A 15 m/sec wind at 101.3 kPa and 20 degrees Celeus enter a two-bladed wind turbi

Question

A 15 m/sec wind at 101.3 kPa and 20 degrees Celeus enter a two-bladed wind turbine with a diameter of 15 m. Calculate the following.

(a) The power of the incoming wind

(b) The theoretical maximum power that could be extracted

(c) A reasonable attainable turbine power

(d) The speed in RPM required for part (c)

(e) The torque for part (c)

This is the fom the book:" Alternative Energy Systems and Applications" by B.K. Hodge

chapter 4 problem 1

Please show all the work and formulas I am studying for final exam and need to understand this problem

Explanation / Answer

(a) Density of air, rho= P*M/RT= 1.23kg/m^3 where molar mass of air M=28.97gms, T=293K,R=8.314

Now, the power of the incoming wind P_in = (1/2)*rho*A*v^3=0.366 MW, where A=pi*d^2/4, d=15m, v=15m/s

(b) The theoretical maximum power that could be extracted P_max=0.592*P_in=0.217MW which comes from the betz limit which 59.2% of incoming wind power.

(c) A reasonable attainable turbine power P_out=1/2 of the theoretical maximum power=P_max/2=0.108MW

(d) The speed in RPM, w=(0.79)*(v/R)=1.58rad/s=4.19*10(-3) RPM

(e) Torque=P_out/w=25.6 MN/m

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