Wind vs.Solar

Why not balance out a system with both?
A little PV and a little Wind Genny.
If you got time, collect some information first.
All UL listed electrical gizmo's will have a tag or sticker on them somewhere that states a combination of; watts and volts, or volts and amps. There is a mathmatical procedure called Ohm's law. Simply stated it says, that volts multiplied by amps equals watts, or that watts divided by either volts or amps will equal the resultant amps or volts. But lets not forget about adding time to this procedure.
Example; Lets say the tag on your refrigerator tells you that it consumes 50 watts at 120 vac. That would be, 50/ 120 = .42 amps.
But, if it consumes 50 watts at 12 vdc  
that would be, 50/ 12 = 4.2 amps.
Now, what that tag will not tell us is for just how long of time the fridge will consume that 50 watts in a 24 hour period. I will pick a number from the air, 6. Lets say that the fridge is consuming 50 watts of power for 6 hours out of a 24 hour period. This is where it can get tricky.

I will assume that you are planing to live off grid. If so then you will need batteries, so to know how much battery to get you will need to know all of your electricity needs over a 24 hour period in advance, and you will need to phrase those needs in terms of amp hours at a nominal battery voltage. I will stick with just the fridge for these examples.

We know from the previous example that at 12 vdc nominal the fridge needs 4.2 amps for 6 out of 24 hours. Simply multiply, 6 x 4.2 = 25.2 amp hours. There is one more step to sizing the battery bank. We need to know what number 25.2 is 20% of. To find that you can simply guess, trial and error or just multiply 25.2 x 5 = 126 amp hours of battery capacity at 12 vdc nominal. (we know that 20% is one fifth of the whole.)

Now, to size a PV array to recharge the battery. (A load this small could easily be handle by PV so I will use it only for this example.)
We know the fridge will need a battery with at least 126 amphours of storage at 12 vdc nominal.
Now we need to know the the number of hours of equivilent full rated charge that you can get from a PV array. It gets tricky again!

Hours of Equivilent Full Rated Charge. There are so many variable to this;
The region, the season, how the PV is mounted, its oreintation to our Sun, and last the PV modules rated output. You mentioned Texas and I will assume a flat plate collector, perpendicular to our Sun at noon time on the shortest day (Dec.21) and that there is full unshaded sunlight from 9:00 am to 3:00 pm. All of this considered will give you about 4 hours of equivilent full rated charge.

To replace with PV what the fridge takes away from the battery simply take the 50 watts the fridge consumes and multiply it by the 6 hours that it might run and we have 300 watt hours. Then divide that by the 4 hours of equivilent full rated charge and we have a 75 watt PV array but, not every day will produce the 300 watt hours needed from the PV array because of inclimate weather. So an increase of 30% to 100% would be, well just a smart thing to do. If at all possible I would double the PV array set in this example and round up the battery to 220 amp hours. Between two 75 watt PV modules wired in parallel and the 220 amp hour battery, the fridge in this example should never run out of power. Who knows, there might even be enough power for a small light or two, but the fridge comes first.

I just used the fridge in these examples but the same princibles will apply to other electric loads. Oh, if you plan on using an inverter, there is no differance between vac or vdc when using Ohm's law for these calculations. In other words if the fridge tag says 50 watts at 120 vac that 50 watts still has to be calculated at 12 vdc, you will have to add about 15% to the calculation because the inverter will consume power in order to make the 120 vac from the 12 vdc. In the example above though, that was taken care of by "bumping" up the size of the system.

Oh! dont forget, there must be a charge controller. Disconects as well as fuses and or breakers and the right size and type of wire are the most important part of any electrical system. Too many amps moving through too small of a wire thats not protected by a fuse or breaker for the size of that wire can cause a fire.