Engineering, 15.10.2020 17:01 maddyclark19

Large wind turbines with blade span diameters of over 100 m are available for electric power generation. 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 32 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

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Answer from: nic9837

The wind turbine generates 19297.222 kilowatt-hours of electricity daily.

The wind turbine makes a daily revenue of 1736.75 US dollars.

Explanation:

First, we have to determine the stored energy of wind ( $E_{wind}$ ), measured in Joules, by means of definition of Kinetic Energy:

$E_{wind} = \frac{1}{2}\cdot \dot m_{wind}\cdot \Delta t \cdot v_{wind}^{2}$ (Eq. 1)

Where:

$\dot m_{wind}$ - Mass flow of wind, measured in kilograms per second.

$\Delta t$ - Time in which wind acts in a day, measured in seconds.

$v_{wind}$ - Steady wind speed, measured in meters per second.

By assuming constant mass flow and volume flows and using definitions of mass and volume flows, we expand the expression above:

$E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot \dot V_{air} \cdot \Delta t \cdot v_{wind}^{2}$ (Eq. 1b)

Where:

$\rho_{air}$ - Density of air, measured in kilograms per cubic meter.

$\dot V_{air}$ - Volume flow of air through wind turbine, measured in cubic meters per second.

$E_{wind} = \frac{1}{2}\cdot \rho_{air}\cdot A_{c}\cdot \Delta t\cdot v_{wind}^{3}$ (Eq. 2)

Where $A_{c}$ is the area of the wind flow crossing the turbine, measured in square meters. This area is determined by the following equation:

$A_{c} = \frac{\pi}{4}\cdot D^{2}$ (Eq. 3)

Where is the diameter of the wind turbine blade, measured in meters.

If we know that $\rho_{air} = 1.25\,\frac{kg}{m^{3}}$ , $D = 100\,m$ , $\Delta t = 86400\,s$ and $v_{wind} = 8\,\frac{m}{s}$ , the stored energy of the wind in a day is:

$A_{c} = \frac{\pi}{4}\cdot (100\,m)^{2}$

$A_{c} \approx 7853.982\,m^{2}$

$E_{wind} = \frac{1}{2}\cdot \left(1.25\,\frac{kg}{m^{3}} \right) \cdot (7853.982\,m^{2})\cdot (86400\,s)\cdot \left(8\,\frac{m}{s} \right)^{3}$

$E_{wind} = 2.171\times 10^{11}\,J$

Now, we proceed to determine the quantity of energy from wind being used by the wind turbine in a day ( $E_{turbine}$ ), measured in joules, with the help of the definition of efficiency:

$E_{turbine} = \eta\cdot E_{wind}$ (Eq. 4)

Where $\eta$ is the overall efficiency of the wind turbine, dimensionless.

If we get that $E_{wind} = 2.171\times 10^{11}\,J$ and $\eta = 0.32$ , then the energy is:

$E_{turbine} = 0.32\cdot (2.171\times 10^{11}\,J)$

$E_{turbine} = 6.947\times 10^{10}\,J$

The wind turbine generates $6.947\times 10^{10}$ joules of electricity daily.

A kilowatt-hours equals 3.6 million joules. We calculate the equivalent amount of energy generated by wind turbine in kilowatt-hours:

$E_{turbine} = 6.947\times 10^{10}\,J\times\frac{1\,kWh}{3.6\times 10^{6}\,J}$

$E_{turbine} = 19297.222\,kWh$

The wind turbine generates 19297.222 kilowatt-hours of electricity daily.

Lastly, the revenue generated per day can be found by employing the following:

$C_{rev} = c\cdot E_{turbine}$ (Eq. 5)

Where:

- Unit price, measured in US dollars per kilowatt-hour.

$C_{rev}$ - Revenue generated by the wind turbine in a day, measured in US dollars.

If we know that $c = 0.09\,\frac{USD}{kWh}$ and $E_{turbine} = 19297.222\,kWh$ , then the revenue is:

$C_{rev} = \left(0.09\,\frac{USD}{kWh} \right)\cdot (19297.222\,kWh)$

$C_{rev} = 1736.75\,USD$

The wind turbine makes a daily revenue of 1736.75 US dollars.

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Answer from: Quest

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