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2 years ago

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A wind turbine with a rotor diameter of 40 m produces 90 kW of electrical power when the wind speed is 8 m/s. The density of air

impinging on the turbine is 1.2 kg/m3. What fraction of the kinetic energy of the wind impinging on the turbine is converted to electrical energy

1 answer:

aleksley [76]2 years ago

8 0

Answer:

0.233

Explanation:

Given that

Diameter of rotor, d = 40 m

Power of rotor, P = 90 kW

Speed of the wind, v = 8 m/s

Density of air, p = 1.2 kg/m³

It is a known fact that

KE = ½mv², where mass flow rate, m

m = p.A.v, where Area, A

A = πd²/4

A = (3.142 * 40²)/4

A = 3.142 * 1600/4

A = 3.142 * 400

A = 1256.8 m², substitute for A in the mass flow rate equation

m = p.A.v

m = 1.2 * 1256.8 * 8

m = 12065.28, substitute for m in the KE equation

KE = ½mv²

KE = ½ * 12065.28 * 8²

KE = 12065.28 * 32

KE = 386088.96 W or

KE = 386.1 kW

Fraction of kinetic energy converted to electric energy is

Fraction = Electric Power / Total KE

Fraction = 90 / 386.1

Fraction = 0.233

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Similarly, the distance y the object has traveled is calculated as follows:

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<u>Complete question</u>

The force F required to compress a spring a distance x is given by F– F0 = kx where k is the spring constant and F0 is the preload. Determine the work required to compress a spring whose spring constant is k= 200 lbf/in a distance of one inch starting from its free length where F0 = 0 lbf. Express your answer in both lbf-ft and Btu.

Solution

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F= F₀× kx where F₀ initial force, k is spring constant and x is the deflection from original point of the spring.

$W=\int\limits^2_1 {} \, Fds \\\\\\W=\int\limits^2_1( {F_0+kx} \,) dx \\\\\\W=\int\limits^a_b {kx} \, dx ; F_0=0\\\\\\W=k\int\limits^2_1 {x} \, dx \\\\\\W=k*\frac{1}{2} (x_2^{2}-x_1^{2} )\\\\\\W=200*\frac{1}{2} (1^2-0)\\\\\\W=100.lbf.in\\\\$

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