Assignment 1 Sola 5053

Assignment 1

Question 1
a)

Location 1
11 July 1200 UTC W Moderate
11 July 1800 UTC NW Gentle
12 July 0000 UTC W Moderate
12 July 1200 UTC NE Strong
13 July 1200 UTC NE Strong

Location 2
11 July 1200 UTC N Moderate
11 July 1800 UTC NW Gentle
12 July 0000 UTC NW Moderate
12 July 1200 UTC W Gentle
13 July 1200 UTC SW Gentle

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