Imagine you are a pipeline company trying to sell a new project to a shipper of
ID: 1767152 • Letter: I
Question
Imagine you are a pipeline company trying to sell a new project to a shipper of oil for a new refinery.
Pipeline Length: 266 miles (16,853,760 inches)
The refinery is scheduled to take 300,000 barrels per day of Bakken Crude at typical temperatures. Temperature: 60 degrees Fahrenheit. Flow rate = 33687.5 in3/s. Kinematic viscosity: 3.337 centistokes or 0.0051723603 in2/s. Specific Gravity = 0.7
Calculate Reynold's number.
Friction Factor: 0.0275.
Calculate Pressure Drop.
Pipeline flows entirely through class I locations except for 10 miles located in a class IV location. You have the option of bypassing the class IV location by running the pipeline an additional 50 miles.
Pipe Diameter: 12 inches X-42 Schedule 20
Calculate Maximum Pressure
The inlet pressure of pumping stations should not fall below 400 psi.
Set distance between pumping stations optimally.
Calculate horsepower requirements
Explanation / Answer
Pipeline Length: 266 miles (16,853,760 inches)
Flow rate,Q = 33687.5 in3/s
Kinematic viscosity=0.0051723603 in2/s
Friction Factor: 0.0275
Pipe diameter, d=12 in
Therefore Area of pipe, A= pi X d2 / 4=3.142 X 144/4=113.112 sq in
Flow rate = Area x Velocity
Velcoty, V =Flow rate/Area= 33687.5/113.112 = 297.824 in per sec
Reynolds Number, R = V x D/(Kinematic viscosity) = 297.824 x 12/0.0051723603 = 690959.47
To find Pressure drop,
Friction Factor, f: 0.0275
Pipeline Length, L: 16,853,760 in
Acceleration due to gravity, g = 392.4 in/s2
Head loss h1= f L V2/(2 g D) = 0.0275 x 16,853,760 x 297.8242 /( 2 x 392.4 x 12) = 4365245.196 in
Density of oil = Specific Gravity x density of water =0.7 x 1000= 700 kg/m3 = 0.0252891 lb/in3
Therefore pressure drop = density x g xh1= 0.0252891 x 392.4 x 4365245.196 = 43318261.19 psi
After By- Passing the the oil through Class IV section oil needs to travel extra 50 miles(3168000 in),
Therefore pressure drop will be( By performing the same presure drop calculation by changing L= 16,853,760 + 3168000 =20021760 in)
Head loss h2 = 0.0275 x 20021760 x 298.0142 /(2 x 392.4 x 12) = 5192398.766 in
Total pressure drop = density x g x h2= 0.0252891 x 392.4 x 5192398.766= 51526472.35 psi
Therefore Maximum pressure in pipe should be after by-passing is 51526472.35 psi
As we know that the total pressure should be 51526472.35 psi and inlet pressure of pumping stations should not fall below 400 psi,
then total number of pumping station required is 51526472.35/400 = 128816.18 or 128817
The distance between each pumping station is 20021760/128817 =155.428 in
Total Horse power required is P= Density x g x Q x h2 = 0.0252891 x 392.4 x 33687.5 x 5192398.766= 1.735798 x 1012 in2 lb/s3 = 508.42 M watt = 681802.45 Hp (1Hp = 745.7 W)
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