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Reciprocating compressor
Reciprocating compressors compress gases via pistons moving inside cylinders. They have high pressure pulsations and can surge if not regulated properly. Flow rate in reciprocating compressors is regulated through variable clearance volumes, variable speed, bypass valves, or suction valve unloaders. Calculating intermediate pressures in multi-stage reciprocating compressors involves determining the common pressure ratio between stages using the overall pressure ratio and number of stages.
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Overview of reciprocating compressors is introduced, highlighting its importance in applications.
Comparison of compressor types: reciprocating, centrifugal, axial and their performance metrics such as pressure and efficiency.
Detailed components of reciprocating compressors including crosshead, valves, piston, and crank mechanism.
Introduction to the power end component of reciprocating compressors, crucial for energy transmission.
Details on gas end mechanisms and the importance of fixed clearance volume in compressor operation.
Explains variable clearance volume mechanisms that improve efficiency and flow rate regulation.
Methods for flow rate regulation in reciprocating compressors including variable clearance, speed, and unloaders.
Mathematical representation of total clearance volume equation prior to operating conditions.
Mathematical representation of total clearance volume equation after operating conditions.
Illustration of the effect of clearance volume on suction and discharge valve operations during compressor cycles.
Equations presenting volumetric efficiency calculations based on pressure and clearance volume.
Relationship between clearance volume changes and suction valve operations impacting volumetric efficiency.
Explanation of suction valve unloaders and their role in compressor efficiency.
Insights on stuffing box mechanics, rider rings, and pressure sealing components.
Discussion on tangential rings and radial rings as essential sealing elements.
Concept of intermediate pressures and their significance in multi-stage reciprocating compressors.
Detailed example calculation of intermediate pressures for a 4-stage reciprocating air compressor.
Calculation of intermediate pressures for a 3-stage reciprocating gas compressor, illustrating practical application.
Another example detailing pressure calculations for a 3-stage gas compressor for practical understanding.
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Reciprocating compressor
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- 2. CompressorsCompressors HIGH LOWV. HIGH NONOYES YESNO YES LOW HIGH V. HIGH YES NO NO Pressure PPressure P Flow Rate QFlow Rate Q S .R .VS .R .V EfficiencyEfficiency Maint. costMaint. cost PulsationPulsation SurgingSurging Reciprocating Centrifugal Axial V. HIGH LOW LOW LOW HIGH V. HIGH
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- 9. RECIPROCATING COMPRESSORS: FLOW RATEREGULATION RECIPROCATING COMPRESSORS: FLOW RATE REGULATION • VARIABLE CLEARANCE • VARIABLE SPEED • BYPASS • SUCTION VALVE UNLOADERS
- 10. CF CR CF =C CR = C 2 3 1 3 • VARIABLE CLEARANCE• VARIABLE CLEARANCE C = Total Clearance Volume = CF + CR Before operatingBefore operating
- 11. After operatingAfter operating CFCR CF = C CR = C 1 2 1 2
- 12. Point g suctionvalves open C = Clearance volume C v b a c g Pd Ps c1M EFFECT OF CLEARANCE VOLUME EFFECT OF CLEARANCE VOLUME AtAt Point a suction valves closed Point b discharge valves open Point c discharge valves closed
- 13. Compression ratio Pressure ratio)ABSOLUTE( Pd P s = Volumetric efficiency = Va g V cylinder a M a g = = V cylinder Vb M C v b a c g Pd Ps c1M b b M a M =
- 14. C v Pd Ps c1 M b a c g Volumetric efficiency= Va g V cylinder a M a g =
- 15. C v Pd Ps M b a c g Volumetric efficiency = Vag V cylinder a M a g =
- 16. v Pd Ps M C g a bc Volumetric efficiency= Va g V cylinder a M a g =
- 17. If C Increasedsuction valves will open at ( g( C v a c g Pd Ps b M EV DECREASES Q DECREASES Volumetric efficiency = Va g V cylinder a M a g =
- 18. If C Decreasedsuction valves will open at ( d( C v a c bPd Ps M g EV INCREASES Q INCREASES Volumetric efficiency = Va g V cylinder a M a g =
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- 24. Example 1 4Stages ReciprocatingAir comp has outlet pressure P4 = 80 barg what are the intermediate pressures P0 0barg 1st stage P1 2nd stage P2 3rd stage P3 P4 80 barg 4th stage
- 25. P0 0barg P1 P2 P3P4 80 barg 1st stage 2nd stage 3rd stage 4th stage R= P1 P0 P4 P3 = P3 P2 = P2 P1 = Assume R is The equal Pressure Ratio between stages. R= 4 P4 P0 RT= P4 P0 BUT RTR = 4 AND RT= P4 P0 R = RT 4 Generally R = RT N
- 26. P4 = R* P3 = 3 * 27 = 81 bara = 80 barg P0 0barg P1 P2 P3 P4 80 barg 1st stage 2nd stage 3rd stage 4th stage RT = P4 / P0 = (80 +1) / (0 +1) = 81/ 1 = 81 N 4 R = RT = 81 = 3 P1 = R * P0 = 3 * 1 = 3 bara = 2 barg P2 = R * P1 = 3 * 3 = 9 bara = 8 barg P3 = R * P2 = 3 * 9 = 27 bara = 26 barg
- 27. Example 2 3Stages ReciprocatingGas compressor has Outlet pressure P3 = 127 barg Inlet pressure P0 = 1 barg what are the intermediate pressures P0 1barg 1st stage P1 2nd stage P2 3rd stage P3 127barg
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- 29. P3 = R* P2 = 4 * 32 = 128 bara = 127 bar g 3 R = RT = 64 = 4 = )127+1( )1+1( = 128 2 = 64 P1 = R * P0 = 4 * 2 = 8 bara = 7 bar g P2 = R * P1 = 4 * 8 = 32 bara = 31 bar g P3 P0 3 R=RT =
- 30. Example 2 3Stages Gascompressor has Outlet pressure = 255 bar g Inlet pressure = 3 bar g What are the intermediate pressures P0 3barg 1st stage P1 2nd stage P2 3rd stage P3 255barg
- 31. P0 3barg 1st stage P1 2nd stage P2 3rd stage P3 255barg P3 =R * P2 = 4 * 64 = 256 bara = 255 bar g 3 R = RT = 64 = 4 = )255+1( )3+1( = 256 4 = 64 P1 = R * P0 = 4 * 4 = 16 bara = 15 bar g P2 = R * P1 = 4 * 16 = 64 bara = 63 bar g P3 P0 3 R=RT =
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