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Nookie1986 [14]
2 years ago
6

At a certain location, wind is blowing steadily at 5 mph. Suppose that the mass density of air is 0.0796 lbm/ft3 and determine t

he radius of a wind turbine with a power generation potential of 1000 kW. Express your answer in feet. Also, determine how the power generation potential (PGP) scales with wind speed, that is, find a such that PGP = Cva, where C is a constant.
Engineering
1 answer:
nlexa [21]2 years ago
5 0

Answer:

The radius of a wind turbine is 691.1 ft

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

Explanation:

Given;

power generation potential (PGP) = 1000 kW

Wind speed = 5 mph = 2.2352 m/s

Density of air = 0.0796 lbm/ft³ = 1.275 kg/m³

Radius of the wind turbine r = ?

Wind energy per unit mass of air, e = E/m = 0.5 v² = (0.5)(2.2352)²

Wind energy per unit mass of air = 2.517 J/kg

PGP = mass flow rate * energy per unit mass

PGP = ρ*A*V*e

PGP = \rho *\frac{\pi r^2}{2} *V*e  \\\\r^2 = \frac{2*PGP}{\rho*\pi *V*e} , r=\sqrt{ \frac{2*PGP}{\rho*\pi *V*e}} = \sqrt{ \frac{2*10^6}{1.275*\pi *2.235*2.517}}

r = 210.64 m = 691.1 ft

Thus, the radius of a wind turbine is 691.1 ft

PGP = CVᵃ

For best design of wind turbine Betz limit (c) is taken between (0.35 - 0.45)

Let C = 0.4

PGP = Cvᵃ

take log of both sides

ln(PGP) = a*ln(CV)

a = ln(PGP)/ln(CV)

a = ln(1000)/ln(0.4 *2.2352) = 7.73

The power generation potential (PGP) scales with speed at the rate of 7.73 kW.s/m

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Answer:

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Above equation can be written as:

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(\frac{r}{R} )^{4}+\frac{(1-(\frac{r}{R} )^{2})}{ln\frac{r}{R}}^{2})

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Part B:(d/D=0.01)

DeltaV=\frac{\pi*R^{4}*DeltaP}{8*u*l}(1-(0.01)^{4}+\frac{(1-(0.01 )^{2})}{ln0.01}^{2})

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DeltaV percent=\frac{1-0.783}{1}*100

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