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Eduardwww [97]
3 years ago
10

Windmills slow the air and cause it to fill a larger channel as it passes through the blades. Consider a circular windmill with

a D1 = 8m diameter rotor in a V1 = 12m/s wind. The wind speed after the windmill is measured at 8 m/s. (a) (7 pt) Determine the diameter of the wind channel downstream from the rotor (D2). (b) (8 pt) Calculate the power produced by this windmill in kW. Assume air is incompressible and use air density of 1.2kg/m3 .
Engineering
1 answer:
Scilla [17]3 years ago
8 0

Answer:

DIAMETER  = 9.797 m

POWER = \dot W = 28.6 kW

Explanation:

Given data:

circular windmill diamter D1 = 8m

v1 = 12 m/s

wind speed = 8 m/s

we know that specific volume is given as

v =\frac{RT}{P}

  where v is specific volume of air

considering air pressure is 100 kPa and temperature 20 degree celcius

v =  \frac{0.287\times 293}{100}

v = 0.8409 m^3/ kg

from continuity equation

A_1 V_1 = A_2 V_2

\frac{\pi}{4}D_1^2 V_1 = \frac{\pi}{4}D_1^2 V_2

D_2 = D_1 \sqrt{\frac{V_1}{V_2}}

D_2 = 8 \times \sqrt{\frac{12}{8}}

D_2 = 9.797 m

mass flow rate is given as

\dot m = \frac{A_1 V_1}{v} = \frac{\pi 8^2\times 12}{4\times 0.8049}

\dot m = 717.309 kg/s

the power produced \dot W = \dot m \frac{ V_1^2 - V_2^2}{2} = 717.3009 [\frac{12^2 - 8^2}{2} \times \frac{1 kJ/kg}{1000 m^2/s^2}]

\dot W = 28.6 kW

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A motorist enters a freeway at 25 mi/h and accelerates uniformly to 65 mi/h. From the odometer in the car, the motorist knows th
Helga [31]

Answer:

a) 2.2 m/s² b) 8 s

Explanation:

a) Assuming that the acceleration is constant, we can use any of the kinematic equations to solve the question.

As we don´t know the time needed to accelerate, we can use the following equation:

vf2 – vo2 = 2*a*∆x

At first, we can convert the values of vf, vo and ∆x, to SI units, as follows:

vf = 65 mi/h* (1,605 m / 1mi) * (1h/3,600 sec) = 29 m/s

vo = 25 mi/h *(1,605 m / 1mi) * (1h/3,600 sec) = 11.2 m/s

∆x = 0.1 mi*(1,605 m / 1mi) = 160.5 m

Replacing these values in (1), and solving for a, we have:

a = (29 m/s – 11.2 m/s) / 321 m = 2.2 m/s2

b) In order to obtain the time needed to reach to 65 mi/h, we can rearrange the equation for the definition of acceleration, as follows:

vf = vo + at  

Replacing by the values already known for vo, vf and a, and solving for t, we get:

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5 0
3 years ago
Please answer the questions !
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120

Explanation:

6 0
3 years ago
In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
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Answer:

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Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0
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