Graduate student Kozin A.A.

South Ural State University

Wind potential`s calculation of the village Nepryakhin (Chelyabinsk region)

The calculation was carried out in accordance with the draft of the recreation center, which is located in the village Nepryakhin. To determine the possibility of using wind power plants in this area it necessary to analyze its wind potential.

For the analysis of the data were taken from the site www.wunderground.com for weather station, located at the airport "Balandino", Chelyabinsk. Since this weather station is located in a wind zone Nepryahino obtained values of wind speed can be used for further calculations.

Average wind speed according to the annual observations:

1. In 2009 - 3.74 m/s;

2. In 2010 - 3.75 m/s;

3. In 2011 - 3.7 m/s.

On the basis of annual values of wind speed shall make an assessment of parameters of the random variable with Pearson. To do this, set up the statistical distribution of the sample (Table 1).

Table 1

Wind speed, m / s	Number of days
0,0	1
0,6	5
0,8	13
1,4	18
1,7	24
2,2	36
2,8	26
3,1	37
3,6	50
3,9	28
4,4	37
5,0	19
5,3	17
5,8	11
6,4	16
6,7	7
7,2	6
7,5	6
8,1	3
8,6	3
8,9	2
Total:	365

The sample size 365 days.

The scope of the sample8,9 m/s.

Thus, the number of classes

;

9.

the length of the sub-interval

;

.

Lets construct a histogram of relative frequencies form the following table:

Table 2

Number of interval i	A partial interval	The sum of the frequencies	The relative frequency	The density of the relative frequency
1	0-1	19	0,05	0,05
2	1-2	42	0,12	0,12
3	2-3	62	0,17	0,17
4	3-4	115	0,32	0,32
5	4-5	56	0,15	0,15
6	5-6	28	0,08	0,08
7	6-7	23	0,06	0,06
8	7-8	12	0,03	0,03
9	8-9	8	0,02	0,02

We find an unbiased estimate of the expectation, that is, the sample mean:

;

.

To find unbiased estimates of the variance, skewness and kurtosis parameters, coefficient of variation, we construct the table:

Table 3

Number of interval i	The boundaries of the interval		Mid-range		ni
	xi	xi+1
1	0	1	0,5	-3,2	19	-60,8	194,56	-622,59	1992,29
2	1	2	1,5	-2,2	42	-92,4	203,28	-447,22	983,88
3	2	3	2,5	-1,2	62	-74,4	89,28	-107,14	128,56
4	3	4	3,5	-0,2	115	-23	4,6	-0,92	0,18
5	4	5	4,5	0,8	56	44,8	35,84	28,67	22,94
6	5	6	5,5	1,8	28	50,4	90,72	163,30	293,93
7	6	7	6,5	2,8	23	64,4	180,32	504,90	1413,71
8	7	8	7,5	3,8	12	45,6	173,28	658,46	2502,16
9	8	9	8,5	4,8	8	38,4	184,32	884,74	4246,73
∑							1160,2	1067,20	11590,39

Thus, the

,

.

,

.

,

.

,

.

,

.

,

.

,

.

For the Pearson test the hypothesis of normal distribution of random variable X - wind speed, it is necessary to calculate the theoretical frequency

, and ,

Set up the following table:

Table 4

№
1	0	1	-3,7	-2,7	-2,07	-1,51
2	1	2	-2,7	-1,7	-1,51	-0,95
3	2	3	-1,7	-0,7	-0,95	-0,39
4	3	4	-0,7	0,3	-0,39	0,17
5	4	5	0,3	1,3	0,17	0,73
6	5	6	1,3	2,3	0,73	1,28
7	6	7	2,3	3,3	1,28	1,84
8	7	8	3,3	4,3	1,84	2,40
9	8	9	4,3	5,3	2,40	2,96

Using a table of values of the Laplace function, we obtain:

Table 5

№
1	-0,4808	-0,4345	0,0463	16,8995
2	-0,4345	-0,3289	0,1056	38,544
3	-0,3289	-0,1517	0,1772	64,678
4	-0,1517	0,0675	0,2192	80,008
5	0,0675	0,2673	0,1998	72,927
6	0,2673	0,3997	0,1324	48,326
7	0,3997	0,4671	0,0674	24,601
8	0,4671	0,4918	0,0247	9,0155
9	0,4918	0,4985	0,0067	2,4455

Intervals containing numerically small empirical frequencies (n_i <20, the intervals 1, 8, 9), should be merged, and the corresponding frequencies add up.

Thus to the obtain the following table:

Table 6

i
1	42	38,544	3,456	11,944	0,310
2	62	64,678	-2,678	7,172	0,111
3	115	80,008	34,992	1224,440	15,304
4	56	72,927	-16,927	286,523	3,929
5	28	48,326	-20,326	413,146	8,549
6	23	24,601	-1,601	2,563	0,104
7	39	25,915	13,085	171,217	6,607
∑					38,914

In the table we find the critical points of distribution (0,05; 7), as significance level adopted , and the number of degrees of freedom k = 7 – 3 (after the union of intervals equal to the number 7).

As , there is reason to reject the hypothesis of normal distribution of the population, but the average wind speed, median and modal values are in the same range of wind speeds, which is a sign of a normal distribution

To plot the density of a normal distribution with parameters and S, complete the following table.

For a normal distribution with parameters, the probability density. Using the table of function values, we obtain:

Table 6

i
1	0,5	-3,2	-1,80	0,0790	0,0444
2	1,5	-2,2	-1,24	0,1849	0,1039
3	2,5	-1,2	-0,67	0,3870	0,2174
4	3,5	-0,2	-0,11	0,3965	0,2228
5	4,5	0,8	0,45	0,3605	0,2025
6	5,5	1,8	1,01	0,2396	0,1346
7	6,5	2,8	1,57	0,1163	0,0653
8	7,5	3,8	2,13	0,0413	0,0232
9	8,5	4,8	2,70	0,0104	0,0058

According to the obtained values of a plot of the probability density

Figure 1 - Graph of the probability density

The confidence interval for , as so we get t_y = 1,960.

Thus, (3,522; 3,888) – confidence interval for .

The confidence interval for we get (1,627; 1,945).

Thus, 95% could be argued that the random values of wind speed with. Nepryakhin will be in the range of 3.522 m/c to 3.888 m/c.

These speeds are considered a minimum for the horizontal-axis wind turbines and their use impractical, but there are no current vertical-axis wind turbines at such a rate can already generate electricity, which is the basis for further work on the project supply of recreation with a group of wind turbines.

Bibliography:

1. www.wunderground.com.

2. Zabeyvorota V.I., Volokhova K.I. Mathematics in Economics (Elements of mathematical statistics). Textbook. UrSEI, Chelyabinsk, 1998.

3. Saplin L.A. Research and study on the use of non-traditional and renewable energy sources in the energy supply of the Chelyabinsk region. Report on research work. CHGAU, Chelyabinsk, 1998