Electricity from the Sun, Wind and Water
Units of measure:
Any time we deal with the subject of energy, power, pressure or heat, various definitions are required to make meaningful calculations and comparisons. This section explains the meaning of the 'SI' or metric units, and compares them to the imperial or US units.
Every attempt has been made to ensure accuracy on these figures. In all your calculations, be sure the result makes sense, and that you are not getting a result that seems out of proportion to the input numbers. In dealing with energy transfer, be sure to factor in losses due to in-efficiencies. This is very important when comparing costs from the BTU's produced from burning fossil fuels to KWh / BTU's achieved from electrical heat sources.
Joules - kilowatts - BTU's
Definition of a BTU
Energy required to heat one pound of water by 1 degree F.
10 BTU = 10 pounds (1 imperial gallon ) heated by 1 degree F.
Power and Heat flow
Power and Heat flow
Volumes and conversions.
1 PSI = 6894.76 Newtons / sq. m = 6894.76 Pascal
Absolute Pressure = psia = Gauge pressure (psig) + 14.7
Chemical content of fuels
1 ft^3 of natural gas = 1020 BTU (chemical energy constant)
1 gallon (US) # 2 fuel oil = 140,000 BTU
1 gallon (US) propane = 91,200 BTU
Heat required to raise 1 gm. water by 1 degree C.
Kilocalorie (C) = 1000 calories. C = food calorie
Some practical fuel cost examples:
is sold by its heating value in 'Giga Joules', not by its volume because the heating ability per unit volume can vary. Since it is a commodity, its price can vary.
is sold by the 'kilowatt hour'. In BC, the average cost for 1 kwh is currently $0.065, December 2010 price and ignoring the two tier price structure. That is 6.5 cents for 1,000 watts for one hour - or 100 watts for ten hours. So to run a typical 1500 watt electric heater for one hour it costs (1.5kW X 6.5 cents, or about 10 cents an hour)
Some definitions you have to understand for this to make sense:
|Wire Guage||Ohms / 1000 ft. Cu||Ampacity (see note *)||Ohms / 1000 ft. Al||Ampacity (see note *)|
For a DC system, or single phase AC, there will be two wires carrying current, positive and negative, or hot and neutral for AC. Calculate the total resistance of both wire runs, (there and back). Values given are for 1000 feet, so use the ratio of your distance to 1000 feet * the resistance indicated per 1000 feet.
Example: for 350 feet from the generator to the battery, it would be 700 feet total wire. Say you are using # 6 which has a resistance of 0.403 ohms per 1000 feet. Your resistance would be 700 / 1000 * 0.403 = 0.282 ohms
Determine the voltage drop by multiplying the resistance you come up with by the anticipated current, (amps in the wire). (Amps = power in watts / system voltage) This voltage will subtract from the generator voltage, so in the above example, if the generator is producing some DC voltage, and we expect 20 amps at 26 volts at the load, that means there is 20 amps flowing through the 0.282 ohms of wire which causes a voltage drop of ( 0.282 * 20 = 5.64 volts drop)
Depending on the generator, it will either spin faster and produce a higher voltage to overcome this loss, and still provide the 20 amps at the load, or it may get maxed out unable to produce more voltage, and the voltage drop will limit the ability to push the current to the load. (Note, voltage is like pressure which pushes the Amps. Amps are like the volume of electricity in the wire, hence the name 'current').