# Assignment 1 Sola 5053

Topics: Wind power, Wind, Wind turbine Pages: 12 (3076 words) Published: May 20, 2013
Assignment 1

Question 1
a)

Location 1
11 July 1200 UTCWModerate
11 July 1800 UTCNWGentle
12 July 0000 UTCWModerate
12 July 1200 UTCNEStrong
13 July 1200 UTCNEStrong

Location 2
11 July 1200 UTCNModerate
11 July 1800 UTCNWGentle
12 July 0000 UTCNWModerate
12 July 1200 UTCWGentle
13 July 1200 UTCSWGentle

b)
I) The wind direction variation is not great and the speeds do not chande too much either. That would be great for wind generation regardless the low speeds of the wind. II) Great variations in the wind direction that goes from N to NW and then to SW are not good for wind generation as the turbines do not have the capacity to change directions that much. The wind direction changes due to the cold front that pass trough that área. The low speed of winds alos does not help in the generation.

Question 2
a)
I)
Location 1
To calculate the pressure at 110m we use the expression: z=Po× EXP-zH So, substituting the values we have P110=101.5 × EXP-1108000
And as result we got 100.1kPa

To calculate the temperature at 110 we use: Tz=To-Γ(z-zo)

So, substituting the values we have T110=285.15-100×0.005
And as result we got 284.65 x10³

To calculate the density of the air we use: ρ=3.4836×PT

So, substituting the values we have ρ=3.4836×100.114284.65
And as result we got 1.225

Location 2

In location 2 we use the same expressions
So, for pressure at 110m we have : P1100=101.5 × EXP-11009000 As result we got: 89.8 x 10³Pa

For temperature at 110m we have : T1100=285.15-100×0.005
As result we got: 289.1 x 10³

For the density of air we have : ρ=3.4857×89.82289.15
As result we got: 1.082

II)
In the location 1 the direction is SW and in the location 2 the direction is SE

III)
Location 1
To calculate the coriolis fator we use the expression:f=2Ω×sin∅

Substituting the values we have f=1.458×10-4×sin(410)
And as result we got 0.11 x 10-4 (1s)
To calculate the air density at see level we use the expression:ρ=3.4857×PT Substituting the values we have:ρ=3.4857×101.5285.15
And as result we got: 1.24

To calculate the wind speed near surface we use the expression:U=1ρ×f×∆P∆D Where
∆P=400
∆D=200
Substituting the values we haveU=11.24×0.11 x 10-4×400200×103 And as result we got: 14.4ms

Location 2

In location 2 we use the same expressions
So, for coriolis fator we have : f=1.458×10-4×sin(310)
As result we got: 0.11 x 10-4 (1s)

For air density we have : ρ=3.4857×PT
As result we got: 1.24

For the wind speed near surface we have :
∆P=400
∆D=280
So, U=11.09×0.11 x 10-4×400280×103
As result we got: 11.7ms

IV)
The location 1 can be classified as near gale (number 7) and location 2 can be classified as strong breeze (number 6). V) To calculate the power we use the expression: P=0.5×ρ×v3

So for location 1 we have: 0.5*1.225*14.443=1.85kWm2

And for location 2 we have: 0.5*1.082*11.733=0.87kWm2

b)
The profile of the terrain can change the velocity of the wind, due to its roughness the changes from each terrain. If the velocity od the wind is changed, the power will also change. The wind speed can influence in the power because it is related exponencially with it. Small variances in speed can cause a huge difference in the power generated. The topography of the site can interfeer as well because it can create obstacles for the wind or turbulance, changing the generated power. The pressure can also affect the power, the higher the pression is the best for the wind power potencial because the air density i salso greater.

c)
This evaluation is not suficiente because we have to take in account others elements that are no related to the power generated, like the proximity to roads and transmission lines. The demografic density of the site, the presence of national parks and problems with the society like noise and visual polution.

Question 3

In the winter the the high...