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Ronch [10]
3 years ago
5

At a certain location, wind is blowing steadily at 10 m/s. Determine the mechanical energy of air per unit mass and the power ge

neration potential of a wind turbine with 70-m-diameter blades at that location. Also determine the actual electric power generation assuming an overall efficiency of the wind turbine to be 30 percent. Take the air density to be 1.25 kg/m^3.
Engineering
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

<em>a) 50 J/kg</em>

<em>b) 721 67 KW</em>

<em></em>

Explanation:

The velocity of the wind v = 10 m/s

diameter of the blades d = 70 m

efficiency of the turbine η = 30%

density of air ρ = 1.25 kg/m^3

The area of the blade A = \pi d^2/4

A = \frac{3.142 * 70^2}{4} = 3848.95 m^2

The mechanical energy air per unit mass is gives as

e = v^2/2 = \frac{10^2}{2} = <em>50 J/kg</em>

<em></em>

Theoretical Power of the turbine P = ρAve

where

ρ is the density of air

A is the area of the blade

v is the velocity of the wind

e is the energy per unit mass

substituting values, we have

P = 1.25 x 3848.95 x 10 x 50 = 2405593.75 W

Actual power = ηP

where η is the efficiency of the turbine

P is the theoretical power of the turbine

Actual power = 0.3 x 2405593.75 = 721678.1 W

==> <em>721 67 KW</em>

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When you see a street with white markings only, what kind of street is it?
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2 years ago
There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter
weqwewe [10]

Answer:

u_e = 9.3 * 10^-8 J / m^3  ( 2 sig. fig)

Explanation:

Given:

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- permittivity of free space (electric constant) e_o =  8.854 *10^-12 s^4 A^2 / m^3 kg

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Solution:

- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:

                                        u_e = 0.5*e_o * E^2

- Plug in the values given:

                                        u_e = 0.5*8.854 *10^-12 *145^2

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3 years ago
The dam cross section is an equilateral triangle, with a side length, L, of 50 m. Its width into the paper, b, is 100 m. The dam
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Answer:

Explanation:

In an equilateral trinagle the center of mass is at 1/3 of the height and horizontally centered.

We can consider that the weigth applies a torque of T = W*b/2 on the right corner, being W the weight and b the base of the triangle.

The weigth depends on the size and specific gravity.

W = 1/2 * b * h * L * SG

Then

Teq = 1/2 * b * h * L * SG * b / 2

Teq = 1/4 * b^2 * h * L * SG

The water would apply a torque of elements of pressure integrated over the area and multiplied by the height at which they are apllied:

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The term sin(30) is because of the slope of the wall

The pressure of water is:

p(y) = SGw * (h - y)

Then:

T1 = \int\limits^h_0 {SGw * (h-y) * sin(30) * L * (h-y)} \, dy

T1 = \int\limits^h_0 {SGw * sin(30) * L * (h-y)^2} \, dy

T1 = SGw * sin(30) * L * \int\limits^h_0 {(h-y)^2} \, dy

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1/4 * b^3 * cos(30) * L * SG  > 1/3 * SGw * sin(30) * L * b^3 * (cos(30))^3

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For the dam to hold, it should have a specific gravity of at leas half the specific gravity of water.

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