1 / 13

The Iowa State University AGC Student Chapter Presents:

The Iowa State University AGC Student Chapter Presents:. Common Math Problems in Wind Turbine Construction (“Good Grief, There’s A Lot of Trig In Wind Turbine Construction.”). Problem #1 - Base- ics

pomona
Download Presentation

The Iowa State University AGC Student Chapter Presents:

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. The Iowa State University AGC Student Chapter Presents: Common Math Problems in Wind Turbine Construction (“Good Grief, There’s A Lot of Trig In Wind Turbine Construction.”)

  2. Problem #1 - Base-ics What is the total height of a wind turbine tower if the foundation base is 16 meters in diameter, the top of the tower is 10 meters in diameter and the tower sides make a 2.45 degree angle with the vertical? top of tower foundation base

  3. Solution (Problem #1) – Base-ics • Wind turbine bases are actually round. Imagine a really tall, thin cone with a small portion of the top cut off. • The diagram below represents what the base might look like of you were to cut it directly down the middle. You end up with a tall, thin trapezoid. The height of the trapezoid represents the height of the base. 10 m top of tower h foundation base 16 m

  4. Solution (Problem #1) – Base-ics • We think it’s easiest to break the trapezoid into 2 triangles and 1 rectangle. • The trangle’s height is the same as the trapezoids, but the base is now 3 m. 10 m top of tower h h foundation base 3 m 10 m 3 m 3 m 16 m

  5. Solution (Problem #1) – Base-ics • We are told that the sides make a 2.45 degree angle with the vertical. • This means that the smallest angle in our triangle is 2.45°. • We can make the assumption that the angle between the ground around the base and the vertical is 90°. • Now that we know three pieces of info about the triangle, we can figure out what the height is using some simple trigonometry. “the vertical” 2.45° 2.45° h h 90° 90° 3 m 3 m

  6. Solution (Problem #1) – Base-ics • We simply apply one of the basic 90-degree-angle relationships (soh, cah, toa) and solve for our unknown height, h. • We chose tangent (toa) because it didn’t require that we solve for an other lengths or angles than we already had. • Turns out that our turbine base is 70.1 meters high, that’s 230 ft! 2.45° h 90° 3 m

  7. Problem #2 – Cutting Edge Given a wind turbine as seen in the picture below and the given dimensions, answer following questions: Dimensions: Blade length: 138 feet Blade Rotational Speed: 200 ft/sec What is the angle between any two of the blades? What is the rotor diameter? How many feet does the tip of the blade cover in one rotation? How much surface area will one blade cover in one minute? Rotor Diameter Blade Angle Blade Length

  8. Solution (Problem #2) – Cutting Edge Given a wind turbine as seen in the picture below and the given dimensions, answer following questions: What is the angle between any two of the blades? We can assume that the angle between each blade is equal. Since the angel between all three blades obviously add up to a full circle (We even drew it for you!), and you know that a full circle means 360°, we simple divide 360° by 3 angles or blades and get 120°. Rotor Diameter Blade Angle Blade Length

  9. Solution (Problem #2) – Cutting Edge Given a wind turbine as seen in the picture below and the given dimensions, answer following questions: Dimensions: Blade length: 138 feet What is the rotor diameter? The cute little diagram on the right tells us that the rotor diameter is the maximum width of the area swept by the blades. This is basically just the diameter of the circle shown. Each blade length is about half the diameter of the circle, so we just multiply the given blade length by 2 and fine that the rotor diameter is about 276 ft. For perspective, that is just short of the length of your football field. The longest manufactured blade is about 200 ft long with a possible rotor diameter of 400 ft! Rotor Diameter Blade Angle Blade Length

  10. Solution (Problem #2) – Cutting Edge Given a wind turbine as seen in the picture below and the given dimensions, answer following questions: How many feet does the tip of the blade cover in one rotation? The distance covered by the tip of any blade is simply the circumference of the circle shown. Using the rotor diameter we just solved for (276 ft) we use the C = 2πR = πD. Recall that π (pi) = 3.1415 C = π(276 ft) = 867.1 ft Rotor Diameter Blade Angle Blade Length

  11. Solution (Problem #2) – Cutting Edge Given a wind turbine as seen in the picture below and the given dimensions, answer following questions: Dimensions: Blade length: 138 feet Blade Rotational Speed: 200 ft/sec How much surface area will one blade cover in one minute? This one is a bit tricky. Here is what we know so far. Rotor Diameter (D) = 276 ft Circumference (C) = 867.1 ft We should start by figuring out how many feet the tip of each blade will travel in one minute. One minute equals 60 seconds and we know that speed = distance / time, so if we multiply the rotational speed by the circumference (total distance each tip travels in one full circle) we will have the total distance the tip travels in one minute. 200 ft/sec x 60 sec = 12000 ft. This is obviously more than the length of one rotation, so we will divide 12000 ft by 867.1 ft to get 13.8. This represents how many full circles the tip makes in one minute. We would then multiply 13.8 by the total area of one circle (A = π(D/2)2). A = π(276/2) 2 = 59828.5 ft2 x 13.8 = 825,633.2 ft2 Rotor Diameter Blade Angle Blade Length

  12. Problem #3 – Things that go boom What will the angle of the main boom be when the cell is placed on top of the tower if the boom truck is 20 m away from the base? Dimensions: Base Height: 70 m Line Length : 10 m Line Length boom cell (Nacelle) base

  13. Problem #3 – Things that go boom What will the angle of the main boom be when the cell is placed on top of the tower if the boom truck is 20 m away from the base? Dimensions: Base Height: 70 m Line Length : 10 m We first see that the boom of the truck and the base of the turbine for a right triangle. Thus all we need are two more pieces of information. We are told that the truck is 20 m away from the base; that’s one piece of info. We are also told that the height of the base is 70 m and the line length is 10 m. To complete the right side of the triangle, we must add these two together to get 80 m. We will just use the same bit if trig from problem #1 and find that the angle of the boom must be 82.9°. This is the smallest angle that the boom can be at to avoid hitting the base with the cell. Please note that the truck is not to scale. If it were, it would be almost 60 ft high! Line Length cell (Nacelle) base boom 10 m

More Related