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Compressor

1) A compressor takes in air at atmospheric pressure and compresses it, delivering it at a higher pressure to a storage vessel. 2) Reciprocating compressors use pistons driven by a crankshaft to compress air in cylinders. As the piston moves, the air is compressed and discharged from the cylinder. 3) The work required to compress air depends on the pressure and volume changes. Actual work done is greater than theoretical isothermal work due to heat transfer during compression.

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Compressor Engr Bilal Wahla
Air Compressors COMPRESSOR – A device which takes a definite quantity of fluid ( usually gas, and most often air ) and deliver it at a required pressure. Air Compressor – 1) Takes in atmospheric air, 2) Compresses it, and 3) Delivers it to a storage vessel ( i.e. Reservoir ). Compression requires Work to be done on the gas, Compressor must be driven by some sort of Prime Mover ( i.e. Engine )
Classification Air Compressors Reciprocating Rotary Single – acting Double - Acting No. of Sides of Piston in operation No. of Stages for Compression Centrifugal Single – stage Multi - stage
Reciprocating Compressor - Working 2. Principle of Operation  Fig. shows single-acting piston actions in the cylinder of a reciprocating compressor.  The piston is driven by a crank shaft via a connecting rod.  At the top of the cylinder are a suction valve and a discharge valve.  A reciprocating compressor usually has two, three, four, or six cylinders in it.
Reciprocating Compressor - Working
Reciprocating Compressor – Equation for Work Volume Pressure P1 P2 V1V2 3 2 2”2’ 4 1 (Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP  Operations : 4 – 1 : Volume V1 of air aspirated into Compressor, at P1 and T1. 1 – 2 : Air compressed according to PVn = Const. from P1 to P2. → Temp increase from T1 to T2. 2 – 3 : Compressed air at P2 and V2 with temperature T2 is delivered.
Reciprocating Compressor – Equation for Work During Compression, due to the excess temperature above surrounding, the air will exchange the heat to the surrounding.  Compression Index, n is always less than γ, the adiabatic index. As Compressor is a work consuming device, every effort is desired to reduce the work. Work done = Area under P-V curve  1 – 2” : Adiabatic Compression = Max. Work.  1 – 2 : Polytropic Compression  1 – 2’ : Isothermal Compression = Min. Work.
Reciprocating Compressor – Equation for Work Thus, comparison between the Isothermal Work and the Actual Work is important. Isothermal Efficiency, ηiso = Isothermal Work Actual Work Thus, more the Isothermal Efficiency, more the actual compression approaches to the Isothermal Compression. P1 P2 V1V2 3 2 2”2’ 4 1(Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP  Actual Work = Wact = Area 4-1-2-3-4 Wact = Area (4-1) – Area (1-2) – Area (2-3)                           1 1 1 2211 2211 1122 2211 22 1122 11 n VPVP VPVP n VPVP VPVP VP n VPVP VP
                                  11 22 11 2211 2211 1 1 1 1 1 1 VP VP VP n n VPVP n n VPVP n Wiso Reciprocating Compressor – Equation for Work P1 P2 V1V2 3 2 2”2’ 4 1(Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP  Now, n nn P P V V VPVP /1 2 1 1 2 2211                               n iso P P P P VP n n W /1 2 1 1 2 11 1 1
                                             n n iso P P P P VP n n P P P P VP n n W /1 1 2 1 2 11 /1 2 1 1 2 11 1 1 1 1 Reciprocating Compressor – Equation for Work P1 P2 V1V2 3 2 2”2’ 4 1(Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP                           n n iso P P VP n n W 1 1 2 11 1 1 The solution of this equation is always negative. This shows that Work is done ON the Compressor.                          n n iso P P mRT n n W 1 1 2 1 1 1Delivery Temperature, n n P P TT 1 1 2 12        
Reciprocating Compressor – Equation for Work P1 P2 V1V4 6 2 5 1 CVP n  3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc Clearance Volume : Volume that remains inside the cylinder after the piston reaches the end of its inward stroke. Thus, Effective Stroke Volume = V1 – V4 Actual Work = Wact = Area 1-2-3-4 Wact = Area (5-1-2-6) – Area (5-4-3-6)
Reciprocating Compressor – Equation for Work                                         n m n m act P P VP n n P P VP n n W 1 1 2 41 1 1 2 11 1 1 1 1                         n act P P P P VVP n n W /1 2 1 1 2 411 1 1                                        n m n m act P P VP n n P P VP n n W 1 4 3 44 1 1 2 11 1 1 1 1 P1 P2 V1V4 6 2 5 1 CVP n  3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc But, P4 = P1 and P3 = P2
Reciprocating Compressor – Volumetric Efficiency Volumetric Efficiency : Ratio of free air delivered to the displacement of the compressor. Ratio of Effective Swept Volume to Swept Volume. Volumetric Efficiency = Effective Swept Volume Swept Volume V1 – V4 V1 – V3 = Vc Vs = = γ Clearance Volume Swept Volume Clearance Ratio = Presence of Clearance Volume Volumetric Efficiency less than 1. ( 60 – 85 % ) P1 P2 V1V4 6 2 5 1 CVP n  3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc ( 4 – 10 % )
Reciprocating Compressor – Volumetric Efficiency ↑ Pr. Ratio ↑ Effect of Clearance Volume ….Clearance air expansion through greater volume before intake   Cylinder bore and stroke is fixed. Effective Swept Volume (V1 – V4) ↓ with ↑ Pr. Ratio ↓ Volumetric Efficiency                    3 4 31 3 31 3 3 3 31 4 31 3 31 4 31 3 31 4331 31 41 1 1 1 V V VV V VV V V V VV V VV V VV V VV V VV VVVV VV VV vol                     P1 P2 V1V4 6 2 5 1 3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc
Reciprocating Compressor – Volumetric Efficiency                                                  11 11 11 11 /1 4 3 /1 4 3 31 3 4 3 31 3 4 3 31 3 n vol n vol vol vol P P P P VV V V V VV V V V VV V     P1 P2 V1V4 6 2 5 1 3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc
Reciprocating Compressor – Actual P-V Diagram P1 P2 2 1 3 4 Valve Bounce Intake Depression Atmospheric Pressure Receiver Pressure 1-2-3-4-1 : Theoretical P-V Diagram. At 4, inlet valve does not open due to : 1. There must be a pressure difference across the valve to open. 2. Inlet valve inertia. Pr. Drop continues till sufficient level for valve to force its seat. Some valve bounce is set (wavy line). Eventually, the pressure sets down at a level lower than atmospheric pressure. This negative pressure difference is known as Intake Depression. Similar situation appears at 2, i.e. at the start of the delivery. Pressure rise, followed by valve bounce and then pressure settles at a level higher than the delivery pressure level. Air delivery to a tank / receiver, hence, generally known as Receiver Pressure.
Reciprocating Compressor – F.A.D. Free Air Delivery (F.A.D.) : If the volume of the air compressor is reduced to atmospheric temperature and pressure, this volume of air is called FAD (m3/min) Delivered mass of air = intake mass of air     2 322 1 411 T VVP T VVP T VP t tt     If clearance volume is neglected Where KCT mKNP t t 28815 /325.101 0 2   2 22 1 11 T VP T VP T VP t tt 
Reciprocating Compressor – Multistage High Pressure required by Single – Stage :  1. Requires heavy working parts. 2. Has to accommodate high pressure ratios. 3. Increased balancing problems. 4. High Torque fluctuations. 5. Requires heavy Flywheel installations. This demands for MULTI – STAGING…!!
Reciprocating Compressor – Multistage Series arrangement of cylinders, in which the compressed air from earlier cylinder (i.e. discharge) becomes the intake air for the next cylinder (i.e. inlet). Intercooler : Compressed air is cooled between cylinders. L.P. = Low Pressure I.P. = Intermediate Pressure H.P. = High Pressure L.P. Cylinder I.P. Cylinder H.P. Cylinder Intercooler Intercooler Air Intake Air Delivery
Reciprocating Compressor – Multistage Intake Pr. P1 or Ps Delivery Pr. P3 or Pd 3 2 9 5 4 1 CVP n  8 7 6 Intermediate Pr. P2 CVP  Without Intercooling Perfect Intercooling L.P. H.P. Volume Overall Pr. Range : P1 – P3 Single – stage cycle : 8-1-5-6 Without Intercooling : L.P. : 8-1-4-7 H.P. : 7-4-5-6 With Intercooling : L.P. : 8-1-4-7 H.P. : 7-2-3-6 Perfect Intercooling : After initial compression in L.P. cylinder, air is cooled in the Intercooler to its original temperature, before entering H.P. cylinder i.e. T2 = T1 OR Points 1 and 2 are on SAME Isothermal line.
Reciprocating Compressor – Multistage Ideal Conditions for Multi – Stage Compressors : A. Single – Stage Compressor : CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. Single – stage cycle : 8-1-5-6                    1 1 5 11 1 1 n n P P VP n n W Delivery Temperature, n n P P TT 1 1 5 15        
Reciprocating Compressor – Multistage CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. B. Two – Stage Compressor (Without Intercooling) : Without Intercooling : L.P. : 8-1-4-7 H.P. : 7-4-5-6                                         n n n n P P VP n n P P VP n n W 1 4 5 44 1 1 4 11 1 1 1 1 This is SAME as that of Work done in Single – Stage. Delivery Temperature also remains SAME. Without Intercooling 
Reciprocating Compressor – Multistage CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. C. Two – Stage Compressor (With Perfect Intercooling) : With Intercooling : L.P. : 8-1-4-7-8 H.P. : 7-2-3-6-7                                         n n n n P P VP n n P P VP n n W 1 2 3 22 1 1 4 11 1 1 1 1 Delivery Temperature, 12 1 2 3 1 1 2 3 23 , TTas P P T P P TT n n n n              
Reciprocating Compressor – Multistage CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. C. Two – Stage Compressor (With Perfect Intercooling) : With Intercooling : L.P. : 8-1-4-7-8 H.P. : 7-2-3-6-7                           n n n n P P P P VP n n W 1 2 3 1 1 2 11 2 1 Now, T2 = T1 P2V2 = P1V1 Also P4 = P2 Shaded Area 2-4-5-3-2 : Work Saving due to Intercooler…!!
Reciprocating Compressor – Multistage Condition for Min. Work : CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. Intermediate Pr. P2 → P1 : Area 2-4-5-3-2 → 0 Intermediate Pr. P2 → P3 : Area 2-4-5-3-2 → 0  There is an Optimum P2 for which Area 2-4-5-3-2 is maximum, i.e. Work is minimum…!!                           n n n n P P P P VP n n W 1 2 3 1 1 2 11 2 1 0 2 1 2 3 1 1 2 2                          dP P P P P d dP dW n n n n For min. Work,
Reciprocating Compressor – Multistage Condition for Min. Work :         0 111 1 1 2 1 3 1 1 21 1                                            n n n n n n n n P n n PP n n P 0 2 1 2 3 1 1 2 2                          dP P P P P d dP dW n n n n                       n n n n n PP P P 1 3112 2 /1 2    31 2 2 PPP  2 3 1 2 312 P P P P ORPPP  CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P.
Reciprocating Compressor – Multistage P2 obtained with this condition (Pr. Ratio per stage is equal) is the Ideal Intermediate Pr. Which, with Perfect Intercooling, gives Minimum Work, Wmin.                         n n P PP VP n n W 1 1 2/1 31 11 1 1 2                     n n P P VP n n W 1 1 2 11 1 1 2                      n n P P VP n n W 2 1 1 3 11 1 1 2 Equal Work per cylinder…!!
Reciprocating Compressor – Efficiency Isothermal work done / cycle = Area of P – V Diagram = P1V1 loge(P2/P1) Isothermal Power = P1V1 loge(P2/P1) N 60 X 1000 kW Indicated Power : Power obtained from the actual indicator card taken during a test on the compressor. Compressor Efficiency = Isothermal Power Indicated Power Isothermal Efficiency = Isothermal Power Shaft Power NOTE : Shaft Power = Brake Power required to drive the Compressor.
Reciprocating Compressor – Efficiency Adiabatic Efficiency : Ratio of Power required to drive the Compressor; compared with the area of the hypothetical Indicator Diagram; assuming Adiabatic Compression. CompressorthedrivetorequiredPowerBrake P P VP adiabatic                          1 1 2 11 1 1 Mechanical Efficiency : Ratio of mechanical output to mechanical input. Mechanical Efficiency, ηmech = Indicated Power Shaft Power
Reciprocating Compressor – Efficiency How to Increase Isothermal Efficiency ? A. Spray Injection : Assimilation of water into the compressor cylinder towards the compression stroke. Object is to cool the air for next operation. Demerits : 1. Requires special gear for injection. 2. Injected water interferes with the cylinder lubrication. 3. Damage to cylinder walls and valves. 4. Water must be separated before delivery of air. B. Water Jacketing : Circulating water around the cylinder to help for cooling the air during compression.
Reciprocating Compressor – Efficiency How to Increase Isothermal Efficiency ? C. Inter – Cooling : For high speed and high Pr. Ratio compressors. Compressed air from earlier stage is cooled to its original temperature before passing it to the next stage. D. External Fins : For small capacity compressors, fins on external surfaces are useful. E. Cylinder Proportions : Short stroke and large bore provides much greater surface for cooling. Cylinder head surface is far more effective than barrel surface.
Reciprocating Compressor – Efficiency Clearance Volume : Consists of two spaces. 1. Space between cylinder end & the piston to allow for wear. 2. Space for reception of valves. High – class H.P. compressors : Clearance Vol. = 3 % of Swept Vol. : Lead (Pb) fuse wire used to measure the gap between cylinder end and piston. Low – grade L.P. compressors : Clearance Vol. = 6 % of Swept Vol. : Flattened ball of putty used to measure the gap between cylinder end and piston. Effect of Clearance Vol. : Vol. taken in per stroke < Swept Vol. ↑ Size of compressor ↑ Power to drive compressor. 
P1 P2 V1V4 6 2 5 1 3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V4=Vs Total Volume, V1 Clearance Volume, V3=Vc Reciprocating Compressor – Work Done                                        n n n n P P VP n n P P VP n n W 1 4 3 44 1 1 2 11 1 1 1 1 Assumption : Compression and Expansion follow same Law. Work / cycle = Area 1-2-3-4-1 P3 = P2 and P4 = P1                                         n n a n n P P VP n n P P VVP n n W 1 1 2 1 1 1 2 411 1 1 1)( 1
P1 P2 V1V4 6 2 5 1 3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V4=Vs Total Volume, V1 Clearance Volume, V3=Vc Reciprocating Compressor – Work Done                     n n P P TRm n n W 1 1 2 11 1 1 m1 is the actual mass of air delivered. Work done / kg of air delivered :                     n n P P TR n n W 1 1 2 1 1 1

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Compressor

  • 1.
  • 2.
    Air Compressors COMPRESSOR –A device which takes a definite quantity of fluid ( usually gas, and most often air ) and deliver it at a required pressure. Air Compressor – 1) Takes in atmospheric air, 2) Compresses it, and 3) Delivers it to a storage vessel ( i.e. Reservoir ). Compression requires Work to be done on the gas, Compressor must be driven by some sort of Prime Mover ( i.e. Engine )
  • 3.
    Classification Air Compressors Reciprocating Rotary Single– acting Double - Acting No. of Sides of Piston in operation No. of Stages for Compression Centrifugal Single – stage Multi - stage
  • 4.
    Reciprocating Compressor -Working 2. Principle of Operation  Fig. shows single-acting piston actions in the cylinder of a reciprocating compressor.  The piston is driven by a crank shaft via a connecting rod.  At the top of the cylinder are a suction valve and a discharge valve.  A reciprocating compressor usually has two, three, four, or six cylinders in it.
  • 5.
  • 6.
    Reciprocating Compressor –Equation for Work Volume Pressure P1 P2 V1V2 3 2 2”2’ 4 1 (Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP  Operations : 4 – 1 : Volume V1 of air aspirated into Compressor, at P1 and T1. 1 – 2 : Air compressed according to PVn = Const. from P1 to P2. → Temp increase from T1 to T2. 2 – 3 : Compressed air at P2 and V2 with temperature T2 is delivered.
  • 7.
    Reciprocating Compressor –Equation for Work During Compression, due to the excess temperature above surrounding, the air will exchange the heat to the surrounding.  Compression Index, n is always less than γ, the adiabatic index. As Compressor is a work consuming device, every effort is desired to reduce the work. Work done = Area under P-V curve  1 – 2” : Adiabatic Compression = Max. Work.  1 – 2 : Polytropic Compression  1 – 2’ : Isothermal Compression = Min. Work.
  • 8.
    Reciprocating Compressor –Equation for Work Thus, comparison between the Isothermal Work and the Actual Work is important. Isothermal Efficiency, ηiso = Isothermal Work Actual Work Thus, more the Isothermal Efficiency, more the actual compression approaches to the Isothermal Compression. P1 P2 V1V2 3 2 2”2’ 4 1(Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP  Actual Work = Wact = Area 4-1-2-3-4 Wact = Area (4-1) – Area (1-2) – Area (2-3)                           1 1 1 2211 2211 1122 2211 22 1122 11 n VPVP VPVP n VPVP VPVP VP n VPVP VP
  • 9.
                                      11 22 11 2211 2211 1 1 1 1 1 1 VP VP VP n n VPVP n n VPVP n Wiso ReciprocatingCompressor – Equation for Work P1 P2 V1V2 3 2 2”2’ 4 1(Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP  Now, n nn P P V V VPVP /1 2 1 1 2 2211                               n iso P P P P VP n n W /1 2 1 1 2 11 1 1
  • 10.
                                                 n n iso P P P P VP n n P P P P VP n n W /1 1 2 1 2 11 /1 2 1 1 2 11 1 1 1 1 Reciprocating Compressor– Equation for Work P1 P2 V1V2 3 2 2”2’ 4 1(Polytropic) (Adiabatic) (Isothermal) CVP n  CVP  CVP                           n n iso P P VP n n W 1 1 2 11 1 1 The solution of this equation is always negative. This shows that Work is done ON the Compressor.                          n n iso P P mRT n n W 1 1 2 1 1 1Delivery Temperature, n n P P TT 1 1 2 12        
  • 11.
    Reciprocating Compressor –Equation for Work P1 P2 V1V4 6 2 5 1 CVP n  3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc Clearance Volume : Volume that remains inside the cylinder after the piston reaches the end of its inward stroke. Thus, Effective Stroke Volume = V1 – V4 Actual Work = Wact = Area 1-2-3-4 Wact = Area (5-1-2-6) – Area (5-4-3-6)
  • 12.
    Reciprocating Compressor –Equation for Work                                         n m n m act P P VP n n P P VP n n W 1 1 2 41 1 1 2 11 1 1 1 1                         n act P P P P VVP n n W /1 2 1 1 2 411 1 1                                        n m n m act P P VP n n P P VP n n W 1 4 3 44 1 1 2 11 1 1 1 1 P1 P2 V1V4 6 2 5 1 CVP n  3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc But, P4 = P1 and P3 = P2
  • 13.
    Reciprocating Compressor –Volumetric Efficiency Volumetric Efficiency : Ratio of free air delivered to the displacement of the compressor. Ratio of Effective Swept Volume to Swept Volume. Volumetric Efficiency = Effective Swept Volume Swept Volume V1 – V4 V1 – V3 = Vc Vs = = γ Clearance Volume Swept Volume Clearance Ratio = Presence of Clearance Volume Volumetric Efficiency less than 1. ( 60 – 85 % ) P1 P2 V1V4 6 2 5 1 CVP n  3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc ( 4 – 10 % )
  • 14.
    Reciprocating Compressor –Volumetric Efficiency ↑ Pr. Ratio ↑ Effect of Clearance Volume ….Clearance air expansion through greater volume before intake   Cylinder bore and stroke is fixed. Effective Swept Volume (V1 – V4) ↓ with ↑ Pr. Ratio ↓ Volumetric Efficiency                    3 4 31 3 31 3 3 3 31 4 31 3 31 4 31 3 31 4331 31 41 1 1 1 V V VV V VV V V V VV V VV V VV V VV V VV VVVV VV VV vol                     P1 P2 V1V4 6 2 5 1 3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc
  • 15.
    Reciprocating Compressor –Volumetric Efficiency                                                  11 11 11 11 /1 4 3 /1 4 3 31 3 4 3 31 3 4 3 31 3 n vol n vol vol vol P P P P VV V V V VV V V V VV V     P1 P2 V1V4 6 2 5 1 3 4 V3 Effective Swept Volume, V1-V4 Swept Volume, V1-V3=Vs Total Volume, V1 Clearance Volume, V3=Vc
  • 16.
    Reciprocating Compressor –Actual P-V Diagram P1 P2 2 1 3 4 Valve Bounce Intake Depression Atmospheric Pressure Receiver Pressure 1-2-3-4-1 : Theoretical P-V Diagram. At 4, inlet valve does not open due to : 1. There must be a pressure difference across the valve to open. 2. Inlet valve inertia. Pr. Drop continues till sufficient level for valve to force its seat. Some valve bounce is set (wavy line). Eventually, the pressure sets down at a level lower than atmospheric pressure. This negative pressure difference is known as Intake Depression. Similar situation appears at 2, i.e. at the start of the delivery. Pressure rise, followed by valve bounce and then pressure settles at a level higher than the delivery pressure level. Air delivery to a tank / receiver, hence, generally known as Receiver Pressure.
  • 17.
    Reciprocating Compressor –F.A.D. Free Air Delivery (F.A.D.) : If the volume of the air compressor is reduced to atmospheric temperature and pressure, this volume of air is called FAD (m3/min) Delivered mass of air = intake mass of air     2 322 1 411 T VVP T VVP T VP t tt     If clearance volume is neglected Where KCT mKNP t t 28815 /325.101 0 2   2 22 1 11 T VP T VP T VP t tt 
  • 18.
    Reciprocating Compressor –Multistage High Pressure required by Single – Stage :  1. Requires heavy working parts. 2. Has to accommodate high pressure ratios. 3. Increased balancing problems. 4. High Torque fluctuations. 5. Requires heavy Flywheel installations. This demands for MULTI – STAGING…!!
  • 19.
    Reciprocating Compressor –Multistage Series arrangement of cylinders, in which the compressed air from earlier cylinder (i.e. discharge) becomes the intake air for the next cylinder (i.e. inlet). Intercooler : Compressed air is cooled between cylinders. L.P. = Low Pressure I.P. = Intermediate Pressure H.P. = High Pressure L.P. Cylinder I.P. Cylinder H.P. Cylinder Intercooler Intercooler Air Intake Air Delivery
  • 20.
    Reciprocating Compressor –Multistage Intake Pr. P1 or Ps Delivery Pr. P3 or Pd 3 2 9 5 4 1 CVP n  8 7 6 Intermediate Pr. P2 CVP  Without Intercooling Perfect Intercooling L.P. H.P. Volume Overall Pr. Range : P1 – P3 Single – stage cycle : 8-1-5-6 Without Intercooling : L.P. : 8-1-4-7 H.P. : 7-4-5-6 With Intercooling : L.P. : 8-1-4-7 H.P. : 7-2-3-6 Perfect Intercooling : After initial compression in L.P. cylinder, air is cooled in the Intercooler to its original temperature, before entering H.P. cylinder i.e. T2 = T1 OR Points 1 and 2 are on SAME Isothermal line.
  • 21.
    Reciprocating Compressor –Multistage Ideal Conditions for Multi – Stage Compressors : A. Single – Stage Compressor : CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. Single – stage cycle : 8-1-5-6                    1 1 5 11 1 1 n n P P VP n n W Delivery Temperature, n n P P TT 1 1 5 15        
  • 22.
    Reciprocating Compressor –Multistage CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. B. Two – Stage Compressor (Without Intercooling) : Without Intercooling : L.P. : 8-1-4-7 H.P. : 7-4-5-6                                         n n n n P P VP n n P P VP n n W 1 4 5 44 1 1 4 11 1 1 1 1 This is SAME as that of Work done in Single – Stage. Delivery Temperature also remains SAME. Without Intercooling 
  • 23.
    Reciprocating Compressor –Multistage CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. C. Two – Stage Compressor (With Perfect Intercooling) : With Intercooling : L.P. : 8-1-4-7-8 H.P. : 7-2-3-6-7                                         n n n n P P VP n n P P VP n n W 1 2 3 22 1 1 4 11 1 1 1 1 Delivery Temperature, 12 1 2 3 1 1 2 3 23 , TTas P P T P P TT n n n n              
  • 24.
    Reciprocating Compressor –Multistage CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. C. Two – Stage Compressor (With Perfect Intercooling) : With Intercooling : L.P. : 8-1-4-7-8 H.P. : 7-2-3-6-7                           n n n n P P P P VP n n W 1 2 3 1 1 2 11 2 1 Now, T2 = T1 P2V2 = P1V1 Also P4 = P2 Shaded Area 2-4-5-3-2 : Work Saving due to Intercooler…!!
  • 25.
    Reciprocating Compressor –Multistage Condition for Min. Work : CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P. Intermediate Pr. P2 → P1 : Area 2-4-5-3-2 → 0 Intermediate Pr. P2 → P3 : Area 2-4-5-3-2 → 0  There is an Optimum P2 for which Area 2-4-5-3-2 is maximum, i.e. Work is minimum…!!                           n n n n P P P P VP n n W 1 2 3 1 1 2 11 2 1 0 2 1 2 3 1 1 2 2                          dP P P P P d dP dW n n n n For min. Work,
  • 26.
    Reciprocating Compressor –Multistage Condition for Min. Work :         0 111 1 1 2 1 3 1 1 21 1                                            n n n n n n n n P n n PP n n P 0 2 1 2 3 1 1 2 2                          dP P P P P d dP dW n n n n                       n n n n n PP P P 1 3112 2 /1 2    31 2 2 PPP  2 3 1 2 312 P P P P ORPPP  CVP  3 2 9 5 4 1 CVP n  8 7 6 L.P. H.P.
  • 27.
    Reciprocating Compressor –Multistage P2 obtained with this condition (Pr. Ratio per stage is equal) is the Ideal Intermediate Pr. Which, with Perfect Intercooling, gives Minimum Work, Wmin.                         n n P PP VP n n W 1 1 2/1 31 11 1 1 2                     n n P P VP n n W 1 1 2 11 1 1 2                      n n P P VP n n W 2 1 1 3 11 1 1 2 Equal Work per cylinder…!!
  • 28.
    Reciprocating Compressor –Efficiency Isothermal work done / cycle = Area of P – V Diagram = P1V1 loge(P2/P1) Isothermal Power = P1V1 loge(P2/P1) N 60 X 1000 kW Indicated Power : Power obtained from the actual indicator card taken during a test on the compressor. Compressor Efficiency = Isothermal Power Indicated Power Isothermal Efficiency = Isothermal Power Shaft Power NOTE : Shaft Power = Brake Power required to drive the Compressor.
  • 29.
    Reciprocating Compressor –Efficiency Adiabatic Efficiency : Ratio of Power required to drive the Compressor; compared with the area of the hypothetical Indicator Diagram; assuming Adiabatic Compression. CompressorthedrivetorequiredPowerBrake P P VP adiabatic                          1 1 2 11 1 1 Mechanical Efficiency : Ratio of mechanical output to mechanical input. Mechanical Efficiency, ηmech = Indicated Power Shaft Power
  • 30.
    Reciprocating Compressor –Efficiency How to Increase Isothermal Efficiency ? A. Spray Injection : Assimilation of water into the compressor cylinder towards the compression stroke. Object is to cool the air for next operation. Demerits : 1. Requires special gear for injection. 2. Injected water interferes with the cylinder lubrication. 3. Damage to cylinder walls and valves. 4. Water must be separated before delivery of air. B. Water Jacketing : Circulating water around the cylinder to help for cooling the air during compression.
  • 31.
    Reciprocating Compressor –Efficiency How to Increase Isothermal Efficiency ? C. Inter – Cooling : For high speed and high Pr. Ratio compressors. Compressed air from earlier stage is cooled to its original temperature before passing it to the next stage. D. External Fins : For small capacity compressors, fins on external surfaces are useful. E. Cylinder Proportions : Short stroke and large bore provides much greater surface for cooling. Cylinder head surface is far more effective than barrel surface.
  • 32.
    Reciprocating Compressor –Efficiency Clearance Volume : Consists of two spaces. 1. Space between cylinder end & the piston to allow for wear. 2. Space for reception of valves. High – class H.P. compressors : Clearance Vol. = 3 % of Swept Vol. : Lead (Pb) fuse wire used to measure the gap between cylinder end and piston. Low – grade L.P. compressors : Clearance Vol. = 6 % of Swept Vol. : Flattened ball of putty used to measure the gap between cylinder end and piston. Effect of Clearance Vol. : Vol. taken in per stroke < Swept Vol. ↑ Size of compressor ↑ Power to drive compressor. 
  • 33.
    P1 P2 V1V4 6 2 5 1 3 4 V3 EffectiveSwept Volume, V1-V4 Swept Volume, V1-V4=Vs Total Volume, V1 Clearance Volume, V3=Vc Reciprocating Compressor – Work Done                                        n n n n P P VP n n P P VP n n W 1 4 3 44 1 1 2 11 1 1 1 1 Assumption : Compression and Expansion follow same Law. Work / cycle = Area 1-2-3-4-1 P3 = P2 and P4 = P1                                         n n a n n P P VP n n P P VVP n n W 1 1 2 1 1 1 2 411 1 1 1)( 1
  • 34.
    P1 P2 V1V4 6 2 5 1 3 4 V3 EffectiveSwept Volume, V1-V4 Swept Volume, V1-V4=Vs Total Volume, V1 Clearance Volume, V3=Vc Reciprocating Compressor – Work Done                     n n P P TRm n n W 1 1 2 11 1 1 m1 is the actual mass of air delivered. Work done / kg of air delivered :                     n n P P TR n n W 1 1 2 1 1 1