ALASKA PIPELINE PROJECT

This is the official repo of the Alaska pipeline project made for the course of fluid mechanics.

ASSUMPTIONS

1. The pipeline is straight and has a shape as per the terrain.
2. Bernoulli's Theorem holds for the flow.
3. Temperature is maintained at 30 $^\circ$C.
4. The pipeline is always full. (No two-phase flow)

Implementation

STRAIGHT-LEVEL PIPELINE WITH INFINITE THICKNESS

Case 1

Diameter=12.7mm $\left(\frac{1}{2}\right)$ inch

VFL=1.3 million barrels per day output from the pipeline=$1.310^6158.987$ liters/day=$206.6831*10^6$ liters/day=$2392.166$ liters/second=2.393 m^3^/s

Velocity Of Flow=18890.584 m/s

:::danger This velocity is quite large and cannot be used in real-life applications. So this diameter of the pipe is unrealistic and irrelevant. ::: Let's now assume the velocity to be 2 m/s and find the diameter of the pipe.

Case 2

Velocity of Flow = 2 m/s (Assumed)

VFL = 2.393 m^3^/s

Diameter of Pipe (calculated) = 48.6 inch $\approx$ 48 inch $\approx$ 1.219 m

$Re=\frac{\rhoVD}{\mu}$ = $\frac{86021.219}{5.64*10^{-3}}$ = 371751.773

Hence, the flow is turbulent. Let's find the fD value. $f_D=0.046R_e^{-0.2}$ $f_D=0.046371751.773^{-0.2}$ $f_D=3.5736*10^{-3}$

Lets calculate friction head as function of length $h_f=4*\frac{f}{g}\frac{\Delta x}{D}\frac{V^2}{2}$ $h_f=4*\frac{3.573610^{-3}}{9.81}\frac{\Delta x}{1.219}\frac{(2)^2}{2}$ $h_f=2.366610^{-3} \Delta x$ meter

Let's simulate this value for our pipeline length of 800 miles i.e. 1287 KM Let's assume the line has only one pump at the start of the line. The graph is the result obtained from using Bernoulli's equation for pressure drop along the pipeline $P=P_{initial}-h_f*\rho*g$ The Profile found is as follows:

Now let's consider the addition of additional pumps to keep pressure above 1 atm in pipes. We will add a pump when the pressure drops below 1.2 atm. By using the above correction and the equation $P=P_{initial}-h_f*\rhog+W_{pump} \rho*g$ We get the following graph

Hence this completes our model of the level pipe system.

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STRAIGHT PIPELINE ELEVATED WITH INFINITE THICKNESS

Now let's consider our path has two mountains in our path. Our new terrain data is shown in the graph below. Now, Let's solve Bernoulli's equation for this process $$\Delta P+\rho* g*\Delta* Z=- h_f*\rho* g$$$$P=P_{initial}- h_f*\rho* g-\rho* g*\Delta Z$$Let's use this equation and terrain data and plot the graph between pressure inside pipe over span of pipeline. The Graph obtained is shown below. Now let's consider the addition of additional pumps to keep pressure above 1 atm in pipes. We will add a pump when the pressure drops below 1.2 atm. By usingthe above correction and the equation $$\Delta P+\rho* g*\Delta* Z=W_{pump}* \rhog- h_f\rho* g$$$$P=P_{initial}- h_f*\rho* g-\rho* g*\Delta Z+W_{pump}* \rho*g$$The graph obtained is as follows

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STRAIGHT PIPELINE ELEVATED ON ACTUAL TERRAIN WITH INFINITE THICKNESS

Now let's take the actual elevation data The Elevation profile is shown below

Now using the formula $$P=P_{initial}- h_f*\rho* g-\rho* g*\Delta Z+W_{pump}* \rho*g$$The graph obtained is as follows With this our model of a pipeline with infinite thickness is complete. Now in the succeeding section, we'll work on the pipeline with finite thickness

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STRAIGHT LEVEL PIPELINE WITH FINITE THICKNESS

Now let's consider the pipe as it is in real life. For this model, we will consider pipes of two pipe thicknesses 0.462" and 0.562" and three grades of steel Grade 60, Grade 65, and Grade 70. The design tensile stress data is as follows

Grade	Tensile Stress(MPa)
60	420
65	551.581
70	585.5

The thickness of the Pipe is given by the formula $t=\frac{PD}{2\sigma_{tensile}}$ Rearranging the equation, $P_{max}=\frac{t_{max}2\sigma_{tensile}}{D}$ Using the above equation we get the following data

Steel Grade	Maximum Pressure (atm) for 0.462" thick pipe	Maximum Pressure (atm) for 0.562" thick pipe
60	79.80	97.08
65	104.1	127.49
70	111.255	135.33

But, it is standard practice to take 30% excess pressure for safety purposes. By considering 30% less pressure from the above values we get the following data.

Steel Grade	Maximum Pressure (atm) for 0.462" thick pipe	Maximum Pressure (atm) for 0.562" thick pipe
60	61.39	74.68
65	80.623	98.07
70	85.581	104.103

So let's develop a model considering Straight pipe with 0.462" thickness. For pressure reduction, we'll add an orifice which will make a pressure drop of 200 psi i.e. 13.61 atm. Considering the following assumptions the graphs obtained are shown below

Grade 60 Pipe 0.462" thick pipe

Grade 65 Pipe 0.462" thick pipe

Grade 70 Pipe 0.462" thick pipe

Grade 60 Pipe 0.562" thick pipe

Grade 65 Pipe 0.562" thick pipe

Grade 70 Pipe 0.562" thick pipe