1 / 85

Reciprocating Gas Compressors

Reciprocating Gas Compressors. 10psi. 100psi. Gas Wellhead Gas Pipeline. P ounds per S quare I nch. 10psi. 105psi. Discharge of Compressor. Suction of Compressor. Applications for Gas compressors. Gas Injection. Gas Lift. Gas Gathering.

quinn-tran
Download Presentation

Reciprocating Gas Compressors

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Reciprocating Gas Compressors

  2. 10psi 100psi Gas Wellhead Gas Pipeline Pounds per Square Inch

  3. 10psi 105psi Discharge of Compressor Suction of Compressor

  4. Applications for Gas compressors

  5. Gas Injection

  6. Gas Lift

  7. Gas Gathering

  8. Coal Seam Gas Evacuation

  9. Gas Storage Summer open closed closed open

  10. Gas Storage Winter closed open open closed

  11. Station 1 Station 2 Gas Transmission

  12. WATER STEAM BOILER GENERATOR TURBINEENGINE Gas Turbine Fuel Booster Co-generation

  13. First Stage WHY ? Third Stage Second Stage

  14. The size of the cylinder is determined by the pressure and quantity of the gas to be compressed. Larger diameter cylinders are for lower pressures. Gas, at a lower pressure,occupies more space.

  15. The size of the cylinder is determined by the pressure and quantity of the gas to be compressed. Smaller diameter cylinders are for higher pressures. Gas, at a higher pressure, occupies less space.

  16. CYLINDER BORE 200 psig 10 “ 350 psig

  17. INLET VALVES DISCHARGE VALVES

  18. 205 PSI 200 PSI Differential pressure opens the inlet valves.

  19. 200 PSI 200 PSI Springs inside the valve close the valve.

  20. 200 PSI 205 PSI Differential pressure holds the inlet valve closed.

  21. 355 PSI 350 PSI Differential pressure opens the discharge valves.

  22. 350 PSI 350 PSI Springs inside the valve close the valve.

  23. 350psi 355 PSI Differential pressure holds the discharge valves closed.

  24. DAMPING SPRINGS DAMPING PLATE CLOSING SPRINGS SEAT PLATE Valve internals

  25. Suction or Discharge Valve ????????? Valve internals

  26. What causes the pressure changes in the cylinder bore?

  27. ~x r x length of bore Swept Volume 2

  28. PISTON PISTON ROD

  29. “Head End” also called the “Outboard end” “Crank End” also called the “Inboard end”

  30. Changing the volume of an area has a direct affect on What?? Pressure Temperature

  31. As the Piston Travels toward the head end the pressures will do what?

  32. Head End will Rise Crank End will fall

  33. What Happens to this Gas ? This Space is referred to as “Inherent Clearance”

  34. Re-expansion Must Occur

  35. Rod Load • Rod load is a measurement of total forces that act upon the “running Gear” of a compressor during the normal compression cycle • The term “Rod Load “ does not mean only the “Rod” is affected or will ultimately fail. • Two Types of Rod Load • Rod Load Compression • Rod Load Tension

  36. As the Piston Travels toward the head Which type of rod load is most effected ? Rod load Compression

  37. As the Piston Travels toward the crank Which type of rod load is most effected ? Rod load Tension

  38. How Do We Calculate “Rod load ? Remember that Pressure is measured in: Pounds Per Square Inch e.g... 200 P.S.I. PSIG AND PSIA

  39. P.ounds P.er S.quare I.nch A 10 Inch Circle ….. AREA Has 78.54 S.I.

  40. P.ounds P.er S.quare inch A 10 Inch Circle ….. AREA = D2x .7854 Has 78.54 S.I. 10 x 10 x .7854

  41. Pounds Per Square Inch Absolute D2x .7854 A 10 Inch Circle ….. AREA = Has 78.54 S.I. Or: 10 x 10 x .7854 Answer Should Be :16,862.54 pounds of force Always “round” up two digits SO…… Absolute Pressure x 78.54 S.I. 200 or 214.7 x 78.54 S.I.

  42. 7.00” = 7.⅛ “ = 11⅓= 14⅞= 5 11/32= 22.43 6 9/16 = Let ‘s Practice Calculate the Areas of the cylinders below 40.08 38.54 Remember (D2x.7854)

  43. 200 psig Rod Load Compression 10 “ RLC = (AHE x Pd) – ( ACE x Ps) 350 psig

  44. 200 psig Ps Rod Load Compression ACE AHE 10 “ Pd RLC = (AHE x Pd) – ( ACE x Ps) 350 psig

  45. 200 psig Assume that The Rod is 2.5 ‘’ Dia. Rod Load Compression ACE AHE 10 “ Will the AHE and ACE be the same ? RLC = (AHE x Pd) – ( ACE x Ps) 350 psig

  46. 200 psig Assume that The Rod is 2.5 ‘’ Dia. Rod Load Compression ACE AHE 10 “ Area of the Rod (Ar) is 4.90 S.I The AHE is 78.54 S.I ACE = (AHE – Ar) ACE = (78.54 – 4.09) 350 psig ACE = (74.45 S.I.)

  47. What We Know So Far • The ACE = 74.45 (10.0’”Dia. – The 2.5 “ Rod Dia) • The AHE = 78.54 (10.00” Dia) • The Ar = 4.09 (2.5” Dia.) • The Suction Pressure = 214.7 PSIA • 200 + 14.7 • The Discharge Pressure = 364.7 • 350 + 14.7

More Related