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Strike441 [17]
3 years ago
10

A wind turbine rotates at 20 rev/min. If its power output is 2.0 MW, the torque produced on the turbine from the wind is

Engineering
1 answer:
Hoochie [10]3 years ago
4 0

Answer:

The torque torque produced on the turbine from the wind is approximately 955 kN·m

Explanation:

The number of revolution per minute of the turbine = 20 rev/min

The power output of the turbine = 2.0 MW

The power transmitted to a shaft equation is given as follows;

Power , \ P = \dfrac{2 \cdot \pi \cdot N \cdot T}{60}

Where;

P = The power transmitted to a turbine shaft = 2.0 MW

N = The number or revolutions per minute = 20 rev/min

T = The torque produced on the turbine by the wind

Therefore;

Torque , \ T = \dfrac{P  \cdot 60}{2 \cdot \pi \cdot N } = \dfrac{2.0 \times 60}{2 \times \pi \times 20}  = \dfrac{3}{\pi } \ MN \cdot m

The torque torque produced on the turbine from the wind = 3/π MN·m ≈ 0.955 MN·m = 955 kN·m.

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A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T
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Answer:

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Explanation:

given data

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solution

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R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

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R min = 28.173 ohm

and

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A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
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Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

<u>Part (A)</u> The stress on the steel wire;

δ = F/A

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