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LAMPIRAN 3 PERHITUNGAN SPESIFIKASI PERALATAN



ABSORBER (AB-01) Fungsi



: Memisahkan produk Metil Klorida dari sisa reaktan keluaran Reaktor-02



Tipe



: Packed Tower



Gambar



:



AB-01



Gambar L.3.1. Absorber-01 (AB-01) Kondisi Operasi : -



Tekanan



= 1,8 atm



-



Temperatur = 58 oC



-



Gas Masuk Laju alir massa, G = 14.347,763 kg/jam = 3,985 kg/s Densitas,  pada 58 oC (331 K) ρcampuran dihitung menggunakan rumus : 1 ρ campuran







Xi ρi



didapatkan ρGas = 1,247 kg/m3 = 0,078 lb/ft3



159



160



Viskositas Gas, G Viskositas masing-masing komponen dicari menggunakan grafik 2-32, Perry’s Chemical Engineers’ Handbook. Didapatkan : Senyawa HCl MeCl DME MeOH H2O Jumlah



m,kg 73,011 7999,209 4950,005 47,896 1277,630 14347,763



Xi μi, cP 0,005 0,016 0,345 0,012 0,003 0,011 0,089 0,011 0,557 0,011 1  



n, mol 2,000 158,400 107,609 1,497 70,979 340,486



Viskositas campuran menggunakan rumus : campuran =  Xi . i didapatkan μGas = 0,11 cp = 1,116 kg/ms Diffusivitas Gas, DG = 1,48 x 10-4 m2/s BMAVG = 30,647 kg/kmol -



Liquid Masuk Laju alir massa, L1



= 11.882,886 kg/jam = 3,301 kg/s



Viskositas Liquid, L



= 0,85 cp = 8,5 x 10-4 kg/ms



Densitas Liquid, L



= 995,467 kg/m3



Diffusivitas Liquid, DL = 2,04 x 10-8 m2/s BM



= 18,015 kg/kmol



Dari Tabel 6.4 Mass-Transfer Operations, Robert E Treyball dipilih : -



Jenis Packing = Ceramic Raschig Ring



-



Nominal size = 2 in = 50 mm



Dari Tabel 6.3, 6.4 dan 6.5 didapatkan : Wall Thickness Cf CD 



= 6 mm = 65 = 135,6 = 0,74



161



= 92 m2/m3 = 34,03 =0 = 0,362 = 0,0725



ap m n p ds 1.



Menentukan Diameter tower, Dt Lihat Grafik 6.34 Mass-Transfer Operations-Robert E Treyball L' G'



=



 ρG   ρL  ρG



  



0,5



(Treyball, halaman 195)



 1,247 kg/m 3  3 3  995,467 kg/m  1,247 kg/m



3,301 kg/s 3,985 kg/s



  



0,5



= 0, 029 L' Dengan menarik garis lurus nilai ' G



 ρG   ρL  ρG



0,5



  = 0, 029 ke garis pressure drop 



gas pada 200 (N/m2)/m, maka didapat ordinat = 0, 05 0,1



G 2 Cf μ L J = 0, 05 ρG  ρL  ρG g c  0,05 ρ G  ρ L  ρ G  g c  G'    0,1 Cf μ L J  



0.5



dimana, J = 1 dan gc = 1



 0,05 x 1,247  995,467  1,247  1 G'    0 ,1 165 x 8,5 x10  4 1  











G’ = 1,391 kg/m2s Tower Cross Sectional Area, A A



3,985 kg/s G  = 2,866 m2 G' 1,391 kg/m 2 s



Diameter Tower, Dt 4A Dt     π 



0,5



 4 x 2,866    3,14  



Jari-jari, r = 0,955 m



0,5



 1,911



m



0 .5



162



2.



Menentukan Hold up Untuk Liquid Sc L 



μL ρL DL



Sc L 



8,5.10 4 995,467 x 2,04 x 10 -8



(Treyball,halaman 205)



Sc L  41,856 m



Untuk Gas Sc G 



μG ρG DG



Sc G 



(Treyball,halaman 205)



1,116 x 10 -4 1,247 x 1,48 x 10 - 4



Sc G  0,0605



L’ =



3,301 kg/s = 1,152 kg/m2s 2,866 m 2



Log L’ = 0,0613 Dari Tabel 6.5, untuk L < 0,012, maka :  = 1,508 x ds0,376 = 1,508 (0,0725)0,376 = 0,562 Φ LSW  Φ LSW 



2,47  10 4  d s  1,21 2,47  10 4



 0,0725 1,21



 LSW  0,0059



Φ LTW 



 2,09  10  737,5  L' 6



ds 2



β



163



Φ LTW 



 2,09  10  737,5  1,152 6



0,562



 0,0725 2



Φ LTW  0,0176



LOW = LTW - LSW LOW = 0,0176 – 0,0059 LOW = 0,0117 9,57 L'0,57 μ 0,13  σ  H   0,43 ρ L 0,84  2,024 L' 1  0,073 



9,57 1,152



0,57



0,1737 0,262logL'



8,5  10 



4 0,13



 0,0773  H   0,43 995,467  0,84 2,024 1,152  1  0,073 







0,1737 0,262log1,152







H  0,0043 LO = LOW – H LO = 0,0117 – 0,0043 LO = 0,0074



Φ LS 



0,0486  μ L



0,02



ds1,21ρ L



 σ 0,99



0,37



0,0486  8,5 10  4    0,0773   0,0725 1,21  995,467 0,37 0,02



Φ LS



0,99



Φ LS  0,0062



Lt = LO + LS Lt = 0,0074 + 0,0062 Lt = 0,0136



(Pers. 6.69, Treyball)



164



3.



Menentukan Interfacial Area Dari Tabel 6.4 n



 808 G'   L'p aAW = m  0,5  ρ  G  0



 808  1,3905  0,362   1,1516 0,5 1,2467  



aAW = 34,03 



aAW = 35,8146 m2/m3 aA = a AW



Φ LO Φ LOW



(Pers. 6.73, Treyball) 0,0074



aA = 35,8146 m2/m3 0,0117 aA = 22,5523 m2/m3 4.



Operating Void Space dalam packing ε = 0,74



ε LO  ε  Φ LT



(Pers. 6.71, Treyball)



ε LO  0,74  0,0136 ε LO  0,7264



5.



Koefisien Fase Gas, FG



FG S CG G



2/3



 dsG '   1,195   μ G 1  ε LO  



G



0,36



(Pers. 6.70, Treyball)



G' 1,3905  = 0,0454 BM 30,6468



maka, FG  0,0605 0,0454



2/3



  0,0725  1,3905  1,195  -4  0,1116 x 10 1  0,7264  



FG = 0,0212



6.



Koefisien Fase Liquid, KL



0,36



165



 ds L'  K L ds  25,1  DL  μL 



0,45



S CL



0,5



K L 0,0725  0,0725  1,152   25,1  8 -4 2,04  10  8,5 x 10 



(Pers. 6.72, Treyball) 0,45



 41,856  0,5



KL = 3,6 x 10-4 kmol/m2s C



ρL 995,467   55,3037 BM 18,015



FL = KL x C = 3,6 x 10-4 kmol/m2s x 55,3037 = 0,0199 7.



Koefisien Volumetrik FG x aA = 0,0212 x 22,5523 = 0,4776 kmol/m3s FL x aA = 0,0199 x 22,5523 = 0,4491 kmol/m3s



8.



Tinggi Transfer Unit, Htog H tg  L



G 0,0454   0,0949 FG  a A 0,4776



L' 1,152   0,0639 BM AVG 18,015



H tl 



L 0,0639   0,1425 FL  a A 0,4491



Pada T = 580C, Tekanan parsial HCl = 3,52 atm Pt = 1,8 atm m



P 3,52   1,9556 atm Pt 1,8



H tog  H tg 



m G H tl L



H tog  0,0949 



1,9556  0,0454  0,1425 0,0639



H tog  0,2926



Number of Transfer Unit, Ntog



(Pers. 8.54, Treyball)



166



N tog 







y1  y 2 y  y* m







dimana : y1 = Fraksi mol HCl dalam fase gas pada bottom kolom



= 0,0058



y2 = Fraksi mol HCl dalam fase gas pada top kolom



= 0,0001



x1 = Fraksi mol HCl dalam fase liquid pada bottom kolom = 0,0029 x2 = Fraksi mol HCl dalam fase liquid pada top kolom



=0



y1* = m . x1 = 1,9556 x 0,0029 = 0,0057 y2* = m . x2 = 1,9556 x 0 = 0 N tog 



y1  y 2   * *  yy 1  yy   y  y* 1   ln   y  y*   2  







N tog 



     







  2    



0,0057  0       0,0058  0,0057 1   0,0001  0  2      0,0058  0,0057 1    ln      0,0001  0  2 



N tog  48,7172



9.



Tinggi Packing,z Z = Htog x Ntog Z = 0,2926 x 48,7172 Z = 14,2541 m



10.



Pressure Drop Pressure Drop untuk packing yang terbasahi dengan tinggi (z) = 14,2541 m P1 = P x Z = 200 (N/m2)/m x 14,2541 m = 2850,8137 N/m2 ΔP2 G' 2  CD z ρG



167



ΔP2 1,3905 2  135,6 14,25 1,2467



ΔP2  2997,7534 N/m2



Total Pressure Drop : ΔP  ΔP1  ΔP2 ΔP  2850,8137  2997,7534



ΔP  5848,5671 N/m2 ΔP  0,0577 atm 11.



Tebal Dinding, t t



P.r C S E  0,6P



(Tabel 4,halaman 537, Peters)



Tekanan Design (P)



= 1,8 atm = 26,46 psi



Jari-jari



= 0,9554 m



Working Stress yang diizinkan (S) = 18700 psi (Hlmn 538, Peters) Korosi yang diizinkan (C)



= 0,003175 m (Tabel 23.2, Peters)



Efisiensi Pengelasan (E)



= 0,85 (Hlmn 638, Coulson)



t=



26,46 . 0,9554 + 0,003175 18700 × 0,85 0,6 × 26,46



t = 0,0048 m OD = 2 t + D OD = 2 (0,0048) + 1,9108 OD = 1,9204 m Summary Tipe Tekanan Temperatur Diameter Tebal Dinding Tinggi Packing Pressure Drop



: : : : : : :



Packed Tower 1,8 atm 58 oC 1,9108 m 0,0048 m 14,2541 m 0,0577 atm



168



COMPRESSOR-01 (CP-01) Fungsi



: Mengalirkan dan menaikkan tekanan dari top KOD-01 menuju R-01



Tipe



: Centrifugal Multi Stage Compressor



Jumlah



: 1 buah



Gambar :



169



Gambar L.3.2. Compressor-01 (CP-01) Data Design Laju alir, W



: 9866,8582 kg/jam = 362,5445 lb/min



Tekanan masuk, Pin



: 1,5 atm = 3174,325 lbf/ft2



Tekanan keluar, Pout



: 2,5 atm = 5290,541 lbf/ft2



Temperatur masuk, Tin



: 110 oC = 383 K



Densitas, ρ



:



Dari Tabel 2-30 Perry’s Chemical Engineer’s Handbook diketahui : Senyawa H2O ρH O = 2 ρ H 2O =



A 4,391



A



(



B 1 + (1 - T / C)



(



D



B 0,2487



C 647,13



BM 18 36,5



Xi 0,63 0,37 1



D 0,2534



)



4,391



0,2487 1 + (1 - 383 / 647,13)



0,2534



)



= 53,5120 kmol/m3 = 963,2154 kg/m3 ρHCl



= 12,246 kmol/m3 = 446,9790 kg/m3



Senyawa H2O HCl Jumlah



m, kg/jam 6216,1206 3650,7375 9866,8582



ρcamp = 772,2079 kg/m3 = 48,2088 lb/ft3 BMcamp = 24,8450 kg/kmol



ρ, kg/m3 963,2154 446,9790



ρi. Xi 606,8257 165,3822 772,2079



BM . Xi 11,3400 13,5050 24,8450



170



1. Kapasitas Kompressor Volume, V = =



W ρ 362,5445 lb/min 48,2088 lb/ft3



= 7,5203 ft3/min Safety Factor = 10 % Maka, kapasitas kompressor, V = 1,1 x V = 1,1 x 7,5203 ft3/min = 8,2723 ft3/min 2. Jumlah stage, Ns Cp H2O = 4,37002 + 0,1127634 T – 0,0003197621 T2 + 0,0000003139008 T3 Cp HCl = 4,232994 + 0,215979 T – 0,0013482160 T2 + 0,0000027081060 T3



Senyawa H2O HCl Jumlah



m, kg/jam 6216,1206 3650,7375 9866,8582



BM 18 36,5



n, kmol 345,3400 100,0202 445,3602



Xi 0,63 0,37 1



Cp, kkal/kmol.K 18,2884 41,3244



Cpi. Xi 14,1812 9,2807 23,4619



Cpcamp = 23,4619 kkal/kmol.K Cvcamp = Cpcamp – R , dimana R = 1,985765 kkal/kmol.K = 23,4619 kkal/kmol.K - 1,985765 kkal/kmol.K = 21,4761 kkal/kmol.K k =



Cp 23,4619 kkal/kmol.K = = 1,0925 Cv 21,4761 kkal/kmol.K



k Wm = RT1 ( k- 1



( P2 / P1 ) k -1 k - 1



)



1,0925 Wm = 1,985765 kkal/kmol.K × 382 × ( 1,0925 1



= 397,0284 kkal/kmol = 555665,6307 ft-lb/lmol



( 2,5 atm / 1,5 atm)



1,0925-1 1,0925 - 1 )



171



Ns = Ns =



Wm BM × 10000 ft - lb/lmol



555665,6307 ft - lb/lmol 24,8450 lb/lbmol × 10000 ft - lb / lb



= 2,2365 ≈ 3 stage 3. Temperatur keluaran gas kompressor, Tout P Tout = ( 2 P ) k - 1 k 1 Tout = (



2,5 atm



1,0925 - 1 1,0925 1,5 atm )



= 388,56 K = 115,56 oC 4. Tenaga yang dibutuhkan, P hp =



3,03.10 - 5 k Ns P1 V ( P2 / P1 ) k -1 k - 1 k -1



hp =



3,03.10 - 5 1,0925 × 3 × 3174,325 lbf/ft 2 × 8,2723 ft 2 /min × (2,5 / 1,5)1,0925-1 1,0925 - 1 1,0925-1



)



(



)



(



= 0,4094 hp Diketahui friction loss max 3 % Power yang dibutuhkan P =



0,4094 hp = 0,4220 hp 0,97



5. Panas yang dihasilkan CP-01 Tin



= 383 K



Tout



= 388,56 K



ΔT



= 5,56 K



Senyawa H2O HCl Jumlah



m, kg/jam 6216,1206 3650,7375 9866,8582



BM 18 36,5



6. Jumlah air pendingin yang dibutuhkan T air in



= 28 oC



mol, n 345,3400 100,0202 445,3602



ΔT.Cp 4,9871 5,3926



Q, kkal/jam 1722,2513 539,3646 2261,6159



172



T air out = 48 oC = 20 oC



ΔT m air =



Q duty CP 01 Cp × ΔT



m air =



2261,6159 kkal/jam 1 kkal / kg o C × 20 o C



= 113,0808 kg/jam



COMPRESSOR-02 (CP-02) Fungsi



: Mengalirkan dan menaikkan tekanan dari top KOD-02 menuju R-01



Tipe



: Centrifugal Multi Stage Compressor



Jumlah



: 1 buah



Gambar :



Gambar L.3.3. Compressor-02 (CP-02) Data Design



173



Laju alir, W



: 4480,9052 kg/jam = 164,6449 lb/min



Tekanan masuk, Pin



: 1,5 atm = 3174,325 lbf/ft2



Tekanan keluar, Pout



: 2,5 atm = 5290,541 lbf/ft2



Temperatur masuk, Tin



: 110 oC = 383 K



Berat Molekul, BM



: 32 kg/kmol = 32 lb/lbmol



Densitas, ρ



:



Dari Tabel 2-30 Perry’s Chemical Engineer’s Handbook diketahui : Senyawa CH3OH ρ CH OH = 3 ρ CH 3OH =



A 2,288



B 0,2685



C 512,65



D 0,2453



A



D) ( B 1 + (1 - T / C) 2,288



0,2534 ) ( 0,2685 1 + (1 - 383 / 512,65)



= 21,7822 kmol/m3 = 392,0787 kg/m3 = 24,4774 lb/ft3 7. Kapasitas Kompressor Volume, V = =



W ρ 164,6449 lb/min 24,4774 lb/ft3



= 6,7264 ft3/min Safety Factor = 10 % Maka, kapasitas kompressor, V = 1,1 x V = 1,1 x 6,7264ft3/min = 7,3990 ft3/min 8. Jumlah stage, Ns Cp CH3OH = -61,68195 + 0,8020923 T – -0,00277988 T2 + 3,356.10-6 T3 Cp = 26,2974 kkal/kmol.K Cv = Cp – R , dimana R = 1,985765 kkal/kmol.K



174



= 26,2974 kkal/kmol.K - 1,985765 kkal/kmol.K = 24,3116 kkal/kmol.K k =



Cp 26,2974 kkal/kmol.K = = 1,0817 Cv 24,3116 kkal/kmol.K



Wm = k RT1 ( k 1



( P2 / P1 ) k -1 k - 1



)



1,0817 Wm = 1,985765 kkal/kmol.K × 382 × ( 1,0817 -1



( 2,5 atm / 1,5 atm)



1,0817 -1



1,0817 - 1 )



= 396,0977 kkal/kmol = 554363,1128 ft-lb/lmol Ns = Ns =



Wm BM × 10000 ft - lb/lmol



554363,1128 ft - lb/lmol 32 lb/lbmol × 10000 ft - lb / lb



= 2,2648 ≈ 3 stage 9. Temperatur keluaran gas kompressor, Tout P Tout = ( 2 P ) k - 1 k 1 Tout = (



2,5 atm



1,0817 - 1 1,0817 1,5 atm )



= 387,96 K = 114,96 oC 10. Tenaga yang dibutuhkan, P hp =



3,03.10 - 5 k Ns P1 V ( P2 / P1 ) k -1 k - 1 k -1



hp =



3,03.10 - 5 1,0925 × 3 × 3174,325 lbf/ft 2 × 8,2723 ft 2 /min × ( 2,5 / 1,5)1,0817-1 1,0817 - 1 1,0817-1



(



(



= 0,3659 hp Diketahui friction loss max 3 % Power yang dibutuhkan P =



)



0,3659 hp = 0,3772 hp 0,97



)



175



11. Panas yang dihasilkan CP-02 Tin



= 383 K



Tout



= 388,56 K



ΔT



= 5,56 K



Senyawa CH3OH



m, kg/jam 4480,9052



BM 32



12. Jumlah air pendingin yang dibutuhkan T air in



= 28 oC



T air out = 48 oC ΔT



= 20 oC



m air =



Q duty CP 02 Cp × ΔT



m air =



8090,0492 kkal/jam 1 kkal / kg o C × 20 o C



= 404,5025 kg/jam



mol, n ΔT.Cp 140,0283 -57,7744



Q, kkal/jam -8090,0492



176



COMPRESSOR-03 (CP-03) Fungsi



: Menaikkan tekanan produk top Absorber sebelum dimasukkan kedalam tangki penyimpanan T-03



Tipe



: Centrifugal Multi Stage Compressor



Jumlah



: 1 buah



Gambar :



Gambar L.3.4. Compressor-03 (CP-03) Data Design Laju alir, W



: 5000,0317 kg/jam = 183,7195 lb/min



Tekanan masuk, Pin



: 1,8 atm = 3809,19 lbf/ft2



Tekanan keluar, Pout



: 10 atm = 21162,17 lbf/ft2



Temperatur masuk, Tin



: 30 oC = 303 K



Densitas, ρ



:



Dari Tabel 2-30 Perry’s Chemical Engineer’s Handbook diketahui : senyawa HCl



A 3.342



B 0.2729



C 324.65



D 0.3217



177



CH2Cl CH3OCH3  HCl 



 HCl 



1.817 1.5693



0.25877 0.2679



416.25 400.1



0.2833 0.2882



A  1  ( 1 - T / C ) D   B



3,342  1  ( 1 - 383 / 324,65 )0,3217   0,2729 



= 5,2175 kmol/m3 = 190,4376054 kg/m3



 CH 2 Cl 



1,817  1  ( 1 - 383 / 416,25 )0,2833   0 , 25877 



= 2,95711291 kmol/m3 = 149,334202 kg/m3



CH 3OCH 3 



1,5693  1  ( 1 - 383 / 400,1) 0,2882   0 , 2679 



= 2,51484819 kmol/m3 = 115,6830167 kg/m3



Senyawa m, kg/jam HCl 2.1304 CH2Cl 4950.0051 CH3OCH3 47.89624887 Jumlah 5000.0317



BM 36.5 50.5 46



Xi 0.000426077 0.989994734 0.009579189 1



ρcamp = 149,0293642 kg/m3 = 9,303868 lb/ft3 BMcamp = 50,0103 kg/kmol 1. Kapasitas Kompressor Volume, V = =



W ρ 183,7195 lb/min 9,303868 lb/ft3



= 19,7466 ft3/min



ρ,kg/m3 190.4376 149.3342 115.683



ρi. Xi 0.0811 147.8401 1.1081 147.9212



BM . Xi 0.0156 49.9947 0.4406 50.0103



178



Safety Factor = 10 % Maka, kapasitas kompressor, V = 1,1 x V = 1,1 x 19,7466 ft3/min = 21,7212 ft3/min 2. Jumlah stage, Ns Cp HCl = 4,232994 + 0,215979 T – 0,0013482160 T2 + 0,00000271 T3 Cp CH2Cl = 1,909831 + 0,1907184 T – 0,000837771 T2 + 0,000001317238 T3 Cp CH3OCH3 = 9,335364 + 0,1929287 T – 0,000883945 T2 + 0,00000151118 T3 Senyawa HCl CH2Cl CH3OCH3 Jumlah



m, kg/jam 2.1304 4950.0051 47.8963 5000.0317



BM 36.5 50.5 46



mol, n Xi Cp Cp.xi 0.0584 0.000589 -5.9907 -0.0035 98.0199 0.988906 19.4257 19.2102 1.0412 0.010505 28.6769 0.3012 99.1195 1 19.5079



Cpcamp = 19,5079 kkal/kmol.K Cvcamp = Cpcamp – R , dimana R = 1,985765 kkal/kmol.K = 19,5079 kkal/kmol.K - 1,985765 kkal/kmol.K = 17,5221 kkal/kmol.K k 



Cp 19,5079 kkal / kmol.K   1,1133 Cv 17,5221 kkal / kmol.K



Wm = k RT1 ( k- 1



Wm =



( P2 / P1 ) k -1 k - 1



1,1133 1 1,1133 1,1133 1  1,985765kkal / kmol.K  10 / 1,8    1,1133  1



= 1127,2981kkal/kmol = 1577723,020 ft-lb/lmol Ns =



)



Wm BM × 10000 ft - lb/lmol



1577723,020 ft  lb / lmol



Ns = 50,0103 x 1000 ft  lb / lmol



179



= 3,1548 ≈ 4 stage



3. Temperatur keluaran gas kompressor, Tout P Tout = ( 2 P ) k - 1 k 1 1,1133  1 1,1133



 Tout = 10 atm  1 , 8 atm   = 316,5152 K = 43,5152 oC



4. Tenaga yang dibutuhkan, P hp =



3,03.10 - 5 k Ns P1 V ( P2 / P1 ) k -1 k - 1 k -1



(



3, 03.10 -5 1,0925 4 1,1133-1



hp 



)



 21162,17 lbf/ft 2  19,7466 ft 2 /min   (10 / 1,8)1,1133-1 1,1133 - 1



= 4,3942 hp Diketahui friction loss max 3 % Power yang dibutuhkan P=



4,3942 hp = 4,5301 hp 0,97



5. Panas yang dihasilkan CP-03 Tin



= 303 K



Tout



= 316,52 K



ΔT



= 13,52 K



Senyawa HCl CH2Cl CH3OCH3 Jumlah



m, kg/jam BM 2.1304 36.5 4950.0051 50.5 47.8962 46 5000.0317  



mol, n ΔT.Cp 0.0584 5.6960 98.0199 4.3377 1.0412 11.7851 99.11949  



Q, kkal/jam 0.3325 425.1769 12.2709 437.7803



180



6. Jumlah air pendingin yang dibutuhkan T air in



= 28 oC



T air out = 48 oC = 20 oC



ΔT



Qduty CP  03



mair mair



=



=



Cp x T



437.7803 kkal / jam 1 kkal / kg O C x 20 OC



= 21,8890 kg/jam



181



CONDENSER – 01 (CD-01) Fungsi



: Mengkondensasikan uap HCl dari top Stripper



Type



: Double Pipe Heat Exchanger



Bahan



: Stainless Steel



Gambar



:



Return bend



Gland



Gland



Gland



Tee



Return Head



Gambar L.3.5. Condenser-01 (CD-01) Fluida Panas



: campuran uap keluar top stripper



Wt



= 82,6729



kg/jam = 182,2623



lb/jam



T1



= 69,96



o



= 157,93



o



T2



= 30



o



= 86



o



Fluida Dingin



C C



F F



: Cooling water



W2



= 42,7660



kg/jam = 94,2829



lb/jam



t1



= 28



o



= 82,4



o



t2



= 48



o



= 118,4



o



C C



F F



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern. 1) Beban panas PH – 01 Q =



855,3209



kkal/hr = 3394,1879 Btu/hr



182



2) Menghitung ΔT



No 1 2 3



Fluida Panas 157,93 86



LMTD (ΔT)



Temperatur Tinggi Temperatur Rendah Selisih



=



ΔT2 - ΔT1 ln(ΔT2 /ΔT1 )



=



39,53 - 3,6 ln(39,53 / 3,6)



Fluida Dingin 118,4 82,4



Selisih 39,53 3,6 -35,93



= 15,011 oF 3) Ta dan ta Ta = ½ (157,93 + 86) = 121,964 oF ta = ½ (118,4 + 82,4) = 100 oF Dari table 8, Kern, UD = 200 hingga 500 Btu / jam ft2 oF a) Asumsi UD A



= 250 Btu / jam ft2 oF



=



Q (U D .T )



=



3394,1879 Btu / jam (250 Btu / jam.ft 2 .o F × 15,011 o F)



= 0,9044 ft2 b) Karena A< 200 ft2, maka dipilih HE dengan jenis Double Pipe Heat Exchanger, dengan klasifikasi sebagai berikut (dari table 11 Kern) :



183



No 1 2 3 4 5



Annulus 1 40 1,049 in 1.320 in 0,344 ft2



Data Pipa IPS SN IDp ODp a’



4) Annulus : produk keluar reaktor b) Flow area, aa D2



= 1,049 in = 0,0874 ft



D1



= 1,320 in = 0,0700 ft



aa



=



 ( D2  D1 ) 4 2



2



2 2 = 3,14(0,0874 - 0,0700 ) 4



= 0,00215 ft2 Diameter Equivalent, De 2



De



D  D1 = 2 D1 =



2



0,0874 2 - 0,0700 2 0,0700



= 0,0392 ft c) Laju alir massa, Ga Ga



= W / aa =



182,2623 lb / jam 0,00215 ft 2



= 84685,984 lb / jam ft2 d) Bilangan Reynold, Rea Pada



Tc



= 121,964



o



F = 49,98 oC



Perhitungan viskositas campuran :



Inner 0,5 40 0,622 in 0,840 in 0,220 ft2



184



Dari fig 2-32, Perry Chemical Engineers Handbook Senyawa



Massa, Kg/jam 70,8807 11,7922 14347,7634



HCl H2O Jumlah



Fraksi, Xi 0,857 0,143 1,000



μ, cP 0,0150 0,0106



μi x Xi, cP 0,0129 0,0015 0,0144



μ = 0,0144cp = 0,0348 lb/ft jam (fig.2-32, Perry) Rea = De . Ga / μ =



0,0392



ft × 84685,984 lb / ft 2 hr 0,0348 lb / ft. jam



= 9,54.104 JH = 250 Perhitungan kapasitas panas, Cp : tabel B.2 , Felder, Elementary Principles of chemical Processes) Senyawa H2O HCl



a 33,46 29,13



b.102 0,668 -0,1341



c. 106 7,6 0,9715



d. 109 -3,593 -4,335



Cp = a + bT + cT 2 + dT 3



Cp ( H 2 O) = 33,46 + 0,668.10- 2 × 49,98 + 0,7604.10-5 × 49,98 2 + - 3,593.10- 9 × 49,98 3 = 33,7904 J / mol.o C



= 1,8772 J/g.oC = 0,4488 Btu / lb oF



Cp( HCl) = 29,13 + - 0,1341.10- 2 × 49,98 + 0,9715.10- 5 × 49,98 2 + - 4,335.10- 9 × 49,98 3 = 29,1942 J / mol.o C



= 0,7998 J/g.oC = 0,1912 Btu / lb oF



Perhitungan konduktivitas termal, k : 10,4 th edition ) k = μ (Cp + ( MW )  ( pers Coulson 6 10,4 k ( H 2 O) = 0,0106 (1,8772 J/g.oC + ( 18 g / mol)



= 0,0260 W/m.oC = 0,0151 lb/ft hr



185



10,4 k ( HCl) = 0,0150 (0,7998 J/g.oC + ( 36,5 g / mol)



= 0,0163 W/m.oC = 0,0094 lb/ft hr Senyawa



Massa, Xi Cp, Kg/jam Btu / lb oF HCl 70,8807 0,857 0,1912 H2O 11,7922 0,143 0,4488 Jumlah 14347,7634 1,000 Cp(campuran) = 0,9536 Btu / lb oF



Xi.Cp 0,6858 0,2678 0,9536



k(campuran) = 0,0177 lb/ft hr k e) ho = jH D ( e



=



250 ×



c.μ



1/ 3 ( μ 0,14 k) μw )



0,0177 lb/ft hr 0,9536 Btu / lb oF . 0,0348 lb/ft jam 1 / 3 ( ) 0,0392 ft 0,0177 lb/ft hr



= 59,9342 Btu/hr.ft2.oF 5) Inner Pipe : Fluida Dingin b) Flow area, ap Dp ap



= 0,622 in = 0,0518 ft =



 ID p 2 4



2 2 = 3,14 × 0,0518 ft 4



= 0,0021 ft2 c) Laju alir massa, Gp Gp



W



= a p =



94,2829 lb / jam 0,0021 ft 2



= 44703,847 lb / jam ft2 d) Perhitungan μ, Cp, k campuran Perhitungan viskositas campuran : T =100 oF



k, lb/ft hr 0,0094 0,0151



Xi.k 0,0140 0,0037 0,0177



186



Dari fiq 2-32 ,Perry didapat : μH2O = 0,7 lb/ft.hr ρH2O = 992,594 kg/m3 = 61,9674 lb/ft3 (tabel 2-92, Perry) Cp(H 2 O) = 75,4 J/mol.oC = 4,4353 J/g.oC = 3,97 Btu / lb oF k = (3,56 / 105 ) × Cp × (ρ 4 / BM)1 / 3  ( pers Coulson 6 th edition )



k H 2 O = (3,56 / 10 5 ) × 4,4353 J/g.oC × (992,594 4 (kg/m3) 4 / 18)1 / 3



= 0,5965 W/m.oC = 0,3452 lb/ft hr e) Bilangan Reynold, Rep Pada



o



tc



= 100



μ



= 0,7 lb/ft. jam



Rep



= =



F



G p ID p



 44703,847 lb / jam ft2 × 0,0518 ft 0,7 lb/ft. jam



= 3310,2135 f) jH



= 13



k g) hi = jH D ( e



=



13 ×



c.μ



1/ 3 ( μ 0,14 k) μw )



0,3452 lb/ft hr 3,97 Btu/lb. oF . 0,7 lb/ft. jam 1 / 3 ( ) 0,0518 ft 0,3452 lb/ft hr



= 299,8524 Btu/hr.ft2.oF h) Koefisien perpindahan panas, hio Untuk kondensasi steam



: hio = =



hi × D p OD inner 299,8524



….(Kern hal.164) Btu/hr.ft2 .oF × 0,0518 ft 0,07 ft



= 222,0335 Btu/hr.ft2.oF i) Clean everaal Coefficient, Uc Uc



=



h io x h o h io + h o 222,0335



= 222,0335



Btu/hr.ft2.oF x 59,9342 Btu/hr.ft2.oF Btu/hr.ft2.oF + 59,9342 Btu/hr.ft2.oF



187



= 47,1948 Btu / jam ft2 oF j) Design overall Coefficient, UD 1 / UD



= 1 / U c + Rd



Rd



= 0,002



1/UD



=



47,1948



1 + 0,002 Btu / jam ft2 oF



= 0,0232 jam.ft2.oF/Btu = 43,1243 Btu / jam ft2 oF



UD



k) Required Surface, A A



= =



Q U D x T 3394,1879 43,1243



Btu/hr Btu / jam ft2 o F x 15,011 o F



= 5,2432 ft2 l) Required Length, L L



= A / a”



a”



= 0,22 ft



L



=



…(Tabel 11 Kern)



5,2432 ft2 0,622 ft



= 23,8326 ft Diambil panjang 1 hairpin



= 2 x 12 ft



Jumlah hairpin yang diperlukan =



23,8326 24 ft



ft



=1



Actual Length



= 1 x 24 ft = 24 ft



Actual Surface



= 24 ft x 0,22 ft = 5,28 ft2



Actual Design Coefficient, Ud Ud



Q



= A .T act



188



3394,1879 Btu/hr 5,28 ft 2 × 15,011 oF



=



= 42,8234 Btu / hr. ft2 oF Rd



UD - Ud UD × Ud



=



43,1243



= 43,1243



Btu / jam ft2 oF - 42,8234 Btu / hr. ft2 oF Btu / jam ft2 oF × 42,8234 Btu / hr. ft2 oF



= 0,00016 hr ft2 oF 6) Pressure Drop b) Annulus : Fluida Panas 1) De’



= (D2 – D1) = (0,0874 – 0,070) ft = 0,0174 ft



Rea



= De . Ga / μ =



0,0174



ft × 84685,984 lb / jam ft2 0,0348 lb/ft jam



= 42420 f



=



0,0035 +



=



0,0035 +



0,264 0,42 R ea



….(Pers 3.47b, Kern)



0,264 (42420 4 ) 0,42



= 0,0065 Densitas pada Tc = 121,964 oF = 49,98 oC = 322,98 K : Densitas HCl (tabel 2-30, Perry’s Chemical Engineers Handbook) : ρ H2O = 280,699 kg/m3 Densitas H2O (dari tabel 2-28, Perry’s Chemical Engineers Handbook) : ρ H2O = 988,037 kg/m3, maka densitas campuran adalah :



189



Senyawa



Massa, Kg/jam 70,8807 11,7922 14347,7634



HCl H2O Jumlah



Fraksi, Xi 0,857 0,143 1,000



ρ, kg/m3 280,699 988,037



ρi x Xi, kg/m3 240,6611 140,9307 381,5918



ρ campuran = 381,5918 kg/m3 = 23,8227 lb/ft3 2



2) ΔFa



4. f .G a L = 2.g . 2 De =



4 × 0,0065 × 84685,984 lb / jam ft2 2 × 24 ft 2 × 4,18.108 ft / hr 2 × ( 23,8227 lb/ft3) 2× 0,1372 ft



= 0,5421 ft 3) Va



= Ga / 3600 . ρ =



…(hal 115 Kern)



84685,984 lb /det ft2 3600 × 23,8227 lb/ft3



= 0,9875 ft/det Fl



=n



V2 2.g



=1x



…(hal. 112 Kern)



0,9875 9 ft/det 2 2 × 32,2 ft / det 2



= 0,0151 ft 4) ΔPa



=



(ΔFa + FL ).ρ 144



=



(0,5421 ft + 0,0151 ft ).23,8227 lb/ft3 144



…(hal. 114 Kern)



= 0,0922 psi c) Inner Pipe : Fluida Panas 1) Rep f



= 3310,2135 =



0,0035 +



=



0,0035 +



0,264 0,42 R ea 0,264



3310,2135 0,42



…(Pers. 3.47b Kern)



190



= 0,0122 Pada t = 100 oF,



ρ = 61,9906 lb/ft3 2



2) ΔFp



= =



4. f .G p L 2 . g . 2 .D 4 × 0,0122 × 44703,847 lb / jam ft2 2 × 24 ft 2 × 4,18.108 × 61,9906 lb/ft3 2 × 0,0174 ft



= 0,0000000189 ft 3) ΔP



= ( ΔFp . ρ)/144 =



0,0000000189 ft × 61,9906 lb/ft3 144



= 8,15.10-9 psi



hio = 222,0335



0,0922 10



SUMMARY houtside Uc = 47,1948 Btu / jam ft2 oF UD = 43,1243 Btu / hr. ft2 oF Rd calculated = 0,00016 hr ft2 oF Rd required = 0,002 hr ft2 oF Calculated ΔP, psi Allowable ΔP, psi



ho = 59,9342



8,15.10-9 10



191



COOLER – 01 (C-01) Fungsi



: Menurunkan temperatur produk keluaran reaktor



Type



: Double Pipe Heat Exchanger



Bahan



: Stainless Steel



Gambar



:



Return bend



Gland



Gland



Gland



Tee



Return Head



Gambar L.3.6. Cooler-01 (C-01) Fluida Panas



: Campuran produk keluar reaktor



Wt



= 14347,7634 kg/jam = 31631,37



T1



= 125



o



= 257



o



T2



= 58



o



= 136,4



o



Fluida Dingin



C C



lb/jam F F



: Cooling water



W2



= 16609,1102 kg/jam = 36616,78



t1



= 28



o



= 82,4



o



t2



= 48



o



= 118,4



o



C C



lb/jam F F



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern.



192



1) Beban panas PH – 01 Q =



334477,238



kkal/hr = 1327313,048 Btu/hr



2) Menghitung ΔT



No 1 2 3



Fluida Panas 257 136,4



LMTD (ΔT)



Temperatur Tinggi Temperatur Rendah Selisih



=



ΔT2 - ΔT1 ln(ΔT2 /ΔT1 )



=



138,6 - 54 ln(138,6 / 54)



Fluida Dingin 118,4 82,4



Selisih 138,6 54 -84,6



= 89,85 oF



3) Ta dan ta Ta = ½ (257 + 136,4) = 196,7 oF ta = ½ (118,4 + 82,4) = 100 oF Dari table 8, Kern, UD = 50 hingga 150 Btu / jam ft2 oF a) Asumsi UD A



= 100 Btu / jam ft2 oF



=



Q (U D .T )



=



1327313,048 Btu / jam 2 o (100 Btu / jam.ft . F × 89,85 o F)



= 147,7224 ft2 b) Karena A< 200 ft2, maka dipilih HE dengan jenis Double Pipe Heat Exchanger, dengan klasifikasi sebagai berikut (dari table 11 Kern) :



193



No 1 2 3 4 5



Annulus 4 40 4,026 in 4,500 in 1,178 ft2



Data Pipa IPS SN IDp ODp a’



4) Annulus : produk keluar reaktor a) Flow area, aa D2



= 4,026 in = 0,3355 ft



D1



= 2,380 in = 0,1983 ft



aa



 ( D2  D1 ) = 4 2



2



2 2 = 3,14(0,3355 - 0,1983 ) 4



= 0,057 ft2 Diameter Equivalent, De 2



De



D  D1 = 2 D1 =



2



0,3355 2 - 0,1983 2 0,1983



= 0,3692 ft d) Laju alir massa, Ga Ga



= W / aa =



31631,37 lb / jam 0,057 ft 2



= 550293,04 lb / jam ft2 e) Bilangan Reynold, Rea



Inner 2 40 2,067 in 2,380 in 0,622 ft2



194



Pada



Tc



= 196,7



o



F



Perhitungan viskositas campuran : Dari fig 2-32, Perry Chemical Engineers Handbook Senyawa HCl CH3Cl CH3OCH3 CH3OH H2O Jumlah



Massa, Kg/jam 73,0111 4950,0051 47,8962 1277,6301 7999,7634 14347,7634



Fraksi, Xi 0,0051 0,3450 0,0033 0,0890 0,5575 1,0000



μ, cP 0,0175 0,0130 0,0116 0,0120 0,0125



μ = 0,01265 cp = 0,0306 lb/ft jam (fig.2-32, Perry) Rea = De . Ga / μ =



0,3692 ft × 550293,04 lb / ft 2 hr 0,0306 lb / ft. jam



= 6,639.106 f) ho = 1500 Btu/hr.ft2oF 5) Inner Pipe : Fluida Dingin a) Flow area, ap Dp



= 2,067 in = 0,1723 ft



ap



 ID p = 4



2



2 2 = 3,14 × 0,1723 ft 4



2



= 0,0233 ft



b) Laju alir massa, Gp Gp



W



= a p



μi x Xi, cP 0,00089 0,04485 0,00004 0,01069 0,00696 0,01265



195



36616,78 lb / jam 0,0233 ft 2



=



= 1572142,802 lb / jam ft2 c) Perhitungan μ, Cp, k campuran Perhitungan viskositas campuran : T =100 oF Dari fiq 2-32 ,Perry didapat : μH2O = 0,7 lb/ft.hr ρH2O = 992,594 kg/m3 = 61,9674 lb/ft3 (tabel 2-92, Perry) Cp(H 2 O) = 75,4 J/mol.oC = 4,4353 J/g.oC = 3,97 Btu / lb oF k = (3,56 / 105 ) × Cp × (ρ 4 / BM)1 / 3  ( pers Coulson 6 th edition )



k H 2 O = (3,56 / 10 5 ) × 4,4353 J/g.oC × (992,594 4 (kg/m3) 4 / 18)1 / 3



= 0,2059 W/m.oC = 0,1192 lb/ft hr d) Bilangan Reynold, Rep Pada



o



tc



= 100



μ



= 0,7 lb/ft. jam



Rep



= =



F



G p ID p



 1572142,802 lb / jam ft2 × 0,1723 ft 0,7 lb/ft. jam



= 386859,4253 e) jH



= 730



k c . μ )1 / 3 ( μ 0,14 f) hi = jH D ( k μw ) e



=



730 ×



0,1192 lb/ft hr 3,97 Btu/lb. oF . 0,7 lb/ft. jam 1 / 3 ( ) 0,1723 ft 0,1192 lb/ft hr



= 5066,8368 Btu/hr.ft2.oF g) Koefisien perpindahan panas, hio Untuk kondensasi steam



: hio = =



hi × D p OD inner



….(Kern hal.164)



5066,8368 Btu/hr.ft2.oF × 0,1723 ft 0,1983 ft



= 4400,4839 Btu/hr.ft2.oF



196



h) Clean everaal Coefficient, Uc Uc



=



h io x h o h io + h o 4400,4839



Btu/hr.ft2 .oF x 1500 Btu/hr.ft2 .oF Btu/hr.ft2 .oF + 1500 Btu/hr.ft2 .oF



= 4400,4839



= 1118,6753 Btu / jam ft2 oF i) Design overall Coefficient, UD 1 / UD



= 1 / U c + Rd



Rd



= 0,002



1/UD



=



1 + 0,002 1118,6753 Btu / jam ft2 oF



= 0,0029 jam.ft2.oF/Btu = 345,5527 Btu / jam ft2 oF



UD



j) Required Surface, A A



=



Q U D x T



=



1327313,048 Btu/hr 345,5527 Btu / jam ft2 o F x 89,85 o F



= 42,7496 ft2 k) Required Length, L L



= A / a”



a”



= 0,622 ft



L



=



…(Tabel 11 Kern)



42,7496 ft2 0,622 ft



= 68,7292 ft Diambil panjang 1 hairpin



= 2 x 12 ft



Jumlah hairpin yang diperlukan = Actual Length



68,7292 ft 24 ft



= 2 x 24 ft = 72 ft



=3



197



= 72 ft x 0,622 ft = 44,784 ft2



Actual Surface



Actual Design Coefficient, Ud Q



Ud



= A .T act =



1327313,048 Btu/hr 44,784 ft 2 × 89,85 oF



= 329,8552 Btu / hr. ft2 oF Rd



=



UD UD



Ud Ud



345,5527



Btu / jam ft2 oF - 329,8552 Btu / hr. ft2 oF



= 345,5527 Btu / jam ft2 oF × 329,8552 Btu / hr. ft2 oF = 0,00014 hr ft2 oF 6) Pressure Drop l) Annulus : Fluida Panas 1) De’



= (D2 – D1) = (0,3355 – 0,1983) ft = 0,1372 ft



Rea



= De . Ga / μ =



0,1372 ft × 550293,04 lb / jam ft2 0,0306 lb/ft jam



= 2466414,221 f



=



0,0035 +



=



0,0035 +



0,264 0,42 R ea



0,264 (2466414,2214 ) 0,42



= 0,0004 Densitas pada Tc = 196,7 oF = 91,5 oC = 364,5 K : Dari tabel 2-30, Perry’s Chemical Engineers Handbook



….(Pers 3.47b, Kern)



198



Senyawa



A



B



C



D



BM



HCl CH3Cl CH3OCH3 CH3OH



3,342 1,817 1,5693 2,288



0,2729 0,2588 0,2679 0,2685



324,65 416,25 400,1 512,64



0,3217 0,2883 0,2882 0,2453



36,5 50,5 46 32



ρ, kmol/m3 11,6574 7,3873 6,0658 9,4698



ρ, kg/m3 425,4936 272,0596 279,0291 303,0343



Densitas H2O (dari tabel 2-28, Perry’s Chemical Engineers Handbook) : ρ H2O = 964,3080 kg/m3, maka densitas campuran adalah :



Senyawa HCl CH3Cl CH3OCH3 CH3OH H2O Jumlah



Massa, Kg/jam 73,0111 4950,0051 47,8962 1277,6301 7999,7634 14347,7634



Fraksi, Xi 0,0051 0,3450 0,0033 0,0890 0,5575 1,0000



ρ, kg/m3 425,4936 373,0596 279,0291 303,0343 964,3080



ρi x Xi, kg/m3 2,1652 128,7063 0,9315 26,9844 537,6247 696,4121



ρ campuran = 964,3080 kg/m3 = 43,4768 lb/ft3 2



2) ΔFa



4. f .G a L = 2.g . 2 De =



4 × 0,0004 × 550293,04 lb / jam ft2 2 × 72 ft 2 × 4,18.108 ft / hr 2 × 43,4768 lb/ft3 2× 0,1372 ft



= 1,6278 ft



3) Va



= Ga / 3600 . ρ =



…(hal 115 Kern)



550293,04 lb /det ft2 3600 × 43,4768 lb/ft3



= 3,5159 ft/det Fl



=n



V2 2.g



…(hal. 112 Kern)



199



=3x



3,5159 ft/det 2 2 × 32,2 ft / det 2



= 0,5758 ft 4) ΔPa



=



(ΔFa + FL ).ρ 144



=



(1,6278 ft + 0,5758 ft ).43,4768 lb/ft3 144



…(hal. 114 Kern)



= 0,6653 psi m) Inner Pipe : Fluida Panas 1) Rep f



= 386859,4253 =



0,0035 +



=



0,0035 +



0,264 0,42 R ea



…(Pers. 3.47b Kern)



0,264 386859,425 30,42



= 0,0047 Pada t = 100 oF,



ρ = 61,9906 lb/ft3 2



2) ΔFp



= =



4. f .G p L 2 . g . 2 .D 4 × 0,0047 × 1572142,802 lb / jam ft2 2 × 72 ft 2 × 4,18.108 × 61,9906 lb/ft32 × 0,0874 ft



= 0,0033 ft 3) ΔP



= ( ΔFp . ρ)/144 =



0,0033 ft × 61,9906 lb/ft3 144



= 0,0014 psi



hio = 2110,4821



SUMMARY houtside Uc = 1118,6753 Btu / jam ft2 oF UD = 345,5527 Btu / hr. ft2 oF Rd calculated = 0,00014 hr ft2 oF



ho = 8,1016



200



Rd required = 0,002 hr ft2 oF Calculated ΔP, psi Allowable ΔP, psi



0,6653 10



0,0014 10



HEATER – 01 (H-01) Fungsi



: Memanaskan produk keluaran bottom absorber



Type



: Double Pipe Heat Exchanger



Bahan



: Stainless Steel



Gambar



: Gambar L.3.7. Heater-01 (H-01)



Return bend



Gland



Gland



Gland



Tee



Return Head



Fluida Panas



: Saturated Steam



Wt



= 273,1055



kg/jam = 602,0937



lb/jam



T1



= 150



o



= 302



o



T2



= 150



o



= 302



o



Fluida Dingin



C C



F F



: Bahan baku methanol



W2



= 21230,6174 kg/jam = 46805,4437 lb/jam



t1



= 58



o



= 136,4



o



t2



= 75



o



= 167



o



C C



F F



201



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern. 1) Beban panas H – 01 Q =



137850,8063



kkal/hr = 547036,2494 Btu/hr



2) Menghitung ΔT



No 1 2 3



Fluida Panas 302 302



LMTD (ΔT)



Temperatur Tinggi Temperatur Rendah Selisih



=



ΔT2 ΔT1 ln(ΔT2 /ΔT1 )



=



135 - 165,6 ln(135/165,6)



Fluida Dingin 167 136,4



= 149,948 oF



3) Ta dan ta Ta = ½ (302 + 302) = 302 oF ta = ½ (167 + 136,4) = 151,7 oF Dari table 8, Kern, UD = 100 hingga 200 Btu / jam ft2 oF Trial UD a)



: Steam & Heavy Organik



Asumsi UD A



=



= 150 Btu / jam ft2 oF



Q (U D .T )



Selisih 135 165,6 30,6



202



=



547036,2494 Btu/hr (150 Btu / jam.ft 2 .o F × 149,948 o F)



= 24,3212 ft2 b) Karena A< 200 ft2, maka dipilih HE dengan jenis Double Pipe Heat Exchanger, dengan klasifikasi sebagai berikut (dari table 11 Kern) : No 1 2 3 4 5



Annulus 2 40 2,067 in 2,380 in 0,622 ft2



Data Pipa IPS SN IDp ODp a’



Inner 1 40 1,049 in 1,320 in 0,344 ft2



4) Annulus : Fluida panas, steam a) Flow area, aa D2



= 1,61 in = 0,1342 ft



D1



= 1,320 in = 0,1100 ft



aa



=



 ( D2 2  D1 2 ) = 4



3,14(0,1342 2 - 0,1100 2 ) = 4



Diameter Equivalent, De 2



De



D  D1 = 2 D1 =



2



0,1342 2 - 0,1100 2 0,1100



= 0,054 ft b) Laju alir massa, Ga Ga



= W / aa =



602,0937 lb / jam 0,0046 ft 2



= 129984,43 lb / jam ft2



0,0046 ft2



203



c) Bilangan Reynold, Rea Pada



o



Tc



=



302



μ



=



0,0144 cp



Rea



= De . Ga / μ =



F = 0,03484 lb/ft jam (fig.2-32, Perry)



0,054 ft × 129984,43 lb / ft 2 hr 0,03484 lb / ft. jam



= 2,002.105 d) jH



= 440



e) Pada



Tc



=



302



k



=



0,0208 lb/ft hr



…(Tabel 5, Kern)



Cp



=



0,4597 Btu / lb oF



…(Fig 4, Kern)



o



F = 150 oC



…(Fig 24, Kern)



Perhitungan Cp : tabel B.2 , Felder, Elementary Principles of chemical Processes) b.102 0,668



a 33,46



c. 105 0,7604



d. 109 -3,593



Cp = a + bT + cT 2 + dT 3 Cp = 33,46 + 0,668.10- 2 × 150 + 0,7604.10- 5 × 150 2 + - 3,593.10- 9 × 1503 Cp = 34,623 J / mol.o C



= 1,9233 J/g.oC = 0,4597 Btu / lb oF



Perhitungan nilai konduktivitas termal, k : 10,4 th edition ) k = μ (Cp + ( MW )  ( pers Coulson 6 10,4 k = 0,0144cp (1,9233 J/g.oC + ( 18 g / mol)



k = 0,036 W/m.oC = 0,0208 lb/ft hr k c . μ )1 / 3 ( μ 0,14 ho = jH D ( k μw ) e



=



440 ×



0,0208 lb/ft hr 0,4597 Btu / lb oF . 0,03484 lb/ft jam 1 / 3 ( ) 0,054 ft 0,036 lb/ft hr



204



= 156,58 Btu/hr.ft2.oF 5) Inner Pipe : Fluida Dingin a) Flow area, ap Dp



= 1,049 in = 0,0874 ft



ap



=



 ID p 4



2



2 2 = 3,14 × 0,0874 ft 4



= 0,006 ft2 b) Laju alir massa, Gp Gp



W



= a p =



46805,4437 lb / jam 0,006 ft 2



= 7802579,9117 lb / jam ft2 T = 152 oF = 66,5 oC Perhitungan viskositas campuran : Dari fiq 2-32 ,Perry didapat : μCH3OH = 0,36 cP = 0,8709 lb/ft.hr μH2O = 0,38 cP = 0,9193 lb/ft.hr μHCl = 0,45 cP = 1,0886 lb/ft.hr



Senyawa HCl CH3OH H2O Jumlah



Massa, Kg/jam 70,8807 1277,6301 19882,1066 21230,6174



Fraksi, Xi 0.003 0,936 0,060 1,000



μ, lb/ft.hr 1,0886 0,8709 0,9193



μi x Xi, lb/ft.hr 0,0524 0,8609 0,0036 0,9169



205



Perhitungan kapasitas panas campuran : tabel B.2 , Felder, Elementary Principles of chemical Processes) Senyawa H2O HCl CH3OH



a 75,4 29,13 42,93



b.102 -0,1341 8,301



c. 106 0,9715 -37,72



d. 109 -4,335 100,8



Cp = a + bT + cT 2 + dT 3



Cp(H 2 O) = 75,4 J/mol.oC = 4,4353 J/g.oC = 1,0601 Btu / lb oF Cp( HCl) = 29,13 + - 0,1341.10 - 2 × 66,5 + 0,9715.10 - 5 × 66,5 2 + - 4,335.10 - 9 × 66,5 3 = 29,0825 J / mol.o C



= 0,7968 J/g.oC = 0,1904 Btu / lb oF



Cp(CH 3OH ) = 42,93 + 8,301.10 - 2 × 66,5 = 46,8117 J / mol.o C



+ - 37,72.10 - 5 × 66,5 2 + 100,8.10 - 9 × 66,5 3



= 1,4629 J/g.oC = 0,3496 Btu / lb oF



Perhitungsn densitas campuran : ρ(H2O) = 979,736 kg/m3 = 61,1649 lb/ft3 (tabel 2-28, Perry Chemical Engineer Handbook) ρ(HCl) = 446,979 kg/m 3 = 27,9048 lb/ft3 (tabel 2-30, Perry Chemical Engineer Handbook) densitas methanol : P



: 1 atm = 1,01325 bar



T



: 66,5 oC = 339,5 K



Tc



: 512,64 K



Pc



: 80,97 bar



Zc



: 0,224



R



: 0,08314 L.bar/mol.K



Vs =



2/7) R.Tc Zc (1+ (1 T / Tc ) Pc



Vs =



2/7) 0,08314 L bar/mol.K × 512,64 K 0,224(1+ (1 339,5 K / 512,64 K ) 80,97 bar



206



= 0,03936 mol/L ρ(CH3OH) = 813,0135 kg/m3 = 50,7562 lb/ft3



Massa, Kg/jam HCl 70,8807 H2O 1277,6301 CH3OH 19882,1066 Jumlah 21230,6174 ρ(campuran) = 60,4273lb/ft3 Senyawa



Xi 0.003 0,936 0,060 1,000



ρ, Kg/m3 27,9048 61,1649 50,7562



Xi.ρi 0,0932 57,2797 3,0544 60,4273



Perhitungan konduktivitas termal, k : k = (3,56 / 10 5 ) × Cp × ((ρ 4 / BM)1 / 3 )  ( pers Coulson 6 th edition )



k H 2 O = (3,56 / 105 ) × 4,4353 × ((61,16494 / 18)1 / 3 )  (pers Coulson 6 th edition )



= 0,02059 W/m.oC = 0,1192 lb/ft hr k HCl = (3,56 / 105 ) × 0,7968 × ((27,90484 / 36,5)1 / 3 )  (pers Coulson 6 th edition)



= 0,0163 W/m.oC = 0,0094 lb/ft hr k CH 3OH = (3,56 / 105 ) × 1,4629



× ((50,7562 4 / 32)1 / 3 )  (pers Coulson 6 th edition )



= 0,1245 W/m.oC = 0,0720 lb/ft hr Senyawa HCl H2O CH3OH Jumlah



Massa, Kg/jam 70,8807 1277,6301 19882,1066 21230,6174



Xi 0.003 0,936 0,060 1,000



Cp, Btu / lb oF 0,1904 1,0601 0,3496



Cp(campuran) = 1,0143 Btu / lb oF k(campuran) = 0,11598 lb/ft hr c) Bilangan Reynold, Rep Pada



o



tc



= 152



F



μ



= 0,9169 lb/ft. jam



Xi.Cp 0,0006 0,9927 0,0210 1,0143



k, lb/ft hr 0,0094 0,1192 0,0720



Xi.k 0,00006 0,00433 0,11159 0,11598



207



Rep



= =



G p ID p



 7802579,9117 lb / jam ft2 × 0,006 ft 0,9169 lb/ft. jam



= 743852,7278 JH = 1000 k d) hi = jH D ( e



= 1000 ×



c.μ



1/ 3 ( μ 0,14 k) μw )



0,11598 lb/ft hr 1,0143 Btu / lb oF . 0,9169 lb/ft jam 1 / 3 ( ) 0,0874 ft 0,11598 lb/ft hr



= 2655,7067 Btu/hr.ft2.oF e) Koefisien perpindahan panas, hio Untuk kondensasi steam



: hio = =



hi × D p OD inner



….(Kern hal.164)



2655,7067 Btu/hr.ft2.oF × 0,0874 ft 0,1100 ft



= 2110,4821 Btu/hr.ft2.oF



f) Clean everaal Coefficient, Uc Uc



=



h io x h o h io + h o 2110,4821



Btu/hr.ft2 .oF x 156,58 Btu/hr.ft2 .oF



= 2110,4821 Btu/hr.ft2 .oF + 156,58 Btu/hr.ft2 .oF = 145,77 Btu / jam ft2 oF g) Design overall Coefficient, UD 1 / UD



= 1 / U c + Rd



208



Rd



= 0,002



1/UD



=



1 + 0,002 145,77 Btu / jam ft2 oF



= 0,0088 jam.ft2.oF/Btu = 112,87 Btu / jam ft2 oF



UD



h) Required Surface, A A



=



Q U D x T



=



547036,2494 Btu/hr 112,87 Btu / jam ft2 o F x 149,948 o F



= 32,32 ft2 i) Required Length, L L



= A / a”



a”



= 0,344 ft



L



=



…(Tabel 11 Kern)



32,32 ft2 0,344 ft



= 93,96 ft Diambil panjang 1 hairpin



= 2 x 12 ft



Jumlah hairpin yang diperlukan =



93,96 ft 24 ft



=4



Actual Length



= 4 x 24 ft = 96 ft



Actual Surface



= 96 ft x 0,344 ft = 33,024 ft2



Actual Design Coefficient, Ud Ud



Q



= A .ΔT act =



547036,2494 Btu/hr 33,024 ft 2 × 149,948oF



= 110,47 Btu / hr. ft2 oF -



209



Rd



=



UD - Ud UD × Ud 112,87 Btu / jam ft2 oF - 110,47 Btu / hr. ft2 oF



= 112,87 Btu / jam ft2 oF × 110,47 Btu / hr. ft2 oF = 0,0002 hr ft2 oF 6) Pressure Drop j) Annulus : Fluida Panas 1) De’



= (D2 – D1) = (0,1342– 0,1100) ft = 0,0242 ft



Rea



= De . Ga / μ =



0,0242 ft ×129984,43 lb / jam ft2 0,03484 lb/ft jam



= 9,017.104 f



=



0,0035 +



=



0,0035 +



0,264 0,42 R ea



….(Pers 3.47b, Kern)



0,264 (9,017.10 4 ) 0,42



= 0,0057 Pada Tc = 302 oF, ρsteam = 955,4261 lb/ft3 = 59,6470 lb/ft3 (Tabel 2.30, Perry) 2



2) ΔFa



4. f .G a L = 2.g . 2 De =



4 × 0,0057 × 129984,43 lb / jam ft2 2 × 1344 ft 2 × 4,18.108 ft / hr 2 × 59,6470 lb/ft3 2× 0,0242 ft



= 0,514 ft 3) Va



= Ga / 3600 . ρ



…(hal 115 Kern)



210



=



129984,43 lb /det ft2 3600 × 59,6470 lb/ft3



= 0,605 ft/det Fl



=n =4



V2 2. g



…(hal. 112 Kern)



0,605 ft/det 2 2 × 32,2 ft / det 2



= 0,023 ft 4) ΔPa



=



(ΔFa + FL ).ρ 144



=



(0,514 ft + 0,023 ft ).59,6470 lb/ft3 144



…(hal. 114 Kern)



= 0,22 psi k) Inner Pipe : Fluida Panas 1) Rep f



= 743852,7278 =



0,0035 +



=



0,0035 +



0,264 0,42 R ea



…(Pers. 3.47b Kern)



0, 264 743852,7278 0,42



= 0,0044 Pada t = 152 oF,



ρ = 60,4273 lb/ft3 2



2) ΔFp



= =



4. f .G p L 2 . g . 2 .D 4 × 0,0044 × 7802579,9117 lb / jam ft2 2 × 96 ft 2 × 4,18.108 × (60,4273 lb/ft3) 2 × 0,087 ft



= 0,0139 ft 3) ΔP



= ( ΔFp . ρ)/144 =



0,0139 ft × 60,4273 lb/ft3 144



= 0,0058 psi



211



hio = 2110,4821



0,22 10



SUMMARY houtside Uc = 145,77 Btu / jam ft2 oF UD = 112,87Btu / hr. ft2 oF Rd calculated = 0,0002 hr ft2 oF Rd required = 0,002 hr ft2 oF Calculated ΔP, psi Allowable ΔP, psi



ho = 156,58



0,0058 10



KNOCK OUT DRUM -01 (KOD-01) Fungsi : Memisahkan fase gas dan liquid produk keluaran Vaporizer -01 Type : Liquid Knock Out Drum (Empty) Vertical Jumlah : 1 buah Bahan : Carbon Steel Gambar :



KOD



Gambar L.3.8. Knock Out Drum (KOD-01) Kondisi operasi Tekanan, P



: 1,5 atm



Temperatur, T : 110 oC Karakteristik fluida Fase gas Laju alir gas, wg = 9866,858 kg/jam



212



kg/m3



Densitas, ρg



= 2,2494



Volumetrik flowrate



= 4386,4399 m3/jam = 154900,0136 ft3/jam



Fase liquid Laju alir gas, wL = 2466,7145 kg/jam Densitas, ρL



= 1631,3521 kg/m3



Volumetrik flowrate



= 1,5121 m3/jam = 49,9200 ft3/jam







Kecepatan uap, U U = 0,14



U = 0,14



ρ ( L) ρg (



1



1631,3521 ) 2,2494



1



U = 3,7676







Diameter drum, D D=



Q v 1/ 4 π U



D=



4386, 4399 × 0, 25 × 3,14 × 3,7676 / 3600



D = 11,2809 ft / sec D = 3,4385 m







Tinggi space liquid Liquid Hold Up = 5-20 meni Liquid Hold Up yang dipilih adalah 10 menit L liq = L liq =



QL π



2 4D 49,9200



3,14 × 0,25 × 11,2809 2 L liq = 0,4997 ft







Tinggi total drum Vapor space minimum = 5,5 ft



213



L = Lliq + 5,5 ft L = 0,4997 ft + 5,5 ft = 5,9997 ft = 1,8702 m 



Kapasitas drum, Vd Vd = 4 π D 2 3 Vd = 4 × 3,14 × 3,4385 2 3 Vd = 170,2003 m 3 = 37585,3400 gallons







Tebal dinding drum, t t =



SE



P.R +C 0,6.P



.........(tabel. 4 hal 537, Peters)



P



= tekanan design



= 1,5 atm = 22,044 psi



R



= jari –jari drum



= 1,7192 m



S



= working stress yang diizinkan



= 13700 psi (tabel 4 hal 538,Peters)



E



= efisiensi pengelasan



= 0,85



C



= korosi yang diizinkan



= 0,003175



22,044 × 1,7192 + 0,003175 13700 × 0,85 0,6 × 22,044 t = 0,006433 m t =



OD = 2 t + D = (2 x 0,006433) + 3,4285 = 3,4513 m



Summary Diameter drum = 3,4285 m Tinggi drum = 1,8702 m Tebal dinging drum = 0,006433 m Kapasitas drum = 170,2003 m3



(hal. 636 Coulson) (table.23.2, Perry)



214



KNOCK OUT DRUM -02 (KOD-02) Fungsi



: Memisahkan fase gas dan liquid produk keluaran Vaporizer -02



Type



: Liquid Knock Out Drum (Empty) Vertical



Jumlah



: 1 buah



Bahan



: Carbon Steel



Gambar



:



KOD



Gambar L.3.9. Knock Our Drum (KOD-02) Kondisi operasi Tekanan, P



: 1,5 atm



Temperatur, T : 110 oC Karakteristik fluida Fase gas Laju alir gas, wg



= 4480,9052 kg/jam



Densitas, ρg



= 1,5625



Volumetrik flowrate



= 2867,7793 m3/jam = 101270,9775 ft3/jam



Fase liquid



kg/m3



215



Laju alir gas, wL



= 1120,226 kg/jam



Densitas, ρL



= 400 kg/m3



Volumetrik flowrate



= 2,8006 m3/jam = 92,4589 ft3/jam







Kecepatan uap, U U = 0,14



ρ ( L) ρg



1



U = 0,14



400 ( 1,5625



)



1



U = 2,2356







Diameter drum, D D=



Q v 1/ 4 π U



D=



101270,9775 × 0,25 × 3,14 × 2,2356 / 3600



D = 7,0263 ft / sec D = 2,1416 m







Tinggi space liquid Liquid Hold Up = 5-20 meni Liquid Hold Up yang dipilih adalah 10 menit L liq = L liq =



QL π



2 4D 92,4589



3,14 × 0,25 × 7,0263 2



L liq = 2,3858 ft







Tinggi total drum Vapor space minimum = 5,5 ft L = Lliq + 5,5 ft L = 2,3858 ft + 5,5 ft = 7,8858 ft = 2,4582 m







Kapasitas drum, Vd



216



Vd = 4 π D 2 3 4 Vd = × 3,14 × 2,1416 2 3 Vd = 41,1249 m 3 = 9081,6051 gallons







Tebal dinding drum, t t =



SE



P.R +C 0,6.P



.........(tabel. 4 hal 537, Peters)



P



= tekanan design



= 1,5 atm = 22,044 psi



R



= jari –jari drum



= 1,0708 m



S



= working stress yang diizinkan



= 13700 psi (tabel 4 hal 538,Peters)



E



= efisiensi pengelasan



= 0,85



C



= korosi yang diizinkan



= 0,003175



(hal. 636 Coulson) (table.23.2, Perry)



22,044 × 1,0708 + 0,003175 13700 × 0,85 0,6 × 22,044 t = 0,005204 m t =



OD = 2 t + D = (2 x 0,005204) + 2,1416 = 2,1520 m



Summary Diameter drum = 2,1416 m Tinggi drum = 2,4582 m Tebal dinging drum = 0,005204 m Kapasitas drum = 41,1249 m3 POMPA – 01 (P-01) Fungsi



: Untuk mengalirkan HCl dari tanki menuju Vaporizer-01



Tipe



: Pompa sentrifugal



217



Gambar



:



Gambar L.3.10. Pompa-01 (P-01) A. Kondisi Operasi Temperatur, T



= 52,93 oC



= 127,27 oF



Densitas, ρ



= 1324,011 kg/m3



= 82,655 lb/ft3



Viskositas, μ



= 0,364 cp



= 0,881 lb/ft jam



Laju alir, W



= 12333,5727 kg/jam = 452,231 lb/menit



Tekanan uap, Pv



= 88,617 lbf/ft2



B. Menentukan Ukuran Pipa a) Volume pompa, Vf Vf



W



=  =



452,231 lb / menit 82,655 lb / ft 3



= 6,018 ft3 /menit = 45,018 gpm b) Volumetrik flowrate, qf qf



=



ft 3 / menit s 60 menit



6,018



= 0,1003 ft3/det c) Menentukan Diameter Optimum, Dopt Dopt



= 3,9 x qf 0,45 x ρ0,13



(Pers 15 hal. 496, Peters)



= 3,9 x ( 0,1003)0,45 x ( 39,10 lb/ft3 )0,13 = 2,460 in



218



d) Ukuran Pipa  Suction pipe SN



= 40



IPS



=



3



in



L



=



4



m



= 13,123



ft



ID



=



3,068



in



=



0,255



ft



OD



=



3,500



in



=



0,291



ft



a



=



7,380



in2



=



0,051



ft2



 Discharge Pipe SN



= 40



IPS



=



3



in



L



=



4



m



= 13,123



ft



ID



=



3,068



in



=



0,255



ft



OD



=



3,500



in



=



0,291



ft



a



=



7,380



in2



=



0,051



ft2



C. Perhitungan pada Suction 1) Menentukan Suction Friction Loss a) Suction Velocity, υ υ



=



υ



=



= =



qf a



qf a



0,1003 ft 3 / det 0,051 ft 2



1,9730 ft/det



219



= 7102,6705 ft/jam Velocity head



=



(1,9730 ft / det ) 2 υ2 = = gc 32,174 lb / ft 3



0,121 ft.lb f / lb m



b) Bilangan Reynlod, Re Re



=



D 



=



82,655 lb / ft 3 x 7102,6705 ft / jam x 0,255 ft 0,881 lb / ftjam



169717,217 Meterial yang digunakan adalah commercial steel pipe Rough factor, έ



= 0,00015



έ/D



= 0,00059



(Fig 14-1, Peters)



Fanning friction factor, f Pada Re



= 169717,217 didapat f



= 0,005 (Fig.14-1, Peters)



c) Skin friction loss, Hfs Digunakan 1 elbow 90 o std dan 1 gate valve Le/D = 1 elbow 90 0 std dan 1 gate valve = 32 + 7 = 39 L



= L + Le x ID Suction



L



= 13,123 ft + 39 x 0,255 ft = 23,0543 ft



Hfs



=



2 fL v 2 x D gc



=



2 x 0,005 x 23,0543 ft (0,121 ft / det) 2 x 0,255 ft 32,174 lb / ft 3



= 0,110 ft lbf / lbm



…(Pers 5.64, Mc Cabe)



=



220



d) Sudden Contraction Friction Loss, Hfc 1 v2 x 2 g c



Hfc



= Kc .



α



= 1 ( untuk aliran turbulen )



Hfc



= Kc . 0,5 ( v2 / gc )



…(Pers. 5.71, Mc Cabe)



Kc



= 0,4 ( 1- Sb/Sa)



…(Pers. 5.65, Mc Cabe)



Sb



= Luas penampang melintang upstream b



Sa



= Luas penampang melintang downstream a



Dimana Sb> Sa, maka : Kc



= 0,4 ( 1 – 0 )



Hfc



= 0,4 x 0,5 x 0,121 ft lbf / lbm = 0,024 ft lbf/lbm



e) Fitting + Valve Friction Loss, Hff Hff



= Kf x ( v2 / 2 gc )



Dimana



:



Kf



= gate valve + elbow



(Pers II.7, Syarifudin Ismail)



= 1 gate valve + 4 elbow 90o



(Tabel II.7, Syarifudin Ismail)



= 0,2 + 0,9 * 4 = 3,8 Hff



= 3,8 x 0,5 x ( 0,121 ft lbf / lbm )



224



= 0,230 ft lbf /lbm f) Total Discharge Friction Loss, Hfd Hfd



= Hfs



+



Hfc



+



Hff



= (1,226 + 0,024 + 0,230) ft lbf / lbm = 0,480 ft lbf / lbm = 0,275 psi 2) Discharge Pressure Zc = 0



m



=0



ft



Zd = 8



m



= 26,246



ft



Static Suction Head (SH) = 26,246 ft Pd = 1



atm



= 2117,3762 ft2 lbf / lbm



=14,7 psi



g/gc = 1 lbf /lbm v =0 Pc



Pd ρ



+



g (Zc gc



Pc Pd g = + (Zd ρ ρ gc



Zd ) +



(v c



Zc ) +



vd )2 = Hf 2g α



(v d



vc )2 2gα



Hf



Pc 2117 ,3762 ft 2 lb f / lb m = + 26,246 ft (1 lb f / lb m ) ρ 82,665 lb / ft 3 Pc = 52,343 ft lb f / lb m ρ Pc = 4326,431 lb f / ft 2 Pc = 30,044 psi



3) Discharge Head, Hd Hd



=



Total disch arg e presure lb / ft 3



x 144in 2 / ft 2



0 ,275 ft lb f / lb m



225



Hd



=



30,044 psi 82,655 lb / ft 3



x 144 in 2 / ft 2



= 52,343 ft. lbf / lbm E. Menghitung Tenaga Pompa 1) Diffrential Pressure (Total ΔP) a. Suction pressure



= 20,238



psi



b. Discharge pressure



= 30,044



psi



9,806



psi



2) Total Head a. Suction head



= 35,259



ft



b. Discharge head



= 52,343



ft



17,084



ft



3) Efisiensi Pompa Kapasitas pompa



= 45,018



gpm



Dari gambar 14-37 Peters diperoleh harga ή



=



E p1  E p 2 2



= 46



%



4) Brake Horse Power (BHP) Persamaan Bernoulli



:



P v 2  Z   H f  gc



Ws



=



Ws



= 34,168 ft lbf / lbm



BHP



= BHP 



=



Ws x x q f ( gpm) 7,481gal / ft x 550 ftlbf / sHp x (60s / min) x 3



34,168 ft lb f / lb m x 82,655 lb / ft 3 x 45,018 gpm 7,481 gal / ft 3 x 550 ftlb f / s Hp x (60s / min) x 46 %



= 1,120 HP



226



5) Tenaga Pompa (MHP) Dari gambar 14-38 Peters diperoleh ή



= 80



MHP



=



:



%



1,120 HP HP 80%



= 1,399 HP Dipilih pompa : Power



= 2 HP



Tipe



= Sentrifugal



Jumlah



= 2 buah ( 1 cadangan )



POMPA – 02(P-02) Fungsi



: Untuk mengalirkan metanol dari tanki menuju vaporizer



Tipe



: Pompa sentrifugal



Gambar



:



Gambar L.3.11. Pompa-02 (P-02) A. Kondisi Operasi Temperatur, T



= 39,36 oC



= 102,848 oF



Densitas, ρ



= 851,611 kg/m3



= 56,163 lb/ft3



Viskositas, μ



= 0,535 cp



= 1,294 lb/ft jam



Laju alir, W



= 4480,9052



Tekanan uap, Pv



=0,048 lbf / ft2



B. Menentukan Ukuran Pipa



kg/jam = 164,3 lb/menit



227



a) Volume pompa, Vf Vf



W



=  =



164,3 lb / menit 56,163 lb / ft 3



= 3,4 ft3 /menit = 0,129 gpm



b) Volumetrik flowrate, qf qf



3,4 ft 3 / menit = s 60 menit



= 0,057 ft3/det c) Menentukan Diameter Optimum, Dopt Dopt



= 3,9 x qf 0,45 x ρ0,13



(Pers 15 hal. 496, Peters)



= 3,9 x ( 0,057)0,45 x ( 56,163 lb/ft3 )0,13 = 1,796 in



d) Ukuran Pipa  Suction pipe SN



= 40



IPS



=



2,5 in



L



=



3



m



= 9,842



ft



ID



=



2,469



in



= 0,205



ft



OD



=



2,880



in



= 0,239



ft



a



=



4,790



in2



= 0,033



ft2



228



 Discharge Pipe SN



= 40



IPS



=



2,5



in



L



=



6



m



= 19,685



ft



ID



=



2,469



in



=



0,205



ft



OD



=



2,880



in



=



0,239



ft



a



=



4,790



in2



=



0,033



ft2



C. Perhitungan pada Suction 1). Menentukan Suction Friction Loss a. Suction Velocity, υ υ



= = =



Velocity head



qf a 0,057 ft 3 / det 0,033ft 2



1,717



ft/det



= 6181,357



ft/jam



=



(1,717 ft / det ) 2 υ2 = = gc 76,460 lb / ft 3



0,092 ft.lb f / lb m



b. Bilangan Reynlod, Re Re



=



D 



=



53,163 lb / ft 3 x 6181,357 ft / jam x 0,205 ft 1,294 lb / ftjam



52026,740 Meterial yang digunakan adalah commercial steel pipe Rough factor, έ



= 0,00015



έ/D



= 0,00073



(Fig 14-1, Peters)



=



229



Fanning friction factor, f Pada Re c.



= 52026,740 didapat f



= 0,0047 (Fig.14-1, Peters)



Skin friction loss, Hfs Digunakan 1 elbow 90 o std dan 1 gate valve Le/D = 1 elbow 90 0 std dan 1 gate valve = 32 + 7 = 39 L



= L + Le x ID Suction



L



= 9,842 ft + 39 x 0,205 ft = 17,8346 ft



Hfs



2 fL v 2 x = D gc



=



…(Pers 5.64, Mc Cabe)



2 x 0,0047 x 17,8346 ft (1,717 ft / det) 2 x 0,205 ft 32,174 lb / ft 3



= 0,075 ft lbf / lbm



d. Sudden Contraction Friction Loss, Hfc 1 v2 x 2 g c



Hfc



= Kc .



α



= 1 ( untuk aliran turbulen )



Hfc



= Kc . 0,5 ( v2 / gc )



…(Pers. 5.71, Mc Cabe)



Kc



= 0,4 ( 1- Sb/Sa)



…(Pers. 5.65, Mc Cabe)



Sb



= Luas penampang melintang upstream b



Sa



= Luas penampang melintang downstream a



Dimana Sb> Sa, maka : Kc



= 0,4 ( 1 – 0 )



Hfc



= 0,4 x 0,5 x 0,092 ft lbf / lbm = 0,018 ft lbf/lbm



e. Fitting + Valve Friction Loss, Hff Hff



= Kf x ( v2 / 2 gc )



Dimana



:



Kf



= gate valve + elbow



(Pers II.7, Syarifudin Ismail)



= 1 gate valve + 2 elbow 90o



(Tabel II.7, Syarifudin Ismail)



= 0,2 + 0,9 * 2 =2 Hff



= 2 x 0,5 x ( 0,092 ft lbf / lbm ) = 0,092 ft lbf /lbm



f. Total Discharge Friction Loss, Hfsuc Hfd



= Hfs



+



Hfc



+



Hff



= (0,144 + 0,018 + 0,092) ft lbf / lbm = 0,254 ft lbf / lbm = 0,094 psi 2). Discharge Pressure Zc = 0



m



=0



ft



Zd = 15



m



= 49,212



ft



234



Static Suction Head (SH) = 49,212 ft Pd = 1



atm



= 2117,3762 ft2 lbf / lbm



=14,7 psi



g/gc = 1 lbf /lbm v =0 Pc



Pd ρ



+



g (Zc gc



Zd ) +



Pc Pd g = + (Zd ρ ρ gc



(v c



Zc ) +



vd )2 = Hf 2gα



(v d



vc )2 2gα



Hf



Pc 2117 ,3762 ft 2 lb f / lb m = + 49,212 ft (1 lb f / lb m ) ρ 53,163 lb / ft 3 Pc = 89,294 ft lb f / lb m ρ Pc = 4747,106 lb f / ft 2 Pc = 32,966 psi



0,254ft lb f / lb m



3). Discharge Head, Hd Hd



Total disch arg e presure = lb / ft 3



Hd



=



32,966 psi 53,163 lb / ft 3



x 144in 2 / ft 2



x 144 in 2 / ft 2



= 89,294 ft. lbf / lbm E. Menghitung Tenaga Pompa 1). Diffrential Pressure (Total ΔP) a. Suction pressure



= 18,284



psi



b. Discharge pressure



= 32,966



psi



14,681



psi



2). Total Head a. Suction head



= 49,527



ft



b. Discharge head



= 89,294



ft



39,767



ft



235



3). Efisiensi Pompa Kapasitas pompa



= 25,429



gpm



Dari gambar 14-37 Peters diperoleh harga ή



=



E p1  E p 2 2



= 35



%



4). Brake Horse Power (BHP) Persamaan Bernoulli



:



Ws



P v 2   Z   H f =  gc



Ws



= 79,534 ft lbf / lbm



BHP



=



=



BHP =



Ws xρ x q f (gpm) 3 7,481gal / ft x 550ftlbf / sHp x (60s / min) xη



79,534 ft lb f / lb m x 53,163 lb / ft 3 x 25,429 gpm 7,481 gal / ft 3 x 550 ftlb f / s Hp x (60s / min) x 35%



= 1,244 HP 5). Tenaga Pompa (MHP) Dari gambar 14-38 Peters diperoleh ή



= 80



MHP



=



:



%



1,244 HP 80%



= 1,555 HP Dipilih pompa :



POMPA – 03 (P-03)



Power



= 2 HP



Tipe



= Sentrifugal



Jumlah



= 2 buah ( 1 cadangan )



236



Fungsi



: Untuk mengalirkan produk keluar bottom absorber menuju stripper.



Tipe



: Pompa sentrifugal



Gambar



:



Gambar L.3.12. Pompa-03 (P-03) A. Kondisi Operasi Temperatur, T



= 58 oC



= 136,4 oF



Densitas, ρ



= 972,866 kg/m3



= 60,734 lb/ft3



Viskositas, μ



= 0,492 cp



= 1,189



Laju alir, W



= 21230,617 kg/jam = 780,091 lb/menit



Tekanan uap, Pv



= 378,81 lbf / ft2



lb/ft jam



B. Menentukan Ukuran Pipa a. Volume pompa, Vf Vf



W



=  =



780,091 lb / menit 60,734 lb / ft 3



= 14,129 ft3 /menit = 105,683 gpm b. Volumetrik flowrate, qf qf



=



ft 3 / menit s 60 menit



14,129



= 0,235 ft3/det c. Menentukan Diameter Optimum, Dopt Dopt



= 3,9 x qf 0,45 x ρ0,13



(Pers 15 hal. 496, Peters)



= 3,9 x ( 0,235 )0,45 x ( 60,734 lb/ft3 )0,13



237



= 3,470 in d. Ukuran Pipa  Suction pipe SN



= 40



IPS



=



4,5



in



L



=



5



m



=



16,404



ft



ID



=



4,026



in



=



0,334



ft



OD



=



4,500



in



=



0,374



ft



a



=



12,700



in2



=



0,087



ft2



 Discharge Pipe SN



= 40



IPS



=



4,5



in



L



=



8



m



= 26,246



ft



ID



=



4,026



in



=



0,334



ft



OD



=



4,500



in



=



0,374



ft



a



= 12,700



in2



=



0,087



ft2



C. Perhitungan pada Suction 1) Menentukan Suction Friction Loss a. Suction Velocity, υ υ



= = =



qf a 0,235 ft 3 / det 0,087 ft 2



2,691 ft/det



= 9689,399



ft/jam



238



Velocity head



=



( 2,691 ft / det ) 2 υ2 = = gc 32,174 lb / ft 3



0,225 ft.lb f / lb m



b. Bilangan Reynlod, Re Re



=



D 60,734 lb / ft 3 x 9689,399 ft / jam x 0,334 ft =  1,189 lb / ftjam



= 165340,1778 Meterial yang digunakan adalah commercial steel pipe Rough factor, έ



= 0,00015



έ/D



= 0,00045



(Fig 14-1, Peters)



Fanning friction factor, f Pada Re



= 165340,1778



didapat f



=



0,0065



(Fig.14-1,



Peters) c. Skin friction loss, Hfs Digunakan 1 elbow 90 o std dan 1 gate valve Le/D = 1 elbow 90 0 std dan 1 gate valve = 32 + 7 = 39 L



= L + Le x ID Suction



L



= 16,404 ft + 39 x 0,334 ft = 29,4361 ft



Hfs



=



2 fL v 2 x D gc



=



2 x 0,00045x 29,4361 ft ( 2,691 ft / det) 2 x 0,334 ft 32,174 lb / ft 3



= 0,258 ft lbf / lbm d. Sudden Contraction Friction Loss, Hfc Hfc



1 v2 x = Kc . 2 g c



…(Pers 5.64, Mc Cabe)



239



α



= 1 ( untuk aliran turbulen )



Hfc



= Kc . 0,5 ( v2 / gc )



…(Pers. 5.71, Mc Cabe)



Kc



= 0,4 ( 1- Sb/Sa)



…(Pers. 5.65, Mc Cabe)



Sb



= Luas penampang melintang upstream b



Sa



= Luas penampang melintang downstream a



Dimana Sb> Sa, maka : Kc



= 0,4 ( 1 – 0 )



Hfc



= 0,4 x 0,5 x 0,462 ft lbf / lbm = 0,045 ft lbf/lbm



e. Fitting + Valve Friction Loss, Hff Hff



= Kf x ( v2 / 2 gc )



Dimana



:



Kf



= gate valve + elbow



(Pers II.7, Syarifudin Ismail)



= 1 gate valve + 4 elbow 90o



(Tabel II.7, Syarifudin Ismail)



= 0,2 + 0,9 * 4 = 3,8 Hff



= 3,8 x 0,5 x ( 0,225 ft lbf / lbm ) = 0,307 ft lbf /lbm



f. Total Discharge Friction Loss, Hfsuc Hfd



= Hfs



+



Hfc



+



Hff



= (0,462 + 0,045 + 0,307) ft lbf / lbm



243



= 0,934 ft lbf / lbm = 0,394 psi 2) Discharge Pressure Zc = 0



m



=0



ft



Zd = 8



m



= 26,264



ft



Static Suction Head (SH) = 26,264 ft Pd = 2



atm



= 4234,7524 ft2 lbf / lbm



= 29,4 psi



g/gc = 1 lbf /lbm v =0 Pc



Pd ρ



+



g (Zc gc



Pc Pd g = + (Zd ρ ρ gc



Zd ) +



(v c



Zc ) +



vd ) 2 = Hf 2gα



(v d



vc )2 2gα



Hf



Pc 4234,7524 ft 2 lb f / lb m = + 26,246 ft (1 lb f / lb m ) ρ 60,734 lb / ft 3 Pc = 96,907 ft lb f / lb m ρ Pc = 5885,553 lb f / ft 2 Pc = 40,872 psi



0,394 ft lb f / lb m



3) Discharge Head, Hd Hd



Total disch arg e presure = lb / ft 3



Hd



=



40,872 psi 60,734 lb / ft 3



x 144in 2 / ft 2



x 144 in 2 / ft 2



= 96,906 ft. lbf / lbm E. Menghitung Tenaga Pompa 1) Diffrential Pressure (Total ΔP) a. Suction pressure



= 18,675



psi



244



b. Discharge pressure



= 40,872



psi



22,197



psi



2) Total Head a. Suction head



=



44,278



ft



b. Discharge head



=



96,906



ft



52,628



ft



3) Efisiensi Pompa Kapasitas pompa



= 105,683



gpm



Dari gambar 14-37 Peters diperoleh harga ή



=



E p1  E p 2 2



= 52



%



4) Brake Horse Power (BHP) Persamaan Bernoulli



:



P v 2  Z   H f  gc



Ws



=



Ws



= 70,393 ft lbf / lbm



BHP



= BHP 



=



Ws x x q f ( gpm) 7,481gal / ft 3 x 550 ftlbf / sHp x (60s / min) x



70,393 ft lb f / lb m x 60,734 lb / ft 3 x 105,683 gpm 7,481 gal / ft 3 x 550 ftlb f / s Hp x (60s / min) x 52%



= 3,520 HP 5) Tenaga Pompa (MHP) Dari gambar 14-38 Peters diperoleh



: ή = 83



%



245



MHP



=



3,52 HP HP 83%



= 4,240 HP Dipilih pompa : Power



= 5 HP



Tipe



= Sentrifugal



Jumlah



= 2 buah ( 1 cadangan



POMPA – 04 (P-04) Fungsi



: Untuk mengalirkan campuran keluaran top Stripper



Tipe



: Pompa sentrifugal



Gambar



:



Gambar L.3.13. Pompa-04 (P-04) A. Kondisi Operasi Temperatur, T



= 69,96 oC



= 158 oF



Densitas



= 522,690 kg/m3



= 32,630 lb/ft3



Viskositas campuran



= 0,353 cp



= 0,854 lb/ft jam



Laju alir, W



= 82,673 kg/jam



= 3,038 lb/menit



Tekanan uap, Pv



= 650,446 lbf / ft2



B. Menentukan Ukuran Pipa 1) Volume pompa, Vf Vf



W



=  =



3,038 lb / menit 32,630 lb / ft 3



= 0,102 ft3 /menit = 0,766 gpm



246



2) Volumetrik flowrate, qf qf



0,102 ft 3 / menit = s 60 menit



= 0,002 ft3/det 3) Menentukan Diameter Optimum, Dopt Dopt



= 3,9 x qf 0,45 x ρ0,13



(Pers 15 hal. 496, Peters)



= 3,9 x ( 0,002 )0,45 x ( 32,630 lb/ft3 )0,13 = 0,349 in 4) Ukuran Pipa  Suction pipe SN



= 40



IPS



= 0,25



in



L



=



4



m



= 13,123



ft



ID



=



0,364



in



=



0,030



ft



OD



=



0,540



in



=



0,045



ft



a



=



0,104



in2



=



0,001



ft2



 Discharge Pipe SN



= 40



IPS



=



0,25



in



L



=



5



m



= 16,404



ft



ID



=



0,364



in



=



0,030



ft



OD



=



0,540



in



=



0,045



ft



a



=



0,104



in2



=



0,001



ft2



C. Perhitungan pada Suction



247



1) Menentukan Suction Friction Loss a. Suction Velocity, υ υ



= =



qf a 0,002 ft 3 / det 0,104 ft 2



=



2,382 ft/det



= 8575,077 ft/jam Velocity head



=



( 2,382 ft / det ) 2 υ2 = = gc 32,174 lb / ft 3



0,176 ft.lb f / lb m



b. Bilangan Reynlod, Re Re



=



D 32,630 lb / ft 3 x 8575,077 ft / jam x 0,03ft = = 9904,035  0,854 lb / ftjam



Meterial yang digunakan adalah commercial steel pipe Rough factor, έ



= 0,00015



έ/D



= 0,00496



(Fig 14-1, Peters)



Fanning friction factor, f Pada Re



= 9904,035



didapat f



= 0,0098 (Fig.14-1, Peters)



c. Skin friction loss, Hfs Digunakan 1 elbow 90 o std dan 1 gate valve Le/D = 1 elbow 90 0 std dan 1 gate valve = 32 + 7 = 39 L



= L + Le x ID Suction



L



= 13,123 ft + 39 x 0,03 ft = 14,3015 ft



Hfs



=



2 fL v 2 x D gc



…(Pers 5.64, Mc Cabe)



248



=



2 x 0,0098 x 14,3015 ft ( 2,382 ft / det) 2 x 0,03 ft 32,174 lb / ft 3



= 1,636 ft lbf /



lbm d. Sudden Contraction Friction Loss, Hfc 1 v2 x 2 g c



Hfc



= Kc .



α



= 1 ( untuk aliran turbulen )



Hfc



= Kc . 0,5 ( v2 / gc )



…(Pers. 5.71, Mc Cabe)



Kc



= 0,4 ( 1- Sb/Sa)



…(Pers. 5.65, Mc Cabe)



Sb



= Luas penampang melintang upstream b



Sa



= Luas penampang melintang downstream a



Dimana Sb> Sa, maka : Kc



= 0,4 ( 1 – 0 )



Hfc



= 0,4 x 0,5 x 0,176 ft lbf / lbm = 0,035 ft lbf/lbm



e. Fitting + Valve Friction Loss, Hff Hff



= Kf x ( v2 / 2 gc )



Dimana



:



Kf



= gate valve + elbow



(Pers II.7, Syarifudin Ismail)



= 1 gate valve + 2 elbow 90o



(Tabel II.7, Syarifudin Ismail)



= 0,2 + 0,9 * 2 = 2,0 Hff



= 2,0 x 0,5 x ( 0,176 ft lbf / lbm ) = 0,176 ft lbf /lbm



252



f. Total Discharge Friction Loss, Hfsuc Hfd



= Hfs



+



Hfc



+



Hff



= (2,122 + 0,035 + 0,176) ft lbf / lbm = 2,334 ft lbf / lbm = 0,529 psi 2) Discharge Pressure Zc = 0



m



=0



ft



Zd = 8



m



= 26,246



ft



Static Suction Head (SH) = 26,246 ft Pd = 1



atm



= 2117,3762 ft2 lbf / lbm



= 14,7 psi



g/gc = 1 lbf /lbm v =0 Pc



Pd ρ



+



g (Zc gc



Pc Pd g = + (Zd ρ ρ gc



Zd ) +



(v c



Zc ) +



vd ) 2 = Hf 2gα



(v d



vc )2 2gα



Hf



Pc 2117,3762 ft 2 lb f / lb m = + 26,246 ft (1 lb f / lb m ) 0,529ft lb f / lb m ρ 32,630 lb / ft 3 Pc = 93,470 ft lb f / lb m ρ Pc = 3049,958 lb f / ft 2 Pc = 21,180 psi



3)



Discharge Head, Hd Hd



=



Hd



=



Total disch arg e presure lb / ft 3 21,180 psi 32,630 lb / ft 3



x 144in 2 / ft 2



x 144 in 2 / ft 2



= 93,469 ft. lbf / lbm



253



E. Menghitung Tenaga Pompa 1)



2)



3)



Diffrential Pressure (Total ΔP) a. Suction pressure



= 16,533



psi



b. Discharge pressure



= 21,180



psi



4,647



psi



Total Head a. Suction head



=



72,963



ft



b. Discharge head



=



93,469



ft



20,506



ft



Efisiensi Pompa Kapasitas pompa



= 0,766 gpm



Dari gambar 14-37 Peters diperoleh harga ή



=



E p1  E p 2 2



= 3,82 % 4) Brake Horse Power (BHP) Persamaan Bernoulli



:



P v 2  Z   H f  gc



Ws



=



Ws



= 41,012 ft lbf / lbm



BHP



= BHP 



Ws x x q f ( gpm) 7,481gal / ft x 550 ftlbf / sHp x (60s / min) x 3



41,012 ft lb f / lb m x 32,630 lb / ft 3 x 0,766 gpm = 7,481 gal / ft 3 x 550 ftlb f / s Hp x (60s / min) x 3,82%



= 0,109 HP



254



5) Tenaga Pompa (MHP) Dari gambar 14-38 Peters diperoleh MHP



=



: ή = 76



%



0,109 HP HP 76%



= 0,143 HP Dipilih pompa : Power



= 1 HP



Tipe



= Sentrifugal



Jumlah



= 2 buah ( 1 cadangan )



POMPA – 05 (P-05) Fungsi



: Untuk mengalirkan aliran campuran keluaran bawah Stripper menuju IPAL



Tipe



: Pompa sentrifugal



Gambar



:



Gambar L.3.14. Pompa-05 (P-05) A. Kondisi Operasi Temperatur, T



= 99,39 oC



= 211 oF



Densitas, ρ



= 918,982 kg/m3



= 57,370 lb/ft3



Viskositas campuran



= 0,263 cp



= 0,636 lb/ft jam



Laju alir, W



= 21147,944 kg/jam = 777,053 lb/menit



Tekanan uap, Pv



= 2043,245 lbf / ft2



B. Menentukan Ukuran Pipa a) Volume pompa, Vf



255



Vf



W



=  =



777,053 lb / menit 57,370 lb / ft 3



= 14,899 ft3 /menit = 111,445 gpm



b) Volumetrik flowrate, qf qf



=



ft 3 / menit s 60 menit



14,899



= 0,248 ft3/det c) Menentukan Diameter Optimum, Dopt Dopt



= 3,9 x qf 0,45 x ρ0,13



(Pers 15 hal. 496, Peters)



= 3,9 x ( 0,248 )0,45 x (57,370 lb/ft3 )0,13 = 3,527 in d) Ukuran Pipa  Suction pipe SN



= 40



IPS



= 24 in



L



=



3



m



=



9,842



ft



ID



=



4,026



in



=



0,334



ft



OD



=



4,500



in



=



0,374



ft



a



=



12,70



in2



=



0,087



ft2



 Discharge Pipe



256



SN



= 40



IPS



=



4



in



L



=



5



m



= 16,404



ft



ID



=



4,026



in



=



0,334



ft



OD



=



4,500



in



=



0,374



ft



a



=



12,70



in2



=



0,088



ft2



C. Perhitungan pada Suction 1) Menentukan Suction Friction Loss a. Suction Velocity, υ υ



= = =



qf a 0,248 ft 3 / det 0,087 ft 2



2,838



ft/det



= 10217,589 ft/jam Velocity head



=



( 2,838 ft / det ) 2 υ2 = = gc 32,174 lb / ft 3



0,250 ft.lb f / lb m



b) Bilangan Reynlod, Re Re =



D 57,370 lb / ft 3 x 10217,589 ft / jamx 0,334ft = = 308010,890  0,636 lb / ftjam



Meterial yang digunakan adalah commercial steel pipe Rough factor, έ



= 0,00015



έ/D



= 0,00045



(Fig 14-1, Peters)



Fanning friction factor, f Pada Re



= 308010,890 didapat f



c) Skin friction loss, Hfs



= 0,00440 (Fig.14-1, Peters)



257



Digunakan 1 elbow 90 o std dan 1 gate valve Le/D = 1 elbow 90 0 std dan 1 gate valve = 32 + 7 = 39 L



= L + Le x ID Suction



L



= 9,842 ft + 39 x 0,334 ft = 22,8746 ft



Hfs



=



2 fL v 2 x D gc



=



2 x 0,00045 x 22,8746 ft ( 2,838 ft / det) 2 x 0,334 ft 32,174 lb / ft 3



…(Pers 5.64, Mc Cabe)



= 0,151 ft lbf / lbm d) Sudden Contraction Friction Loss, Hfc 1 v2 x 2 g c



Hfc



= Kc .



α



= 1 ( untuk aliran turbulen )



Hfc



= Kc . 0,5 ( v2 / gc )



…(Pers. 5.71, Mc Cabe)



Kc



= 0,4 ( 1- Sb/Sa)



…(Pers. 5.65, Mc Cabe)



Sb



= Luas penampang melintang upstream b



Sa



= Luas penampang melintang downstream a



Dimana Sb> Sa, maka : Kc



= 0,4 ( 1 – 0 )



Hfc



= 0,4 x 0,5 x 0,249 ft lbf / lbm = 0,05 ft lbf/lbm



e)



Fitting + Valve Friction Loss, Hff Hff



= Kf x ( v2 / 2 gc )



Dimana



:



Kf



= gate valve + elbow



(Pers II.7, Syarifudin Ismail)



= 1 gate valve + 2 elbow 90o (Tabel II.7, Syarifudin Ismail)



261



= 0,2 + 0,9 * 2 =2 Hff



= 2 x 0,5 x ( 0,249 ft lbf / lbm ) = 0,249 ft lbf /lbm



f)



Total Discharge Friction Loss, Hfsuc Hfd



= Hfs



+



Hfc



+



Hff



= (0,264 + 0,05 + 0,249) ft lbf / lbm = 0,563 ft lbf / lbm = 0,224 psi 2). Discharge Pressure Zc = 0



m



=0



ft



Zd = 4



m



= 13,123



ft



Static Suction Head (SH) = 13,123 ft Pd = 1



atm



= 14,7 psi



= 2117,3762 ft2 lbf / lbm



g/gc = 1 lbf /lbm v =0 Pc



Pd ρ



+



g (Zc gc



Zd ) +



(v c



vd )2 = Hf 2gα



(v d v c ) 2 Pc Pd g = + (Zd Zc ) + Hf ρ ρ gc 2gα Pc 2117 ,762 ft 2 lb f / lb m = + 13,1236ft (1 lb f / lb m ) ρ 57,370 lb / ft 3 Pc = 50,593 ft lb f / lb m ρ Pc = 2902,552 lb f / ft 2 Pc = 20,156 psi



3). Discharge Head, Hd



0,224 ft lb f / lb m



262



Hd



=



Hd



=



Total disch arg e presure lb / ft 3 20,156 psi 57,370 lb / ft 3



x 144in 2 / ft 2



x 144 in 2 / ft 2



= 50,593 ft E. Menghitung Tenaga Pompa 1). Diffrential Pressure (Total ΔP) a. Suction pressure



= 15,876



psi



b. Discharge pressure



= 20,156



psi



4,280



psi



2). Total Head a. Suction head



= 39,849



ft



b. Discharge head



= 50,593



ft



10,744



ft



3). Efisiensi Pompa Kapasitas pompa



= 111,445 gpm



Dari gambar 14-37 Peters diperoleh harga ή



=



E p1  E p 2 2



= 55



%



4). Brake Horse Power (BHP) Persamaan Bernoulli



:



P v 2  Z   H f  gc



Ws



=



Ws



= 21,488 ft lbf / lbm



BHP



= BHP 



Ws x x q f ( gpm) 7,481gal / ft x 550 ftlbf / sHp x (60s / min) x 3



263



21,488 ft lb f / lb m x 57,370 lb / ft 3 x 111,445 gpm = 7,481 gal / ft 3 x 550 ftlb f / s Hp x (60s / min) x 55%



= 1,012 HP 5). Tenaga Pompa (MHP) Dari gambar 14-38 Peters diperoleh ή



= 80



MHP



=



:



%



1,012 HP HP 80%



= 1,265 HP Dipilih pompa : Power



= 2 HP



Tipe



= Sentrifugal



Jumlah



= 2 buah ( 1 cadangan )



POMPA – 06 (P-06) Fungsi



: Untuk mengalirkan H2O



Tipe



: Pompa sentrifugal



Gambar



:



Gambar L.3.15. Pompa-06 (P-06) A. Kondisi Operasi Temperatur, T



= 30 oC



= 86 oF



Densitas, ρ



= 995,647 kg/m3



= 65,156 lb/ft3



Viskositas campuran



= 0,850 cp



= 2,056 lb/ft jam



Laju alir, W



= 108,8965 kg/jam



= 4,001 lb/menit



Tekanan uap, Pv



= 88,566 lbf / ft2



B. Menentukan Ukuran Pipa



264



a). Volume pompa, Vf Vf



W



=  =



4,001 lb / menit 65,156 lb / ft 3



= 0,071 ft3 /menit = 0,530 gpm b). Volumetrik flowrate, qf qf



=



ft 3 / menit s 60 menit



0,071



= 0,00118 ft3/det c). Menentukan Diameter Optimum, Dopt Dopt



= 3,9 x qf 0,45 x ρ0,13



(Pers 15 hal. 496, Peters)



= 3,9 x (0,00118)0,45 x (65,156 lb/ft3 )0,13 = 0,321 in d). Ukuran Pipa  Suction pipe SN



= 40



IPS



= 0,25



in



L



=



4



m



=



12,123



ft



ID



=



0,364



in



=



0,030



ft



OD



=



0,540



in



=



0,045



ft



a



=



0,104



in2



=



0,001



ft2



 Discharge Pipe SN



= 40



IPS



=



0,25



in



265



L



=



5



m



=



16,404



ft



ID



=



0,364



in



=



0,030



ft



OD



=



0,540



in



=



0,045



ft



a



=



0,104



in2



=



0,001



ft2



C. Perhitungan pada Suction 1). Menentukan Suction Friction Loss a. Suction Velocity, υ υ



= =



qf a 0,00118 ft 3 / det 0,001 ft 2



=



Velocity head



1,647



ft/det



= 5929,626



ft/jam



=



(1,647 ft / det ) 2 υ2 = = gc 32,174 lb / ft 3



0,084 ft.lb f / lb m



b. Bilangan Reynlod, Re Re =



D 65,156 lb / ft 3 x 5929,626 ft / jam x 0,030ft = = 5415,250  2,056 lb / ftjam



Meterial yang digunakan adalah commercial steel pipe Rough factor, έ



= 0,00015



έ/D



= 0,000496



(Fig 14-1, Peters)



Fanning friction factor, f Pada Re



= 5415,250



didapat f



c. Skin friction loss, Hfs Digunakan 1 elbow 90 o std dan 1 gate valve Le/D = 1 elbow 90 0 std dan 1 gate valve = 32 + 7 = 39 L



= L + Le x ID Suction



= 0,012 (Fig.14-1, Peters)



266



L



= 13,123 ft + 39 x 0,030 ft = 14,3015 ft



Hfs



=



2 fL v 2 x D gc



=



2 x 0,012 x 14,3015 ft (1,647 ft / det) 2 x 0,030 ft 32,174 lb / ft 3



…(Pers 5.64, Mc Cabe)



= 0,958 ft lbf / lbm d. Sudden Contraction Friction Loss, Hfc Hfc



1 v2 x = Kc . 2 g c



α



= 1 ( untuk aliran turbulen )



Hfc



= Kc . 0,5 ( v2 / gc )



…(Pers. 5.71, Mc Cabe)



Kc



= 0,4 ( 1- Sb/Sa)



…(Pers. 5.65, Mc Cabe)



Sb



= Luas penampang melintang upstream b



Sa



= Luas penampang melintang downstream a



Dimana Sb> Sa, maka : Kc



= 0,4 ( 1 – 0 )



Hfc



= 0,4 x 0,5 x 0,084 ft lbf / lbm = 0,017 ft lbf/lbm



270



e. Fitting + Valve Friction Loss, Hff Hff



= Kf x ( v2 / 2 gc )



Dimana



:



Kf



= gate valve + elbow



(Pers II.7, Syarifudin Ismail)



= 1 gate valve + 2 elbow 90o (Tabel II.7, Syarifudin Ismail) = 0,2 + 0,9 * 2 =2 Hff



= 2 x 0,5 x ( 0,084 ft lbf / lbm ) = 0,084 ft lbf /lbm



f. Total Discharge Friction Loss, Hfsuc Hfd



= Hfs



+



Hfc



+



Hff



= (1,243 + 0,017 + 0,084) ft lbf / lbm = 1,344 ft lbf / lbm = 0,580 psi 2). Discharge Pressure Zc = 0



m



=0



ft



Zd = 3



m



= 9,842



ft



Static Suction Head (SH) = 9,842 ft Pd = 1



atm



g/gc = 1 lbf /lbm v =0



= 14,7 psi



= 2117,3762 ft2 lbf / lbm



271



(v c - v d ) 2 Pc - Pd g + (Zc - Zd ) + = Hf ρ gc 2gα (v d - v c ) 2 Pc Pd g = + (Zd - Zc ) + - Hf ρ ρ gc 2gα Pc 2117,762 ft 2 lb f / lb m = + 9,842ft (1 lb f / lb m ) - 0,084 ft lb f / lb m ρ 65,156 lb / ft 3 Pc = 45,251 ft lb f / lb m ρ Pc = 2812,661 lb f / ft 2 Pc = 19,532 psi



3). Discharge Head, Hd Hd



=



Hd



=



Total disch arg e presure lb / ft 3 19,532 psi 65,156 lb / ft 3



x 144in 2 / ft 2



x 144 in 2 / ft 2



= 45,251 ft E. Menghitung Tenaga Pompa 1). Diffrential Pressure (Total ΔP) a. Suction pressure



= 18,511



psi



b. Discharge pressure



= 19,532



psi



1,021



psi



2). Total Head a. Suction head



= 42,886



ft



b. Discharge head



= 45,251



ft



2,365



ft



3). Efisiensi Pompa Kapasitas pompa



= 0,530 gpm



Dari gambar 14-37 Peters diperoleh harga



272



ή



=



E p1  E p 2 2



= 0,15 % 4). Brake Horse Power (BHP) Persamaan Bernoulli



:



P v 2  Z   H f  gc



Ws



=



Ws



= 21,488 ft lbf / lbm



BHP



= BHP  =



Ws x x q f ( gpm) 7,481gal / ft x 550 ftlbf / sHp x (60s / min) x 3



21,488 ft lb f / lb m x 57,370 lb / ft 3 x 111,445 gpm 7,481 gal / ft 3 x 550 ftlb f / s Hp x (60s / min) x 0,15%



= 0,4205 HP 5). Tenaga Pompa (MHP) Dari gambar 14-38 Peters diperoleh ή



= 79



MHP



=



:



%



0,4205 HP HP 79%



= 0,5323 HP Dipilih pompa : Power



= 1 HP



Tipe



= Sentrifugal



Jumlah



= 2 buah ( 1 cadangan )



PREHEATER – 01 (PH-01) Fungsi



: Menaikkan temperatur bahan baku` Reaktor-01



Type



: Shell and Tube Heat Exchanger



HCl



sebelum



diinput



ke



273



Bahan



: Stainless Steel



Gambar



:



Gambar L.3.16. Heater-01 (H-01) Fluida Panas



: Saturated Steam



Wt



= 170,3323



kg/jam = 375,5180



lb/jam



T1



= 150



o



= 302



o



T2



= 150



o



= 302



o



Fluida Dingin



C C



F F



: HCl



W2



= 9866,8582 kg/jam = 21752,6729 lb/jam



t1



= 110



o



= 230



o



t2



= 125



o



= 257



o



C C



F F



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern. 1) Beban panas PH – 01 Q =



85975,7418



2) Menghitung ΔT



kkal/hr = 341179,3415 Btu/hr



274



No 1 2 3



Fluida Panas 302 302



LMTD (ΔT)



Temperatur Tinggi Temperatur Rendah Selisih



=



T2 - T1 ln(T2 /T1 )



=



45 - 72 ln(45/ 72)



= 57,511 oF



S =



t 2  t1 T1  t1 257  230



= 302  230 = 0,375 Ft = 0,71 ∆t = LMTD x Ft = 57,511 x 0,71 = 40,8328 3) Ta dan ta Ta = ½ (302 + 302) = 302 oF ta = ½ (257 + 230) = 243,5 oF Dari table 8, Kern, UD = 50 Btu / jam ft2 oF Trial UD c)



: Steam & Heavy Organik



Asumsi UD A



=



= 25 Btu / jam ft2 oF



Q (U D .T )



Fluida Dingin 257 230



Selisih 45 72 27



275



=



341179 ,3415 Btu / jam ( 25 Btu / jam. ft 2 .o F  57,511 o F )



= 334,2212 ft2 d) Karena A> 200 ft2, maka dipilih HE dengan jenis Shell and Tube Heat Exchanger, dengan klasifikasi sebagai berikut (dari table 11 Kern) : ID baffle Pitch C'



Shell side 13.25 in 5 in 1.25 in 0.25 in



Nt L OD a" pass BWG ID



Tube side 66 buah 15 ft 1 in 0.693 in2 2 18 in 0.902 in



A. Fluida Dingin 1. Flow area dalam tube (a’t) Total flow area (at)



= 0,693 inch2 = Nt x a’t / 144 x n =



66 x 0,693 144 x 2



= 0,1588 ft3 2. Laju alir,



Gt = =



W at 21752,6729 0,1588



= 136970,8 lb/ hr. ft2 Perhitungan viskositas campuran :



(Tabel10.Kern)



276



Dari fiq 2-32 ,Perry didapat : μH2O = 0,145 cP = 0,3508 lb/ft.hr μHCl = 0,19 cP = 0,4596 lb/ft,hr Perhitungan kapasitas panas, Cp : tabel B.2 , Felder, Elementary Principles of chemical Processes) Senyawa H2O HCl



b.102 0,668 -0,1341



a 33,46 29,13



c. 106 7,6 0,9715



d. 109 -3,593 -4,335



Cp = a + bT + cT 2 + dT 3



Cp(H 2 O) = 33,46 + 0,668.10 - 2 × 117,5 + 0,7604.10 -5 × 117,5 2 + - 3,593.10 -9 × 117,5 3 = 34,3441 J / mol.o C



= 1,9080 J/g.oC = 0,455 Btu / lb oF



Cp(HCl) = 29,13 + - 0,1341.10- 2 × 117,5 + 0,9715.10- 5 × 117,5 2 + - 4,335.10- 9 × 117,5 3 = 29,0995 J / mol.o C



= 1,7972 J/g.oC = 0,1905 Btu / lb oF



Perhitungan konduktivitas termal, k : 10,4 th edition ) k = μ (Cp + ( MW )  ( pers Coulson 6 10, 4 k ( H 2 O) = 0,0144 (1,9080 J/g.oC + ( 18 g / mol)



= 0,3604 W/m.oC = 0,2086 lb/ft hr 10,4 k (HCl) = 0,0144 (1,7972 J/g.oC + ( 36,5 g / mol)



= 1,5418 W/m.oC = 0,8922 lb/ft hr Senyawa H2O HCl Jumlah



Massa, kg 3650,7375 6216,1206 9866,8581



Xi 0,37 0,63 1



Cp, Btu / lb oF 0,455 0,1905



Cp(campuran) = 0,3603 Btu / lb oF k(campuran) = 0,4615 lb/ft hr μ(campuran) = ∑ X i .μ i = 0,3910 lb/ft.hr



Xi.Cp 0,0705 0,2898 0,3603



k, lb/ft hr 0,2086 0,8922



Xi.k 0,1314 0,3301 0,4615



277



3. Bilangan Reynold, Ret Pada



Tc



= 243,5 oF



μ



= 0,3603 lb/ft jam



Ret



= =



De .G a



 0,902 x 136970,8 0,3910



= 26327,27 15



4. Dengan L/D



= 0,902 x 12 = 199,5565



5. Pada



 c.     k 



, diperoleh



jH



= 85



tc



= 243,5 oF



Cp



= 0,3603 Btu/lb. oF



k



= 0,4615 Btu/ft. oF. jam



1



3



=



(Fig 24. Kern)



0,3603 x 0,3910 0,4615



1



3



= 0,6733 1/ 3



6. hi



 k   Cp .   = jH     D  k 



    w



  



0 ,14



Koreksi viskositas diabaikan karena tidak significant, maka didapat : hi



= 351,4287 Btu / hr. ft2 oF



hio



= hi 



 ID    OD   0,902    1 



= 351,4287 



= 316,9887 Btu / hr. ft2 oF B. Fluida panas (steam)



: Shell Side



1. Flow area pada shell (as)



278



as



= =



 ID



x C" x B  (144 Pt )



13,25 x 0,25 x 5 144 x 1,25



= 0,0920 ft2 2. Laju alir massa dalam shell, Gs Gs



W



= a s =



375,5180 0,0920



= 4081,101 lb / jam. ft2 3. Condensate loading per linier foot (G”) G”



=



W L.Nt 2 / 3



=



375,5180 15 x 66 2 / 3



=



1,5329 lb/jam. ft = D x Gs / µ



4. Bilangan Reynold, Res Pada



Tc



= 302 oF



Cp



= 0,1904 Btu/lb.oF



k



= 0,2194 Btu/lb.oF



μ



=



 c.     k 



1



0,0348 lb/ft . jam



3



=



 0,1904 x 0,0348    0,2194  



= 0,3115 De



=



Res=



= =



0,55 inch = 0,0458 ft GS D



 4081,101 x 0,0458 0,0348



= 5369,385



1



3



279



jH = 40



(Fig. 28 Kern)



5. Koefisien Perpindahan Panas, ho Koreksi viskositas diabaikan karena tidak significant, maka diperoleh : ho



= jH . (k/D). (cμ/k)1/3



(Pers. 6.28 Kern)



40 x 0,2194 x 0,3115 0,0458



=



= 59,6581 Btu / jam ft2 oF 6. Clean Overall Coefficient, Uc Uc =



hio x ho hio  ho 316,9887 x 59,6581



= 316,9887 + 59,6581



(Pers. 6.38 Kern)



= 50,2087 Btu / jam ft2 oF 7. Dirt Factor, Rd Rd



=



Uc UD 50,2087 - 25 = 50,2087 x 25 U c .U D



= 0,02 4) PRESSURE DROP Tube Side 1. Untuk NRe Factor friksi



2. ΔPt



= 28575,93 = 0,0002



s



= 0,4685



Фt



= 0,3401 f Gt 2 Ln = x D e s t



=



0,0002 x 136970,8 2 x 15 x 2 5,22 x 1010 x 0,0458 x 0,4685 x 0,3401



= 0,2953 psi



(Fig 26, Kern)



280



3. V2 / 2g



= 0,07 = ( 4n / s ) ( V2 / 2g )



ΔPr



=



4 x 2 x 0,07 0,4685



= 1,1953 psi 4. ΔPT



= ΔPt + ΔPr = 0,2953 + 1,1953 = 1,4905 psi



Shell Side 1. Faktor Friksi Re



= 5369,385



f



= 0,0023



2. Number of cross, N+1



(Fig 29, Kern)



(N + 1)



= 12 L / B



(Pers. 7.43 Kern)



= (12 x 15)/ 5 = 36 Ds



= ID / 12 = 13,25 / 12 = 1,1042 ft



s



= 0,94 2



ΔPs



= =



f G s D f ( N  1) x De S  s 0,0023 x 4081,1012 x 1,1042 x 36 5,22 x 1010 x 0,0458 x 0,94



= 0,000677 psi



hio = 316,9887



SUMMARY houtside =



ho = 59,6581



281



Uc = 50,2087 Btu/jam ft2 oF UD = 25 Btu/jam ft2 oF Rd calculated = 0,02 1,4905 10



Calculated ΔP, psi Allowable ΔP, psi



0,000677 10



PREHEATER – 02 (PH-02) Fungsi



: Memanaskan bahan baku metanol



Type



: Double Pipe Heat Exchanger



Bahan



: Stainless Steel



Gambar



:



Return bend



Gland



Gland



Gland



Tee



Return Head



Gambar L.3.17. Preheater-02 (PH-02) Fluida Panas



: Saturated Steam



Wt



= 36,0041



kg/jam = 79,3756



lb/jam



T1



= 150



o



= 302



o



T2



= 150



o



= 302



o



Fluida Dingin



C C



F F



: Bahan baku methanol



W2



= 4480,9052 kg/jam = 9878,6932 lb/jam



t1



= 110



o



C



= 230



o



F



282



t2



o



= 125



C



= 266



o



F



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern. 1) Beban panas PH – 02 Q =



18354,9646



kkal/hr = 72838,3915 Btu/hr



2) Menghitung ΔT



No 1 2 3



Fluida Panas 302 302



LMTD (ΔT)



Temperatur Tinggi Temperatur Rendah Selisih



=



ΔT2 ΔT1 ln(ΔT2 /ΔT1 )



=



45 - 72 ln(45/ 72)



Fluida Dingin 257 230



= 57,511 oF 3) Ta dan ta Ta = ½ (302 + 302) = 302 oF ta = ½ (257 + 230) = 244 oF Dari table 8, Kern, UD Trial UD e)



: Steam – gases (5 – 50 Btu / jam ft2 oF)



Asumsi UD A



=



= 40 Btu / jam ft2 oF



Q (U D .T )



Selisih 45 72 27



283



=



72838,3915 Btu/hr ( 40 Btu / jam.ft 2 .o F × 57,511 o F)



= 31,6628 ft2 f) Karena A< 200 ft2, maka dipilih HE dengan jenis Double Pipe Heat Exchanger, dengan klasifikasi sebagai berikut (dari table 11 Kern) :



No 1 2 3 4 5



Annulus 4 40 4,026 in 4,500 in 1,178 ft2



Data Pipa IPS SN IDp ODp a’



4) Annulus : Fluida panas, steam a) Flow area, aa D2



= 4,026 in = 0,3355 ft



D1



= 3,500 in = 0,2917 ft



aa



=



 ( D2 2  D1 2 ) 4



2 2 = 3,14(0,3355 - 0,2917 ) 4



= 0,0216 ft2 Diameter Equivalent, De 2



De



D  D1 = 2 D1 =



2



0,3355 2 - 0,2917 2 0,2917



= 0,0943 ft b) Laju alir massa, Ga Ga



= W / aa



Inner 3 40 3,068 in 3,500 in 0,917 ft2



284



=



79,3756 lb / jam 0,0216 ft 2



= 3678,1536 lb / jam ft2 c) Bilangan Reynold, Rea Pada



o



Tc



=



302



μ



=



0,0144 cp



Rea



= De . Ga / μ =



F = 0,03484 lb/ft jam (fig.2-32, Perry)



0,0943 ft × 3678,1536 lb / ft 2 hr 0,03484 lb / ft. jam



= 9,952.103 d) jH



= 40



e) Pada



Tc



=



302



k



=



0,0208 lb/ft hr



…(Tabel 5, Kern)



Cp



=



0,4597 Btu / lb oF



…(Fig 4, Kern)



o



F = 150 oC



…(Fig 24, Kern)



Perhitungan Cp : tabel B.2 , Felder, Elementary Principles of chemical Processes) b.102 0,668



a 33,46



c. 105 0,7604



d. 109 -3,593



Cp = a + bT + cT 2 + dT 3 Cp = 33,46 + 0,668.10- 2 × 150 + 0,7604.10- 5 × 150 2 + - 3,593.10- 9 × 1503 Cp = 34,623 J / mol.o C



= 1,9233 J/g.oC = 0,4597 Btu / lb oF



Perhitungan nilai konduktivitas termal, k : 10,4 th edition ) k = μ (Cp + ( MW )  ( pers Coulson 6 10,4 k = 0,0144cp (1,9233 J/g.oC + ( 18 g / mol)



k = 0,036 W/m.oC = 0,0208 lb/ft hr k c . μ )1 / 3 ( μ 0,14 ho = jH D ( k μw ) e



285



=



40 ×



0,0208lb/ft hr 0,4597Btu / lb oF . 0,03484 lb/ft jam 1 / 3 ( ) 0,0943 ft 0,036lb/ft hr



= 8,1016 Btu/hr.ft2.oF 5) Inner Pipe : Fluida Dingin a) Flow area, ap Dp ap



= 3,068 in = 0,2557 ft =



 ID p 2 4



2 2 = 3,14 / 0,2557 ft 4



= 0,0513 ft2 b) Laju alir massa, Gp Gp



W



= a p =



9878,6932 lb / jam 0,0513 ft 2



lb / jam ft2



= 192522,5595



Dari fiq 2-32 ,Perry didapat : μCH3OH = 0,0137 cP = 0,0331 lb/ft.hr c) Bilangan Reynold, Rep Pada



o



tc



= 244



μ



= 0,3910 lb/ft. jam



Rep



= =



F



G p ID p



 192522,5595 lb / jam ft2 × 0,2557 ft 0,0331 lb/ft. jam



= 1485126,3208



286



d) hi = 1500 Btu/hr.ft2.oF e) Koefisien perpindahan panas, hio Untuk kondensasi steam



: hio = =



hi × D p OD inner



….(Kern hal.164)



1500 Btu/hr.ft2 .oF × 0,2557 ft 0,2917 ft



= 1314,8571 Btu/hr.ft2.oF f) Clean everaal Coefficient, Uc Uc



=



h io x h o h io + h o 1314,8571 Btu/hr.ft2 .oF x 8,1016 Btu/hr.ft2 .oF



= 1314,8571 Btu/hr.ft2 .oF + 8,1016 Btu/hr.ft2 .oF = 8,0520 Btu / jam ft2 oF g) Design overall Coefficient, UD 1 / UD



= 1 / U c + Rd



Rd



= 0,002



1/UD



=



1 + 0,002 8,0520 Btu / jam ft2 oF



= 0,1262 jam.ft2.oF/Btu = 7,9244 Btu / jam ft2 oF



UD



h) Required Surface, A A



Q



= U x T D =



72838,3915 Btu/hr 7,9244 Btu / jam ft2 o F x 57,511 o F



= 159,8254 ft2



287



i) Required Length, L L



= A / a”



a”



= 0,917 ft



L



=



…(Tabel 11 Kern)



159,8254 ft2 0,917 ft



= 174,2917 ft Diambil panjang 1 hairpin



= 2 x 12 ft



Jumlah hairpin yang diperlukan =



174,2917 ft 24 ft



=8



Actual Length



= 8 x 24 ft = 192 ft



Actual Surface



= 192 ft x 0,917 ft = 176,064 ft2



Actual Design Coefficient, Ud Q



Ud



= A .T act =



72838,3915 Btu/hr 176,064 ft 2 × 57,511 oF



= 7,1935 Btu / hr. ft2 oF Rd



=



UD UD



Ud Ud



7,9244 Btu / jam ft2 oF - 7,1935 Btu / hr. ft2 oF



= 7,9244 Btu / jam ft2 oF × 7,1935 Btu / hr. ft2 oF = 0,013 hr ft2 oF 6) Pressure Drop a) Annulus : Fluida Panas 1) De’



= (D2 – D1) = (0,3355 – 0,2917) ft = 0,0438 ft



Rea



= De . Ga / μ =



0,0438 ft ×3678,1536lb / jam ft2 0,03484 lb/ft jam



288



= 4,628.103 f



=



0,0035 +



=



0,0035 +



0,264 0,42 R ea



….(Pers 3.47b, Kern)



0,264 (4,628.103 ) 0,42



= 0,0111 Pada Tc = 302 oF, ρsteam = 955,4261 lb/ft3 = 59,6470 lb/ft3 (Tabel 2.30, Perry) 2



2) ΔFa



4. f .G a L = 2.g . 2 De =



4 × 0,0111 × 3678,1536 lb / jam ft2 2 × 192 ft 2 × 4,18.108 ft / hr 2 × 59,6470 lb/ft3 2× 0,0438 ft



= 0,00089 ft 3) Va



= Ga / 3600 . ρ =



…(hal 115 Kern)



3678,1536 lb /det ft2 3600 × 59,6470 lb/ft3



= 0,017 ft/det Fl



=n =8



V2 2. g



…(hal. 112 Kern)



0,017 ft/det 2 2 × 32,2 ft / det 2



= 0,00003645 ft 4) ΔPa



=



(ΔFa + FL ).ρ 144



=



(0,00089 ft + 0,00003645 ft ).59,6470 lb/ft3 144



= 0,000382 psi b) Inner Pipe : Fluida Panas 1) Rep



= 1485126,321



…(hal. 114 Kern)



289



f



=



0,0035 +



=



0,0035 +



0,264 0,42 R ea



…(Pers. 3.47b Kern)



0,264 1485126,3210,42



= 0,004175 Pada t = 117,5 oF,



ρ = 58,57 lb/ft3



2



2) ΔFp



= =



4. f .G p L 2 . g . 2 .D 4 × 0,004175 × 192522,5595 lb / jam ft2 2 × 4512 ft 2 × 4,18.108 × 58,57 lb/ft3 2 × 0,2557 ft



= 0,00373 ft 3) ΔP



= ( ΔFp . ρ)/144 =



0,00373 ft × 58,57 lb/ft3 144



= 0,000625 psi hio = 1314,8571



0,000382 10



SUMMARY houtside Uc = 8,0520 Btu / jam ft2 oF UD = 7,9244 Btu / hr. ft2 oF Rd calculated = 0,013 hr ft2 oF Rd required = 0,002 hr ft2 oF Calculated ΔP, psi Allowable ΔP, psi



ho = 8,1016



0,000625 10



290



REAKTOR (R-01) Fungsi



:



Tempat mereaksikan Metanol dan Asam Klorida untuk menghasilkan Metil Klorida.



Tipe



:



Operasi :



Multi Tubular Fixed Bed Reactor Kontinyu



Gambar :



R-01



Gambar L.3.18. Reaktor-01 (R-01) Kondisi Operasi :  Tekanan, P



= 2,5 atm



(US Patent 6.111.153)



291



 Temperatur, T



= 125 oC



(US Patent 6.111.153)



 Konversi HCl, X



= 86 %



(US Patent 6.111.153)



 Residence Time,  = 10 s



(US Patent 5.196.618)



 Laju Alir Massa, W = 13765,2457 kg/jam = 3823,6794 gr/s  Densitas,  pada 125 oC (398,15 K) ρ dihitung dengan persamaan : A



ρ B



  T D  1  1     C  



di mana, ρ dalam kmol/m3 dan T dalam K



,



Dari Tabel 2-30 Perry’s Chemical Engineers’ Handbook diketahui : Senyawa CH3OH HCl H2O



A 2,288 3,342



B 0,2685 0,2729



C 512,64 324,65



D 0,2453 0,3217



5,459



0,30542



647,13



0,081



Senyawa CH3OH HCl



m, kg 3898,3876 3678,2449



H2O Total



6278,2819 13854,9144



n, mol 32 36,5



Xi 0,2832 0,2652



ρi, kg/m3 677,8514 446,9790



18



0,4516 1,0000



964,4141



ρcampuran dihitung menggunakan rumus : 1 ρ campuran







Xi ρi



didapatkan ρcampuran = 746,0270 kg/m3 = 46,5743 lb/ft3 -



Viskositas,  Viskositas Campuran, campuran dihitung berdasarkan persamaan : campuran =  Xi . i



292



Data viskositas masing-masing senyawa didapatkan dari gambar 2-32 Perry’s Chemical Engineers’ Handbook. Senyawa CH3OH



m, kg 3898,3876



Xi 0,2832



μi, cP 0,0140



HCl H2O



3678,2449 6278,2819 13854,9144



0,2652 0,4516



0,0190 0,0135



Total



1,0000



didapatkan campuran = 0,0151 cp = 1,51.10-5 kg/ms -



Reaksi : CH3OH(g)



+



HCl(g)



CH3OH(g)



CH3Cl(g)



+



H2O (g)



CH3OCH3(g)



Data Katalis : 



Nama Katalis = Zinc Chloride dengan support Karbon Aktif







Diameter Katalis, dp = 5 mm Dari Perry’s Chemical Engineers’ Handbook diketahui range diameter katalis untuk reaktor Multi Tubular Fixed Bed adalah 3 – 5 mm.







Porositas, 



= 0,4







Densitas, c



= 1732 kg/m3







Turtuosity,  = 3,0







Constriction Factor,  = 0,7 Volumetric Flowrate, qf qf 



W ρ 13765,2457 kg/jam



= 746,0270 kg/m3 = 18,4514 m3/jam = 0,005125 m3/s



293



Konsentrasi Umpan FAO = mol umpan HCl =



3678,2449 kg/jam 36,5 kg/kmol



FAO



CAO = q f



=



= 100,0202 kmol/jam



100,0202 kmol/jam = 5,4207 kmol/m3 18,4514 m 3 /jam



FBO = mol umpan CH3OH =



3898,3876 kg/jam 32 kg/kmol



FBO



CBO = q f



=



= 121,8246 kmol/jam



121,8246 kmol/jam = 6,6025 kmol/m3 18,4514 m 3 /jam



Konstanta Laju Reaksi, ko Pada reaktor ini, terjadi reaksi fase gas dengan menggunakan katalis solid berbentuk bola berpori. Katalis yang digunakan adalah ZnCl2 dengan support Karbon Aktif. Dalam hal ini, Karbon Aktif berperan sebagai bola berpori. Reaksi yang terjadi antar reaktan dengan bantuan katalis ZnCl 2 berlangsung di dalam pori-pori Karbon Aktif ini. Tahapan yang terjadi untuk reaksi jenis ini adalah : 1. Difusi Eksternal Reaktan 2. Difusi Internal Reaktan 3. Adsorpsi reaktan A pada permukaan katalis 4. Reaksi pada permukaan katalis ( A



B)



5. Desorbsi produk pada permukaan katalis 6. Difusi Internal produk 7. Difusi eksternal produk r



Gambar katalis pellet dengan pori Large pore



Large openings Compressed porous powder



Small pores Single large catalyst pellet



Model of pore structure



294



A



B



7



A



B



External diffusion



7



1



B 2



2 6



6 3



Internal diffusion



5 4



A-B Catalytic surface



Tahapan Reaksi Gas Katalitik



Menghitung nilai Diffusivitas HCl dalam CH3OH



D HCl CH 3OH



1/2    1   T 3/2  1   M HCl M CH 3OH      0,0018583  2  p t σ HCl CH 3OH Ω HCl CH 3OH   



dimana : D HCl CH 3OH = Bulk Diffusivity, cm2/s



T



= Temperatur, K



M HCl M CH 3OH



= Berat molekul HCl dan CH3OH



       



(Pers. 11-11, J.M. Smith)



295



Pt



= Tekanan total, atm



σ HCl CH 3OH



= Konstanta Lennard-Jones



Ω HCl CH 3OH = Integral Colision



Dari Tabel E.1 , Transport Phenomena- R.Byron Bird Senyawa HCl CH3OH



σ, Ǻ 4,374 3,585



σ HClCH3OH 







ε/k,K 378 507



1 σ HCl  σ CH3OH 2







(Pers. 17.3-18, R. Byron



Bird) σ HCl CH 3OH 



1  4,374  3,585 2



σ HCl CH 3OH  3,9795



ε HCl CH OH 3



k



(Pers. 17.3-19, R. Byron



 (ε HCl ε CH 3OH )1/2



Bird) ε HCl CH OH 3



k ε HCl  CH 3 OH k



 ( 378  507)1/2  437,774



Pada temperatur 125 oC = 398,15 K kT ε HCl CH OH 3



=



398,15 = 437,774 0,91



Dari Tabel E.2 , Transport Phenomena - R.Byron Bird, untuk ε didapatkan harga Ω Jadi,



HCl  CH 3 OH



= 1,517



kT HCl  CH 3 OH



= 1,13



296



1 1 1/2 + 36,5 32 2,5 ( 3,9795) 2 1,517



( 398,15) 3/2 D HCl CH 3OH = 0,0018583



D HCl CH 3OH = 0,0595 cm2/s



Luas permukaan terluar partikel solid per satuan volume gas, ac ac 



=



6 1    dp



(Fogler, Hal.766)



61  0,4  5.10 3



= 720 m2/m3 Menghitung Superficial Velocity, U Untuk menghitung U, diperlukan data spesifikasi tube. dk  0,15 dT dT =



(JM Smith, hal. 571)



dk 0,50 cm = = 3,33 cm = 1,3123 in 0,15 0,15



Dari Perry’s Chemical Engineers’ Handbook Bab 23 hal. 53 diketahui bahwa diameter tube reaktor maximal untuk katalis dengan range diameter 3-5 mm adalah 2 in. Berikut adalah data spesifikasi tube IPS 2 in diambil dari Tabel. 11, Kern IPS



= 2 in



Sch. No.



= 40



OD



= 2,38 in = 0,0605 m



ID



= 2,067 in = 0,0525 m



a”



= 3,35 in2 = 0,0021 m2



Lt



= 10 ft



U 



qo a"



= 3,048 m



297



U=



0,005125 m 3 /s 0,0021 m 2



= 2,3714 m/s



Menghitung Faktor Perpindahan Massa (jD) jD



0,455  d p U       μ 



0,455 jD = 0,4



0,407



0,407



5 × 10 -3 × 2,215 1,51 x 10 -5



j D  0,0755



Menghitung Koefisien Transfer Massa Reaktan ke permukaan Katalis kg 



jD G  ρ D    ρ  μ 



2/3



(Syarifuddin Ismail, hal. 181)



dimana : kg



= Koefisien Perpindahan Massa , cm/s



ρ



= Densitas fluida, gr/cm3



G



= Laju alir linear massa, gr/s



μ



= Viskositas, poise



D



= Diffusivitas Bulk katalis dalam fluida, cm2/s



kg =



0,0755 × 3823,6794 0,746



(



0,746 × 0,0595 ) 1,51 x 10 - 5



2/3



k g = 79356,3443 cm/s



kg = 793,563443 m/s Menghitung Luas permukaan eksternal Katalis S ext 



6 ρc d p



S ext =



6 3 1732 kg/m 5 x 10 - 3 m



S ext  0,6928 m2/kg



298



Menghitung Luas permukaan Internal Katalis S int 



ac ρc



Sint =



(Fogler, eq.11-55) 720 m -1



1732 kg/m 3



S int  0,4157



dengan dp = 5 mm dan Rpori = 2,5 10-3 m De 



D AB  σ τ



De =



0,0595 cm 2 /s × 0,4 × 3,9795 10



D e  0,00947 cm2/s



Menghitung Laju reaksi spesifik Nilai k dihitung dengan menggunakan persamaan Arrhenius : τ=



10 =



1 k



C Ao 1 ( ln ( - X A ) - ln ( 1 - X A ) ) C Ao C Bo 1C Bo



1 k



5,4207 1 ( ln ( - 0,86 ) - ln ( 1 - 0,86) ) 5,4207 6,6025 16,6025



k = 0,904 m3/kmol.s Modulus Thiele, s Φ S2



 k S ext ρ C C AO  R  De 



  



0,5



Φ S 2 = 2,5 × 10



(Fogler, Hal.750)



3



Φ S2  168,7933



Menghitung Effectiveness Factor



0,904 × 0,6928 ×1732×5, 4207 0,000000947



0,5



299



3  Φ1cothΦ1  1 Φ12



η



(Pers.12.32, Fogler)



Karena nilai S2 besar (S2>20) maka, η



3 Φ S2



η=



3 η  0,0178 168,7933



Menghitung Koefisien Transfer Massa Pada permukaan katalis : μ



v = ρ =



1,51 x 10 - 5 kg/ms 746,0270 kg/m 3



= 2,02 x 10-8 m2/s N RE  N RE =



U dp



(Fogler, hal. 633)



1    v 2,3714 × 5 × 10 - 3 ( 1 - 0,4) 2,02 x 10 -8



N RE  976340,3498



Schmidst Number, Sc Sc 



v D AB



Sc =



2,02 × 10 -8 = 0,0595.10 - 4



(Fogler, hal. 633) 0,0034



Sh = NRe1/2 Sc1/3



(Pers.10-65, Fogler)



Sh = (976340,3498)1/2 (0,0034)1/3 Sh = 148,6107 ks 



1    D AB   d p



 S h  



(Fogler, hal. 633)



300



ks =



1



0,0595.10- 4 148,6107 5 × 10 -3



0,4 0,4



k s  0,265



ko 



1 1 1 1 1    k g S ext k g Sint η k ads k s



(Syarifuddin Ismail, hal.187)



Reaksi yang terjadi pada reaktor ini adalah reaksi gas katalitik. Katalis yang digunakan berbentuk pellet. Dengan adanya penggunaan katalis pada reaksi ini, kads akan menjadi sangat kecil. Dengan demikian, konstanta laju reaksi, global rate, akan ditentukan oleh koefisien transfer massa k g dan ks. Jadi persamaan diatas, direduksi menjadi : ko 



ko =



1 1 1 1   k g S ext k g S int η k s 1 1 1 1 + + 793,563443 × 0,6928 793,563443 × 0,4157 × 0,0178 0,265



k o  0,254



Menghitung Volume Tube Reaktor, VTR



VTR  FAO



Xf



dX A



Xi



R



 r



dimana : VTR = Volume Tube Reaktor FAO = Umpan Masuk Xf



= Konversi = 0,998



- rR = Laju reaksi -rR = ko [(CA.CB) + (CA)] di mana : CBo = CAo , XA + XB = 1 -rR = ko.[CAo(1 – XA).CBo(1 – XB) + CAo(1 – XA)]



301



-rR = ko [CAo(1 – XA).CAoXA + CAo(1 – XA)] -rR = ko [CAo2 XA(1 – XA) + CAo(1 – XA)] maka : X



f VTR dX A  FAO Xi k o C Ao 2 X A (1  X A )  C Ao (1 - X A )







VTR 







X  FAo  f dX A   k o C Ao  X i C Ao X A 1  X A   1  X A  



VTR



X  FAo  f dX A    k o C Ao  Xi C Ao X A  C Ao X A 2  1  X A 



VTR



F  Ao k o C Ao



X f  dX A   2  X i  C Ao X A   C Ao  1 X A  1



VTR



F  Ao k o C Ao



X f  dX A    Xi   C Ao X A - 1  X A  1 



VTR



X  FAo  f dX A    k o C Ao  Xi  C Ao X A  1 1 - X A  



Untuk menyelesaikan integral di atas, digunakan penjabaran menjadi pecahan parsial (faktor linear). integral di atas diubah ke persamaan berikut : Xf



X



X



f f dX A A dX A B dX A   X  C Ao X A  1 1 - X A  X  C Ao X A  1 X 1 - X A  i i i



A(1 – XA) + B(CAoXA + 1) = 1 A – AXA + BCAoXA + B = 1 (BCAo - A)XA + (A + B) = 1 dari persamaan di atas diketahui : BCAo – A = 0 A+B=1



302



didapatkan A =



C Ao 1  C Ao 1



B = 1 C Ao Sehingga persamaan awal dapat ditulis sebagai berikut :



VTR



F  Ao k o C Ao



Xf  C Ao  dX A      X i  1  C Ao   C Ao X A  1



X  f  1  dX A         X i  1  C Ao  1 - X A  



VTR 



FAo k o C Ao



 C Ao  1    1  X X   ln  C Ao X A  1 X f    ln 1 - X A  X f   i i  1  C Ao   1  C Ao  C Ao  



VTR 



FAo k o C Ao



 1   1  C Ao



VTR = (



100,0202



((



1



0,254) ( 5,4207) 1 + 5,4207



) × ln [ ( 5,4207) ( 0,86) + 1] - (



1 1 + 5,4207



  1 X  ln  C Ao X A  1 X f   i   1  C Ao



) × ln ( 1- 0,86) )



VTR  1,31045 m3



Safety Factor 20% : VTR = 1,2 x 1,31045 m3 VTR = 1,5725 m3 Menentukan Volume Katalis, Vk Vk = (1-Ф) VTR Vk = (1 – 0,4) 1,5725 m3 Vk = 0,9435 m3 Menentukan Berat Katalis, Wk Wk = ρ k  Vk Wk = 1732 kg/m3 x 0,9435 m3 Wk = 1634,1859 kg



Jumlah Tube, NT



  X  ln 1 - X A  X f  i  



303



Volume 1 buah Tube, VT VT 



π ID 2 L 4



VT =



3,14 ( 0,0525) 2 3,048 4



VT  0,0066 m3



NT 



VTR VT



NT =



1,5725 m 3 0,0066 m 3



N T  238,43 ≈ 239 tube



Diameter Shell Reaktor, Ds Tube disusun di dalam shell secara Triangular pitch dengan alasan : 1. Susunan Tube lebih kuat 2. Lebih mudah dibersihkan secara kimiawi 3. Koefisien perpindahan panas lebih baik A



Pt n



B Clearance, C’ =



C



OD 0,0605 m   0,03025 m 2 2



Jarak Antar Tube, PT PT  OD  C' PT  0,0605 + 0,03025 PT  0,09075 m



Luas Triangular Pitch, A



304











A



1 PT PT sin60 0 2



A



1 0,09075 0,09075 sin 60 0 2











A  0,00365 m2



Free Area, Af Af = A -



π  OD  2 8



A f  0,00365 



3,14  0,0605 2 8



A f  0,002213 m2



Total Free Volume, Vf Vf = Af . NT . L Vf = 0,002213 m2 x 239 x 3,048 m Vf = 1,6098 m3 Volume Shell, Vs Vs = Vf + VTR Vs = 1,6098 + 1,5725 Vs = 3,1824 m3 Area Shell, As As 



Vs L



As =



3,1824 m 3 3,048 m



A s  1,0441 m2



Diameter Shell, Ds



305



 N  2  A  4  Ds   T  π   Ds = (



0,5



239 × 2 × 0,00365 × 4 ) 3,14



0,5



Ds  1,4879 m



Tinggi Head Reaktor, Hs Head Reaktor berbentuk ellipsoidal Hs = 0,25 x Ds Hs = 0,25 x 1,4879 m Hs = 0,3720 m Tinggi Reaktor Total, HR HR = L + (2 x Hs) HR = 3,048 m + 2 (0,3720 m) HR = 3,7919 m Volume Head Reaktor, VHR π 3 VHR  2   Ds    24  VHR = 2



3,14 ( 1,4879 ) 3 24



VHR = 0,8619 m 3



Volume Total Reaktor, VR VR = VTR + VHR VR = 1,5725 m3 + 0,8619 m3 VR = 2,4344 m3 Tebal Dinding Shell Reaktor, t t



Pr C 2 S E  0,6 P



Tekanan Operasi (P)



(Peters, Tabel4, Hal.537) = 2,5 atm = 36,74 psi



306



Diameter Shell (Ds)



= 1,4879 m



Working Stress (S)



= 18700 psi



Efisiensi Pengelasan



= 0,85



Korosi yang diijinkan (C) = 0,003175 m t=



36,74 × 1,4879 × 0,5 + 0,003175 2 × 18700 × 0,85 - 0,6 × 36,74



= 0,004035 m OD = Ds + 2 t = 1,4879 m + 2 x 0,004035 m OD = 1,4959 m Pressure Drop, P dP G  dL ρ g c d p



  1    1501    μ  3    1,75G  dp     



(Pers.4.22, Fogler)



dimana : G adalah superficial mass velocity (kg/m2 s) G=xu G=



 qf ρ   a"N



G=



746,0270 kg/m 3



T



   



0,005125 m 3 /s 0,0021 m 2 × 239



G = 7,42 kg/m2s dP 7,42 1 - 0,4 = dL 746,0270 × 1 × 0,005 0,43



dP  15,822 N/m3 dL



dimana, Lo = 0 L = 3,048 m ∆P = 15,822 N/m3 (3,048 – 0) ∆P = 48,23 N/m3



150( 1 - 0,4) 1,51 x 10 - 5 + 1,75 × 7,42 0,005



307



∆P = 0,004823 atm Pout = (2,5 – 0,004823) atm Pout = 2,4952 atm Summary Tipe Tekanan Temperatur Diameter Tebal Dinding Pressure Drop



: Fixed Bed Multi Tubular : 2,5 atm : 125 oC : 1,4879 m : 0,004035 m : 0,004823 atm



REAKTOR (R-02) Fungsi



:



Tempat mereaksikan Metanol dan Asam Klorida untuk menghasilkan Metil Klorida.



Tipe



:



Operasi : Gambar :



Multi Tubular Fixed Bed Reactor Kontinyu



308



R-01



Gambar L.3.19. Reaktor-02 (R-02) Kondisi Operasi :  Tekanan, P



= 2,5 atm



(US Patent 6.111.153)



 Temperatur, T



= 125 oC



(US Patent 6.111.153)



 Konversi HCl, X



= 86 %



(US Patent 6.111.153)



 Residence Time,  = 10 s



(US Patent 5.196.618)



 Laju Alir Massa, W = 14347,7635 kg/jam = 3985,4898 gr/s  Densitas,  pada 125 oC (398,15 K) ρ dihitung dengan persamaan : A



ρ B



  T D  1  1     C  



,



di mana, ρ dalam kmol/m3 dan T dalam K



Dari Tabel 2-30 Perry’s Chemical Engineers’ Handbook diketahui : Senyawa CH3OH HCl H2O CH3Cl CH3O CH3 Senyawa CH3OH HCl



A 2,288 3,342 5,459 1,817 1,5693 m, kg



B 0,2685 0,2729 0,30542 0,25877 0,2679



1669,8707 511,1033



C D 512,64 0,2453 324,65 0,3217 647,13 0,081 416,25 0,2883 400,1 0,2882 n, mol Xi 32 0,1164 36,5 0,0356



ρi, kg/m3 677,8514 446,9790



H2O CH3Cl



7780,8805 42,0314



18 50,5



0,5423 0,0029



946,4141 360,1479



CH3O CH3 Total



4343,8776 14347,7635



46



0,3028 1,0000



613,8388



309



ρcampuran dihitung menggunakan rumus : 1 ρ campuran







Xi ρi



didapatkan ρcampuran = 804,7210 kg/m3 = 50,2385 lb/ft3 -



Viskositas,  Viskositas Campuran, campuran dihitung berdasarkan persamaan : campuran =  Xi . i Data viskositas masing-masing senyawa didapatkan dari gambar 2-32 Perry’s Chemical Engineers’ Handbook. Senyawa CH3OH



m, kg 1669,8707



Xi 0,1164



μi, cP 0,0140



HCl H2O



511,1033 7780,8805



0,0356 0,5423



0,0190 0,0135



CH3Cl CH3O CH3



42,0314 4343,8776 14347,7635



0,0029 0,3028



0,0140 0,0126



Total



1,0000



didapatkan campuran = 0,0096 cp = 9,63.10-6 kg/ms -



Reaksi : CH3OH(g) CH3OH(g)



+



HCl(g)



CH3Cl(g) CH3OCH3(g)



Data Katalis : 



Nama Katalis = Zinc Chloride dengan support Karbon Aktif



+



H2O (g)



310







Diameter Katalis, dp = 5 mm Dari Perry’s Chemical Engineers’ Handbook diketahui range diameter katalis untuk reaktor Multi Tubular Fixed Bed adalah 3 – 5 mm.







Porositas, 



= 0,4







Densitas, c



= 1732 kg/m3







Turtuosity,  = 3,0







Constriction Factor,  = 0,7 Volumetric Flowrate, qf qf 



=



W ρ 14347,7635 kg/jam 804,7210 kg/m3



= 17,8295 m3/jam = 0,00495 m3/s Konsentrasi Umpan FAO = mol umpan HCl =



511,1033 kg/jam 36,5 kg/kmol



FAO



CAO = q f



=



= 14,0028 kmol/jam



14,0028 kmol/jam = 0,7854 kmol/m3 17,8295 m 3 /jam



FBO = mol umpan CH3OH =



1669,8707 kg/jam 32 kg/kmol



FBO



CBO = q f



=



= 52,1835 kmol/jam



52,1835 kmol/jam = 2,9268 kmol/m3 17,8295 m 3 /jam



Konstanta Laju Reaksi, ko



311



Pada reaktor ini, terjadi reaksi fase gas dengan menggunakan katalis solid berbentuk bola berpori. Katalis yang digunakan adalah ZnCl2 dengan support Karbon Aktif. Dalam hal ini, Karbon Aktif berperan sebagai bola berpori. Reaksi yang terjadi antar reaktan dengan bantuan katalis ZnCl 2 berlangsung di dalam pori-pori Karbon Aktif ini. Tahapan yang terjadi untuk reaksi jenis ini adalah : 1. Difusi Eksternal Reaktan 2. Difusi Internal Reaktan 3. Adsorpsi reaktan A pada permukaan katalis 4. Reaksi pada permukaan katalis ( A



B)



5. Desorbsi produk pada permukaan katalis 6. Difusi Internal produk 7. Difusi eksternal produk



r



Gambar katalis pellet dengan pori Large pore



Large openings Compressed porous powder



Small pores Single large catalyst pellet



Model of pore structure



312



A



B



7



A



B



External diffusion



7



1



B 2



2 6



6 3



Internal diffusion



5 4



A-B Catalytic surface



Tahapan Reaksi Gas Katalitik



Menghitung nilai Diffusivitas HCl dalam CH3OH



D HCl CH 3OH



1/2    1   T 3/2  1   M HCl M CH 3OH      0,0018583  2  p t σ HCl CH 3OH Ω HCl CH 3OH   



       



(Pers. 11-11, J.M. Smith)



dimana : D HCl CH 3OH = Bulk Diffusivity, cm2/s



T



= Temperatur, K



M HCl M CH 3OH



= Berat molekul HCl dan CH3OH



Pt



= Tekanan total, atm



σ HCl CH 3OH



= Konstanta Lennard-Jones



Ω HCl CH 3OH = Integral Colision



Dari Tabel E.1 , Transport Phenomena- R.Byron Bird Senyawa HCl CH3OH σ HClCH3OH 



Bird)



σ, Ǻ 4,374 3,585







ε/k,K 378 507



1 σ HCl  σ CH3OH 2







(Pers. 17.3-18, R. Byron



313



σ HCl CH 3OH 



1  4,374  3,585 2



σ HCl CH 3OH  3,9795



ε HCl CH OH 3



k



(Pers. 17.3-19, R. Byron



 (ε HCl ε CH 3OH )1/2



Bird) ε HCl CH OH 3



k ε HCl  CH k



3 OH



 ( 378  507)1/2  437,774



Pada temperatur 125 oC = 398,15 K kT 398,15 = = ε HCl CH OH 437,774 0,91 3



Dari Tabel E.2 , Transport Phenomena - R.Byron Bird, untuk ε didapatkan harga Ω



HCl  CH 3 OH



kT



= 1,13



HCl  CH 3 OH



= 1,517



Jadi, 1 1 1/2 + 36,5 32 2,5 ( 3,9795) 2 1,517



( 398,15) 3/2 D HCl CH 3OH = 0,0018583



D HCl CH 3OH = 0,0595 cm2/s



Luas permukaan terluar partikel solid per satuan volume gas, ac ac 



6 1    dp



(Fogler, Hal.766)



314



61  0,4  5.10 3



=



= 720 m2/m3 Menghitung Superficial Velocity, U Untuk menghitung U, diperlukan data spesifikasi tube. dk  0,15 dT dT =



(JM Smith, hal. 571)



dk 0,50 cm = = 3,33 cm = 1,3123 in 0,15 0,15



Dari Perry’s Chemical Engineers’ Handbook Bab 23 hal. 53 diketahui bahwa diameter tube reaktor maximal untuk katalis dengan range diameter 3-5 mm adalah 2 in. Berikut adalah data spesifikasi tube IPS 2 in diambil dari Tabel. 11, Kern IPS



= 2 in



Sch. No.



= 40



OD



= 2,38 in = 0,0605 m



ID



= 2,067 in = 0,0525 m



a”



= 3,35 in2 = 0,0021 m2



Lt



= 8 ft



U  U=



= 2,4384 m



qo a"



0,00495 m 3 /s 0,0021 m 2



= 2,2915 m/s



Menghitung Faktor Perpindahan Massa (jD) jD



0,455  d p U       μ 



jD =



0,407



- 0,407 0,455 5 × 10 - 3 × 2,2915 ( ) 0,4 9,63 x 10 - 6



315



j D  0,0637



Menghitung Koefisien Transfer Massa Reaktan ke permukaan Katalis kg 



jD G  ρ D    ρ  μ 



2/3



(Syarifuddin Ismail, hal. 181)



dimana : kg



= Koefisien Perpindahan Massa , cm/s



ρ



= Densitas fluida, gr/cm3



G



= Laju alir linear massa, gr/s



μ



= Viskositas, poise



D



= Diffusivitas Bulk katalis dalam fluida, cm2/s



0,0637 × 3823,6794 kg = 0,804721



0,804721× 0,0595 ( ) 9,63 x 10 - 6



2/3



k g = 91925,89 cm/s



kg = 919,2589 m/s Menghitung Luas permukaan eksternal Katalis S ext 



6 ρc d p



S ext =



6 3 1732 kg/m 5 x 10 - 3 m



S ext  0,6928 m2/kg



Menghitung Luas permukaan Internal Katalis S int  Sint =



ac ρc



(Fogler, eq.11-55) 720 m -1



1732 kg/m 3



S int  0,4157



dengan dp = 5 mm dan Rpori = 2,5 10-3 m



316



De 



D AB  σ τ



De =



0,0595 cm 2 /s × 0,4 × 3,9795 10



D e  0,00947 cm2/s



Menghitung Laju reaksi spesifik Nilai k dihitung dengan menggunakan persamaan Arrhenius : τ=



10 =



1 k



C Ao 1 ( ln ( - X A ) - ln ( 1 - X A ) ) C Ao C Bo 1C Bo



1 k



1 0,7854 12,9268



( ln (



0,7854 - 0,86 ) - ln ( 1 - 0,86) ) 2,9268



k = 0,0683 m3/kmol.s Modulus Thiele, s Φ S2



 k S ext ρ C C AO  R  De 



  



0,5



Φ S 2 = 2,5 × 10



(Fogler, Hal.750)



-3



0,0683 × 0,6928 ×1732× 0,7854 0,000000947



Φ S2  46,3976



Menghitung Effectiveness Factor η



3  Φ1cothΦ1  1 Φ12



Karena nilai S2 besar (S2>20) maka, η



3 Φ S2



η=



3 46,3976



η  0,065



(Pers.12.32, Fogler)



0,5



317



Menghitung Koefisien Transfer Massa Pada permukaan katalis : μ



v = ρ =



9,63 x 10 - 6 kg/ms 804,7210 kg/m 3



= 1,2 x 10-8 m2/s N RE 



U dp



(Fogler, hal. 633)



1    v



N RE =



2,2915 × 5 × 10 -3 ( 1 - 0,4) 1,2 x 10-8



N RE  1596170,988



Schmidst Number, Sc Sc 



v D AB



Sc =



1,2 × 10 -8 = 0,0595.10 - 4



(Fogler, hal. 633) 0,002



Sh = NRe1/2 Sc1/3



(Pers.10-65, Fogler)



Sh = (1596170,988)1/2 (0,002)1/3 Sh = 159,4654 ks 



ks =



1    D AB   d p 1



0,4 0,4



 S h  



(Fogler, hal. 633)



0,0595.10- 4 159,4654 5 × 10 - 3



k s  0,2846



ko 



1 k g S ext



1 1 1 1    k g Sint η k ads k s



(Syarifuddin Ismail, hal.187)



318



Reaksi yang terjadi pada reaktor ini adalah reaksi gas katalitik. Katalis yang digunakan berbentuk pellet. Dengan adanya penggunaan katalis pada reaksi ini, kads akan menjadi sangat kecil. Dengan demikian, konstanta laju reaksi, global rate, akan ditentukan oleh koefisien transfer massa k g dan ks. Jadi persamaan diatas, direduksi menjadi : ko 



1 1 1 1   k g S ext k g S int η k s 1



ko =



1 919,2589



× 0,6928



+



1 1 + 919,2589 × 0,4157 × 0,06465 0,2846



k o  0,281



Menghitung Volume Tube Reaktor, VTR



VTR  FAO



Xf



dX A



Xi



R



 r



dimana : VTR = Volume Tube Reaktor FAO = Umpan Masuk Xf



= Konversi = 0,998



- rR = Laju reaksi -rR = ko [(CA.CB) + (CA)] di mana : CBo = CAo , XA + XB = 1 -rR = ko.[CAo(1 – XA).CBo(1 – XB) + CAo(1 – XA)] -rR = ko [CAo(1 – XA).CAoXA + CAo(1 – XA)] -rR = ko [CAo2 XA(1 – XA) + CAo(1 – XA)] maka : X



f VTR dX A  2 FAO Xi k o C Ao X A (1  X A )  C Ao (1 - X A )











319



X f  dX A    X i C Ao X A 1  X A   1  X A  



VTR



F  Ao k o C Ao



VTR



X  FAo  f dX A    k o C Ao  Xi C Ao X A  C Ao X A 2  1  X A 



VTR



F  Ao k o C Ao



VTR



X  FAo  f dX A    k o C Ao  Xi   C Ao X A - 1  X A  1 



VTR



X  FAo  f dX A    k o C Ao  Xi  C Ao X A  1 1 - X A  



X f  dX A   2  X i  C Ao X A   C Ao  1 X A  1



Untuk menyelesaikan integral di atas, digunakan penjabaran menjadi pecahan parsial (faktor linear). integral di atas diubah ke persamaan berikut : Xf



X



X



f f dX A A dX A B dX A   X  C Ao X A  1 1 - X A  X  C Ao X A  1 X 1 - X A  i i i



A(1 – XA) + B(CAoXA + 1) = 1 A – AXA + BCAoXA + B = 1 (BCAo - A)XA + (A + B) = 1 dari persamaan di atas diketahui : BCAo – A = 0 A+B=1 didapatkan A =



C Ao 1  C Ao 1



B = 1 C Ao Sehingga persamaan awal dapat ditulis sebagai berikut :



320



VTR



X FAo  f  C Ao  dX A      k o C Ao  X i  1  C Ao   C Ao X A  1



X  f  1  dX A          1  C 1 X Ao  A    Xi 



VTR 



FAo k o C Ao



 C Ao  1    1  X X   ln  C Ao X A  1 X f    ln 1 - X A  X f   i i  1  C Ao   1  C Ao  C Ao  



VTR 



FAo k o C Ao



 1   1  C Ao



VTR = (



14,0028



((



1



0,281) ( 0,7854) 1 + 0,7854



) × ln [ ( 0,7854) ( 0,86) + 1] - (



1 1 + 0,7854



  1 X  ln C Ao X A  1 X f   i   1  C Ao



) × ln ( 1 - 0,86) )



VTR  0,6937 m3



Safety Factor 20% : VTR = 1,2 x 0,6937 m3 = 0,6937 m3 Menentukan Volume Katalis, Vk Vk = (1-Ф) VTR Vk = (1 – 0,4) 0,6937 m3 Vk = 0,416 m3 Menentukan Berat Katalis, Wk Wk = ρ k  Vk Wk = 1732 kg/m3 x 0,416 m3 Wk = 720,9044 kg Jumlah Tube, NT Volume 1 buah Tube, VT VT 



π ID 2 L 4



VT =



3,14 ( 0,0525) 2 2,4384 4



VT  0,0053 m3



  X  ln 1 - X A  X f  i  



321



NT 



VTR VT



NT =



1,5725 m 3 0,0053 m 3



N T  131,48 ≈ 132 tube



Diameter Shell Reaktor, Ds Tube disusun di dalam shell secara Triangular pitch dengan alasan : 1. Susunan Tube lebih kuat 2. Lebih mudah dibersihkan secara kimiawi 3. Koefisien perpindahan panas lebih baik



A



Pt n C



B Clearance, C’ =



OD 0,0605 m   0,03025 m 2 2



Jarak Antar Tube, PT PT  OD  C' PT  0,0605 + 0,03025 PT  0,09075 m



Luas Triangular Pitch, A











A



1 PT PT sin60 0 2



A



1 0,09075 0,09075 sin 60 0 2











322



A  0,00365 m2



Free Area, Af Af = A -



π  OD  2 8



A f  0,00365 



3,14  0,0605 2 8



A f  0,002213 m2



Total Free Volume, Vf Vf = Af . NT . L Vf = 0,002213 m2 x 132 x 2,4384 m Vf = 0,7084 m3 Volume Shell, Vs Vs = Vf + VTR Vs = 0,7084 + 0,6937 Vs = 1,4022 m3 Area Shell, As As 



Vs L



As =



1,4022 m 3 2,4384 m



A s  0,5750 m2



Diameter Shell, Ds  N  2  A  4  Ds   T  π   Ds = (



0,5



132 × 2 × 0,00365 × 4 ) 3,14



0,5



323



Ds  1,1049 m



Tinggi Head Reaktor, Hs Head Reaktor berbentuk ellipsoidal Hs = 0,25 x Ds Hs = 0,25 x 1,1049 m Hs = 0,2762 m Tinggi Reaktor Total, HR HR = L + (2 x Hs) HR = 2,4384 m + 2 (0,2762 m) HR = 2,9908 m Volume Head Reaktor, VHR π 3 VHR  2   Ds    24 



VHR = 2



3,14 ( 1,1049 ) 3 24



VHR = 0,3529 m 3



Volume Total Reaktor, VR VR = VTR + VHR VR = 0,6937m3 + 0,3529 m3 VR = 1,0466 m3 Tebal Dinding Shell Reaktor, t t



Pr C 2 S E  0,6 P



(Peters, Tabel4, Hal.537)



Tekanan Operasi (P)



= 2,5 atm = 36,74 psi



Diameter Shell (Ds)



= 1,1049 m



Working Stress (S)



= 18700 psi



Efisiensi Pengelasan



= 0,85



Korosi yang diijinkan (C) = 0,003175 m



324



t=



36,74 × 1,1049 × 0,5 + 0,003175 2 × 18700 × 0,85 - 0,6 × 36,74



= 0,0038 m OD = Ds + 2 t = 1,1049 m + 2 x 0,0038 m OD = 1,1125 m Pressure Drop, P dP G  dL ρ g c d p



  1    1501    μ  3    1,75G  dp     



(Pers.4.22, Fogler)



dimana : G adalah superficial mass velocity (kg/m2 s) G=xu G=



 qf ρ   a"N



G=



804,7210 kg/m3



T



   



0,00495 m3/s 0,0021 m 2 × 132



G = 14,0254 kg/m2s dP 14,0254 1 - 0,4 = dL 804,7210 × 1 × 0,005 0,4 3



150( 1 - 0,4) 9,63 x 10 -5 + 1,75 × 14,0254 0,005



dP  51,6961 N/m3 dL



dimana, Lo = 0 L = 2,4384 m ∆P = 51,6961 N/m3 (2,4384 – 0) ∆P = 126,06 N/m3 ∆P = 0,012606 atm Pout = (2,5 – 0,012606) atm Pout = 2,4874 atm Summary



325



Tipe Tekanan Temperatur Diameter Tebal Dinding Pressure Drop



: Fixed Bed Multi Tubular : 2,5 atm : 125 oC : 1,1049 m : 0,0038 m : 0,012606 atm



STRIPPER-01 (ST-01) Fungsi



:



Memisahkan HCl dari campuran produk keluaran bottom Absorber-01



Tipe



:



Packed Tower



Jumlah :



1 buah



Bahan



Stainless Steel



:



Gambar :



ST-01



Gambar L.3.20. Stripper-01 (ST-01) Kondisi Operasi : -



Tekanan



= 1,5 atm



-



Temperatur = 75 oC Gas Masuk -



Laju alir massa, G1



= 502,1258 kg/jam = 0,1395 kg/s



-



Viskositas Gas, G



= 0,0153 cp = 1,53 x 10-5 kg/ms



-



Densitas Gas, G



= 2,3409 kg/m3 = 0,1461 lb/ft3



326



-



Diffusivitas Gas, DG = 3,85 x 10-5 m2/s



-



BMAVG



= 18,015 kg/kmol



Liquid Masuk -



Laju alir massa, L



= 21.230,617 kg/jam = 5,8974 kg/s



-



ρ liquid = 1044,1549 kg/m3



-



Viskositas Liquid, L Viskositas Campuran, campuran dihitung berdasarkan persamaan : campuran =  Xi . i Viskositas masing-masing komponen dicari menggunakan gambar 2-32 Perry’s Chemical Engineer’s Handbook. Senyawa m,kg HCl 70,8807 MeOH 1277,6300 H2O 19882,100 Jumlah 21230,62



Xi 0,0049 0,0900 0,5573 1



μ, cP 1,1 0,32 0,4



mol 1,98 40,7572 448,6493 491,3865



didapatkan campuran = 0,2572 cp = 2,57 x 10-4 kg/m.s -



Diffusivitas Liquid, DL



-



BMAVG = 19,2356 kg/kmol



= 2,89 x 10-10 m2/s



Dari Tabel 6.4 Mass-Transfer Operations-Robert E Treyball dipilih : -



Jenis Packing = Ceramic Raschig Ring



-



Nominal size = 2 in = 50 mm



Dari Tabel 6.3, 6.4 dan 6.5 didapatkan : Wall Thickness = 6 mm Cf



= 65



CD



= 135,6







= 0,74



ap



= 92 m2/m3



327



m



= 34,03



n



=0



p



= 0,362



ds



= 0,0725



1.



Menentukan Diameter tower, Dt Lihat Grafik 6.34 Mass-Transfer Operations-Robert E Treyball L' G'



=



 ρG   ρL  ρG



  



0,5



(Treyball, hlmn 195)



21.230,617 kg/jam  2,3409 kg/m 3  502,1258 kg/jam  1044,1549 kg/m 3  2,3409 kg/m 3



  



0,5



= 2,0042 L' Dengan menarik garis lurus nilai ' G



 ρG   ρL  ρG



0,5



  = 2,0042 ke garis pressure drop 



gas pada 200 (N/m2)/m, maka didapat ordinat = 0,00125 0,1



G 2 Cf μ L J = 0,00125 ρG ρL  ρG gc  0,00125 ρ G  ρ L  ρ G  g c  G'    0,1 Cf μ L J  



0.5



dimana, J = 1 dan gc = 1



 0,00125 x 2,34091044,1549  2,3409  1 G'    0,1 65 x 2,57 x 10 - 4 1  











G’ = 0,3274 kg/m2s Tower Cross Sectional Area, A A



G 0,1395 kg/s  = 0,4260 m2 G' 0,3274 kg/m 2 s



Diameter Tower, Dt 4 A Dt     π 



0,5



 4 x 0,4260    3,14  



Jari-jari (r) = 0,3683 m



0,5







0,7367 m



0.5



328



2.



Menentukan Hold up Untuk Liquid Sc L 



μL ρL DL



Sc L 



2,57.10 4 1044,1549 x 2,89 x 10 -10



(Treyball,hlmn 205)



Sc L  852,326 m



Untuk Gas Sc G 



μG ρG DG



Sc G 



1,53 x 10 -5 2,3409 x 3,85 x 10 -5



(Treyball,hlmn 205)



Sc G  0,1698



L’ =



5,8974 kg/s = 13,8427 kg/m2s 0,4260 m 2



Log L’ = 1,1412 Dari Tabel 6.5, untuk L < 0,012, maka :  = 1,508 x ds0,376 = 1,508 (0,0725)0,376 = 0,5622  LSW   LSW 



2,47  10 4  d s  1, 21 2,47  10 4



 0,0725  1, 21



 LSW  0,0059



Φ LTW 



 2,09  10  737,5  L' 6



ds 2



β



329



Φ LTW 



 2,09  10  737,5  13,8427 6



0,5622



 0,0725 2



 LTW  0,0713



LOW = LTW - LSW LOW = 0,0713 – 0,0059 = 0,0654 H



H



9,57 L'0,57 μ 0,13  σ    0,43 ρ L 0,84  2,024 L' 1  0,073 



975,7 13,8427 



0,57







0,1737 0,262logL'



 2,57  10 



4 0,13



1044,1549  0,84 2,024 13,8427 



0,43



 0,0773     1  0,073 



0,1737 0,262log13,8427







H  0,0031 LO = LOW x H LO = 0,0654 x 0,0031 LO = 0,0623



Φ LS  Φ LS 



0,0486  μ L



0,02



ds1,21ρ L



 σ 0,99



0,37



0,0486   2,57. 10  4 



0,02



  0,0773



0,99



 0,0725 1,21 1044,1549 0,37



 LS  0,0059



Lt = LO + LS



(Pers. 6.69, Treyball)



Lt = 0,0623 + 0,0059 Lt = 0,0682 3.



Menentukan Interfacial Area n



 808 G'   L'p aAW = m  0,5   ρG 



330



0



 808  0,3274   13,8427 0,5  2,3409 



aAW = 34,03 



0,362



aAW = 88,1017 m2/m3 aA = a AW



Φ LO Φ LOW



(Pers. 6.73, Treyball) 0,0623



aA = 88,1017 m2/m3 0,0654 aA = 83,8709 m2/m3 4.



Operating Void Space dalam packing ε = 0,74 ε LO  ε  Φ LT



(Pers. 6.71, Treyball)



ε LO  0,74  0,0682 ε LO  0,6717



5.



Koefisien Fase Gas, FG



FG S CG G G



2/3



 dsG '   1,195   μ G 1  ε LO  



0,36



(Pers. 6.70, Treyball)



G' 0,3274  = 0,0182 BM 18,015



maka, FG  0,1698 0,1395



2/3



  0,0725  0,3274  1,195  -5 1,53 x 10 1  0,6717  



0,36



FG = 0,0061 6. Koefisien Fase Liquid, KL  ds L'  K L ds  25,1  DL  μL 



0,45



S CL



0,5



K L 0,0725  0,0725  13,8427   25,1  10 4 2,89  10  2,57 x 10 



(Pers. 6.72, Treyball) 0,45



 852,326 0,5



331



KL = 1,207 x 10-4 kmol/m2s C



ρL 1044,1549   54,2819 BM 19,2358



FL = KL x C = 1,207 x 10-4 kmol/m2s x 54,2819 = 0,0066 7.



Koefisien Volumetrik FG x aA = 0,0061 x 83,8709 = 0,5106 kmol/m3s FL x aA = 0,0066 x 83,8709 = 0,5494 kmol/m3s



8.



Tinggi Transfer Unit, Htol H tg  L



G 0,0182   0,0356 FG  a A 0,5106



L' 13,8427   0,7196 BM AVG 19,2358



H tl 



L 0,7196   1,3099 FL  a A 0,5494



Pada T = 75 oC, Tekanan parsial HCl = 117,3641 atm P = 1,5 atm m



P 117,3641   78,2428 atm Pt 1,5



H tol  H tl 



L H tg mG



H tol  1,3099 



0,7196  0,0356 78,2428  0,0182



H tol  1,3279



9.



Number of Transfer Unit, Ntol A



L mG 0,7196



A = 78,2428 x 0,0182



(Pers. 8.54, Treyball)



332



A = 0,5053 1/A = 1,9776 dimana : y1 = 0 y2 = 0 x1 = 0,0348 x2 = 0,0349 x 1  y1 /m 0,0348  0/78,2428   0,9977 x 2  y1 /m 0,0349  0/78,2428



Dari grafik 8.20, Mass Transfer Operations, Robert E treyball, didapatkan harga : N tol  3,5



Tinggi Packing,z Z = Htol x Ntol Z = 1,3279 x 3,5 Z = 4,6476 m 10. Pressure Drop Pressure Drop untuk packing yang terbasahi dengan tinggi (z) = 4,6476 m P1 = P x Z = 200 (N/m2)/m x 4,6476 m = 929,5159 N/m2 ΔP2 G' 2  CD z ρG ΔP2 0,3274 2  135,6 4,6476 2,3409



ΔP2  28,8567 N/m2



Total Pressure Drop : ΔP  ΔP1  ΔP2 ΔP  929,5159  28,8567



333



ΔP  958,3726 N/m2 ΔP  0,0095 atm 11. Tebal Dinding, t t



P.r C S E  0,6P



(Tabel 4,hlmn 537, Peters)



Tekanan Design (P)



= 1,5 atm = 22,044 psi



Jari-jari (r)



= 0,3683 m



Working Stress yang diizinkan (S)



= 18700 psi (Hlmn 538, Peters)



Korosi yang diizinkan (C)



= 0,003175 m (Tabel 23.2, Peters)



Efisiensi Pengelasan (E)



= 0,85 (Hlmn 638, Coulson)



t



26,46 . 0,3683  0,003175 18700  0,85  0,6  26,46



t = 0,0038 m OD = 2 t + D OD = 2 (0,0038) + 0,7367 OD = 0,7443 m Summary Tipe Tekanan Temperatur Diameter Tebal Dinding Tinggi Packing Pressure Drop



: : : : : : :



Packed Tower 1,5 atm 75 oC 0,7367 m 0,0038 m 4,6476 m 0,0095 atm



TANKI – 01 Fungsi



: Menampung bahan baku larutan HCl



334



Bentuk



: Silinder vertical dengan tutup ellipsoidal pada bagian atas



Bahan konstruksi



: Carbon steel



Gambar



:



Gambar L.3.21. Tanki-01 (T-01) a. Data : 30 oC



Temperatur, T Tekanan, P



: 1 atm



Laju alir massa, W



: 9866,8582 kg/jam



Densitas campuran , ρ



: 1324,011



Lama persediaan



: 7 hari



kg/m3



b. Kapasitas Tanki, Vt Vt



=



Lajualir x 24 jam x Lama persediaan Densitas



Vt



=



9866,8582 kg / jam x 24 jam x 7 hari 1324,011 kg / m 3



Vt



= 1251,977 m3



Safety factor



= 10 %



Kapasitas tanki (Vt) = (1+0,1) x 1251,977 m3 = 1377,175 m3



c. Diameter Tanki, D  Volume total, Vt Dimana,



H



= tinggi silinder = 3/2 D



335



= ( π / 4 ) . D2 . H



Vt



= ( 3 / 8 )π . D3 D



 =   8 / 3 x



D



=



D



= 10,536 m







= ( π / D3 ) / 24



Ve



=



3,14 10,536 3



x



1 24



= 153,019 m3  Volume silinder, Vs Vt



= Ve + Vs



Vs



= Vt – Ve



Vs



= 1377,175 – 153,019 m3



Vs



= 1224,155 m3



d. Tinggi tanki total, Ht  Tinggi Silinder H



=3/2D = 3 / 2 x 10,536 m = 15,804 m



 Tinggi Elipsoidal, h h



=¼xD



h



= ¼ x 10,536 m



h



= 2,634 m



1/ 3



8 1377,175 1 / 3 x 3 3,14



 Volume Elipsoidal head Ve



Vt   



336



 Tinggi tanki total, Ht Ht



=H+h = 15,804 + 2,634 m = 18,438 m



e. Tebal dinding tanki, t P xR



t



= S x E  0,6 x P  C



P



= Tekanan design



=1



R



= Jari – jari kolom



= 5,268 m



S



= Working stress allowable



= 1272,4551 atm



Ej



= Welding joint efficiency



= 0,85



C



= Tebal korosi yang diijinkan



= 0,003175



t



= 1272,4551



t



= 0,00805 m



t



= 0,805 cm



Dimana : atm



m



1 atm x 5,268 m + 0,003175 m x 0,85 0,6 x 1 atm



Outside diameter (OD)



= ID + 2 t = 10,536 m + 2 . 0,00805 m = 10,552 m



TANKI – 02 (T-02) Fungsi



: Menampung bahan baku larutan metanol



Bentuk



: Silinder vertical dengan tutup ellipsoidal pada bagian atas



Bahan konstruksi



: Carbon steel



337



Gambar



:



Gambar L.3.22. Tanki-02 (T-02) a. Data : 30 oC



Temperatur, T Tekanan, P



: 1 atm



Laju alir massa, W



: 4480,9905



kg/jam



Densitas campuran , ρ



: 851,611 kg/m3



Lama persediaan



: 7 hari



b. Kapasitas Tanki, Vt Vt



=



Laju alir x 24 jam x Lama persediaan Densitas



Vt



=



4480,9905 kg / jam x 24 jam x 7 hari 851,611 kg / m 3



Vt



= 883,963 m3



Safety factor



= 20 %



Kapasitas tanki (Vt) = ( 1 + 0,2 ) x 883,963 m3 = 1060,755 m3



c. Diameter Tanki, D  Volume total, Vt Dimana,



H



= tinggi silinder = 3/2 D



Vt



= ( π / 4 ) . D2 . H = ( 3 / 8 )π . D3



338



D



 =   8 / 3 x



D



=



D



= 9,658 m







Vt   



1/ 3



8 1060,755 1 / 3 x 3 3,14



 Volume Elipsoidal head Ve



= ( π / D3 ) / 24



Ve



=



Ve



= 111,862 m3



3,14 (9,658 ) 3



x



1 24



m3



 Volume silinder, Vs Vt



= Ve + Vs



Vs



= Vt – Ve



Vs



= 1060,755 – 111,862 m3



Vs



= 942,894 m3



d. Tinggi tanki total, Ht  Tinggi Silinder H



=3/2D = 3 / 2 x 9,658 m = 14,487 m



 Tinggi Elipsoidal, h h



=¼xD



h



= ¼ x 9,658 m



h



= 2,414 m



 Tinggi tanki total, Ht Ht



=H+h



339



= 14,487 + 2,414 m = 16,901 m e. Tebal dinding tanki, t P xR



t



= S x E  0,6 x P  C



P



= Tekanan design



=1



R



= Jari – jari kolom



= 4,829 m



S



= Working stress allowable



= 932,2297 atm



Ej



= Welding joint efficiency



= 0,85



C



= Tebal korosi yang diijinkan



= 0,003175 m



t



= 932,2297 x 0,5 0,6 x 1 atm + 0,003175



t



= 0,00927 m



t



= 0,927 cm



Dimana :



1 atm x 4,829 m



Outside diameter (OD)



= ID + 2 t = 9,658 m + ( 2 x 0,00927) = 9,676 m



TANKI – 03 (T-03) Fungsi



: Menampung produk metil klorida



Bentuk



: Spherical tank



Bahan konstruksi



: Stainlees steel



Gambar



:



m



atm



340



Gambar L.3.23. Tanki-03 (T-03) a. Data : 30 oC



Temperatur, T Tekanan, P



: 10 atm



Laju alir massa, W



: 5000,031 kg/jam



Densitas campuran , ρ



: 975,903



Lama persediaan



: 7 hari



kg/m3



b. Kapasitas Tanki, Vt Vt



=



Lajualir x 24 jam x Lama persediaan Densitas



Vt



=



5000,031 kg / jam x 24 jam x 7 hari 975,903 kg / m 3



Vt



= 860,747 m3



Safety factor



= 10 %



Kapasitas tanki (Vt) = ( 1 + 0,1 ) x 860,747 m3 = 946,822



m3



c. Diameter Tanki, D  Volume total, Vt Dimana,



H



= tinggi silinder = 3/2 D



Vt



= ( π / 4 ) . D2 . H = ( 3 / 8 )π . D3



341



D



 =   8 / 3 x



D



=



D



= 9,299 m







Vt   



1/ 3



8 946,822 3 x 3 3,14



 Volume Elipsoidal head Ve



= ( π / D3 ) / 24



Ve



=



3,14 (9,299 ) 3



x



1 24



= 105,202 m3  Volume silinder, Vs Vt



= Ve + Vs



Vs



= Vt – Ve



Vs



= 946,822– 105,202 m3



Vs



= 841,619 m3



d. Tinggi tanki total, Ht  Tinggi Silinder H



=3/2D = 3 / 2 x 9,299 m = 13,948 m



 Tinggi Elipsoidal, h h



=¼xD



h



= ¼ x 9,299 m



h



= 2,325 m



 Tinggi tanki total, Ht Ht



=H+h



342



= 13,948 + 2,325 m = 16,273 m e. Tebal dinding tanki, t P xR



t



= S x E  0,6 x P  C



P



= Tekanan design



=1



R



= Jari – jari kolom



= 4,649 m



S



= Working stress allowable



= 1272,4551 atm



Ej



= Welding joint efficiency



= 0,85



C



= Tebal korosi yang diijinkan



= 0,003175 m



t



= 1272,4551 x 0,5 0,6 x 1 atm + 0,003175 m



t



= 0,6912 m



t



= 69,12 cm



Dimana : atm



1 atm x 4,649 m



Outside diameter (OD)



= ID + 2 t = 9,299 m + ( 2 x 0,6912) m = 10,681 m



TANKI – 04(T-04) Fungsi



: Menampung H2O



Bentuk



: Silinder vertical dengan tutup ellipsoidal pada bagian atas



Bahan konstruksi



: Carbon steel



Gambar



:



343



Gambar L.3.24. Tanki-04 (T-04) a. Data : 30 oC



Temperatur, T Tekanan, P



: 1 atm



Laju alir massa, W



: 108,897 kg/jam



Densitas campuran , ρ



: 1077,406



Lama persediaan



: 7 hari



kg/m3



b. Kapasitas Tanki, Vt Vt



=



Lajualir x 24 jam x Lama persediaan Densitas



Vt



=



108,897 kg / jam x 24 jam x 7 hari 108,897 kg / m 3



Vt



= 16,980 m3



Safety factor



= 10 %



Kapasitas tanki (Vt) = ( 1 + 0,1 ) x 16,980 m3 = 18,678 m3



c. Diameter Tanki, D  Volume total, Vt Dimana,



H



= tinggi silinder = 3/2 D



Vt



= ( π / 4 ) . D2 . H = ( 3 / 8 )π . D3



D



Vt   =   8 / 3 x    



1/ 3



344



8 18,678 x 3 3,14



D



=



D



= 2,513 m



 Volume Elipsoidal head Ve



= ( π / D3 ) / 24



Ve



=



3,14 ( 2,513 ) 3



x



1 24



= 2,075 m3  Volume silinder, Vs Vt



= Ve + Vs



Vs



= Vt – Ve



Vs



= 18,678 – 2,075 m3



Vs



= 16,603 m3



d. Tinggi tanki total, Ht  Tinggi Silinder H



=3/2D = 3 / 2 x 2,513 m = 3,769 m



 Tinggi Elipsoidal, h h



=¼xD



h



= ¼ x 2,513 m



h



= 0,628 m



 Tinggi tanki total, Ht Ht



=H+h = 32,499 + 5,417 m = 4,397 m



3



345



e. Tebal dinding tanki, t P xR



t



= S x E  0,6 x P  C



P



= Tekanan design



=1



R



= Jari – jari kolom



= 1,256 m



S



= Working stress allowable



= 932,2297 atm



Ej



= Welding joint efficiency



= 0,85



C



= Tebal korosi yang diijinkan



= 0,003175 m



t



= 932,2297 x 0,5 0,6 x 1 atm + 0,003175 m



t



= 0,00476 m



t



= 0,476 cm



Dimana : atm



1 atm x 1,256 m



Outside diameter (OD)



= ID + 2 t = 2,513 m + ( 2 x 0,00476) m = 2,522 m



VAPORIZER – 01 (V-01) Fungsi



: Menguapkan bahan baku HCl sebelum diinput ke Reaktor-01



Type



: Shell and Tube Heat Exchanger



Bahan



: Stainless Steel



Gambar



:



346



Gambar L.3.25. Vaporizer-01 (V-01) Fluida Panas



: Saturated Steam



Wt



= 8056,5700 kg/jam = 17761,68



T1



= 150



o



= 302



o



T2



= 150



o



= 302



o



Fluida Dingin



C C



lb/jam F F



: HCl



W2



= 12333,5727 kg/jam = 27190,84



t1



= 52,93



o



= 127,2764



o



t2



= 110



o



= 230



o



C C



lb/jam F F



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern. 1) Beban panas V – 01 Q =



4066578,1444 kkal/hr = 16137487,4487 Btu/hr



2) Menghitung ΔT



No 1 2 3



Fluida Panas 302 302



Temperatur Tinggi Temperatur Rendah Selisih



Fluida Dingin 230 127,2764



Selisih 72 174,72 102,72



347



LMTD (ΔT)



=



ΔT2 - ΔT1 ln(ΔT2 /ΔT1 )



=



72 - 174,72 ln(72 /174,72)



= 116,006 oF S =



t 2  t1 T1  t1 230  127,276



= 302  127,276 = 0,588 Ft = 0,71 ∆t = LMTD x Ft = 116,006 x 0,71 = 82,36



Ta = ½ (302 + 302) = 302 oF ta = ½ (127,276 + 230) = 178,64 oF



Dari table 8, Kern, UD = 100 – 200 Btu/jam.ft2 oF Trial UD a)



: Steam & gas



Asumsi UD A



= 100 Btu / jam ft2 oF



=



Q (U D .T )



=



16137487,448 Btu / jam (100 Btu / jam. ft 2 .o F  116,006 o F )



= 1959,37 ft2



348



b) Karena A > 200 ft2, maka dipilih HE dengan jenis Shell and Tube Heat Exchanger, dengan spesifikasi sebagai berikut (dari table 11 Kern) : ID baffle Pitch C'



Nt L OD a" pass BWG ID



Shell side 13.25 5 1.25 0.25



Tube side 66 buah 15 ft 1 in 0.693 in2 2 18 in 0.902 in



B. Fluida Dingin 1. Flow area dalam tube (a’t) Total flow area (at)



= 0,693 inch2



(Tabel 10. Kern)



= Nt x a’t / 144 x n = 0,1588 ft3



2. Laju alir, Gt =



27190,84 W = 0,1588 = 171213,5 at



lb/ hr. ft2



Perhitungan viskositas campuran : Dari fiq 2-32 ,Perry didapat : μHCl = 0,165 cP = 0,3992 lb/ft.hr μH2O = 0,410 cP = 0,9919 lb/ft.hr μH2O = 0,773 lb/ft.hr Perhitungan kapasitas panas, Cp : T = 81,47 oC tabel B.2 , Felder, Elementary Principles of chemical Processes)



349



Senyawa HCl H2O



b.102 -0,1341 -



a 29,13 75,4



c. 106 0,9715 -



d. 109 -4,335 -



Cp = a + bT + cT 2 + dT 3



Cp( HCl) = 29,13 + - 0,1341.10- 2 × 81,47



+ 0,9715 .10- 5 × 81,47 2 + - 4,335.10- 9 × 81,47 3



= 0,968 J/g.oC = 0,1904 Btu / lb oF Cp(H2O) = 75,4 J/mol.oC = 4,4353 J/g.oC = 1,06 Btu / lb oF Densitas : ρH2O = 977,771 kg/m3 = 61,0420 lb/ft3 (tabel 2-28 Perry’s Chemical Engineer Handbook) ρH2O = 446,979 kg/m3 = 27,9048 lb/ft3 (tabel 2-30 Perry’s Chemical Engineer Handbook) Perhitungan konduktivitas termal, k : ρ4 1 / 3 )  ( pers Coulson 6 th edition ) k = 3,56.10 - 5 × (Cp × ( MW)



k H 2 O = 3,56.10 -5 × (4,4353 J/g.oC × (



(977,771 kg/m3) 4



1/ 3 ) 18 )



= 0,585 W/m.oC = 0,338 lb/ft hr k HCl = 3,56.10 - 5 × (0,968 J/g.oC × (



(446,979 kg/m3) 4



1/ 3 ) 36,5)



= 0,029 W/m.oC = 0,017 lb/ft hr Senyawa



m,kg



Xi



HCl 4563,4219 0,37 H2O 7770,1508 0,63 Jumlah 12333,5727 1,00 Cpcamp = 0,7383 Btu / lb oF



Cp Btu / lb oF 0,19044 1,060061



kcamp = 0,2194 lb/ft hr 3. Bilangan Reynold, Ret Pada



Tc



= 178,6382



o



F



Xi.Cpi Btu / lb oF 0,0705 0,6678 0,7383



k, lb/ft hr 0,017 0,338



Xi.ki lb/ft hr 0,2132 0,0063 0,2194



350



μ



= 0,7726



Ret



= =



lb/ft jam



De .G a



 0,902 x 171213,5 0,773



= 16658,07 15



4. Dengan L/D



= 0,902 x 12 = 199,5565



5. Pada



 c.     k 



, diperoleh



jH



= 180



tc



= 178,6382 oF



Cp



= 0,7383



k



= 0,2194 Btu/ft. oF. jam



1



3



k   D



Btu/lb. oF



0,7383 x 0,773 0,2194



=



 6. hi = jH 



(Fig 24. Kern)



1/ 3



 Cp .      k 



    w



  



1



3



= 1,375



0 ,14



Koreksi viskositas diabaikan karena tidak significant, maka didapat : hi



= 722,4921 Btu / hr. ft2 oF



hio



= hi 



 ID    OD   0,902    1 



= 722,4921 



= 651,6878 Btu / hr. ft2 oF B. Fluida panas (steam) 1.



: Shell Side



Flow area pada shell (as) as



= =



 ID



x C" x B  (144 Pt )



13,25 x 0,25 x 5 144 x 1,25



351



= 0,0920 ft2 2.



Laju alir massa dalam shell, Gs Gs



W



= a s 17761,68



= 0,0920



= 193032,5 lb / jam. ft2 3.



Condensate loading per linier foot (G”) G”



=



W L.Nt 2 / 3



=



17761,68 15 x 66 2 / 3



=



72,5042 lb/jam. lin ft



4. Bilangan Reynold, Res = D x Gs / µ Pada



Tc



= 302 oF



Cp



= 0,1904 Btu/lb.oF



k



= 0,2194 Btu/lb.oF



μ



=



 c.     k 



1



0,0348 lb/ft . jam



3



=



 0,1904 x 0,0348    0,2194  



1



3



= 0,3115 De



=



Res=



= =



0,55 inch = 0,0458 ft GS D



 193032,5 x 0,0458 0,0348



= 253967,3 jH = 90



(Fig. 28 Kern)



5. Koefisien Perpindahan Panas, ho Koreksi viskositas diabaikan karena tidak significant, maka diperoleh :



352



= jH . (k/D). (cμ/k)1/3



ho



=



(Pers. 6.28 Kern)



90 x 0,2194 x 0,3115 0,0458



= 134,2308



Btu / jam ft2 oF



6. Clean Overall Coefficient, Uc Uc =



hio x ho hio  ho 538,1105 x 134,2308



= 538,1105  134,2308



(Pers. 6.38 Kern)



= 107,432 Btu / jam ft2 oF 7. Dirt Factor, Rd Rd



=



Uc UD 107,432  100 = 107,432 x 100 U c .U D



= 0,000692 4) PRESSURE DROP Tube Side 1. Untuk NRe Factor friksi



2. ΔPt



= 16658,07 = 0,00017



s



= 0,4685



Фt



= 0,3401 =



f Gt 2 Ln x D e s t



=



0,00017 x 171231,5 2 x 15 x 2 5,22 x 1010 x 0,4685 x 0,3401



= 0,3922 psi 3. V2 / 2g ΔPr



= 0,07 = ( 4n / s ) ( V2 / 2g ) =



4 x 2 x 0,07 0,4685



(Fig 26, Kern)



353



= 1,1953 psi 4. ΔPT



= ΔPt + ΔPr = 0,3922 + 1,1953 = 1,5874 psi



Shell Side 1. Faktor Friksi Re



= 253967,3



f



= 0,00115



2. Number of cross, N+1



(Fig 29, Kern)



(N + 1)



= 12 L / B



(Pers. 7.43 Kern)



= (12 x 15)/ 5 = 36 Ds



= ID / 12 = 13,25 / 12 = 1,1042 ft



s



= 0,94 2



ΔPs



= =



f G s D f ( N  1) x De S  s 0,00115 x 193032,5 2 x 1,1042 x 36 5,22 x 1010 x 0,0458 x 0,94



= 0,7574 psi



hio = 538,1105



SUMMARY houtside = Uc = 107,432 Btu / jam ft2 oF UD = 100 Btu / jam ft2 oF Rd calculated = 0,000692



ho = 134,2308



354



Rd required = 0,002 Calculated ΔP, psi Allowable ΔP, psi



1,5874 10



0,7574 10



VAPORIZER – 02 (V-02) Fungsi



: Menguapkan bahan baku metanol



Type



: Double Pipe Heat Exchanger



Bahan



: Stainless Steel



Gambar



:



Return bend



Gland



Gland



Gland



Tee



Return Head



Gambar L.3.26. Vaporizer-02 (V-02) Fluida Panas



: Saturated Steam



Wt



= 1620,9659 kg/jam = 3573,6139 lb/jam



T1



= 150



o



= 302



o



T2



= 150



o



= 302



o



Fluida Dingin



C C



F F



: CH3OH



W2



= 5601,1315 kg/jam = 12348,3666 lb/jam



t1



= 39,36



o



= 102,85



o



t2



= 110



o



= 230



o



C C



F F



355



Perhitungan berdasarkan “Process Heat Transfer’, D Q. Kern. 1) Beban panas V – 02 Q =



818187,4584



kkal/hr = 3246830,473 Btu/hr



2) Menghitung ΔT



No 1 2 3



Fluida Panas 302 302



LMTD (ΔT)



Temperatur Tinggi Temperatur Rendah Selisih



=



ΔT2 - ΔT1 ln(ΔT2 /ΔT1 )



=



72 - 199,15 ln(72 /199,15)



Fluida Dingin 230 102,85



= 125,118 oF Ta = ½ (302 + 302) = 302 oF ta = ½ (102,85 + 230) = 166 oF Dari table 8, Kern, UD a) Trial UD



: Steam & gas UD = 200 – 700 Btu / jam ft2 oF



b)



= 400 Btu / jam ft2 oF



Asumsi UD A



=



Q (U D .T )



=



3246830,473 Btu / jam (400 Btu / jam.ft 2 .o F × 125,118 o F)



Selisih 72 199,15 127,15



356



= 103,8006 ft2 c) Karena A< 200 ft2, maka dipilih HE dengan jenis Double Pipe Heat Exchanger, dengan klasifikasi sebagai berikut (dari table 11 Kern) :



No 1 2 3 4 5



Annulus 1 40 1,049 in 1,320 in 0,344 ft2



Data Pipa IPS SN IDp ODp a’



3) Annulus : Fluida panas, steam a) Flow area, aa D2



= 1,049 in = 0,0874 ft



D1



= 0,840 in = 0,0700 ft



aa



=



 ( D2 2  D1 2 ) 4



2 2 = 3,14(0,0874 - 0,0700 ) 4



2



= 0,00215 ft



Diameter Equivalent, De 2



De



D  D1 = 2 D1 =



2



0,0874 2 - 0,0700 2 0,0700



= 0,0392 ft



Inner 0,5 40 0,622 in 0,84 in 0,304 ft2



357



b) Laju alir massa, Ga Ga



= W / aa =



3573,6139 lb / jam 0,00215 ft 2



= 1660436,3 lb / jam ft2 c) Bilangan Reynold, Rea Pada



o



Tc



=



302



μ



=



0,0144 cp



Rea



= De . Ga / μ =



F



0,3692 ft × 1660436,3 lb / ft 2 hr 0,03484 lb / ft. jam



= 1,867.106 d) ho = 1500 Btu/hr.ft2.oF 4) Inner Pipe : Fluida Dingin a) Flow area, ap Dp ap



= 0,622 in = 0,0518 ft =



 ID p 2 4



2 2 = 3,14 × 0,0518 ft 4



= 0,00021 ft2 b) Laju alir massa, Gp Gp



W



= a p =



= 0,03484 lb/ft jam (fig.2-32, Perry)



12348,3666 lb / jam 0,00021 ft 2



= 5854928,3 lb / jam ft2 c) Perhitungan μ, Cp, k campuran Perhitungan viskositas campuran :



358



Dari fiq 2-32 ,Perry didapat : μCH3OH = 0,117 cP = 0,2830 lb/ft.hr Perhitungan kapasitas panas, Cp : T = 74,68 oC tabel B.2 , Felder, Elementary Principles of chemical Processes) Senyawa CH3OH



b.102 7,092



a 21,1523



c. 106 2,507



d. 109 -28,51



Cp = a + bT + cT 2 + dT 3



Cp(CH 3OH) = 21,1523 + 7,092.10- 2 × 74,68 + 2,507 .10- 5 × 74,68 2 + - 28,51.10-9 × 74,68 3



= 0,7281 J/g.oC = 0,1740 Btu / lb oF Perhitungan konduktivitas termal, k : 10,4 th edition ) k = μ (Cp + ( MW )  ( pers Coulson 6 10,4 k (CH 3OH) = 0,117 (0,7281 J/g.oC + ( 32 g / mol)



= 0,0550 W/m.oC = 0,0318 lb/ft hr d) Bilangan Reynold, Rep Pada



o



tc



= 166



μ



= 0,3910 lb/ft. jam



Rep



= =



F



G p ID p



 5854928,3 lb / jam ft2 × 0,0518 ft 0,117 lb/ft. jam



= 1072193,3 hi = 1500 Btu/hr.ft2.oF e) Koefisien perpindahan panas, hio Untuk kondensasi steam



: hio = =



hi × D p OD inner



….(Kern hal.164)



1500 Btu/hr.ft2 .oF × 0,0518 ft 0,07 ft



359



= 1110,71 Btu/hr.ft2.oF f) Clean everaal Coefficient, Uc Uc



=



h io x h o h io + h o 1110,71 Btu/hr.ft2 .oF x 1500 Btu/hr.ft2 .oF



= 1110,71 Btu/hr.ft2.oF + 1500 Btu/hr.ft2.oF = 638,166 Btu / jam ft2 oF g) Design overall Coefficient, UD 1 / UD



= 1 / U c + Rd



Rd



= 0,002



1/UD



=



1 + 0,002 638,166 Btu / jam ft2 oF



= 0,0036 jam.ft2.oF/Btu = 280,35 Btu / jam ft2 oF



UD



h) Required Surface, A A



= =



Q U D x T 280,35



3246830,473 Btu/hr Btu / jam ft2 o F x 125,118 o F



= 92,564 ft2



i) Required Length, L L



= A / a”



a”



= 0,304 ft



L



=



…(Tabel 11 Kern)



92,564 ft2 0,304 ft



= 304,49 ft Diambil panjang 1 hairpin



= 2 x 12 ft



Jumlah hairpin yang diperlukan =



304,49 ft 24 ft



= 13



360



Actual Length



= 13 x 24 ft = 312 ft



Actual Surface



= 312 ft x 0,304 ft = 94,848 ft2



Actual Design Coefficient, Ud Q



Ud



= A .ΔT act =



3246830,473 Btu/hr 94,848 ft 2 × 125,118 oF



= 273,60 Btu / hr. ft2 oF Rd



=



UD - Ud UD × Ud 280,35 Btu / jam ft2 oF - 273,60 Btu / hr. ft2 oF



= 280,35 Btu / jam ft2 oF × 273,60 Btu / hr. ft2 oF = 0,0021 hr ft2 oF 5) Pressure Drop a) Annulus : Fluida Panas 1) De’



= (D2 – D1) = (0,30874 – 0,0700) ft = 0,0174 ft



Rea



= De . Ga / μ =



0,0174 ft × 1660436,3 lb / jam ft2 0,03484 lb/ft jam



= 8,301.106 f



=



0,0035 +



=



0,0035 +



0,264 0,42 R ea



….(Pers 3.47b, Kern)



0,264 (8,301.10 6 ) 0,42



= 0,0044 Pada Tc = 302 oF, ρsteam = 955,4261 lb/ft3 = 59,6470 lb/ft3 (Tabel 2.30, Perry)



361



2) ΔFa



=



4.f .G a 2 L 2.g.ρ 2 D e



=



4 × 0,0044 × 1660436,3 lb / jam ft2 2 × 312 ft 2 × 4,18.108 ft / hr 2 × 59,6470 lb/ft3 2× 0,0174 ft



= 2,90 ft 3) Va



= Ga / 3600 . ρ =



…(hal 115 Kern)



1660436,3 lb /det ft2 3600 × 59,6470 lb/ft3



= 7,73 ft/det



Fl



=n = 13



V2 2. g



…(hal. 112 Kern)



7,73 ft/det 2 2 × 32,2 ft / det 2



= 12,07 ft



4) ΔPa



=



( Fa  FL ). 144



=



(2,90 ft + 12,07 ft ).59,6470 lb/ft3 144



…(hal. 114 Kern)



= 6,20 psi b) Inner Pipe : Fluida dingin 1) Rep f



= 1072193,288 =



0,0035 +



=



0,0035 +



0,264 0,42 R ea 0,264



1072193,288 0, 42



= 0,0043 Pada t = 166 oF,



ρ = 74,68 lb/ft3



…(Pers. 3.47b Kern)



362



2



2) ΔFp



=



4. f .G p L



=



2 . g . 2 .D 4 × 0,0043 × 5854928,3 lb / jam ft2 2 × 312 ft 2 × 4,18.108 × 46,39 lb/ft32 × 0,0174 ft



= 0,0087 ft 3) ΔP



= ( ΔFp . ρ)/144 =



0,0087 ft × 46,39 lb/ft3 144



= 0,0028 psi



hio = 1500



6,20 10



SUMMARY houtside Uc = 638,166 Btu / hr ft2 oF UD = 280,35 Btu / hr. ft2 oF Rd calculated = 0,0021hr ft2 oF Rd required = 0,002 hr ft2 oF Calculated ΔP, psi Allowable ΔP, psi



ho = 1500



0,0028 10