1 / 6

Calculating Oil Pump Arm Height Above 7 Metres Using Sinusoidal Function

This guide explores the oil pump arm height, modeled by the equation H = 6 + 4Sin(πt). We determine the duration the pump arm is above 7 metres in one complete cycle. Four methods are presented: Unit Circle (Method A), Graphical Method (Method C), and Analytical Solutions (Methods B and D). We calculate the angles and corresponding time intervals where the height exceeds 7 metres, providing insights into the sinusoidal behavior of the pump's motion. This is essential for optimizing oil extraction processes.

kaspar
Télécharger la présentation

Calculating Oil Pump Arm Height Above 7 Metres Using Sinusoidal Function

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. Oil Pump Problem The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres How long is the height of the pump arm above 7 metres from the ground in one complete cycle? Method A: unit circle Method B: unit circle Method C: graph Method D: accurate graph Height ‘H’ Notes

  2. A S 0, 2π π T C Home Oil Pump method A The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) Solve 7 = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Solve 7 = 6 + 4Sin(A) 1 = 4Sin(A) RADIANS! Let A = πt How long is the height of the pump arm above 7 metres from the ground in one complete cycle? 0.25 = Sin(A) Positive Sin so Quadrant 1 & 2 Calculate Sin-1 0.25 A = 0.253 A = π–0.253 Quadrant 2 A = 2.889 t =0.253 ÷ π = 0.08 min And t = A ÷π t =2.889 ÷ π = 0.92 min Time = 0.92 – 0.08 = 0.84 min

  3. A S 0, 2π π T C Home Oil Pump method B The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Solve 7 = 6 + 4Sin(πt) 1 = 4Sin(πt) How long is the height of the pump arm above 7 metres from the ground in one complete cycle? RADIANS! 0.25 = Sin(πt) Positive Sin so Quadrant 1 & 2 Calculate Sin-1 0.25 πt= 0.253 1st time = 0.253 ÷π= 0.08 min πt = π–0.253 Quadrant 2 πt = 2.889 Height ‘H’ 2nd time = 0.92 min (2dp) Time = 0.92 – 0.08 = 0.84 min

  4. Home Oil Pump method C The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) Solve 7 = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Solve 7 = 6 + 4Sin(A) 1 = 4Sin(A) RADIANS! Let A = πt How long is the height of the pump arm above 7 metres from the ground in one complete cycle? 0.25 = Sin(A) (A)= Sin-1 0.25 Sketch the graph A = 0.253 Or A = π–0.253 Or A = 2.889 t =0.253 ÷ π = 0.08 min And t = A ÷π t =2.889 ÷ π = 0.92 min Time = 0.92 – 0.08 = 0.84 min

  5. Home Oil Pump method D The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Solve 7 = 6 + 4Sin(πt) 1 = 4Sin(πt) How long is the height of the pump arm above 7 metres from the ground in one complete cycle? RADIANS! 0.25 = Sin(πt) (πt)= Sin-1 0.25 Sketch the graph πt = 0.253 t = 0.08 min (2dp) t = 1 – 0.08 Length of 1 cycle is 2 π÷ π = 2 Midpoint = 1 Height ‘H’ t = 0.92 min Time = 0.92 – 0.08 = 0.84 min

  6. A S 0, 2π π T C Home Oil Pump Notes The height the end of the pump arm is above the ground is given by the equation H = 6 + 4Sin(πt) Solve 7 = 6 + 4Sin(πt) where t = time in minutes And H is the vertical height in metres Solve 7 = 6 + 4Sin(A) 1 = 4Sin(A) RADIANS! Let A = πt How long is the height of the pump arm above 7 metres from the ground in one complete cycle? 0.25 = Sin(A) Positive Sin so Quadrant 1 & 2 Calculate Sin-1 0.25 A = 0.253 A = π–0.253 Quadrant 2 A = 2.889 t =0.253 ÷ π = 0.08 min And t = A ÷π t =2.889 ÷ π = 0.92 min Time = 0.92 – 0.08 = 0.84 min

More Related