Chase Attermann Salt Lake Community College Math 1210 Pipeline Project Spring 2016

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1 Chase Attermann Salt Lake Community College Math 1210 Pipeline Project Spring 2016 Cost Factors as taken from Outline: Cost of drilling through the existing mountain would be a one-time cost of $3,500,000 on top of the normal costs ($500,000 per mile) of 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 $420,000 and will delay the project by 6 months costing the company another $180,000 per month. For any pipeline run across private ground, your company incurs an additional $350,000 per mile cost for right-of-way fees. Costs of Routes: I. Costs of Routes via BLM Ground: A. The cost of the pipeline on BLM ground can be found by multiplying the pipe cost per mile (500,000 dollars/mile) by the length of the pipe in miles. The cost formula below calculates the cost of the pipe in BLM ground :

2 1. Cost of Route Heading West, South, then East on BLM Land: a) In the sketch below, the total number of miles of pipeline is found with the sum of the distances running each direction: D = ( ) = 65 b) Using the above cost formula for pipeline on BLM Land, the cost of the route is calculated as: $ = $32,500, The cost of the route running East through the mountain, then South on BLM Land to the refinery can be calculated using the same cost formula for BLM Land plus the additional costs of installing pipe in the mountain, which is ($3,500,000 + $420,000 + $1,080,000 = $5,000,000)

3 a) From the sketch above, it's found that D = 55. Using the original cost formula plus total additional costs: Cost = ($500000/mile) X (55 miles) + $ = $32,500,000 II. Costs of Routes via Private Land: A. The cost of pipe through private land consists of the right-of-way cost of running pipe through each mile of private land ($350000/mile) plus the original cost of installing each mile of pipe ($500000/mile). Because a route can run partially on private land and partially on BLM ground, the total cost of the route is the cost of running pipe on private land plus the cost of running pipe on BLM ground. From this, the new formula for the total cost of running pipe is as follows: 1. Shortest Distance through Private Ground to Refinery: a) The pipeline running straight from the well to the refinery does not run on BLM land, so D = 0, and only the length of the pipe in private land is needed. Geometry can be used to find the length of the pipe.

4 b) The figure above confirms that the well is directly 15 miles North of the BLM border running to the refinery, and the length of private ground east of the well is the length of private ground west of the well (5 miles) subtracted from the total length of the private land (45 miles), which is 40 miles. A pipeline running directly from the well to the refinery connects these two perpendicular distances, which can be analyzed as the right triangle shown below. The pythagorean theorem can be applied to find the length of the pipeline: The length of the pipe L = miles c) Plugging L into the total cost function gives: Cost = ($500000/mile) X (0 miles) + ($850000/mile) X (42.72miles) = $36,312, The cost of running a pipeline South across private land then East to the refinery can be found using the total cost function. a) Using the same dimensions of the triangle found previously, the pipeline s route will follow the two perpendicular sides running South then East to the Refinery. b) The sketch to the right shows the pipeline running 15 miles South through private land (L) and then 40 miles through BLM ground (D).

5 c) Using the total cost function: Cost = $850000(15) + $500000(40) = $32,750,000 III. Optimal Pipeline A. Given the area that lies between the two previous route options through private land, an optimal route for pipeline can be determined to minimize cost. 1. The cost of the optimal route can be found with the same total cost formula that was previously constructed: Cost = ($500000/mile X D) + ($ X L ) Although the distances of the pipes in private ground (L and D) are unknown, a function can be formed to show where the cost is increasing and decreasing using calculus. Because there are two variables in the formula, substitution must be used to express the cost in terms of one variable. It can be seen from the sketch above that any combination of x and D will have the sum of 40, substitution may be used to express D in terms of x: (D = 40 - X).

6 2 2 2 Using the pythagorean theorem once again: ( L = x + 15 ) which gives: 2. Which gives the cost function in terms of x: 3. The derivative of C(x) set equal to zero, d/dx C(x) = 0, will give the value of x where the cost stops decreasing: a) After calculating, C (x) = 0 when x = miles. The minimum value of x can be observed with the graph of the cost function below.

7 4. Next, the final variables are calculated: D = 40 - (10.911) = miles L = = miles 5. Plugging the lengths back into the total cost formula: Cost = $500000(29.089) + $850000(18.549) = $30,310,795.31, which is the minimum cost that the pipeline route could be. Below is the final sketch of the optimal pipeline: Conclusion: The optimal route for pipeline determined by the derivative of the cost function set equal to zero will greatly reduce the cost of implementing pipeline. Starting at the Well, the pipeline should run south-east about miles until it reaches BLM Ground, where it will then run straight East about miles to the Refinery. The total cost will be about $30,310,795.31, saving over $2,000,000 compared to the previous routes that ranged from the next cheapest, $32,500,000 to the most expensive, $36,312,000.

8 Reflection Throughout this semester of learning calculus, I often start to see how engineered systems in my day to day life have been accomplished, particularly after engaging in the subject either while driving home from class or after time spent studying the material. I see a lot of possible application in how roadways are constructed with the least amount of concrete and asphalt possible, or the cheapest way to construct an overpass highway with the maximum possible slope for cargo trucks to be able to accelerate over. Optimization problems like this have shown me what an amazing tool calculus is, and how it gives us the ability to solve a problem with formulas and algebra and calculate a more accurate result than ever could have been attained by solving it with a time-extensive trial-and-error process. This is where I really see the greatest gift of calculus, in that without it, optimization problems would still be a matter of trial and error and iterating until an approximation is achieved. As an aspiring materials science engineer, it is apparent that calculus will be my best companion in the field. Deciding on the material for any kind of shape, volume, surface area an the densities within them will all be greatly dependent on the costs of constructing them and I feel maximums and minimums will be used almost every day in my profession. Strengths and overall practicality of the materials will rely heavily on calculus also. Some materials are stronger as other elements are added, and often times there is a maximum amount of element added to give strength until it starts to become too hard, brittle, and weak. Similarly in adding elements to materials, the flux of atoms through a surface area of material is equal to the diffusion coefficient multiplied by the concentration gradient (rate of change of concentration with time), otherwise known from calculus as the derivative of concentration as a function of time, or. dc dx

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