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from the oil drum
back-of-the-envelope calculations
Kyle Schuant
the energy E required to compress air at 25C is,

E = 110,000 x ln (P1/P2) /m3/mol

There are about 45mol air in 1m3, so,

E = 110,000 x ln (P1/P2) /m3

This howstuffworks article tells us that an air car tank might have 300lt at 4,561psi, which is 29,999,087.707 - call it 30,000 kPa. Atmospheric pressure is 101.3kPa. 300lt at 30,000kPa will be 90,000lt at atmospheric pressure, or 90m3. And so we get,

E = 110,000 x ln (30,000 / 101.3) x 90
= 110,000 x 5.69 x 90
= 56,331,000J
which is 15.6kWhr

However, a company which supplies air compressors tells us that "Most systems typically waste 25 to 50 percent of the energy required to generate compressed air that actually provides useful work."

Let's be optimistic and assume that with lots of air cars zooming around, service stations will buy the most efficient (expensive) compressors. So we get just a 25% loss. This brings us to 20.9kWhr.

Let's round it up to 21kWhr to refill the tank. Again, this isn't the air car referred to in the article, but it gives us an idea of the order of magnitude.

21kWhr to travel 200km.

A regular small city car gets about 10km/lt. Petrol costs about $1.30/lt, and causes 2.32kg CO2e/lt. So to go 200km in a regular car would cost $26 and cause 46.4kg CO2e in emissions.

Electricity from coal cost $0.1355/kWh and 1.21kg CO2e/kWh, so the 200km journey would cost $2.85 and cause 34.9kg CO2e in emissions.

Electricity from wind costs $0.19/kWh and causes 0.04kg CO2e/kWh. So the 200km journey would cost $3.99 and cause 0.84kg CO2e in emissions.

The average Australian car is driven 15,000km annually. That'd be 75 refills, or 1,575kWh energy in all. That's not bad when the average household uses 6,000kWhr annually.

OK, the numbers: suppose you use 1000 gallons of oil to heat your home (that’s how most of us heat here in the Northeast despite the fact that we don’t like refineries). At $3.00/gallon (my guess for next winter), that costs $3000 (duh). According to my favorite government spreadsheet, there are 138,690 BTUs in each gallon of No. 2 fuel oil so you’re buying about 140 million BTUs to keep you warm. However, because even a good furnace is only 78% efficient, only 108 million of those BTUs do you any good.

If you were to create all the useful BTUs with conventional electric heat, you’d need to buy about 32,000 kilowatt-hours (3412 BTUs per kWh). At the $.16/kWh we’ll be paying here next winter, that’s $5000 dollars. Stick with oil!

But, according to the same spreadsheet, geothermal is 3.3 times as efficient as conventional heat (because you’re just pumping up what you need). With geothermal you’ll need less than 10,000 kWhs and pay about $1500; you save 50% compared to oil!

According to Excel, the present value of $1500 per year over 20 years at 6.5% interest (actually your heat pump system should last longer) is a little more than $16,000. If you can get a heat pump system with the capacity you need installed for that amount or less, you’ve got yourself a bargain – no subsidies involved. Actually, if it’s a new installation then you have to take into account what a furnace and fuel tanks would have cost as well. Distributing the heat inside the house is best done with circulating water but can be done with hot air as well. Either you already have a system to do that or you’d need to pay for one anyway. You can also get your domestic hot water from the heat pump and save a little more. If the price of oil goes up faster than electricity, you save more – and vice versa.

Can you get geothermal heat with this capacity installed for this amount? Depends. It depends on whether you have land that’s easily dug down into, a well, or a pond. If your ho

. It has a complete PAHS system including INSULATION/-WATERSHED UMBRELLA

Calculate the cost of heating and cooling a greehouse.

The question is best answered by using an example, but first some formulas: Watts=amps*volts 1 KWH = 3,413 BTU 1500W electric heater produces 5120BTU Propane is 91,600BTU/gallon Q=deltaT*A/R Q=BTU heat requirement per hour delta T is the difference between outside temp and desired inside temp A= the surface are of the greenhouse is square feet, not including the floor; ie, the roof, side walls, and end walls R= the R value of the glazing

Two important points about this last formula...some heating calculators use "heat loss value" (HLV) instead of R-value. Don't get confused, the HLV is just 1/R. The formula using heat loss value is then Q=deltaT*A*HLV. The other important point is that you use a different temperature when calculating your average energy costs over a month than you do when you you calculate what size heater you need. Heater calculations are used to determine how much heat you will need on the coldest night of the year. For this, you use your coldest winter temps, for example the colder end of your zone. For calculating heating costs you need the average temperature for the month.

Here are some typical R-values for common glazings or coverings: 4 mil polyethylene 0.83; 4 mm (5/32") twinwall polycarbonate 1.43; 6 mil polyethylene 0.87; 6 mm (1/4") twinwall polycarbonate 1.54; 6 mil poly double layer (inflated) 1.43; 11 mil woven polyethylene 0.95; 3 mm (1/8") glass (single layer) 0.88; 16 mm (5/8") triplewall polycarbonate 2.5; Polycarbonate/fiberglass (single layer) 0.83

OK, let's say we are contemplating an ACF 10 x 12 Cottage Greenhouse. We first need to calculate the surface area of the greenhouse. Using the diminsions found on the website and some simple geometry, we calculate the surface area of each sidewall to be 55 sq ft, each roof side to be 64.6 sq ft, and each end wall to be 75 sq ft. Therefore the total area (A) is sq ft is (55*2)+(64.6*2)+(75*2)=389.2 sq f

Three thousand feet below the surface, a sandstone aquifer (caverns that now hold water) will be injected with pressurized air, temporarily displacing some of the water. The electricity from wind turbines will power the compressors. A pipe will deliver underground air compressed to 900 to 1,000 pounds per square inch. The compression of millions of cubic feet of air will be scheduled for nights and weekends, when wind power often sells for next to nothing.

Wind parks pay for themselves when demand and electricity rates are higher - during weekdays and on hot summer days. But when electricity is most needed, sometimes the wind isn't blowing.

The stored-energy park would get around that problem by slowly releasing the pressurized air from the aquifer to provide most of the energy needed to turn the blades of a generator otherwise powered by natural gas. Metered valves would control the release of the pressurized air. Similar operations are used now to store natural gas underground across the nation.

The cavern complex would produce 268 megawatts of electricity to be sold to Midwest utilities on the grid. That's enough to turn on the lights in 268,000 homes.

Kent Holst, development director of the stored-energy park, said the plan could transform the economics of wind power.

With the storage park option, the utility owners are expected to be able to store and produce energy at a price equivalent to 6.5 cents per kilowatt hour, then sell the energy at peak times for 8 to 10 cents a kilowatt hour.

The best metric is energy per unit weight, about 120-150 w-hr/kg for current commercial Li-ion rechargeable batteries. We expect that 500-700 whr/kg can be accomplished in the near term, rising to 1200-1500 whr/kg in the longer term (for the engine and its fuel supply).

choices