The energy consumption of any set of appliances that you might want to use can be worked out to give you an idea both of the size of wind turbine required and the size of the batteries. Note that any appliance which uses electricity for heating will have a very high electricity consumption if used for a long period, so you will not if you are wise try to operate an electric cooker from a wind turbine and batteries.
It is not possible to give figures for electricity consumption of various appliances because these will vary from one appliance to another and from one user to another. The best plan is to make your own estimates based on the wattage of the appliance and the number of hours per day that you use it. The wattage is usually written on the appliance somewhere near the maker’s name, but if you are in doubt the local electricity showroom may be able to tell you.
In the case of some appliances, such as refrigerators and freezers which operate on thermostats and do not run all the time, you will have to get an estimate, again from the electricity board. Remember that a fridge or freezer will use less energy if it is in a cool room such as a larder than if it is in a hot kitchen. As a rough guide an average-sized fridge in a normal kitchen will use about 1 kWh a day, and a large freezer will use 3-4kWh a day. In an unheated kitchen a medium-sized fridge, the sort that fits under a worktop, will use 0.5kWh a day. The one way to be certain is to obtain a meter, connect it to the flex of the fridge or other appliances and read it every day for a year.
To calculate how much electricity you use in a day, multiply the wattage of each of your appliances by the number of hours you use them. This will give you the electricity demand in Watt-hours per day (divide by 1000 to give an answer in kWh).
As a very rough guide the BRE estimate that the average domestic use of electricity for appliances and lights but not cooking, space heating or water heating is 840kWh per household per year, or 2300Wh per day. If you wanted to generate this much electricity in a year in the area used in our earlier example where V50 is 4mlsec you would need to work out the size of wind turbine as follows:
A = Ee/10V50 x Cp X Effa x Effg
A is the wind turbine swept area. Ee is the required electrical output in kWh per year. The amount of electricity required should be increased by 10 per cent to allow for inefficiencies and losses in batteries and wiring, and to give a safety margin over your estimated requirements. This will give a figure of about 925kWh from the 840kWh average value quoted earlier.
Cp is the coefficient of performance; Effa is the efficiency of the alternator; Effg is the efficiency of any step-up drive.
Using the 4m/sec value of V50 from our earlier example, and values for the other terms as used earlier, we get the following result:
A = 925/10 x 43 x 0.3 x 0.5 x 0.8 = 12 m2
So the diameter is 3.9 metres.
The power output at the shaft for a wind turbine of this diameter can be found at the appropriate rated speed, 9m/sec in this example, using the earlier formula P=0.00064xAxV3xCp. In this case it will be 1680W; if this figure is multiplied by the efficiency of the alternator and the step-up gearing, the electrical power output can be found, which is about 670W. Since the type of alternator selected will determine the efficiency value used, and you cannot select an alternator until you know the required output in Watts, you may have to do the calculation a few times to work out a reasonably accurate diameter, yearly output and alternator type.
Your average daily electricity requirement can be multiplied by the number of days during which you expect the windspeed to be too low to generate electricity to give the capacity of the battery in kWh or ah. For example, a consumption of 2.4kWh per day for ten windless days would need a battery of 24kWh capacity. This would be 2000ah at 12 volts or l000ah at 24 volts. A two day battery would be only 400ah or 200ah, and might work out cheaper, if you use a small diesel generator to charge it during longer calm periods. A lot depends on the price you have to pay for the batteries.
For homemade wind turbines, battery voltage will be 12 or 24 because these are the only alternator voltages that are cheaply available (from the motor industry). The problem with low voltages is that you must have very thick wires to carry the current. For example, at mains voltage a 60W lightbulb has a current demand of only 60/240=0.25amp. A normal 5amp lighting cable can carry 5/0.25 or twenty bulbs at mains voltage. At 12 volts a 60W bulb draws a current of 60/12=5amps so a 5amp cable can only operate one bulb.
The ampere rating of cables is determined by their cross-sectional area, the thicker the cable the greater the amperage and of course the cost. At 12 volts even the thick 30amp cable used to connect electric cookers could handle only six 60W bulbs. This problem can be reduced by the use of transistorised low voltage fluorescent lights which are made for caravans and which draw much less current than a conventional incandescent lightbulb of the same light output.
For other appliances it is worth considering buying a transistorised inverter which will change 12 volts DC to 240 volts AC, so you can use normal size house wiring and normal mains voltage equipment. If you decide on an inverter be sure that it produces a mains voltage accurate enough to suit the equipment you wish to operate, and remember that any electric motor when starting draws a current about four times the current it uses when running. The inverter must have the capacity to handle the starting load and not just the running load.
You can see from all this that building a wind turbine is only the beginning of your problems, but if you really want to do it do not expect it to work properly at once and be prepared for periods without electricity. You will not be able to run an all-electric house on wind power but you should be able to provide yourself with light and perhaps a radio without too much trouble.