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

Answers: 3

Show answers

Answers

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.

Answer from: Quest

answer: a

a is the answer

Answer from: Quest

answer:

i am not 100% sure but i think it is a.)

dont take my word for it

explanation:

Another question on Engineering

Engineering, 03.07.2019 14:10

Explain the difference laminar and turbulent flow. explain it with the shear stress and the velocity profiles.

Answers: 1

Answer

Engineering, 03.07.2019 14:10

When at a point two solid phase changes to one solid phase on cooling then it is known as a) eutectoid point b) eutectic point c) peritectic point d) peritectoid point

Answers: 3

Answer

Engineering, 03.07.2019 14:10

Amass of 1.5 kg of air at 120 kpa and 24°c is contained in a gas-tight, frictionless piston-cylinder device. the air is now compressed to a final pressure of 720 kpa. during the process, heat is transferred from the air such that the temperature inside the cylinder remains constant. calculate the boundary work input during this process.

Answers: 2

Answer

Engineering, 03.07.2019 14:10

Amass of m 1.5 kg of steam is contained in a closed rigid container. initially the pressure and temperature of the steam are: p 1.5 mpa and t 240°c (superheated state), respectively. then the temperature drops to t2= 100°c as the result of heat transfer to the surroundings. determine: a) quality of the steam at the end of the process, b) heat transfer with the surroundings. for: p1.5 mpa and t 240°c: enthalpy of superheated vapour is 2900 kj/kg, specific volume of superheated vapour is 0. 1483 m/kg, while for t 100°c: enthalpy of saturated liquid water is 419kj/kg, specific volume of saturated liquid water is 0.001043m/kg, enthalpy of saturated vapour is 2676 kj/kg, specific volume of saturated vapour is 1.672 m/kg and pressure is 0.1 mpa.

Answers: 3

Answer

You know the right answer?

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

Questions