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- 1. Air or Gas Compressors: A steady-state, steady flow machine that is used to compressed air or gas to final pressure exceeding 241.25 Kpa gage. Types of Compressor: 1. Centrifugal Compressors: For low pressure and high capacity applications. 2. Rotary Compressors: For medium pressure and low capacity application. 3. Reciprocating Compressors: For high pressure and low capacity application. Uses of compressed air: 1. Operation of small engines 2. Pneumatic tools 3. Air hoists 4. Industrial cleaning by air blast 5. Tire inflation 6. Paint Spraying
- 2. 7. Air lifting of liquids 8. Manufacture of plastics and other industrial products 9. To supply air in mine tunnels 10. Other specialized industrial applications Analysis of Centrifugal and Rotary Type 1 Suction 2 Discharge W (Work) Assumption KE = 0 PE = 0 For a compressor, work is done on the system W = h - Q <ul><li>For Isentropic Compression: PV k = C </li></ul><ul><li>Q = 0 </li></ul><ul><li>W = h </li></ul>P V P 2 P 1 PV k = C
- 3. Where: m – mass flow rate in kg/sec C p – constant pressure specific heat in KJ/kg- C or KJ/kg- K
- 4. 2. Polytropic compression: PV n = C P V P 2 P 1 PV n = C
- 5. 3. Isothermal Compression: PV = C Analysis of Reciprocating Type Compressor (Piston-in-cylinder type): piston Valves cylinder Piston rod P V P 2 P 1 PV = C
- 6. Pressure-Volume Diagram (PV) HE – head end CE – Crank end L – length of stroke P 1 – suction pressure P 2 – discharge pressure V 1’ – volume flow rate at intake V D – displacement volume CV D – clearance volume CV D = V 3 V D L HE CE P V 1 2 3 4 P 2 P 1 V 1’ V D CV D
- 7. 1. Isentropic Compression: PV k = C Where: V1’ – volume flow rate at intake, m 3 /sec m – mass flow rate corresponding V 1’ P 1 – suction pressure, Kpa P 2 – discharge pressure, Kpa T 1 – suction temperature, K T 2 – discharge temperature, K W – work, KW 2. Polytropic Compression: PV n = C
- 8. 3. Isothermal Compression: PV = C Percent Clearance : Ratio of the clearance volume to the displacement volume. Note: For compressor design values of C ranges from 3 to 10 percent.
- 9. Pressure Ratio: Ratio of the discharge pressure to suction pressure. Volumetric Efficiency: Ratio of the volume flow rate at intake to the displacement volume. 1. For Isentropic Compression and Expansion process: PV k = C
- 10. 2. For Polytropic Compression and Expansion process: PV n = C 2. For Isothermal Compression and Expansion process: PV = C Actual Volumetric Efficiency : Ratio of the actual volume of air drawn in by the compressor to the displacement volume.
- 11. For an air compressor handling ambient air where pressure drop and heating of air occurs due to fluid friction and irreversibilities of fluid flow, less amount of air is being drawn by the cylinder. The actual volumetric efficiency is: Where: P O – ambient air pressure in Kpa T O – ambient air temperature in K Displacement Volume: Volume of air occupying the highest stroke L of the piston within the cylinder. The length of stroke L is the dis- tance from the HE (head end) to the CE (crank end).
- 12. <ul><li>1. For Single Acting : For single acting compressor, the piston works on </li></ul><ul><ul><li>one side (at head end only) of the cylinder. </li></ul></ul><ul><li>2. For Double Acting : For double acting compressor, the piston works on </li></ul><ul><ul><li>both sides (at head end & crank end) of the cylinder. </li></ul></ul>a. Without considering the volume of the piston rod. b. Considering the volume of the piston rod.
- 13. Where: D – diameter of piston in meters d – diameter of piston rod in meters N – no. of RPM n’ – no. of cylinders Piston Speed : It is the linear speed of the piston. Compressor Performance Factor: 1. Compression Efficiency: Ratio of Ideal Work to Indicated Work.
- 14. 2. Mechanical Efficiency: Ratio of Indicated Work to Brake or Shaft Work. 3. Compressor Efficiency: Ratio of Ideal Work to Brake or Shaft Work.
- 15. MULTISTAGE COMPRESSION: Multi staging is simply the compression of air or gas in two or more cylinders in place of a single cylinder compressor. It is used in reciprocating compressors when pressure of 300 KPa and above are desired, in order to: 1) Save power 2) Limit the gas discharge temperature 3) Limit the pressure differential per cylinder 4) Prevent vaporization of lubricating oil and to prevent its ignition if the tem- perature becomes too high. It is a common practice for multi-staging to cool the air or gas between stages of compression in an intercooler, and it is this cooling that affects considerable saving in power.
- 16. For an ideal multistage compressor, with perfect inter-cooling and minimum work, the cylinder were properly designed so that: a) the work at each stage are equal b) the air in the intercooler is cooled back to the initial temperature c) no pressure drop occurs in the intercooler 2 Stage Compressor without pressure drop in the intercooler : 1 2 3 4 Suction Discharge Qx Intercooler 1 st stage 2 nd stage
- 17. Work of 1 st stage cylinder ( W 1 ) : Assuming Polytropic compression on both stages. Work of 2 nd stage cylinder ( W 2 ) : Assuming Polytropic compression on both stages.
- 18. For perfect inter-cooling and minimum work: W 1 = W 2 T 1 = T 3 W = W 1 + W 2 W = 2W 1 P 2 = P 3 = P x therefore P 1 V 1’ = P 3 V 3’ Where: Px – optimum intercooler pressure or interstage pressure P V P 4 P 1 P x 1 4 3 2 5 6 7 8 PV n = C W 1 W 2 S T 4 3 2 1 P 4 P x P 1 Q x
- 19. Then the work W for an ideal 2-stage compressor is: <ul><li>Heat losses calculation: </li></ul><ul><li>Heat loss during compression at 1 st stage cylinder </li></ul><ul><li>Q 1 = mC n (T 2 – T 1 ) </li></ul><ul><li>2. Heat loss during compression at 2 nd stage cylinder </li></ul><ul><li>Q 2 = mCn (T 4 – T 3 ) </li></ul><ul><li>3. Heat loss in the intercooler </li></ul><ul><li>Q x = mC p (T 2 – T 3 ) </li></ul>
- 20. With pressure drop in the intercooler: T 1 T 3 and P 2 P 3 W = W 1 + W 2 P 1 V 1’ P 3 V 3’ 2 Stage Compressor with pressure drop in the intercooler : 1 2 3 4 Suction Discharge Qx Intercooler 1 st stage 2 nd stage
- 21. P V P 4 P 1 P 3 1 4 3 2 5 6 7 8 PV n = C W 1 W 2 S T 4 3 2 1 P 4 P 1 Q x P 2 P 2 P 3 3 Stage Compressor without pressure drop in the intercooler : 1 2 3 4 Suction Discharge Qx LP Intercooler 1 st stage 2 nd stage 3 rd stage 5 6 Qy HP Intercooler
- 22. S T 4 3 2 1 P 6 P x P 1 Q x P y 5 6 Q y For perfect inter-cooling and minimum work: T 1 = T 3 = T 5 P x = P 2 = P 3 W 1 = W 2 = W 3 P y = P 4 = P 5 W = 3W1 P 1 V 1’ = P 3 V 3’ = P 5 V 5’ mRT 1 = mRT 3 = mRT 5 Therefore: r P1 = r P2 = r P3 P V P 6 P 1 P x 1 4 3 2 5 6 7 12 PV n = C W 1 W 2 P y 9 10 11 8 W 3
- 23. Work for each stage: 1 st Stage: 2 nd Stage: 3 rd Stage: Intercooler Pressures:
- 24. Heat Losses during compression : Q 1 = mC n (T 2 – T 1 ) Q 2 = mC n (T 4 – T 3 ) Q 3 = mC n (T 6 – T 5 ) Heat loss in the LP and HP intercoolers: LP Intercooler Qx = mC p (T 2 – T 3 ) HP Intercooler Qy = mC p (T 4 – T 5 ) <ul><li>Note: </li></ul><ul><li>For isentropic compression and expansion process, no heat loss during </li></ul><ul><li>compression. </li></ul><ul><li>For isothermal compression and expansion process, the loss during </li></ul><ul><li>compression is equivalent to the compression work, and no intercooler </li></ul><ul><li>is required. </li></ul>Total Work: W = 3W 1
- 25. For multistage compression with minimum work and perfect inter-cooling and no pressure drop that occurs in the inter-coolers between stages, the following conditions apply: 1. the work at each stage are equal 2. the pressure ratio between stages are equal 3. the air temperature in the inter-coolers are cooled to the original temperature T 1 4. the total work W is equal to Where: s – is the number of stages. Note: For multistage compressor with pressure drop in the intercoolers the equation of W above cannot be applied. The total work is equal to the sum of the work for each stage that is computed separately.

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I have a question. where did you get all these information? because I have to list it as my references.

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