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MA_775_DIABLO [31]

2 years ago

12

A type 3 wind turbine has rated wind speed of 13 m/s. Coefficient of performance of this turbine is 0.3. Calculate the rated pow

er density of the wind that is hitting the turbine. Calculate the mechanical power developed at the shaft connecting rotor and generator. Assume rotor diameter 100 m and air density 1.225 kg/m^3.

1 answer:

Anna [14]2 years ago

8 0

Answer:

Rated power = 1345.66 W/m²

Mechanical power developed = 3169035.1875 W

Explanation:

Wind speed, V = 13 m/s

Coefficient of performance of turbine, $C_p$ = 0.3

Rotor diameter, d = 100 m

or

Radius = 50 m

Air density, ρ = 1.225 kg/m³

Now,

Rated power = $\frac{1}{2}\rho V^3$

or

Rated power = $\frac{1}{2}\times1.225\times13^3$

or

Rated power = 1345.66 W/m²

b) Mechanical power developed = $\frac{1}{2}\rho AV^3C_p$

Here, A is the area of the rotor

or

A = π × 50²

thus,

Mechanical power developed = $\frac{1}{2}\times1.225\times\pi\times50^2\times13^3\times0.3$

or

Mechanical power developed = 3169035.1875 W

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3 years ago

An aluminum block weighing 28 kg initially at 140°C is brought into contact with a block of iron weighing 36 kg at 60°C in an in

Anika [276]

Answer:

Equilibrium Temperature is 382.71 K

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Solution:

As per the question:

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Mass of Iron block, m = 36 kg

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At room temperature

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Now,

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Thus

$T_{e} = 109.71^{\circ}C$ = 273 + 109.71 = 382.71 K

where

$T_{e}$ = Equilibrium temperature

Now,

To calculate the changer in entropy:

$\Delta s = \Delta s_{a} + \Delta s_{i}$

Now,

For Aluminium:

$\Delta s_{a} = MC_{p}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 28\times 0.949\times ln\frac{382.71}{413} = - 2.025\ kJ/K$

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$\Delta s_{i} = mC_{i}ln\frac{T_{e}}{T_{i}}$

$\Delta s_{a} = 36\times 0.45\times ln\frac{382.71}{333} = 2.253\ kJ/K$

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To find the normal stress (σ) in the wall of the basketball we need to use the following equation:

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<u>Where:</u>

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<u>Where:</u>

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I hope it helps you!

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Answer:

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