You should make a wind speed distribution for the hub height with intervals of 1 m/s for the three winter months December, January and February. A wind speed distribution gives the number of hours that the wind is in between 0.5 and 1.5 m/s, in between 1.5 and 2.5 m/s, in between 2.5 and 3.5 m/s and so on. The wind speed should have been measured and averaged for very short periodes like 10 seconds. If you take the average of about 10 minutes, you can get a very different result. This wind speed distribution normally gives one interval with the highest number of hours. Assume this is the interval in between 4.5 and 5.5 m/s. So the average wind speed is 5 m/s for this interval. If this most important interval has an average wind speed lower than 4 m/s, I think that you can better forget wind energy.
Next you chose a certain wind turbine but you must have the Pel-V curve of that wind turbine. For all my HAWT's, I give the calculated Pel-V curves in the design reports but for most VAWT's reliable Pel-V curves are not available. If a Pel-V curve isn't availble, it is impossible to calculate the output for the three winter months. Assume that you chose the VIRYA-5B3 for which the Pel-V curve is given as figure 8 of my public report KD 710. This curve starts at V = 3 m/s. So you can read that Pel = 110 W at V = 3 m/s, 259 W at V = 4 m/s, 506 W at V = 5 m/s, 875 W at V = 6 m/s and so on. Next you multiply the power for a certain wind speed as read from the Pel-V curve with the number of hours of each interval for the same average wind speed and you find the energy generated for that wind speed interval in Whours. The total energy in the three winter months is the sum of the energy of all wind speed intervals.
What you find is the generated electrical energy. If all energy is supplied to a heat pump, you get about a factor four more heat energy depending on the COP-value of the heat pump. But you need a lot of heat even for a well isolated house and therefore you need a big wind turbine especially if the best wind speed interval is gained for a low wind speed. You should also make a calculation of the heat loss of your house to find out how much heat you really need. I expect that with a HAWT with a rotor diameter of 5 m and a tower height of a least 12 m you will get a good result for a well isolated house in a good wind regime. But such a large HAWT is out of the range to design and build by most amateurs. So you should find a grid connected windmill of this size which can be bought somewhere. I can't advise you which brand is best as this requires careful investigation of different brands.
But if you take a VAWT like a Savonious rotor with a Cp value which is less than half of that of a well designed HAWT, you need more than the double swept rotor area. So this results in a Savonious rotor with a diameter and height of about 7 m. For the same wind speed, this Savonious rotor should have a tower of about 8.5 m high. This Savonious rotor will be extremely heavy but also very dangerous as you can't limit the rotational speed at high wind speeds. May be now it will be clear why a Savonious rotor is only a reasonable option if you need only a very little power.
If you take a Darrieus rotor, the maximum Cp is about 0.35 but the required swept rotor area is still larger than for a well design VAWT. A H-Darrieus rotor needs a diameter and a height of about 5 m to get the same power as a HAWT with a rotor diameter of 5 m. But Darrieus rotors have a lot of disadvantages (see report KD 215). It might be possible to solve the starting problem (see KD 601) but the other disadvantages remain. A H-Darrieus rotor is very sensible to vibrations and it is also very difficult to limit the rotational speed and thrust at very high wind speeds. So you should forget this if you don't have the needed skills and money to solve all problems.
