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Lyrx [107]
4 years ago
9

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 80-m-diameter blades at that location. Also determine the actual electric power generation assuming an overall efficiency of 30 percent. Take the air density to be 1.25 kg/m3. Assume that the wind is blowing steadily at a constant uniform velocity and that the efficiency of the wind turbine is independent of the wind speed.
Engineering
1 answer:
valentina_108 [34]4 years ago
3 0

We know that:

Kinetic energy is the only form of  mechanical energy the wind possesses, and it can  be converted to work entirely.

Therefore, the  power potential of the wind is its kinetic energy,  which is V²  /2 per unit mass, and mV² / 2 for a  given mass flow rate:

e (mech) = k e =  V²  /2

e (mech) = (10 m/s)² / 2

To convert the units from m/s to kJ/kg,

e (mech)  = 50 m² /s² x ( 1 kJ/kg / 1000 m² /s²)

e (mech)  = 0.050 kJ/kg

Now,

We also know the relation

m = ρVA

Also A = π D²/4, so above equation becomes:

m = ρV (π D²/4)

m = (1.25 kg/ m³) ( 10 m/s) x ( 3.14 x 80 m x 80 m)/ 4

m = 62,800 kg/s

So,

E (mech) = m x e (mech)

E (mech) =  62,800 kg /s x 0.050 kJ/kg

E (mech) = 3,140 kW

Therefore, the 3140 kW of actual power can be generated by this wind turbine at the given conditions.

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What are hydraulics powered by? Select the best option.
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Answer:

Pressurized Liquid

Explanation:

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How many ticks are in a inch?<br><br><br> A: 12<br> B: 10<br> C: 8<br> D: 16
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Explanation:

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3 years ago
A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr
seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F =  mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625

Weight transfer is given as follows;

W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

F_R × 1.3 = 1.7 × F_F

F_R + F_F = 18000

F_R + \dfrac{1.3 }{1.7} \times  F_R = 18000

F_R = 18000*17/30 = 10200 N

F_F = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel F_R  = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel F_F =  7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

\mu = \dfrac{F}{R} =  \dfrac{11250}{18000} = 0.625

The least coefficient of friction, μ = 0.625.

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