Math 1210 Pipeline Project

The U.S. Interior Secretary recently approved drilling of natural gas wells near Vernal, Utah. Your company has begun drilling and established a high-producing well on BLM ground. They now need to build a pipeline to get the natural gas to their refinery. While running the line directly to the refinery will be the least amount of pipe and shortest distance, it would require running the line across private ground and paying a right-of-way fee. There is a mountain directly east of the well that must be drilled through in order to run the pipeline due east. Your company can build the pipeline around the private ground by going 1 mile directly west and then 5 miles south and finally 21 miles east to the refinery (see figure below). Cost for materials, labor and fees to run the pipeline across BLM ground is $300,000 per mile. For any pipeline run across private ground, your company incurs an additional $200,000 per mile cost for right-of- way fees. Cost of drilling through the existing mountain would be $500,000 on top of the normal costs of the material, labor and fees for the pipeline itself. Also the BLM will require an environmental impact study before allowing you to drill through the mountain. Cost for the study is estimated to be $100,000 and will delay the project by 3 months costing the company another $50,000 per month. Your company has asked you to do the following:

a) Determine the cost of running the pipeline strictly on BLM ground with two different scenarios:

1. heading east through the mountain and then south to the refinery

2. running west, south and then east to the refinery.

b) Determine the cost of running the pipeline the shortest distance (straight line joining well to refinery across the private ground).

c) Determine the cost function for this pipeline for the configuration involving running from the well across the private ground at some angle and intersecting the BLM ground to the south and then running east to the refinery. Use this function to determine the optimal place to run the pipeline to minimize cost. Clearly show all work including sketching the placement of the optimal pipeline. Make it very clear how you use your knowledge of calculus to determine the optimal placement of the pipeline. Draw a graph of this cost function and label the point of minimum cost.

Dear CEO,

The Pipeline Project near Vernal is making progress as we plan out our most cost effective route. When broken down, we have four options:

a) to stay on BLM land. However we count this as two options as there are two ways around the private land:

1. running west, south and then east to the refinery.

With a total of 27 miles at a cost of $300,000 per mile on BLM land, this route will cost $8,100,000 in labor material and fees.

2. heading east through the mountain and then south to the refinery. This option includes $500,000 to drill through the mountain plus approximately $100,000 for an environmental impact study before drilling, setting operations back three months. Operational costs are $50,000 monthly. Although this option is shorter (25 miles from well to refinery), the total cost is $8,250,000.

b) a direct line from well to refinery over private land.

This option cost the most at $10,307,764, figured by an additional $200,000 per mile for right-of-way fees, with just more than 20.6 miles of pipeline. See EQ1

c) minimizing the distance through private land, while still taking a more direct route.

The minimum cost is found by building 16.25 miles from the refinery along BLM land, then building 6.25 miles through private land to the well. For this option, the total cost is $8,000,000. (see EQ2)

Equation list

EQ1: 5 Miles North to South, 20 miles East to West.

5! + 20! = 20.616

EQ2

Cost along x is 300,000x

Remainder of the line d=20-x

Using Pythagorean’s identity to find L

𝐿 = 25 + 20 − 𝑥 !

cost along 𝐿 𝑥 = 500,000 25 + 20 − 𝑥 !

𝐶𝑜𝑠𝑡 𝑥 = 500,000 25 + 20 − 𝑥 ! + 300,000𝑥 with possibilities of 0 ≤ 𝑥 ≤ 20

The derivative of Cost(x) was found as follows:

!!"500,000 ∗ 25 + 20 − 𝑥 !

!! + !

!"300,000𝑥

𝐶𝑜𝑠𝑡! 𝑥 = 250,000 ∗ 25 + 20 − 𝑥 ! !!! ∗ (−2 20 − 𝑥 ) + 300,000

x  d  

θ 5  

L    

𝛽

 

Equation list (cont)

Simplify:

𝐶𝑜𝑠𝑡! 𝑥 =250,000 ∗ −2 20− 𝑥

25+ 20− 𝑥 !+ 300,000

set equal to zero, and solve for x

0 =−500,000(20− 𝑥)

25+ 20− 𝑥 !+ 300,000

300,000 =500,000(20− 𝑥)

25+ 20− 𝑥 !

square both sides

900,000 =2,500,000 20− 𝑥 !

25+ 20− 𝑥 !

Let u=20-x

900,000(25+ 𝑢 !) = 2,500,000 𝑢 !

22,500,000+ 900,000 𝑢 ! = 2,500,000 𝑢 !

22,500,000 = 1,600,000 𝑢 !

𝑢! = 14.0625

𝑢 = 3.75

replace u with 20-x

20− 𝑥 = 3.75

𝑥 = 16.25 miles

insert into cost function

𝐶𝑜𝑠𝑡 16.25 = 500,000 25 + 20 − 16.25 ! + 300,000(16.25)

𝐶𝑜𝑠𝑡 16.25 = $8,000,000

Equation list (cont)

Solve for L:

𝐿 = 25 + 20 − 16.25 !

𝐿 = 6.25  𝑚𝑖𝑙𝑒𝑠

solve for θ

tanθ =3.755

𝜃 = 36.87°  𝑓𝑟𝑜𝑚  𝑑𝑢𝑒  𝑠𝑜𝑢𝑡ℎ

solve for beta

𝛽 = |cos!!3.756.25− 180|

𝛽 = 126.87

 

 

16.25  3.75  

36.9

5  

6.25    

126.9