Neraca Panas Evaporator Dan Barometric Condenser [PDF]

Citation preview

Neraca Panas Di Sekitar Evaporator (EV-01, EV-02)



Tujuan



: menghitung kondisi operasi masing-masing effect, menghitung kebutuhan fresh steam pada effect 1 dan menghitung beban panas pada evaporator.



Fungsi



: memekatkan sirup glukosa hingga konsentrasi 50% berat.



Keterangan : F S V1 V2



: Umpan masuk evaporator : Fresh Steam : Uap keluaran effect 1 : Uap keluaran effect 2 Komponen Air Maltotriosa Maltosa Glukosa







Input (kg/jam) Arus 12 3.808,712 36,425 83,015 1.774,500



L1 L2 C1 C2



: Larutan hasil pemekatan effect 1 : Larutan hasil pemekatan effect 2 : Kondensat effect 1 : Kondensat effect 2 Output (kg/jam) Arus 14 1.893,939 36,425 83,015 1.774,500



Data-data yang diketahui Feed:



xF = 0,332



Effect 3:



P2 = 25,01 kPa



tF = 80oC



t2 = 65oC



F = 5.702,652 kg/jam



L2 = 3.787,879 kg/jam



x2 = 0,5 Steam:



P = 143,27 kPa ts = 110oC λs = 2.230,0 kJ/kg







Neraca massa overall F = C2 + V 2 + L 2 = V1 + V2 + L2 = Vtot + L2 Vtot = F – L2 = 5.702,652 kg/jam – 3.787,879 kg/jam = 1.914,773 kg/jam Asumsi V1 = V2 = V3 = 1/3 Vtot = 1/3 (1.914,773 kg/jam) = 957,387 kg/jam







Neraca massa solute Effect 1 F = V1 + L1 L1 = F – V1 = 5.702,652 kg/jam – 957,387 kg/jam = 4.745,265 kg/jam F xF = L1 xL1 x1 -







=



= 0,399



Effect 2 x2 = 0,5 (spesifikasi yang diinginkan)



BPR masing-masing effect BPR = 1,78 (xi) + 6,22 (xi2) (Geankoplis, 1993, hal. 507)







- BPR1 = 1,78 (0,399) + 6,22 (0,3992) = 1,701 - BPR2 = 1,78 (0,5) + 6,22 (0,52) = 2,445 Cp masing-masing liquid Cp = 4,19 – 2,35xi (Geankoplis, 1993, hal. 507) -







CpF Cp1 Cp2



= 4,19 – 2,35 (0,332) = 3,410 kJ/kg.K = 4,19 – 2,35 (0,399) = 3,252 kJ/kg.K = 4,19 – 2,35 (0,5) = 3,015 kJ/kg.K



Kondisi operasi dan data-data eltalpi masing-masing effect Menebak harga: P1 = 57,8 kPa P2 = 25,01 kPa Effect 1 P1 = 57,8 kPa, maka t1’ = 85oC H1’ = 2.652,0 kJ/kg h1’ = 355,9 kJ/kg t1 = t1’ + BPR1 = 85oC + 1,701 = 65,701oC H1 = H1’ + Cp1 (BPR1)



= 2.652,0 kJ/kg + 3,252 kJ/kg.K (1,701 K) = 2.657,5 kJ/kg h1



= h1’ + Cp1 (BPR1) = 355,9 kJ/kg + 3,252 kJ/kg.K (1,701 K) = 361,4 kJ/kg



∆Teff1 -



= tS – t1 = 110oC – 103.661oC = 23,299oC



Effect 2 P2 = 25,01 kPa, maka t2’ = 65oC H2’ = 2.648,4 kJ/kg h2’ = 272,0 kJ/kg o t2 = t2’ + BPR2 = 65 C + 2,445 = 67,445oC H2 = H2’ + Cp2 (BPR2) = 2.648,4 kJ/kg + 3,015 kJ/kg.K (2,445 K) = 2.655,8 kJ/kg h2



= h2’ + Cp2 (BPR2) = 272,0 kJ/kg + 3,015 kJ/kg.K (2,445 K) = 279,4 kJ/kg



∆Teff2 = t1’ – t2 = 85oC – 67,445oC = 17,555oC 



Neraca panas Effect 1 F hF + S HS = L1 h1 + V1 H1 + S hC1 F CpF (tF – 0) + S λS = (F – V1) h1 + V1 H1 (5.702,652) (3,410) (80) + S (2.230,0) = (5.702,652 – V1) (361,4) + V1 (2.657,5) 2.230,0 S – 2.296,1 V1 = 505.656,257 … (1) Effect 2 L1 h1 + V1 H1 = L2 h2 + V2 H2 + V1 hC2 (F – V1) h1 + V1 H1 = L2 h2 + (Vtot – V1) H2 + V1 h1’ (5.702,652 – V1) (361,4) + V1 (2.657,5) = (3.787,879) (279,4) + (1.914,773 - V 1) (2.655,8) + V1 (355,9) 4.596,0 V1 = 4.082.301,880



…(2)



Dari persamaan (1) dan (2) didapat: V1 = 888,235 kg/jam V2 = 1.026,538 kg/jam S = 1.141,315 kg/jam 



Luas masing-maing effect Diketahui, U1 = 3.123 W/m2.K U2 = 1.987 W/m2.K (Geankoplis, 1993, hal. 507)



A1 =



= 34,979 m2



= 58,609 m2



A2 =



Untuk mengetahui apakah trial berhasil atau tidak, maka harus dilakukan evaluasi luas masing-masing effect terhadap luas rata-rata (AM) AM = = 45,133 m2



AM =



Maka, penyimpangan luas masinng-masing effect dihitung dengan rumus Ai =



x 100%



A1 =



x 100% = 22,498%



A2 =



x 100% = 29,859%



Karena penyimpangan > 10% maka trial harus diulang 



∆Teff revisi ∆Teff1 (rev)



= ∆Teff1 (A1/AM) = 23,299 (34,979/45,133) = 18,057



∆Teff2 (rev)



= ∆Teff2 (A2/AM) = 17,555 (58,609/45,133) = 22,797



Rencanakan : ∆Teff1 = 18,057 oC, ∆Teff2 = 22,797 oC Maka trial P dilakukan pada harga t1’ = 90,242 oC dan t2’ = 65 oC Untuk t1’= 97,374oC, pada steam table: t (oC) P (kPa) 90 70,11 90,242 P1 91 72,81 Dengan interpolasi, didapat: P1 = 70,76 kPa h1’ = 377,9 kJ/kg H1’ = 2.660,5 kJ/kg Untuk t2’= 65oC, didapat: P2 = 25,01 kPa h2’ = 272,0 kJ/kg H2’ = 2.648,4 kJ/kg



h (kJ/kg) 376,9 h1’ 381,1



H (kJ/kg) 2.660,1 H1’ 2.661,7







Neraca massa solute Pada tahap ini data-data yang digunakan sama seperti perhitugan sebelumnya, yaitu: V1 = V2 = 1/2 Vtot = 957,387 kg/jam L1 = 4.745,265 kg/jam, x1 = 0,399 L2 = 3.787,879 kg/jam, x3 = 0,5







Kondisi operasi dan data-data entalpi masing-masing effect Effect 1 P1 = 70,76 kPa, maka t1’ = 90,242oC H1’ = 2.660,5 kJ/kg h1’ = 377,9 kJ/kg o t1 = t1’ + BPR1 = 90,242 C + 1,701 = 91,943oC H1 = H1’ + Cp1 (BPR1) = 2.660,5 kJ/kg + 3,252 kJ/kg.K (1,701 K) = 2.666,0 kJ/kg h1



= h1’ + Cp1 (BPR1) = 377,9 kJ/kg + 3,252 kJ/kg.K (1,701 K) = 383,4 kJ/kg



-



∆Teff1 = tS – t1 = 110oC – 91,943oC = 18,057oC Effect 2 P2 = 25,01 kPa, maka t2’ = 65oC H2’ = 2.648,4 kJ/kg h2’ = 272,0 kJ/kg t2 = t2’ + BPR2 = 65oC + 2,445 = 67,445oC H2 = H2’ + Cp2 (BPR2) = 2.648,4 kJ/kg + 3,015 kJ/kg.K (2,445 K) = 2.655,8 kJ/kg h2



= h2’ + Cp2 (BPR2) = 272,0 kJ/kg + 3,015 kJ/kg.K (2,445 K) = 279,4 kJ/kg



∆Teff2 



= t1’ – t2 = 90,242oC – 67,445oC = 22,797oC



Neraca panas Effect 1 F hF + S HS = L1 h1 + V1 H1 + S hC1 F CpF (tF – 0) + S λS = (F – V1) h1 + V1 H1 (5.702,652) (3,410) (80) + S (2.230,0) = (5.702,652 – V1) (383,4) + V1 (2.666,0) 2.230,0 S – 2.282,6 V1 = 631.201,963 … (3) -



Effect 2 L1 h1 + V1 H1 = L2 h2 + V2 H2 + V1 hC2 (F – V1) h1 + V1 H1 = L2 h2 + (Vtot – V1) H2 + V1 h1’ (5.702,652 – V1) (383,4) + V1 (2.666,0) = (3.787,879) (279,4) + (1.914,773 - V 1) (2.655,8) + V1 (377,9) 4.560,4 V1 = 3.956.756,174



Dari persamaan (3) dan (4) didapat: V1 = 867,628 kg/jam V2 = 1.047,145 kg/jam



…(4)



S = 1.171,133 kg/jam 



Luas masing-masing effect = 46,312 m2



A1 =



= 43,827 m2



A2 =



Untuk mengetahui apakah trial berhasil atau tidak, maka harus dilakukan evaluasi luas masing-masing effect terhadap luas rata-rata (AM) AM = = 44,925 m2



AM =



Maka, penyimpangan luas masinng-masing effect dihitung dengan rumus Ai =







A1 =



x 100% = 3,087%



A2 =



x 100% = 2,445%



Karena penyimpangan < 10% maka trial dianggap berhasil Ekonomi evaporator (Steam economy) η=







x 100%



= 1,635



Neraca panas total Q masuk = Q keluar QF + QS = QC2 + QL2 + QV2 QF QF = F CpF ΔTF = (5.702,652 kg/jam) (3,410 kJ/kg.K) (80 – 0) K = 1.555.468,341 kJ/jam QS QS = S λS = (1.171,133 kg/jam) (2.230,0 kJ/kg) = 2.611.625,533 kJ/jam QC2 QC2 = V1 h1’ = (867,628 kg/jam) (377,9 kJ/kg) = 327.890,021 kJ/jam QL2 -



QL2



-



QV2 QV2



= L2 h2 = (3.787,879 kg/jam) (279,4 kJ/kg) = 1.058.226,042 kJ/jam = V2 H2 = (1.047,145 kg/jam) (2.655,8 kJ/kg) = 2.780.977,812 kJ/jam NERACA PANAS EV-01, EV-02, EV-03 QF QS QC2 QL2 QV2 Total



Input (kJ/jam) 1.555.468,341 2.611.625,533



4.167.093,874



Output (kJ/jam) 327.890,021 1.058.226,042 2.780.977,812 4.167.093,874



Neraca Panas Di Sekitar Barometric Condensor (BC-01)



Tujuan : menentukan kebutuhan air pendingin yang digunakan Fungsi : sebagai system vakum yang bekerja dengan cara menyedot vapor dari EV-02 kemudian mengkondensasikannya Neraca panas di BC-01: Qin = Qout QV2 + QW = QC V2 H2 + W hW = (V3 + W) hCamp Diketahui: V2 = 1.047,145 kg/jam H2 = 2.655,8 kJ/kg tW = Suhu air pendingin masuk, 30oC



hW tC amp hCamp



= 125,7 kJ/kg = Suhu campuran kondensat dan air pendingin, 45oC = 188,4 kJ/kg



QCamp = QV2 + QW Maka, kebutuhan air pendingin:



Maka, QCamp



= (V2 + W) hCamp = (1.047,145 kg/jam + 41.207,268 kg/jam) 188,4 kJ/kg = 7.960.731,410 NERACA PANAS BC-01 QV2 QW QCamp Total



Input (kJ/jam) 2.780.977,812 5.179.753,599 7.960.731,410



Output (kJ/jam) 7.960.731,410 7.960.731,410