Question

# 1. A windmill has a diameter of 3 meters. It converts wind energy to electrical energy...

1. A windmill has a diameter of 3 meters. It converts wind energy to electrical energy at an efficiency of 45 percent of the THEORETICAL MAXIMUM for windmills, when connected to an electrical generator. What is the electric power output at a wind velocity of 10 miles per hour? (Compute your answer in Watts, but do not enter units. If the answer is 50 W, enter 50.)

2. How many 60 watt lightbulbs can be supplied with electricity for the windmill described above? (Note: in reality, bulbs come one by one, so it's a bit absurd to say you can light "1.3 bulbs". But CAPA is a dumb computer, so if your answer comes out to 1.3 bulbs, go ahead and input 1.3, understanding full well that in reality that would mean you could light only 1 bulb fully...)

3. How many 60 watt lightbulbs can be supplied with electricity for the windmill described above? (Note: in reality, bulbs come one by one, so it's a bit absurd to say you can light "1.3 bulbs". But CAPA is a dumb computer, so if your answer comes out to 1.3 bulbs, go ahead and input 1.3, understanding full well that in reality that would mean you could light only 1 bulb fully...)

4. If you tripled the windspeed in part a, by what factor would you increase the number of bulbs you could light up? (If you think you could light up 6 times as many bulbs, input 6) (Note: You can figure this one out even if you didn't get either of the previous two problems right!)

1. Power generated by a windmill is given by, P = 1/2 * v3 * * r2 * Efficiency

First we have to convert the 10 miles/hr to meter/second.

therefore, 10 miles/hr = 4.47 m/s.

Putting the values in the equation,

P = 1/2 * (4.47)3 * 3.14 * (3/2)2 * 0.45

P = 141.97

2. We are getting 141.96 W output power so Number of 60W bulbs supplied by electricity = 141.97 / 60

= 2.36

3. Question 2 and 3 are same.

4. If the velocity of the wind tripled,

then velocity = 3* ( 4.47 m/s)

= 13.41 m/s

Using Power formula, P = 1/2 * v3 * * r2 * Efficiency

= 1/2 * (13.41)3 * 3.14 * (3/2)2 * 0.45

= 3833.37

Number of bulbs which can be light up = 3833.37 / 60

= 63.89

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