--- title: "Wind turbine merits and output: 5.5 m blade" subject: "Power Plant Design" chapter: 1 key: "semester_8/power_plant_design:wind-merits-numerical-5-5m" source: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-merits-numerical-5-5m site: BCT Engineering --- # Wind turbine merits and output: 5.5 m blade Power Plant Design · Chapter 1: Energy Sources and Electric Power Generation - Asked in: 2073 Magh, 6 marks - Video: [Wind Energy | Wind Power | HAWT & VAWT | Calculations](https://www.youtube.com/watch?v=kDK6Da8Crj8) — Engineering & Science Basics - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_8/power_plant_design/wind-merits-numerical-5-5m - Whole chapter: https://www.bctengineer.com/md/semester_8/power_plant_design/chapter-1 ## Question How electricity is generated from Wind-turbine system? What are the merits of wind energy? Estimate the power output from the wind turbine that has blade length of 5.5 m and wind speed of 25 m/sec. Assume that air density is 1.23 kg/m³ and power coefficient is 0.4. ## Short answer Wind turns blades, a gearbox increases speed, and a generator converts mechanical torque into electricity fed through a converter and transformer. Merits are renewable, free fuel, no emissions, low running cost, and modular installation. Using P = 0.5 rho A v^3 Cp with A = pi r^2, output is about 365.3 kW. ## Answer #### Electricity generation from a wind turbine The moving air turns the rotor blades; the rotor converts the kinetic energy of wind into mechanical torque on a shaft. A gearbox steps up the speed, and the generator converts it to electricity, which is fed to the grid through a converter and transformer (or stored in batteries for stand-alone systems). ```text Wind -> Blades/hub -> Low-speed shaft -> Gearbox | Grid <- Transformer <- Converter <- Generator (yaw drive, pitch control and brake on the nacelle) ``` #### Merits of wind energy - Renewable, free fuel, no emissions during operation. - Low running and maintenance cost. - Modular; quick installation; land below can still be used for farming. - Suitable for remote areas and hybrid systems with solar and hydro. #### Power output $$ P = \frac{1}{2}\rho A v^3 C_p, \qquad A = \pi r^2 $$ Given: $r = 5.5$ m (blade length), $v = 25$ m/s, $\rho = 1.23$ kg/m³, $C_p = 0.4$. $$ \begin{aligned} A &= \pi \times 5.5^2 = 95.03\ \text{m}^2 \\ P_w &= 0.5 \times 1.23 \times 95.03 \times 25^3 \\ &= 0.5 \times 1.23 \times 95.03 \times 15625 = 913{,}209\ \text{W} \\ P &= 0.4 \times 913{,}209 = 365{,}284\ \text{W} \end{aligned} $$ **Answer:** Power output $\approx 365.3$ kW. Note: 25 m/s is near the usual cut-out speed of large turbines, so in practice the output would be limited to the rated value by pitch control. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the correct sequence of power flow in a wind turbine system? A. Blades -> generator -> gearbox -> converter B. Blades -> low-speed shaft -> gearbox -> generator -> converter -> transformer -> grid C. Gearbox -> blades -> generator -> grid D. Blades -> transformer -> generator -> gearbox Answer: B. The rotor drives a shaft and gearbox, then the generator, with converter and transformer before the grid. (Key point: Energy conversion chain) 2. What is the purpose of the gearbox? A. Steps up the slow rotor speed for the generator B. Steps down the generator speed C. Stores energy D. Converts AC to DC Answer: A. The rotor turns slowly; the gearbox raises speed for the generator. (Key point: Role of the gearbox) 3. Which is a merit of wind energy? A. Output is constant B. Land below can still be used for farming C. No site limitations D. High running cost Answer: B. Land around turbines can still be farmed, along with low running cost and no emissions. (Key point: Merit of wind) 4. For r = 5.5 m, v = 25 m/s, rho = 1.23 and Cp = 0.4, the output is about A. 913 kW B. 365 kW C. 182 kW D. 36.5 kW Answer: B. A = 95.03 m^2, wind power 913 kW, times 0.4 = 365.3 kW. 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