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kari74 [83]
3 years ago
12

Large wind turbines with a power capacity of 8 MW and blade span diameters of over 160 m are available for electric power genera

tion. Consider a wind turbine with a blade span diameter of 100 m installed at a site subjected to steady winds at 8 m/s. Taking the overall efficiency of the wind turbine to be 44 percent and the air density to be 1.25 kg/m3, determine the electric power generated by this wind turbine. Also, assuming steady winds of 8 m/s during a 24 hour period, determine the amount of electric energy and the revenue generated per day for a unit price of $0.09/kWh for electricity.
The density of air is given to be rho-1.25 kg/m3

The electric power generated by the wind turbine is ____ kWh.

The amount of electric energy generated is ____`` kWh

The revenue generated per day is $ ____.

Physics
2 answers:
kkurt [141]3 years ago
8 0

Answer:

Electric power generated= 26544kWh

Power generated by the turbine = 1105.98kw

Revenue per day = $2388.96

Explanation: Given

Air density, air=1.25kg/m3

Overall efficiency = 0.44

Diameter of turbine = 100m

Velocity= 8m/s

Unit price of electric power generated =$ 0.09

The expression for electric power generated is the product of overall efficiency and change in kinetic energy of turbine blade

n overall = W/mv^2/2

Rearranging the equation gives

W = noveral × mv2/2

W = overall × (pAv)v^2/2

W = overall × pAv^3/2

W = noverall × pAv^3/2

W = no real × pair ×pi/4 d^2×v^3/2

W = 0.44×1.25×(3.142/4 )× 100^2 × (8^3/2)

W = 1105.98kW

Electric energy power per day = E = Wt = 1105.98× 24hrs = 26544Wh

Revenue generated per day = E × Runit

= 26544 × $0.09 = $ 2388.96

blagie [28]3 years ago
6 0

Explanation:

Below are attachments containing the solution

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Answer

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after time T/4 current reach maximum

 t = \dfrac{T}{4}

 t = \dfrac{2\pi\sqrt{LC}}{4}

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        t = 8.2 x 10⁻⁴ s

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b) using law of conservation

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If the radius of an atom is 60 pm and the radius of the Earth is 6000 km, by how many orders of
kherson [118]

Answer:

1 x 10¹⁷

               

Explanation:

Given data:

        Radius of the earth  = 6000km

        Radius of an atom  = 60pm

Now, how many orders is the radius of the earth larger than an atom

Solution:

To solve this problem, let us express both quantity as the same unit;

         1000m  = 1km

    6000km  = 6000 x 10³m   =  6 x 10⁶m

    60pm;

            1 x 10⁻¹²m  = 1pm

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ollegr [7]

<u>Answer:</u>

0.24 m

<u>Explanation:</u>

Given:

Wave velocity ( v ) = 360 m / sec

Frequency ( f ) = 1500 Hz

We have to calculate wavelength ( λ ):

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v = λ f

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Putting values here we get:

= > λ = 360 / 1500 m

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= > λ = 0.24 m

Hence, wavelength of sound is 0.24 m.

6 0
3 years ago
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