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

Answer 1

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)