1. OIL TRANSPORTATION IN PIPELINE Group leader : Meshary Al-Sebhan

2. 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

3. 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 1 - 2 metres (about 3 to 6 feet).feet

4. Example:

5. Given Data : 1.7 cp μ 28 API 150 psig Psep 0.001 ℰ Solution: 1bbl = 5.6146 cuft 1day = 86400 sec 1ft = 12in A @D=1in. = 0.785 sq.in = 0.005451 sq.ft A @D=2in. = 3.14 sq.in = 0.021806 sq.ft

6. = 55.35799 lbm/cuft P,psi∆ P,psi v,(ft/sec)q,cuft/secf,(chart)NReq,bbl/dayD,in.L,ftsegment 1411.958146.0644.7680.0260.02819277.6074001800A 1265.895946.80513.1130.0710.02453013.42011001800B 319.090101.8454.1720.0910.03433735.813140021200C 217.24567.2455.9600.1300.03348194.01820002400D

7. THP,psi∆ P,psi Leq.,ft(L/D)fittingLv,ft/secq, cuft/secf(chart)NReq,bpdD,in.L,ftWell 325.33108.09216.519816.58.344420.0454890.02533735.870012001 342.9123.823216.519816.53.576180.0194950.0314458.230012002 1391.3125.38316.519816.57.152360.038990.02728916.460013003

9. Calculation of pipe dimater

10. 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.

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

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

13. 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 = 0.834 ft = 10 inches

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

16. Effective length calculate by equation Capacity of pipe calculate by equation

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

18. First step Calculate effective length Section one 14 2.662 ÷ le 0.553 = (10 2.661 ÷6 0.553 ) + (12 2.661 ÷ 6 0.553 ) le= 5.3 km Section two 14 2.662 ÷ le 0.553 = (10 2.661 ÷10 0.553 ) + (8 2.661 ÷ 10 0.553 ) le= 22.8 km Total effective length = 5.3+22.8+25 = 53.1 km

19. 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 = 0.003 kg/m-s (2.02×10 -3 Lb m /ft-s)

20. Third step Calculate capacity q=(3.180×0.356 2.661 ÷865 0.446 ×0.003 0.107 )×(1000000÷53100) 0.553 = 0.197 m 3 /s = 107000 bpd

21. Oil flow in parallel pipeline

22. 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

23. 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?

24. 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

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

27. For Section BC: Similarly for the second looped section –

30. Total effective length of system = 10.3936 + 7.6338 + 4.1535 = 22.1809 miles = 117115.1520 ft

31. Calculate ΔP from the Equation: L = 76808.6880 ft q = ((50,000 × 5.615) / (60 × 60 × 24)) = 3.2492 ft 3 /s 49.92 lb m /ft 3 0.00672 lb m /ft s gc = 32.17 lbm× ft / lbf ×s2 PC = 50 psia = 7200 lbf/ft2 d = 10 in. = 0.8333 ft

32. 87721.6535lb f /ft 2 609.1781psi

33. 168243.0051 lb f /ft 2 = 1168.3542 psi

34. The pressure at a = 168243.0051 lb/ft2 The pressure at b = 87721.6535 lb/ft2

35. Oil flow in series pipelines

36. 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.

37. Solution:

39. 43.68 lb m /ft 3 0.008064 lb m /ft s gc = 32.17 lbm× ft / lbf ×s2

40. PB = 300 psia = 43200 lbf/ft2 3.5276 ft3/s

41. 1.9425 ft3/s

42. 0.5321 ft3/s

43. PB = 350 psia = 50400 lbf/ft2

44. 3.1927 ft3/s

45. 2.1316 ft3/s

46. 0.5839 ft3/s

47. PB = 390 psia = 56160 lbf/ft2

48. 2.9023 ft3/s

49. 2.2735 ft3/s

50. 0.6228 ft3/s

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

52. Thank you