Years ago I was flying back from a convention in Montreal in a small airplane and made a fuel stop in Dalhart, TX. This is a small city in the northwest corner of the Texas panhandle. And according to the National Renewable Energy Laboratory, it ought to be an excellent wind resource. I'd like to understand what storage would be necessary, both in terms of power (rate of delivery) and energy (total capacity) to have the wind resource provide reliable base load power. To this end, I'd consider a system whereby the momentary load would be provided by a wind farm if possible, and available energy above that load would be stored and delivered when available wind energy was insufficient. Two things should be kept in mind here: 1) I'm not an electrical engineer; and 2) I'm sure that I'm not the first to carry out such an analysis. Nevertheless, I'll not be deterred from giving it a shot.

My first chore was to understand the wind regime in Dalhart. To do this, I utilized Mathematica's curated WeatherData to download a time series of average daily wind speeds from January 1, 2000 through September 4, 2015. The data is in kilometers/hour and a plot is shown below:

But this data is (I confidently assume) from the Dalhart Municipal Airport (KDHT) and, if taken according to standards, is measured at a height of 10m. But in this day and age, the hub height of a modern turbine of, say, 3MW nameplate capacity is likely to be 110m or even higher. I'm going to base my analysis on the Siemens SWT-3.0-108 turbine. As the designation suggests, this turbine has a nameplate capacity of 3.0 MW and a diameter of 108 meters. I'll put it on a tower yielding a hub height of 120m.

Now, given the wind at 10 meters, what do I do to estimate the wind at 120 meters? The wind gradient has been well-studied and I'm going to use ~v_{w}(h)=v_{10}(\frac{h}{h_{10}})^a~ where ~v_{w}(h)~ is wind velocity at height h meters (here, 120), ~v_{10}~ is the velocity at 10 meters, ~h~ is the hub height in meters (again, 120), ~h_{10}~ is the measurement or reference height (10 meters), and ~a~ is the so-called "Hellman exponent." I'll use~~0.34 ~~0.2, the exponent for "~~neutral air above human inhabited areas~~ tall crops, hedges, and shrubs." Thus, ~(\frac{h}{h_{10}})^a=(\frac{120}{10})^{0.2}=1.644~. Thus, each wind speed in the data above will be multiplied by ~~2.328~~1.644.* Additionally, I've converted the data to meters/second. The resulting plot is below:

This should yield conservative results with respect to rate of energy delivery because power in wind (that is, rate of energy conversion) is proportional to the cube of velocity. The data shown is daily average, and variances above the average have a greater effect than variances below due to the cubic scaling.

Lest the length of this post get out of hand, I'll stop it here. Next up will be determination of capacity factor based on the data above and using the data for the selected turbine. From there, we'll look at intermittency and the storage required to deliver steady power. Finally, we'll look at the land required and the costs so that a Dalhart, TX wind farm can deliver base load power.

Meanwhile...

My first chore was to understand the wind regime in Dalhart. To do this, I utilized Mathematica's curated WeatherData to download a time series of average daily wind speeds from January 1, 2000 through September 4, 2015. The data is in kilometers/hour and a plot is shown below:

But this data is (I confidently assume) from the Dalhart Municipal Airport (KDHT) and, if taken according to standards, is measured at a height of 10m. But in this day and age, the hub height of a modern turbine of, say, 3MW nameplate capacity is likely to be 110m or even higher. I'm going to base my analysis on the Siemens SWT-3.0-108 turbine. As the designation suggests, this turbine has a nameplate capacity of 3.0 MW and a diameter of 108 meters. I'll put it on a tower yielding a hub height of 120m.

Now, given the wind at 10 meters, what do I do to estimate the wind at 120 meters? The wind gradient has been well-studied and I'm going to use ~v_{w}(h)=v_{10}(\frac{h}{h_{10}})^a~ where ~v_{w}(h)~ is wind velocity at height h meters (here, 120), ~v_{10}~ is the velocity at 10 meters, ~h~ is the hub height in meters (again, 120), ~h_{10}~ is the measurement or reference height (10 meters), and ~a~ is the so-called "Hellman exponent." I'll use

This should yield conservative results with respect to rate of energy delivery because power in wind (that is, rate of energy conversion) is proportional to the cube of velocity. The data shown is daily average, and variances above the average have a greater effect than variances below due to the cubic scaling.

Lest the length of this post get out of hand, I'll stop it here. Next up will be determination of capacity factor based on the data above and using the data for the selected turbine. From there, we'll look at intermittency and the storage required to deliver steady power. Finally, we'll look at the land required and the costs so that a Dalhart, TX wind farm can deliver base load power.

Meanwhile...

*Thanks to Michael Tobis for questioning the exponent. Comparing these numbers with a 120 meter wind map yields a better (and more conservative) agreement.

## 4 comments:

This may be too easy, but I'm not sure Dalhart counts as human-inhabited.

Seriously, panhandle towns aren't really very urban.

First... Really? The Stevie Wonder version?

Second... Since when have you let fear of verbosity slow you?

Third... This is interesting and you should let us know when Chapter 2 can be expected.

Fourth... Hi Michael!

Agreed, having been there. It's pretty desolate. Based on Google Earth, it's center pivot irrigation agricultural and grasslands. But there's no "neutral air above uninhabited areas" in the sources I've found. I follow the general reasoning behind the exponent, i.e., the more interference from ground based structures and features and the more stable the flow, the higher the exponent. This makes intuitive sense. But what would you suggest as an exponent? I could feature 0.2, leading to a multiplier of 1.64. Taking the average of the daily averages gives me a number more in line with the map at http://www.tindallcorp.com/120m-map/ so I'll change it. Thanks.

Yes, I like Stevie Wonder's version. Second, people quickly lose interest, especially when there are equations involved. Third, probably next weekend, though I'll try to find time during the week to work on it since I also find it interesting. There is likely wind farm data from the region (lots of wind farms near there) that can simply give me capacity factors but this is more fun and I can compare it to such data if I can find it.

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