Compressor Thermodynamics Rev3

Compressor Thermodynamics

Methods and Alternatives

2 GE Title or job number

11/17/2013

Background

Calculating Volume, SI

The basic equation is 𝑃𝑉 = 𝑛𝑅𝑇𝑍

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 • Note Inlet Pi, Ti

• Calculate Volume (Vi), and Cp, Cv and k=Cp/Cv

Step 2

• Estimate To

• 𝑇𝑜 = 𝑇𝑖𝑃𝑜

𝑃𝑖

𝑘−1

𝑘

Step 3

• Estimate Ideal Work

• 𝑊 ≈1

𝜖

𝑘

𝑘−1

𝑍𝑖𝑅𝑇𝑖

𝑀𝑊

𝑃0

𝑃𝑖

𝑘−1

𝑘− 1 𝑤ℎ𝑒𝑟𝑒 𝑅 = 8.3145

𝑘𝐽

𝑘𝑚𝑜𝑙/𝐾


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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

𝑀𝑊 = 𝑍𝑖𝑀𝑊𝑖

𝑍𝑖


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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

• Needed a convenient model for the fluid

• Little General Access to Process Simulation Tools • Limitations in Equations of State Methods

• BWR 1940, complex solution • RK 1949, limited application to mixtures • NGA 1958, Equilibrium Ratio Data for Computers

Needed an alternative which could be handled using a slide rule

• Ie: Simple log-log relationships • 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 • Efficiency • 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 • Efficiency • Composition


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Compressor Modeling

Classic Compressor Algorithm: Polytropic

• 𝑃𝑉𝑛 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

•𝑃𝑚

𝑇= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡

•𝑃

𝑛−1𝑛 −𝑚

𝑍= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡


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Schultz shortcut: Linearization

𝑘 =𝐶𝑝

𝐶𝑣

hp= polytropic efficiency

h𝑝= 𝑉

𝑑𝑃

𝑑𝐻=

𝑉

𝜕𝐻𝜕𝑃 𝑇

+ 𝐶𝑝𝑑𝑇𝑑𝑃

𝑋 =𝑇

𝑉

𝜕𝑉

𝜕𝑇 𝑃− 1 Z-Factor Charts

𝑌 = −𝑃

𝑉

𝜕𝑉

𝜕𝑃 𝑇


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Schultz Calculation Strategy: Single Phase

Step 1 • Note Inlet e, Pi, Ti

• Calculate Volume (Vi), k= (Cp/Cv)

Step 2

• Estimate Outlet Temperature, To


𝑃𝑖

𝑘−1

𝑘

Step 3

• Estimate Average Pressure, 𝑃

• 𝑃 = 𝑃𝑖𝑃𝑜

𝑃𝑖 𝑜𝑟 𝑃 =

𝑃𝑜+𝑃𝑖

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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Step 4

• Estimate Average Temperature, 𝑇

• 𝑇 =𝑇𝑜+𝑇𝑖

2

Step 5

• Estimate Average Heat Capacity Ratio, 𝑘

• 𝑘 =𝑘𝑖+2𝑘𝑇 ,𝑃 +𝑘𝑜

4

Step 6

• Estimate 𝑋

• 𝑋 =𝑇

𝑉

𝜕𝑉

𝜕𝑇 𝑃− 1 𝑎𝑡 𝑃 , 𝑇


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Next Steps: Same Example


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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Step 7

• Estimate 𝑌

• 𝑌 = −𝑃

𝑉

𝜕𝑉

𝜕𝑃 𝑇𝑎𝑡 𝑃 , 𝑇

Step 8

• Estimate 𝑚

• 𝑚 =

𝑘 −1

𝑘 1

𝜖+𝑋 𝑌

1+𝑋 2

Step 9

• Estimate 𝑛

• 𝑛 =1+𝑋

𝑌 1

𝑘 1

𝜖+𝑋 −

1

𝜖−1


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More Steps


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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Step 10

• Estimate To


𝑃𝑖

𝑚

Step 11

• Estimate Vo

• 𝑉𝑜 = 𝑉𝑖𝑃𝑜

𝑃𝑖

−1

𝑛

Step 12

• Verify n from EOS

• 𝑛 =ln 𝑃𝑜

𝑃𝑖

ln𝑉𝑖

𝑉𝑜


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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


To C 169.35Previous Estimate


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Step 13

• Compare n-values

• Adjust To estimate or subdivide steps and return to step 4

Step 14

• Estimate Work

• 𝑊 ≈1

𝜀

𝑛

𝑛 −1

𝑍𝑖𝑅𝑇𝑖

𝑀𝑊

𝑃0

𝑃𝑖

𝑛 −1

𝑛 − 1 ≈

𝑓

𝜀

𝑛

𝑛 −1

𝑅

𝑀𝑊𝑍𝑜𝑇𝑜 − 𝑍𝑖𝑇𝑖

Step 15 • 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


Classic

Both Schultz Relations

Slight difference between relations Iterate Further to Close

Difference between “Classic” work and Predicted work is related to effect of efficiency on Schultz temperature model


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Next Iteration


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 5 ko - 1.31


kbar - 1.32 kbar - 1.33



Xbar 0.23 Xbar 0.26



Ybar 1.02 Ybar 1.02



Step 10 To C 166.46 C 164.81


Step 12 Zo EOS - 0.99 - 0.99


nbar - 1.38 - 1.37

Step 13 New To C 166.46 C 164.81



Ho J/mol 14958.04 Ho J/mol 14870.88



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After 3 Iterations (Direct Substitution)


Step 2 Po bar 110.00 To C 165.64

Step 3 Pbar bar 62.05 Pbar bar 72.50


Step 5 ko - 1.31


kbar - 1.32 kbar - 1.33



Xbar 0.23 Xbar 0.26



Ybar 1.02 Ybar 1.02



Step 10 To C 166.44 C 164.80


Step 12 Zo EOS - 0.99 - 0.99


nbar - 1.38 - 1.37

Step 13 New To C 166.44 C 164.80



Ho J/mol 14957.01 Ho J/mol 14869.96



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Converged


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



Ho J/mol 14957.07 Ho J/mol 14870.01


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

Compressor Thermodynamics Rev3

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