Reciprocating Compressor - Universitas Diponegoro

33
Reciprocating Compressor Nazaruddin Sinaga Efficiency and Energy Conservation Laboratory Diponegoro University

Transcript of Reciprocating Compressor - Universitas Diponegoro

Page 1: Reciprocating Compressor - Universitas Diponegoro

Reciprocating

Compressor

Nazaruddin Sinaga

Efficiency and Energy Conservation LaboratoryDiponegoro University

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RECIPROCATING COMPRESSORS

RECIPROCATING COMPRESSORS

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CompressorsCompressors

HIGH LOWV. HIGH

NO NOYES

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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Crosshead

Stuffing

Box

Packing

Discharge Valves

Suction Valves

PistonDistance Piece

Con-rodCrank

Piston Rod

Wiper

Rings

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POWER END POWER END

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GASEND

GASEND

FIXED CLEARANCE

VOLUME

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VARIABLECLEARANCE

VOLUME

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RECIPROCATING COMPRESSORS:FLOW RATE REGULATION

RECIPROCATING COMPRESSORS:FLOW RATE REGULATION

• VARIABLE CLEARANCE

• VARIABLE SPEED

• BYPASS

• SUCTION VALVE

UNLOADERS

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CF CR

CF = C

CR = C

23

13

• VARIABLE CLEARANCE• VARIABLE CLEARANCE

C = Total Clearance Volume = CF + CR

Before operating Before operating

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After operating After operating

CF CR

CF = C

CR = C

12

12

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Point g suction valves open

C = Clearance volume

C

v

b

a

c

g

Pd

Ps c1M

EFFECT OF CLEARANCE VOLUME

EFFECT OF CLEARANCE VOLUMEAtAt

Point a suction valves closed

Point b discharge valves open

Point c discharge valves closed

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Compression ratio

Pressure ratio )ABSOLUTE(P d

P s=

Volumetric efficiency =V a g

V cylinder a M

a g=

= V cylinder

V b M

C

vb a

c

g

Pd

Ps c1M

b

b M

a M =

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C

v

Pd

Ps c1M

b

a

c

g

Volumetric efficiency =V a g

V cylinder a M

a g=

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C

v

Pd

PsM

b

a

c

g

Volumetric efficiency =V a g

V cylinder a M

a g=

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v

Pd

Ps M

C

g a

bc

Volumetric efficiency =V a g

V cylinder a M

a g=

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If C Increased suction valves will open at ( g (

C

va

c

g

Pd

Ps

b

M

EV DECREASES

Q DECREASES

Volumetric efficiency =V a g

V cylinder a M

a g=

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If C Decreased suction valves will open at ( d (

C

va

c bPd

Ps M g

EV INCREASES

Q INCREASES

Volumetric efficiency =V a g

V cylinder a M

a g=

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•SUCTION VALVE

UNLOADERS

•SUCTION VALVE

UNLOADERS

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STUFFING BOX

RIDER RINGSRIDER RINGS

PRESSURE RINGS

PRESSURE RINGS

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WIPER RINGS

WIPER RINGS

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T

R

TANGENTIAL RINGST

R RADIAL RINGS

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RECIPROCATING COMPRESSORS

INTERMEDIATEPRESSURES

RECIPROCATING COMPRESSORS

INTERMEDIATEPRESSURES

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Example 1

4 Stages Reciprocating Air comp has outlet pressure P4 = 80 barg

what are the intermediate pressures

P0

0 barg

1st

stage

P1

2nd stage

P2

3rd stage

P3 P4

80 barg

4th stage

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P0

0 barg

P1 P2 P3 P4

80 barg

1st

stage2nd

stage3rd

stage

4th stage

R =P1

P0

P4

P3=

P3

P2=

P2

P1=

Assume

R is The equal Pressure Ratio between stages.

R = 4 P4

P0RT =

P4

P0BUT

RTR =

4

AND RT = P4

P0

R = RT4

Generally

R = RTN

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P4 = R * P3 = 3 * 27 = 81 bara = 80 barg

P0

0 barg

P1 P2 P3 P4

80 barg

1st

stage2nd

stage3rd

stage

4th stage

RT = P4 / P0 = (80 +1) / (0 +1) = 81/ 1 = 81

N 4R = 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

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Example 2

3 Stages Reciprocating Gas compressor has Outlet pressure P3 = 127 barg Inlet pressure P0 = 1 barg

what are the intermediate pressures

P0

1 barg

1st

stage

P1

2nd stage

P2

3rd stage

P3

127 barg

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P0

1 barg

1st

stage

P1

2nd stage

P2

3rd stage

P3

127 barg

Assume

P3

P2=P1

P2=R

P1

P0

=

RTP3

P0

=3

RP3

P0

=BUT

3R

P3

P0

=

R = RT3 3

R = RT

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P3 = R * P2 = 4 * 32 = 128 bara = 127 bar g

3R = RT = 64 = 4

=)127+ 1(

)1+1(=

1282 = 64

P1 = R * P0 = 4 * 2 = 8 bara = 7 bar g

P2 = R * P1 = 4 * 8 = 32 bara = 31 bar g

P3

P0

3R=RT =

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Example 2

3 Stages Gas compressor has Outlet pressure = 255 bar gInlet pressure = 3 bar g

What are the intermediate pressures

P0

3 barg

1st

stage

P1

2nd stage

P2

3rd stage

P3

255 barg

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P0

3 barg

1st

stageP1

2nd stage

P2

3rd stage

P3

255 barg

P3 = R * P2 = 4 * 64 = 256 bara = 255 bar g

3R = RT = 64 = 4

=)255+ 1(

)3+1(=

2564 = 64

P1 = R * P0 = 4 * 4 = 16 bara = 15 bar g

P2 = R * P1 = 4 * 16 = 64 bara = 63 bar g

P3

P0

3R=RT =

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REFERENCES

1. L. Nelik. Centrifugal and Rotary Pumps: Fundamentals with Applications,

CRC Press, Boca Raton, 1999.

2. Sulzer Pumps. Centrifugal Pump Handbook, 3rd edition, Sulzer Pumps

Ltd., Winterthur, Switzerland, Elsevier, 2010.

3. S. L. Dixon and C. A. Hall. Fluid Mechanics and Thermodynamics of

Turbomachinery, 6th Edition, Elsevier Inc., Amsterdam, 2010.

4. M. P. Singh and G. M. Lucas. Blade Design and Analysis for Steam

Turbines, McGraw-Hill Companies, Inc., 2011.

5. M. P. Boyce. Gas Turbine Engineering Handbook, 3rd edition, GPM,

Houston, Texas, 2005.

6. C. Soares. Microturbines: Applications for Distributed Energy Systems,

Elsevier, Amsterdam, 2007.

7. A. S. Leyzerovich. Wet-Steam Turbines for Nuclear Power Plants,

PennWell Corporation, Tulsa, Oklahoma, 2005.

8. S. Yedidiah. Centrifugal Pump User's Guidebook: Problems and Solutions,

Chapman & Hall, New York, 1996.

9. M. L. Adams Jr. Power Plant Centrifugal Pumps Problem Analysis and

Troubleshooting, CRC Press, Boca Raton, 2017.