kfhc tpb reguler zfactor fix

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7/23/2019 Kfhc Tpb Reguler Zfactor Fix http://slidepdf.com/reader/full/kfhc-tpb-reguler-zfactor-fix 1/9 NAMA KELOMPOK : 1. ANDRI RAHMAT W 1401151 2. ILHAM RIFALDI 1401200 3. VALENTINUS DIMAS 1401194 C1 0.775 16.043 12.43333 C2 0.083 30.070 2.49581 C3 0.021 44.097 0.926037 i-C4 0.006 58.124 0.348744 n-C4 0.002 58.124 0.116248 i-C5 0.003 72.151 0.216453 n-C5 0.008 72.151 0.577208 C6 0.001 86.178 0.086178 C7 0.001 100.205 0.100205 N2 0.050 28.013 1.40065 CO2 0.030 44.010 1.3203 H2S 0.020 34.076 0.68152 C1 0.775 16.043 666.4 343.33 266.08075 516.46 C2 0.083 30.070 706.5 549.92 45.64336 58.6395 C3 0.021 44.097 616.0 666.06 13.98726 12.936 i-C4 0.006 58.124 527.9 734.48 4.40688 3.1674 n-C4 0.002 58.124 550.8 765.62 1.53124 1.1016 i-C5 0.003 72.151 490.4 828.63 2.48589 1.4712 n-C5 0.008 72.151 488.6 848.8 6.7904 3.9088 C6 0.001 86.178 436.9 913.60 0.9136 0.4369 C7 0.001 100.205 398.8 972.7 0.9727 0.3988 N2 0.050 28.013 493.1 227.49 11.3745 24.655 CO2 0.030 44.010 1071 547.91 16.4373 32.13 H2S 0.020 34.076 1300 672.45 13.449 26 ( Tci ) ( YiTci ) ( YiPci ) Component Mole Fraction ( Yi )  Apparent Molecular Weight ( Mi ) ( Pci ) KIMIA FISIKA HIDROCARBON 1. For the gas composition given in the following text, determine apparent molecular weight, spesific gravity, pseudo-critical pressure, pseudo critical temperature of gas Component Mole Fraction ( Yi )  Apparent Molecular Weight ( Mi ) ( YiMi ) =   29  =  20.70268 29  = 0.713886 = 1 = 643.242   = 20.70268

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Page 1: Kfhc Tpb Reguler Zfactor Fix

7/23/2019 Kfhc Tpb Reguler Zfactor Fix

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NAMA KELOMPOK : 1. ANDRI RAHMAT W 1401151

2. ILHAM RIFALDI 1401200

3. VALENTINUS DIMAS 1401194

C1 0.775 16.043 12.43333

C2 0.083 30.070 2.49581

C3 0.021 44.097 0.926037

i-C4 0.006 58.124 0.348744

n-C4 0.002 58.124 0.116248i-C5 0.003 72.151 0.216453

n-C5 0.008 72.151 0.577208

C6 0.001 86.178 0.086178

C7 0.001 100.205 0.100205

N2 0.050 28.013 1.40065

CO2 0.030 44.010 1.3203

H2S 0.020 34.076 0.68152

C1 0.775 16.043 666.4 343.33 266.08075 516.46

C2 0.083 30.070 706.5 549.92 45.64336 58.6395

C3 0.021 44.097 616.0 666.06 13.98726 12.936

i-C4 0.006 58.124 527.9 734.48 4.40688 3.1674

n-C4 0.002 58.124 550.8 765.62 1.53124 1.1016

i-C5 0.003 72.151 490.4 828.63 2.48589 1.4712

n-C5 0.008 72.151 488.6 848.8 6.7904 3.9088

C6 0.001 86.178 436.9 913.60 0.9136 0.4369

C7 0.001 100.205 398.8 972.7 0.9727 0.3988

N2 0.050 28.013 493.1 227.49 11.3745 24.655

CO2 0.030 44.010 1071 547.91 16.4373 32.13

H2S 0.020 34.076 1300 672.45 13.449 26

( Tci ) ( YiTci ) ( YiPci )ComponentMole Fraction (

Yi )

 Apparent Molecular

Weight ( Mi )( Pci )

KIMIA FISIKA HIDROCARBON

1. For the gas composition given in the following text, determine apparent molecular weight, spesific

gravity, pseudo-critical pressure, pseudo critical temperature of gas

ComponentMole Fraction (

Yi )

 Apparent Molecular

Weight ( Mi )( YiMi )

=  29   = 20.70268

29   = 0.713886

= 1 = 643.242   = 20.70268

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Known :

 Answer :

2. After you determine Ppc and Tpc, Determine Pseudo-Reduce Pressure and Pseudo-ReduceTemperature . After that find gas compressibility Factor using Standing Kartz Correlation and Beggs

Brill Correlation. ( Standing Kartz Chart is attached )

= 384.07288° = 681.3052

= 384.07288 °   = 681.3052

= 213℉+ 460 = 673° = 3250 = 384.07288 ° = 681.3052

= =   3250

681.3052= 4.77

= =   673°

384.07288°= 1.75

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Beggs Brill Correlation

  = 1.39( − 0.92).−0.36 − 0.10= 1.39(1.75− 0.92).−0.36 1.75 −0.10

= 1.39(0.83).

−0.63−0.10= 1.39 0.911043358 −0.63−0.10= 1.266350268− 0,63− 0.10= 0.536350268

= 0.62− 0.23  +   0.066 − 0.86−0.037  + 0.32

10

= 9( − 1)= 9(1.75 − 1)= 9(0.75)= 6.75

= 0.62−0.23(1.75) 4.77+   0.0661.75−0.86 −0.037 4.77 + 0.32(4.77)

10.

= 0.62−0.23(1.75) 4.77+   0.0661.75−0.86 −0.037 4.77 + 0.32(4.77)

10.

= 1.037475 + 0.845436408+ 0.000670286= 1.883581694

= 0.132 − 0.32 = 0.132− 0.32 1.75= 0.132− 0.32 0.243038049= 0.132− 0.077772176= 0.054227824

= 0.3106− 0.49 + 0.1824

= 0.3106− 0.49 1.75 + 0.1824 1.75 

= 0.3106− 0.8575+ 0.5586= 0.0117

= 10

= 10.

= 1.027306415

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= + 1 −   +

= 0.536350268+ 1−0.5363502682.718.   +0.054227824(4.77)

= 0.536350268+ 0.070762154+ 0.25866672= 0.865779142= 0.87

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