compressor thermodynamics methods and · pdf file8 ge title or job number 11/17/2013...
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Compressor Thermodynamics
Methods and Alternatives
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Background
Calculating Volume, SI
The basic equation is ππ = ππ ππ
Where
Z=Compression factor, (no units)
T=Absolute Temperature
R=0.083145, m3 bar/(mol K) (Volume Units)
P=Absolute pressure, bar
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Process
General Energy Balance
Mixture @ Pi,Ti Output Fluid @ Po,To
F
Work
Heat
1st Law DH=Q+W/e
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Process
General Energy Balance
Pi= 35 bar Ti= 30 oC
F
Work=??
Efficiency = 80%
Po= 110 bar To= ??
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Calculating Discharge Conditions: Ideal Gas Single Phase
Step 1 β’ Note Inlet Pi, Ti
β’ Calculate Volume (Vi), and Cp, Cv and k=Cp/Cv
Step 2
β’ Estimate To
β’ ππ = ππππ
ππ
πβ1
π
Step 3
β’ Estimate Ideal Work
β’ π β1
π
π
πβ1
πππ ππ
ππ
π0
ππ
πβ1
πβ 1 π€βπππ π = 8.3145
ππ½
ππππ/πΎ
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Example: First Calculating Mol Wt
Feed Component MW Zi*mwi
Component mol%
Water 0.00 18.02 0.00
H2S 0.00 34.08 0.00
CO2 0.95 28.01 26.61
N2 8.85 44.01 389.56
Methane 78.96 16.04 1266.70
Ethane 7.75 30.07 233.11
Propane 2.51 44.10 110.78
i-Butane 0.30 58.12 17.35
n-Butane 0.49 58.12 28.50
i-Pentane 0.08 72.15 5.84
n-Pentane 0.07 72.15 5.07
Benzene 0.00 78.11 0.00
Toluene 0.00 92.14 0.00
e-Benzene 0.00 106.17 0.00
o-Xylene 0.00 106.17 0.00
m-Xylene 0.00 Density 106.17 0.00
p-Xylene 0.00 mw kg/m3 106.17 0.00
Hexane 0.02 84.40 670 84.40 1.70
Heptane 0.01 92.60 734 92.60 1.08
Octane 0.00 105.20 760 105.20 0.47
Nonane 0.00 117.70 781 117.70 0.14
Decane 0.00 171.50 800 171.50 0.05
100.00 Stream MW, S 20.87
ππ = πππππ
ππ
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Calculating Ideal Gas Version
Step 1 efficiency 0.80 mw kg/kmol 19.64
Pi bar 35.00 Ti C 60.00
Zi - 0.95 Hi J/mol 10516.77
Vi m /kmol 0.76 k=Cpi/Cvi - 1.33
Step 2 Po bar 110.00 To C 169.35
Ideal Work 178.4 info
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Compressor Modeling
L
The Polytropic Analysis of Centrifugal Compressors
John M. Schultz, Trans. Of the ASME, ASME J. of Engineering for Power, Jan 1962 pp 69-82
The real-gas equations of polytropic analysis are derived in terms of compressibility functions X and Y which supplement the familiar compressibility factor, Z. A polytropic head factor, f, is introduced to adjust test results for deviations from perfect-gas behavior. Functions X and Y are generalized and plotted for gases in corresponding states.
The thermodynamic design and test evaluation of centrifugal compressors is frequently based upon a polytropic analysis employing perfect-gas relations. In many instances real-gas relations would be more accurate, but these are virtually unknown. The purpose of this paper is to derive the real-gas equations of polytropic analysis and to show their application to centrifugal compressor testing and design.
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The problem in 1962
β’ Needed a convenient model for the fluid
β’ Little General Access to Process Simulation Tools β’ Limitations in Equations of State Methods
β’ BWR 1940, complex solution β’ RK 1949, limited application to mixtures β’ NGA 1958, Equilibrium Ratio Data for Computers
Needed an alternative which could be handled using a slide rule
β’ Ie: Simple log-log relationships β’ Utilize Corresponding States Models
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Classic Algorithm Flange to Flange
Hi,Si,ri,CPi,Cvi
V/F
Hos,Sos,Tos
Feed at P,T
Flash at Pi,Ti
Isentropic Flash at Po
Ho=Hi+(Hos-His)/h
Isenthalpic Flash at Po
Ho,So,To,W V/F
Outputs
Inputs β’ Efficiency β’ Composition
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Schultz Algorithm: Wheel by Wheel
Hi,Si,Vi,CPi,Cvi
V/F,X,Y,n,m,
To
Feed at P,T
Flash at Pi,Ti
Calculate To,Po,Vo
Done Ho,So,To,W
Outputs Inputs β’ Efficiency β’ Composition
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Compressor Modeling
Classic Compressor Algorithm: Polytropic
β’ πππ = ππππ π‘πππ‘
β’ππ
π= ππππ π‘πππ‘
β’π
πβ1π βπ
π= ππππ π‘πππ‘
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Schultz shortcut: Linearization
π =πΆπ
πΆπ£
hp= polytropic efficiency
hπ= π
ππ
ππ»=
π
ππ»ππ π
+ πΆπππππ
π =π
π
ππ
ππ πβ 1 Z-Factor Charts
π = βπ
π
ππ
ππ π
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Schultz Calculation Strategy: Single Phase
Step 1 β’ Note Inlet e, Pi, Ti
β’ Calculate Volume (Vi), k= (Cp/Cv)
Step 2
β’ Estimate Outlet Temperature, To
β’ ππ = ππππ
ππ
πβ1
π
Step 3
β’ Estimate Average Pressure, π
β’ π = ππππ
ππ ππ π =
ππ+ππ
2
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Example: Continue from Previous
Step 1 efficiency 0.80 mw kg/kmol 19.64
Pi bar 35.00 Ti C 60.00
Zi - 0.95 Hi J/mol 10516.77
Vi m /kmol 0.76 k=Cpi/Cvi - 1.33
Pbar from Geometric Mean Pbar from Arithmetic Mean
Step 2 Po bar 110.00 To C 169.35
Ideal Work 178.4 info
Step3 Pbar bar 62.05 Pbar bar 72.50
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Schultz Calculation Strategy: Single Phase
Step 4
β’ Estimate Average Temperature, π
β’ π =ππ+ππ
2
Step 5
β’ Estimate Average Heat Capacity Ratio, π
β’ π =ππ+2ππ ,π +ππ
4
Step 6
β’ Estimate π
β’ π =π
π
ππ
ππ πβ 1 ππ‘ π , π
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Next Steps: Same Example
Pbar from Geometric Mean Pbar from Arithmetic Mean
Look at effect of averaging models
Step 4 Tbar C 114.68
Step 5 ko - 1.30
ktbar,pbar - 1.31 ktbar,pbar - 1.33
kbar - 1.32 kbar - 1.32
Step 6 dv/dt m /kmol/bar 0.0016 dv/dt m /kmol/bar 0.0014
T/V K*kmol/ m 770.23 T/V K*kmol/ m 902.98
Xbar 0.22 Xbar 0.25
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Schultz Calculation Strategy: Single Phase
Step 7
β’ Estimate π
β’ π = βπ
π
ππ
ππ πππ‘ π , π
Step 8
β’ Estimate π
β’ π =
π β1
π 1
π+π π
1+π 2
Step 9
β’ Estimate π
β’ π =1+π
π 1
π 1
π+π β
1
πβ1
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More Steps
Pbar from Geometric Mean Pbar from Arithmetic Mean
Step 7 dv/dp m /kmol/K -0.01 dv/dp m /kmol/K -0.01
P/V bar*kmol/ m 123.23 P/V bar*kmol/ m 168.80
Ybar 1.02 Ybar 1.02
Step 8 mbar - 0.24 mbar - 0.24
Step 9 nbar - 1.38 nbar - 1.38
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Schultz Calculation Strategy: Single Phase
Step 10
β’ Estimate To
β’ ππ = ππππ
ππ
π
Step 11
β’ Estimate Vo
β’ ππ = ππππ
ππ
β1
π
Step 12
β’ Verify n from EOS
β’ π =ln ππ
ππ
lnππ
ππ
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Step 10 To C 166.18 C 164.56
Step 11 Vo calc m /kmol 0.33 m /kmol 0.33
Step 12 Zo EOS - 0.99 - 0.99
Vo Eos m /kmol 0.33 m /kmol 0.33
nbar - 1.38 - 1.37
Closing Up First Iteration
Pbar from Geometric Mean Pbar from Arithmetic Mean
To C 169.35Previous Estimate
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Schultz Calculation Strategy: Single Phase
Step 13
β’ Compare n-values
β’ Adjust To estimate or subdivide steps and return to step 4
Step 14
β’ Estimate Work
β’ π β1
π
π
π β1
πππ ππ
ππ
π0
ππ
π β1
π β 1 β
π
π
π
π β1
π
ππππππ β ππππ
Step 15 β’ Done
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Step 13 New To C 166.18 C 164.56
Step 14 Work kJ/kg 226.72 Work kJ/kg 226.18
Work kJ/kg 226.72 Work kJ/kg 226.18
Ho J/mol 14942.93 Ho J/mol 14857.41
Work kJ/kg 225.38 kJ/kg 221.03
End of First Iteration
Pbar from Geometric Mean Pbar from Arithmetic Mean
Classic
Both Schultz Relations
Slight difference between relations Iterate Further to Close
Difference between βClassicβ work and Predicted work is related to effect of efficiency on Schultz temperature model
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Next Iteration
Pbar from Geometric Mean Pbar from Arithmetic Mean
Previous Estimate To C 165.37
Step 2 Po bar 110.00 To C 165.37
Step 3 Pbar bar 62.05 Pbar bar 72.50
Step 4 Tbar C 112.68
Step 5 ko - 1.31
ktbar,pbar - 1.32 ktbar,pbar - 1.33
kbar - 1.32 kbar - 1.33
Step 6 dv/dt m /kmol/bar 0.0016 dv/dt m /kmol/bar 0.0014
T/V K*kmol/ m 771.12 T/V K*kmol/ m 904.17
Xbar 0.23 Xbar 0.26
Step 7 dv/dp m /kmol/K -0.01 dv/dp m /kmol/K -0.01
P/V bar*kmol/ m 124.01 P/V bar*kmol/ m 169.90
Ybar 1.02 Ybar 1.02
Step 8 mbar - 0.24 mbar - 0.24
Step 9 nbar - 1.38 nbar - 1.39
Step 10 To C 166.46 C 164.81
Step 11 Vo calc m /kmol 0.33 m /kmol 0.33
Step 12 Zo EOS - 0.99 - 0.99
Vo Eos m /kmol 0.33 m /kmol 0.33
nbar - 1.38 - 1.37
Step 13 New To C 166.46 C 164.81
Step 14 Work kJ/kg 226.81 Work kJ/kg 226.27
Work kJ/kg 226.81 Work kJ/kg 226.27
Ho J/mol 14958.04 Ho J/mol 14870.88
Work kJ/kg 226.15 kJ/kg 221.71
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After 3 Iterations (Direct Substitution)
Pbar from Geometric Mean Pbar from Arithmetic Mean
Step 2 Po bar 110.00 To C 165.64
Step 3 Pbar bar 62.05 Pbar bar 72.50
Step 4 Tbar C 112.82
Step 5 ko - 1.31
ktbar,pbar - 1.32 ktbar,pbar - 1.33
kbar - 1.32 kbar - 1.33
Step 6 dv/dt m /kmol/bar 0.0016 dv/dt m /kmol/bar 0.0014
T/V K*kmol/ m 771.06 T/V K*kmol/ m 904.09
Xbar 0.23 Xbar 0.26
Step 7 dv/dp m /kmol/K -0.01 dv/dp m /kmol/K -0.01
P/V bar*kmol/ m 123.96 P/V bar*kmol/ m 169.82
Ybar 1.02 Ybar 1.02
Step 8 mbar - 0.24 mbar - 0.24
Step 9 nbar - 1.38 nbar - 1.39
Step 10 To C 166.44 C 164.80
Step 11 Vo calc m /kmol 0.33 m /kmol 0.33
Step 12 Zo EOS - 0.99 - 0.99
Vo Eos m /kmol 0.33 m /kmol 0.33
nbar - 1.38 - 1.37
Step 13 New To C 166.44 C 164.80
Step 14 Work kJ/kg 226.81 Work kJ/kg 226.26
Work kJ/kg 226.81 Work kJ/kg 226.26
Ho J/mol 14957.01 Ho J/mol 14869.96
Work kJ/kg 226.10 kJ/kg 221.67
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Converged
Pbar from Geometric Mean Pbar from Arithmetic Mean
Arithmetic mean seems to agree better with rigorous model, but all are within modeling accuracy. Note that enthalpy match is important for process simulators
Step 13 New To C 166.44 C 164.80
Step 14 Work kJ/kg 226.81 Work kJ/kg 226.26
Work kJ/kg 226.81 Work kJ/kg 226.26
Ho J/mol 14957.07 Ho J/mol 14870.01
Work kJ/kg 226.10 kJ/kg 221.67
Rigorous Model To=164.1oC W=219.7kJ/kg
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If I Integrate Through the Compressor
Choose 7 wheels
Gas
Inlet T Inlet P Discharge P Discharge T Gas Rate Compressibility Gas Density Gas Mol Wt Gas Cp/Cv EnthalpyoC bar bar
oC m3/d Z Tf c,Pf c Tf c,Pf c Tf c,Pf c Change
1 60.00 35.00 41.22 73.8 31.5 0.95 26.11 19.64 1.33 27.28
2 73.76 41.22 48.55 87.9 27.9 0.96 29.47 19.64 1.33 28.60
3 87.92 48.55 57.19 102.5 24.7 0.96 33.24 19.64 1.32 29.97
4 102.48 57.19 67.35 117.4 21.9 0.96 37.49 19.64 1.32 31.42
5 117.43 67.35 79.33 132.8 19.5 0.97 42.24 19.64 1.32 32.94
6 132.77 79.33 93.43 148.5 17.3 0.97 47.56 19.64 1.32 34.55
7 148.49 93.43 110.00 164.5 15.4 0.98 53.49 19.64 1.31 36.18
Sum= 220.95