maybe we should cut the water weight in half, and refill it after a half hour of mowing. Now comes the tricky part of figuring the displacement for this engine. Remember that we're expanding rather than compressing the air in order to develop power. Because the pressure will always be 1 atmosphere (15 psi) or lower, and we never develop any heat, all the parts can be lightweight and not heat resistant. This will save us a lot of weight over comparable cast iron engine construction. And we're going to need every bit! You wouldn't want to mow the grass if it's more than 85F out, and you probably won't need to mow the grass when the weather is much below 70F. So we'll design the engine to operate in this range. Humidity is crucial to this sort of motor, since evaporation provides the power. Since less evaporation is possible at the lower inlet temperature, let's base our design on 75F and 50% humidity for the inlet air. Recall that we need to evaporate enough water to provide 550 BTUs per minute of cooling. At 1,000 BTUs per pound, this means we have to evaporate about a half pound of water every minute. 75F air at 50% humidity can absorb about 2.3 millionths of a pound of water per cubic foot to reach 100% humidity. So we need to pump (0.55 / 2.3) 0.239 million cubic feet of air per minute through our motor to develop 2.5 horsepower. Some large displacement racing V8s pump 600 CFM of air at full throttle. We're going to pump 239,000 CFM to develop 2.5 horsepower! Another strike against a vacuum water motor. So how big does it need to be if we run at 3,000 RPM? We need to pump about (239,000 / 3,000) 80 cubic feet per revolution. (Here we gain a little, since one full cycle occurs each revolution.) Because we can use lightweight, low temperature materials, 80 cubic feet doesn't sound impossible for straight pumping. Unfortunately, this is a vacuum engine; so the pumping effect is reduced by the vacuum ratio needed to flash evaporate the water. The ratio is nearly impossible to calculate. The difficulty arises in the interaction between pressure and temperature, and the nonlinear effect this has on vapor pressure. The best way to figure the ratio would be to build a model engine with a variable vacuum ratio, and find the optimum ratio by experiment. Using a SWAG* based on the vapor pressure of water at 75F (0.435 PSI), I'd guess the vacuum ratio has to be at least 1:34!!! *(SWAG : Scientific, wild-ass, guess) Suddenly a vacuum water motor becomes impractical at this temperature and humidity. Vacuum pistons necessary to sweep a displacement of (34 x 80) 2,720 cubic feet are impractical by anyone's standard. |
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So forget the lawnmower, and consider my last suggestion of
an Air Conditioner system with regenerative bootstrapping from
the condenser heat... Such a system would consist of four parts: a conventional electric air conditioning unit (say one ton for calculation purposes), our vacuum water motor coupled to an electrical generator (say two horsepower for the motor, to run the AC with a little wiggle room for air handling and other overhead), a startup system to provide power until the bootstrap effect brings the water motor up to speed (basically a battery and inverter), and a battery charger to recharge the battery once the water motor is running (a basic battery charger with some control circuitry). Now we're dealing with higher temperatures and lower humidity, and we no longer have to worry about the weight of water, since the system can be plumbed directly into a water supply. Let's start by assuming that the feed air for our motor off the AC condenser will never run less than 100F, and never more than 25% absolute humidity.* The two horsepower specified will require about (44 x 2 / 0.20) 440 BTUs per minute and consume about (440 / 8000) .055 gallons per minute of water (about 3.3 gallons per hour). *(Although the two are similar, I've been using absolute humidity throughout this exercise, rather than the usual relative humidity, because the calculations are easier.) |
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@75F, dew point = 63F @ 50% humidity @100F, dew point = 75F @ 25% humidity Air = .0809 pounds / cubic foot Air = 14.7 PSI at one atmosphere Water = .00133 pounds / cubic foot @75F Water = .002854 pounds / cubic foot @100F Water = 0.435 PSI @75F Water = 0.949 PSI @100F .00005628 pounds water / pound air @75F @100% therefore .0809 x .00005628 = 4.55 micro pounds water / cubic foot air therefore 1000 x 4.55 x 10**-6 / 2 .0023 BTU to change one cubic foot of air from 50% humidity to 100% humidity. (Absolute humidity used instead of relative humidity in order to simplify calculations.) |
© 2005 Jeff L. Anderson