OIL TRANSPORTATION IN PIPELINE Group leader : Meshary Al-Sebhan

Introduction Pipelines are used to move crude oil from the wellhead to gathering and processing facilities and from there to refineries and tanker loading facilities. Crude oil is collected from field gathering systems consisting of pipelines that move oil from the wellhead to storage tanks and treatment facilities where the oil is measured and tested. From the gathering system the crude oil is sent to a pump station where the oil delivered to the pipeline

Pipelines are generally the most economical way to transport large quantities of oil over land they have lowest cost per unit and highest capacity, Although pipelines can be built under the sea. Oil pipelines are made from steel or plastic tubes with inner diameter typically from 10 to 120 cm (about 4 to 48 inches). Most pipelines are buried at a typical depth of about metres (about 3 to 6 feet).feet

Example:

Given Data : 1.7 cp μ 28 API 150 psig Psep ℰ Solution: 1bbl = cuft 1day = sec 1ft = 12in = sq.in = sq.ft = 3.14 sq.in = sq.ft

= lbm/cuft P,psi∆ P,psi v,(ft/sec)q,cuft/secf,(chart)NReq,bbl/dayD,in.L,ftsegment A B C D

THP,psi∆ P,psi Leq.,ft(L/D)fittingLv,ft/secq, cuft/secf(chart)NReq,bpdD,in.L,ftWell

Calculation of pipe dimater

A direct solution for diameter But friction factor depend on Reynolds number (Re) and Re depend on diameter then we can not solve this equation. As we know Large pipe > 8 in. Small pipe < 8 in.

Substituting these equations for “f” into the first equation. Small pipe, Large pipe,

Units Diameter = ft Flow rate = cuft/sec Density=lb/cuft Viscosity=lb/ft.sec Pressure=lb/sq.ft Length= ft

Example: Determine diameter for oil flow rate of 1.46 cuft/s, specific gravity 0.79 A total pressure drop allowance of 72.5 psi over 12.4 miles and viscosity Of 10 cp. Assume large pipe diameter. Solution : d = ft = 10 inches

if we need to rise capacity of pipeline we use : Loop pipeline Parallel pipeline Series pipeline

Effective length calculate by equation Capacity of pipe calculate by equation

Example Calculate capacity of system handling 32 API gravity oil with viscosity of 3 cp and pressure drop equal 1000 kpa (145 psi)

First step Calculate effective length Section one ÷ le = ( ÷ ) + ( ÷ ) le= 5.3 km Section two ÷ le = ( ÷ ) + ( ÷ ) le= 22.8 km Total effective length = = 53.1 km

Second step Find density and viscosity ɣ =0.85 We get ɣ from API ρ= ɣ *1000 = 865 kg /m 3 (54 Lb m /ft 3 ) µ = 3cp÷1000 = kg/m-s (2.02×10 -3 Lb m /ft-s)

Third step Calculate capacity q=(3.180× ÷ × )×( ÷53100) = m 3 /s = bpd

Oil flow in parallel pipeline

Example A pipeline system is composed of two section; AB and BC (see Fig. 1). The former consists of two parallel lines and the later of three parallel lines, of sizes and lengths as indicated in the figure. 50,000 bbl/day are to be transmitted through this system. The oil viscosity is 10 cp. The pressure at C is to be

maintained at 50 psia. The specific gravity of the oil is 0.8 and its temperature is 60 °F. Determine the pressure at A. What would be the pressure at B?

20 mi, 8 in 25 mi, 10 in 30 mi, 12 in 35mi, 8 in 40mi, 10 in a b c Figure.1 series-loop system

solution; For Section AB: Convert the first looped section into an equivalent length, Le, of 10 in. pipe –

For Section BC: Similarly for the second looped section –

Total effective length of system = = miles = ft

Calculate ΔP from the Equation: L = ft q = ((50,000 × 5.615) / (60 × 60 × 24)) = ft 3 /s lb m /ft lb m /ft s gc = lbm× ft / lbf ×s2 PC = 50 psia = 7200 lbf/ft2 d = 10 in. = ft

lb f /ft psi

lb f /ft 2 = psi

The pressure at a = lb/ft2 The pressure at b = lb/ft2

Oil flow in series pipelines

Example Oil of specific gravity 0.7 is being transported from station A to stations C and D. The oil viscosity is 12 cp. A single pipeline of diameter 8 in., length 3 miles, runs from station A to a pipeline junction at B. A 6-in., 2-mile pipeline connects junction B to station C, while a 4-in., 3-mile pipeline connects junction B to D. Given that the pressure at station A (PA) = 600 psia, and that stations C and D are the same pressure (PC, PD) = 30 psia, determine the capacity of the system. Assume that the flowing temperature is 80 °F.

Solution:

43.68 lb m /ft lb m /ft s gc = lbm× ft / lbf ×s2

PB = 300 psia = lbf/ft ft3/s

ft3/s

ft3/s

PB = 350 psia = lbf/ft2

ft3/s

ft3/s

ft3/s

PB = 390 psia = lbf/ft2

ft3/s

ft3/s

ft3/s

The capacity of the system: q = 2.90 ft3/s

Thank you