Lampiran Perhitungan [PDF]

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URAIAN PERHITUNGAN



Berdasarkan Log Sheet tanggal 21 Agustus 2017, didapatkan data sebagai berikut: 



Flow Gas Alam Masuk



= 5369 Nm3







Temperatur Gas Alam



= 50oC







Temperatur Exhaust



= 450oC = 723,15 K







Energi Listrik yang Dihasilkan



= 10,5 MW







% O2 dalam Flue Gas



= 14,2 %



Fraksi Komponen



BM



Heating Value



HV x Fraksi



(BTU/scf)



(BTU/scf)



BM . Fraksi (%)



N2



34.08



0.40



0.14



0.00



0



CO2



44.01



5.52



2.43



0.00



0



CH4



16.04



85.90



13.78



1010.00



867.59



C2H6



30.07



5.85



1.76



1769.80



103.5333



C3H8



44.10



1.32



0.58



2516.20



33.21384



I-C4H10



58.12



0.25



0.15



3252.10



8.13025



N-C4H10



58.12



0.30



0.17



3262.40



9.7872



I-C5H12



72.15



0.12



0.09



4000.90



4.80108



N-C5H10



72.15



0.08



0.06



4008.80



3.20704



C6H14



86.17



0.26



0.22



5065.80



13.17108



100.00



19.37



TOTAL



Nilai GHV =



Sumber: Data Departemen Laboratorium (Tanggal: 21 Agustus 2017)



1043.43379







Perhitungan Laju Alir Gas Alam Laju Alir Gas Alam



= 5369 Nm3/jam : 22,414 Nm3/kmol = 239,537789 kmol x 19,37 kg/kmol = 4639,84697 kg/jam = 189604,6233 scf/jam







Perhitungan Neraca Massa (Material Balance) Diagram Alir Neraca Massa GTG



Gas Turbine Generator



GAS ALAM



CO2 N2



(GTG) 3006-J



H2O O2



UDARA



1. Reaksi yang Terjadi CH4



+



2 O2



→



CO2



+



2 H2O



C2H6



+



7/2 O2



→



2 CO2



+



3 H2O



C3H8



+



5 O2



→



3 CO2



+



4 H2O



I-C4H10



+



13/2 O2



→



4 CO2



+



5 H2O



N-C4H10



+



13/2 O2



→



4 CO2



+



5 H2O



I-C5H12



+



8 O2



→



5 CO2



+



6 H2O



N-C5H10



+



8 O2



→



5 CO2



+



6 H2O



C6H14



+



19/2 O2



→



6 CO2



+



7 H2O



2. Perhitungan Teoritis Basis = 100 kmol Gas Alam Reaksi Pembakaran



O2 Teoritis



CO2 Teoritis



H2O Terbentuk



(Kmol)



(Kmol)



(Kmol)



Kmol Komponen







CH4



85.9



171.8



85.9



171.8



C2H6



5.85



20.475



11.7



17.55



C3H8



1.32



6.6



3.96



5.28



I-C4H10



0.25



1.625



1



1.25



N-C4H10



0.3



1.95



1.2



1.5



I-C5H12



0.12



0.96



0.6



0.72



N-C5H10



0.08



0.64



0.4



0.48



C6H14



0.26



2.47



1.56



1.82



CO2



5.52



-



5.52



-



TOTAL



99.6



206.52



111.84



200.4



Perhitungan Komposisi Udara Masuk Teoritis (100 kmol Gas Alam) a. O2 Teoritis



= 206,52 kmol



b. N2 dalam O2 Teoritis



= (Perbandingan N2 / O2 Udara) x O2



Teoritis = (79/21) x 206,52 kmol = 776,9086 kmol c. Udara Teoritis



= (O2 Teoritis + N2 dalam O2 Teoritis)



= (206,52 + 776,9086) kmol = 983,42856 kmol







Perhitungan Komposisi Udara Masuk Teoritis (239,537789 kmol) a.



N2



=



776,9086 𝑘𝑚𝑜𝑙 100 𝑘𝑚𝑜𝑙



𝑥 230,537789 𝑘𝑚𝑜𝑙



= 1860,9904 kmol b.



O2



= (Perbandingan O2 / N2 Udara) x N2 = (21/79) x 1860,9904 kmol = 494,6937 kmol



c.



N2 in



= 100 % x Fraksi N2 dalam Udara = 100 % x 0,79 = 79 %



d.



O2 in



= 100% x Fraksi O2 dalam Udara = 100 % x 0,21 = 21 %



e.



Analisa Komposisi Flue Gas, O2 = 14,2%



f.



O2 tereaksi



= 21 % – 14,2 % = 6,8 %



g.



% Excess Udara



𝑂2 𝑒𝑥𝑐𝑒𝑠𝑠



=𝑂



2



=



𝑡𝑒𝑟𝑒𝑎𝑘𝑠𝑖



14,2 % 6,8 %



𝑥 100 %



𝑥 100 %



= 208,8235 %



h.



O2 Excess



= 208,8235 % x O2 Teoritis = 2,088235 x 206,52 kmol = 431,2623 kmol



3. Neraca Massa (Material Balance) di GTG o Input Laju Alir Mol Fuel (Bahan Bakar) a. O2 =



=



(𝑂2 𝑇𝑒𝑜𝑟𝑖𝑡𝑖𝑠 + 𝑂2 𝐸𝑥𝑐𝑒𝑠𝑠) 100 𝑘𝑚𝑜𝑙



= 239,537789 kmol



x Laju Alir Mol Fuel



(206,52 𝑘𝑚𝑜𝑙/𝑗𝑎𝑚+ 431,2623 𝑘𝑚𝑜𝑙/𝑗𝑎𝑚) 100 𝑘𝑚𝑜𝑙/𝑗𝑎𝑚



. 239,537789 𝑘𝑚𝑜𝑙/



𝑗𝑎𝑚 = 1527,7296 kmol b. N2 = (79/21) x 1527,7296 kmol = 5747,1734 kmol c. Gas Alam = 239,537789 kmol



o Output a.



CO2



= =



CO2 Dihasilkan 100 𝑘𝑚𝑜𝑙 111,84 kmol 100 𝑘𝑚𝑜𝑙



x Laju Alir Mol Fuel



x 239,537789 kmol



= 267,8991 kmol b.



O2



= O2 input – [



O2 Teoritis 100 𝑘𝑚𝑜𝑙



x Laju Alir Mol Fuel ]



= ( 1527,7296 - [



214,505 100



x 239,537789 ] ) kmol



= 1013,9091 kmol c.



N2



= N2 input + N2 dari Gas Alam = (5747,1734 + (0,4% x 239,537789)) kmol = (5747,1734 kmol + 0,9582 kmol) = 5748.1316 kmol



d. H2O



= =



H2 O Terbentuk 100 𝑘𝑚𝑜𝑙 200,4 𝑘𝑚𝑜𝑙 100 𝑘𝑚𝑜𝑙



x [ Laju Alir Mol Fuel ]



x 239,537789 kmol



= 480,0337 kmol KOMPONEN



INPUT (kmol)



OUTPUT (kmol)



Gas Alam



239,537789



-



CO2



-



267,8991



N2



5747,1734



5748.1316



O2



1527,7296



1013,9091



H2O



-



480,0337



TOTAL



7514,440789



7509,9735



4. Perhitungan Neraca Panas di GTG 



Diagram Alir Neraca Panas di GTG



Q4 Flue Gas to WHB Q1 Gas Alam



Gas Turbine Generator



Q5 FDFan FanFan Q2 Udara Q Q



(GTG) 3006-J



Q5 Listrik



Q3 Reaksi



Q Losses







Q1 Gas Alam (Panas dari Gas Alam) T = 50 oC = 323,15 K , Tref = 27 oC = 300,15 K



Cp N2 = (29) + (0,2199 x 10-2) T + (0,5723 x 10-5) T2 – (2,871 x 10-9) T3 T = 323,15 K Maka, Cp N2 = 30,2114 J/mol K



= 28,6351 BTU /kmol K



Q = m . Cp . (T - Tref) Q = (0,4% x 239,537789 kmol) .30,414 BTU/kmol K .(323,15 K – 300,15 K) Q N2 = 631,0444 BTU



Cp CO2 = 36,11 + (4,233 x 10-2) T – (2,887 x 10-5) T2 + (7,464 x 10-9) T3 T = 323,15 K Maka, Cp CO2 = 47,0260 J/mol K = 44,573 BTU/kmol K



Q = m . Cp . (T - Tref) Q = (5,52% x 239,53778 kmol) .44,573 BTU/kmol K .(323,15 K – 300,15 K) Q CO2 = 13555,26802 BTU



Cp CH4 = 4,5980 + (1,245 x 10-2) T + (2,686 x 10-6) T2 - (2,703 x 10-9) T3 T = 323,15 K Maka, Cp CH4 = 8,82866 Cal/mol K = 35,011524 BTU/kmol K Q = m . Cp . (T - Tref) Q = (85,9% x 239,53778 kmol) .35,012 BTU/kmol K .(323,15 K – 300,15 K) Q CH4 = 165694 BTU



Cp C2H6 = 1,2920 + (4,254 x 10-2) T - (1,657 x 10-6) T2 - (2,08 x 10-9) T3 T = 323,15 K Maka, Cp C2H6 = 13,3787 Cal/mol K



= 53,055403 BTU/kmol K



Q = m . Cp . (T - Tref) Q = (5,85% x 239,53778 kmol) .53,055 BTU/kmol K .(323,15 K – 300,15 K) Q C2H6 = 17099,7 BTU



Cp C3H8 = -1,009 + (7,315 x 10-2) T - (3,789 x 10-5) T2 + (7,678 x 10-9) T3 T = 323,15 K Maka, Cp C3H8 = 18,9318Cal/mol K = 75,0773 BTU/kmol K



Q = m . Cp . (T - Tref) Q = (1,32% x 239,53778 kmol) .75,077 BTU/kmol K .(323,15 K – 300,15 K) Q C3H8 = 5459,9 BTU



Cp N-C4H10 = 2,266 + (7,313 x 10-2) T - (2,647 x 10-5) T2 - (6,74 x 10-10) T3 T = 323,15 K Maka, Cp N-C4H10 = 25,05 Cal/mol K = 99,3398 BTU/kmol K Q = m . Cp . (T - Tref) Q = (0,12% x 239,53778 kmol) .99,398 BTU/kmol K .(323,15 K – 300,15 K) Q N-C4H10 = 1368,25 BTU



Cp I-C4H10 = -0,3320 + (9,189 x 10-2) T - (4,409 x 10-5) T2 + (6,92 x 10-9) T3 T = 323,15 K Maka, Cp I-C4H10 = 24,9915 Cal/mol K = 99,1077 BTU/kmol K Q = m . Cp . (T - Tref) Q = (0,3% x 239,53778 kmol) .99,108 BTU/kmol K .(323,15 K – 300,15 K) Q I-C4H10 = 1638,06 BTU



Cp N-C5H12 = -0,866 + (1,164 x 10-1) T - (6,163 x 10-5) T2 + (1,267 x 10-8) T3 T = 323,15 K Maka, Cp N-C5H12 = 30,7404 Cal/mol K = 121,9063 BTU/kmol K Q = m . Cp . (T - Tref) Q = (0,12% x 239,538 kmol) .121,9063 BTU/kmol K .(323,15 K – 300,15 K) Q N-C5H12 = 805,952 BTU



Cp I-C5H12 = -2,2750 + (1,21 x 10-1) T - (6,519 x 10-5) T2 + (1,367 x 10-8) T3 T = 323,15 K Maka, Cp I-C5H12 = 30,4779 Cal/mol K = 120,87318 BTU/kmol K Q = m . Cp . (T - Tref) Q = (0,08% x 239,538 kmol) .120,8732 BTU/kmol K .(323,15 K – 300,15 K) Q I-C5H12 = 532,748 BTU



Cp C6H14 = -1,054 + (1,39 x 10-1) T - (7,449 x 10-5) T2 + (1,551 x 10-8) T3 T = 323,15 K Maka, Cp C6H14 = 36,6086 Cal/mol K = 145,17728 BTU/kmol K Q = m . Cp . (T - Tref) Q = (0,26% x 239,538 kmol) .145,1773 BTU/kmol K .(323,15 K – 300,15 K)



Q C6H14 = 2079,57 BTU



Jadi, Q1 (Panas Fuel Gas)



= Q N2 + Q CO2 + Q CH4 + Q C2H6 + Q C3H8 + Q N-C4H10 + Q I-C4H10 + Q N-C5H12



+ Q I-C5H12 + Q C6H14



= (631,044 + 13555,24 + 165693,7 + 17099,66 + 5459,895 + 1368,249 + 1638,064



+



805,9523



+



532,748



+



2079,572) BTU = 2,0886 x 105 BTU







Q2 Udara (Panas dari Udara) T = 30 oC = 303,15 K , Tref = 27 oC = 300,15 K



Cp O2 = (29,1) + (1,158 x 10-2) T - (0,6076 x 10-5) T2 + (1,311 x 10-9) T3 T = 303,15 K Maka, Cp O2



= 32,088 J/mol K



= 30,414 BTU/kmol K



Q = m . Cp . (T - Tref) Q = 1527,7296 kmol . 30,414 BTU/kmol K . (303,15 K – 300,15 K) Q O2 = 139393,1042 BTU



Cp N2 = (29) + (0,2199 x 10-2) T + (0,5723 x 10-5) T2 – (2,871 x 10-9) T3 T = 303,15 K Maka, Cp N2 = 30,213 J/mol K



= 28,6367 BTU /kmol K



Q = m . Cp . (T - Tref) Q = 5747,1734 kmol . 28,6367 BTU/kmol K . (303,15 K – 300,15 K) Q N2 = 493740,2414 BTU



Jadi, Q2 (Panas dari Udara) = (139393,1042 + 493740,2414 ) BTU = 6,3313 x 105 BTU







Q3 Reaksi (Panas Pembakaran Gas Alam) Q3



= m . GHV = 189604,6233 scf/jam x 1043,43379 BTU/scf = 19,7840 x 107 BTU







Q4 Flue Gas (Panas sisa menuju WHB) T = 400 oC = 723,15 K , Tref = 27 oC = 300,15 K



Cp O2 = (29,1) + (1,158 x 10-2) T - (0,6076 x 10-5) T2 + (1,311 x 10-9) T3 T = 723,15 K Maka, Cp O2 = 34,7924 J/mol K



= 32,9772 BTU/kmol K



Q = m . Cp . (T - Tref) Q = 1013,9091kmol . 32,9772 BTU/kmol K . (723,15 K – 300,15 K) Q O2 = 12,472 x 106 BTU



Cp N2 = (29) + (0,2199 x 10-2) T + (0,5723 x 10-5) T2 – (2,871 x 10-9) T3 T = 723,15 K Maka, Cp N2 = 30,213 J/mol K



= 28,6367 BTU /kmol K



Q = m . Cp . (T - Tref) Q = 5748,1316 kmol /jam . 28,6367 BTU/kmol K . (723,15 K – 300,15 K) Q N2 = 61,340 x 106 BTU



Cp CO2 = 36,11 + (4,233 x 10-2) T – (2,887 x 10-5) T2 + (7,464 x 10-9) T3 T = 723,15 K Maka, Cp CO2 = 54,53504 J/mol K = 51,670 BTU/kmol K Q = m . Cp . (T - Tref) Q = 267,8991 kmol . 51,670 BTU/kmol K . (723,15 K – 300,15 K) Q CO2 = 5,8553 x 106 BTU



Cp H2O = 7,701 + (4,59 x 10-4) T + (2,521 x 10-6) T2 – (8,59 x 10-10) T3 T = 723,15 K Maka, Cp H2O = 9,026 kCal/kmol K = 35,7941 BTU/kmol K Q = m . Cp . (T - Tref) Q = 480,03337 kmol . 35,7941 BTU/kmol K . (723,15 K – 300,15 K) Q H2O = 7,2681 x 106 BTU



Jadi, Q4 (Panas Flue Gas) BTU



= (12,472+61,34+5,8553 +7,2681) x 106



= 8,69354 x 107 BTU







Q5 Listrik (Listrik yang Diproduksi) Q5



= 10,5 MW



1 MW = 106 Watt 103 W = 0,947832 BTU / s 1h



= 3600 s



Maka, Q5



= (10,5 x 106 W) x



𝟎,𝟗𝟒𝟕𝟖𝟑𝟐 𝑩𝑻𝑼/𝒔 𝟏𝟎𝟑



𝑾



x



𝟑𝟔𝟎𝟎 𝒔 𝟏𝒉



= 3,5828 x 107 BTU / jam







Q6 Force Draft Fan Dari neraca massa, diketahui bahwa V udara masuk = 7274,903 kmol/jam Maka, V



= 7274,903 kmol / jam x 22,414 Nm3 / kmol = 163059,6758 Nm3 / jam = 45,2944 Nm3 / s



Jika diketahui , (Dari data pada tanggal 21 Agustus 2017) P Udara yang dialirkan (P2) = 3 atm = 303975 Pa P Udara lingkungan (P1)



= 1 atm = 101325 Pa



Efisiensi Alat (Design)



= 80 %



n Udara



= 1,4



Maka, P



=



=



𝑃1 . 𝑉 60000



.



𝑛 𝑛−1



𝑃2



. {[ ]



𝑛−1 𝑛



𝑃1



101325 . 45,2944 60000



.



− 1} (KW)



1,4 1,4 − 1



. {[



303975 101325



]



1,4−1 1,4



− 1} (KW)



= 98,7161 KW = 93,5648 BTU / s . 3600 s / jam = 3,3683328 x 105 BTU



5. Menghitung Q Loss Q Loss = (Q1 + Q2 + Q3) – (Q4 + Q5 + Q6) = (2,0886 x 105 + 6,3313 x 105 + 19,7840 x 107) BTU – (8,69354 x 107 + 3,5828 x 107 + 3,3683328 x 105 ) BTU = 19,8682 x 107 BTU – 12,3100 x 107 BTU = 7,5582 x 107 BTU 6. Menghitung Efisiensi Alat GTG Efisiensi



= =



𝑄 𝐼𝑛𝑝𝑢𝑡 – 𝑄 𝐿𝑜𝑠𝑠 𝑄 𝐼𝑛𝑝𝑢𝑡



x 100 %



19,8682 𝑥 107 𝐵𝑇𝑈 – 7,5582 𝑥 107 𝐵𝑇𝑈



= 62 %



19,8682 x 107 BTU



x 100 %