Question

A wind mill of 60 m diameter produces 1200 kW of power
when it receives the wind at speed of 12 m/s having density
of air as 1.19 kg/mº. Find the efficiency of wind mill.​

Answer 1

The efficiency of the wind mill is 96.1%

To Find:

The efficiency of wind mill

Given:

   The diameter of the wind mill is 60 m.

   The wind mill produces 1200 kW of power when the wind is blowing at a speed of 12 m/s.

   The density of the air is 1.19 kg/m^3.

Solution:

To find the efficiency of a wind mill, we need to calculate the power output of the wind mill and the power input of the wind.

The power output of the wind mill can be calculated using the formula:

Power output = (1/2) * ρ * A * v^3

Where:

ρ = density of air (1.19 kg/m^3)

A = area of the blades (π * (d/2)^2)

d = diameter of the wind mill (60 m)

v = wind speed (12 m/s)

So,Power output = (1/2) * 1.19 kg/m^3 * π * (30 m)^2 * 12^3 m^3/s^3 = 1244.57 KW

The power input of the wind can be calculated using the formula:

Power input = 1/2 * ρ * A * v^3

Where:

ρ = density of air (1.19 kg/m^3)

A = area of the blades (π * (d/2)^2)

d = diameter of the wind mill (60 m)

v = wind speed (12 m/s)

So,Power input = 1/2 * 1.19 kg/m^3 * π * (30 m)^2 * 12^3 m^3/s^3 = 1244.57 KW

The efficiency of the wind mill can be calculated using the formula:

Efficiency = (Power output / Power input) * 100

So,

Efficiency = (1200 KW / 1244.57 KW) * 100 = 96.1 %

Therefore, the efficiency of the wind mill is 96.1%.

   

Answer 2

The efficiency of the wind mill is 96.1%

Given:

The diameter of the wind mill is 60 m.

The wind mill produces 1200 kW of power when the wind is blowing at a speed of 12 m/s.

The density of the air is 1.19 kg/m^3.

To find: Efficiency

Solution:

Simplification method simplifying processes complicated in work towards regularity of exertions, costs and opportunity. It removes useless and detrimental (or unneeded) difference and variety.The interpretation and estimate field is individual of the easiest in the mathematical talent division and only efficient estimation can help a contender resolve questions from this division easily.

To find the efficiency of a wind mill, we need to calculate the power output of the wind mill and the power input of the wind.

The power output of the wind mill can be calculated using the formula:

Power output = $(1/2) * ρ * A * v^3$

Where:

ρ = density of air (1.19 kg/m^3)

A = area of the blades (π * (d/2)^2)

d = diameter of the wind mill (60 m)

v = wind speed (12 m/s)

So,Power output =
$(1/2) * 1.19 kg/m^3 * π * (30 m)^2 * 12^3 m^3/s^3 \\= 1244.57 KW$

The power input of the wind can be calculated using the formula:

Power input = $1/2 * ρ * A * v^3$

Where:

ρ = density of air (1.19 kg/m^3)

A = area of the blades (π * (d/2)^2)

d = diameter of the wind mill (60 m)

v = wind speed (12 m/s

So,Power input = 1/2 * 1.19 kg/m^3 * π * (30 m)^2 * 12^3 m^3/s^3 = 1244.57 KW

The efficiency of the wind mill can be calculated using the formula:

Efficiency = (Power output / Power input) * 100

So,

Efficiency =
$(1200 KW / 1244.57 KW) * 100 = \\96.1 %$

Therefore, the efficiency of the wind mill is 96.1%.

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