Neraca Massa [PDF]

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NERACA MASSA A. STASIUN GILINGAN Data analisa stasiun gilingan (13 Agustus 2015) Tebu digiling



= 34030 ku



Nira mentah



= 32832 ku



Imbibisi



= 10250 ku Tabel . Data Analisis Stasiun Gilingan Uraian % brix % pol HK N1



15,91



11,96



75



N2



6,92



4,89



70,6



N3



4,97



3,4



68,3



N4



3,25



2,1



66,1



N5



2,17



1,39



64,1



NM



11,3



8,32



74,1



-



2,73



-



Ampas Perhitungan: N3



N4



N5



I



Tebu



Ampas N3+ Nt2 = N2 NM



N4 + Nt3 = N3



N5 + Nt4 = N4



N5



1) Menghitung N1 NM bnm



N2 bn2



N1 bn1 N1 = =



𝑏.𝑛𝑚 −𝑏.𝑛2 𝑏 .𝑛1−𝑏.𝑛2 11,3−6,92 15,91−6,92



x NM x 32.832 ku



= 15.996,01 ku P n1= =



𝑝.𝑛1 100



x N1



11,96 100



x 15.996,01 ku



= 1.913,123 ku B n1= =



𝑏.𝑛1 100



x N1



15,91 100



x 15.996,01 ku



= 2.544,966 ku



2) Menghitung N3 N2 bn2



N3 bn3 N3 =



Nt2 bn1 𝑏.𝑛1−𝑏.𝑛2 𝑏.𝑛1−𝑏.𝑛3



x N2



N2 = NM – N1 = 32.832 ku - 15.996,01 ku = 16.835,99 ku P n2=



𝑝.𝑛2



=



4,89



100



100



x N2 x 16.835,99 ku



= 823,2797 ku B n2=



𝑏.𝑛2



=



6,92



100



100



x N2 x 16.835,99 ku



= 1.165,05 ku



=



15,91−6,92 15,91−4,97



x 16.835,99 ku



= 13.835,06 ku P n3= =



𝑝.𝑛3 100 3,4 100



x N3 x 16.835,99 ku



= 470,3919 ku B n3=



𝑏.𝑛3



=



4,97



100



100



x N3 x 16.835,99 ku



= 687,6023 ku



3) Menghitung N4 N3 bn3



Nt3 bn1



N4 bn4



N4 =



𝑏.𝑛1−𝑏.𝑛3



=



15,91−4,97



𝑏.𝑛1−𝑏.𝑛4



15,91−3,25



x N3 x 13.835,06 ku



= 11.955,41 ku P n4= =



𝑝.𝑛4 100 2,1 100



x N4 x 11.955,41 ku



= 251,0637 ku



B n4=



𝑏.𝑛4



=



3,25



100



100



x N4 x 11.955,41 ku



= 388,5509 ku



4) Menghitung N5 N4 bn4



Nt4 bn1



N5 bn5 N5 =



𝑏.𝑛1−𝑏.𝑛4



=



15,91−3,25



𝑏.𝑛1−𝑏.𝑛5



15,91−2,17



x N4 x 11.955,41 ku



= 11.015,69 ku P n5=



𝑝.𝑛5



=



1,39



100



100



x N5 x 11.015,69 ku



= 251,0637 ku B n5=



𝑏.𝑛5



=



2,17



100



100



x N5 x 11.015,69 ku



= 239,0404 ku



5) Menghitung Ampas (Bagasse) Neraca massa overall: In – out = akumulasi



,akumulasi= 0



In = out Tebu + Imbibisi



= Nira Mentah Kotor + Ampas = NMK – Fnmk = NMK – (fnmk x NMK)



NM fnmk



= 2%



NM



= NMK – 0,02 NMK = 0,98 NMK



NMK



= 1,0204 NM = 1,0204 x 32.832 ku



NMK



= 33.502,04 ku



Maka, Ampas



= Tebu + Imbibisi – Nira Mentah Kotor = 34.030 + 10.250 – 33.502,04 = 10.778,23 ku



Pa5



𝑝.𝑎



= 100 x Ampas =



2,73 100



x 10.778,23 ku



= 294,2457 ku



B. STASIUN PEMURNIAN Nira Mentah Kotor



I



Nira Tapis



Nira Sulfitir II



Nira Encer



Nira Kotor



Air Pencuci



III



Blotong



Keterangan: I : Sulfitir II : Door Clarifier III : Rotary Vacuum Filter Perhitungan: Ampas = 10.778,23 ku Pa



𝑝.𝑎



2,73 100



= =



𝑝.𝑛𝑚 100 8,32 100



= 64,1 x 100 = 4,23



x 10.778,23 ku



= 294,2457 ku



Ba



x NM



𝑏.𝑎



= 100 x A =



4,23 100



x 10.778,23 ku



= 455,919 ku



x 32.832 ku



= 2.731,6224 ku



𝑝.𝑎



= 𝐻𝐾 𝑛5 x 100 2,73



= 100 x Ampas =



Pnm



ba



Bnm



=



11,3 100



x 32.832 ku



Pt



= Pnm + Pa



= 3.710,016 ku



= 2.731,6224 ku + 294,2457 ku



Bt



= 3.025,8681 ku



= Bnm + Ba = 3.710,016 + 455,919 = 4.165,935 ku



Zktl tebu Ft



= Ft + Fnmk 𝑓𝑎



= Fa = 100 x A =



49,5 – 4,23 100



, fa = zka – ba



x 10.778,23 ku



= 4.879,304 ku Fnmk = 𝑓𝑛𝑚𝑘 x NMK 100 = 2% x 33.502,04 ku = 670,04 ku Maka, Zktl tebu



= 4.879,304 ku + 670,04 ku = 5.549,34 ku



Zk



= Bt + Zktl tebu = 4.165,935 ku + 5.549,34 ku = 9.715,28 ku



1. Neraca Massa di Rotary Vacuum Filter Nira Tapis



Nira Kotor



Air Pencuci



III



Blotong input – output



= akumulasi



input



= output



, akumulasi = 0



Nira Kotor + Air Pencuci



= Nira Tapis + Blotong



Nira tapis



= Nira Kotor + Air Pencuci – Blotong



Nira Kotor



= 15% x Tebu digiling



Air pencuci



= 2,5% x Tebu digiling



Blotong



= 4% x Tebu digiling



Maka, Nira tapis



= =



15+2,5−4 100 13,5 100



x 34.030 ku



x 34.030 ku



= 4.594,05 ku 2. Neraca massa di Sulfitir Nira Mentah Kotor



I



Nira Sulfitir



Nira Tapis



Kadar Maksimal



input – output



= akumulasi



input



= output



, akumulasi = 0



Nira mentah Kotor + Nira Tapis



= Nira sulfitir



Nira sulfitir



= 33.502,04 + 4594,05 = 38.096,1 ku



3. Neraca Massa di Door Clarifier



Nira Sulfitir Nira Kotor



II



Nira Encer input – output



= akumulasi



input



= output



Nira sulfitir



= Nira encer + Nira kotor



Nira Encer



= Nira sulfitir - Nira kotor



, akumulasi = 0



= 38.096,1 – 5.104,5 Nira encer



= 32.991,6 ku



C. STASIUN PENGUAPAN



Uap Bekas Air Jatuhan Nira Encer



Nira Kental Gambar. Diagram alir proses penguapan



Data analisa stasiun pemurnian: Nira encer Brix nira encer



= 32.991,6 ku = 13,82



Brix nira kental = 61,67



Nira Encer



STASIUN PENGUAPAN



Nira Kental



Air Gambar . Perhitungan: 1) Menghitung nira kental Neraca massa komponen gula: Akumulasi = input – output input = out bnira encer x NE = bnira kental x NK



, akumulasi = 0



NK = =



𝑏.𝑛𝑖𝑟𝑎𝑒𝑛𝑐𝑒𝑟 𝑥 𝑁𝐸 𝑏 .𝑛𝑖𝑟𝑎𝑘𝑒𝑛𝑡𝑎𝑙 13,82 𝑥 32.991,6 61,67



= 7.393,2854 ku 2) Menghitung total air yang diuapkan Neraca massa overall stasiun penguapan: Akumulasi = input – output



, akumulasi = 0



input = out Nira encer = Nira Kental + Air Air yang diuapkan = Nira encer – Nira Kental = 32.991,6 ku - 7.393,2854 ku = 25.598,31 ku 3) Menghitung air yang diuapkan di masing-masing badan evaporator Luas pemanasan badan penguap yang beroperasi: (13 Agustus 2015) Badan I



= 1500 m2



Badan II



= 1500 m2



Badan III



= 1100 m2



Badan IV



= 1100 m2



Total



= 5200 m2



a. Badan I 𝐿𝑢𝑎𝑠 𝑒𝑣𝑎𝑝 𝐼



Air yang diuapkan = 𝐿𝑢𝑎𝑠 𝑘𝑒𝑠𝑒𝑙𝑢𝑟𝑢 1500



𝑕𝑎𝑛



x Total air yang diuapkan



= 5200 x 25.598,31 ku



= 7.384,129 ku Nira keluar badan I = nira encer – air diuapkan di badan I = 32.991,6 ku - 7.384,129 ku = 25.607,47 ku b. Badan II 𝐿𝑢𝑎𝑠 𝑒𝑣𝑎𝑝 𝐼𝐼



Air yang diuapkan = 𝐿𝑢𝑎𝑠 𝑘𝑒𝑠𝑒𝑙𝑢𝑟𝑢 𝑕𝑎𝑛 x Total air yang diuapkan 1500



= 5200 x 25.598,31 ku = 7.384,129 ku Nira keluar badan II = Nira keluar badan I – air diuapkan di badan II = 25.607,47 ku - 7.384,129 ku = 18.223,34 ku c. Badan III 𝐿𝑢𝑎𝑠 𝑒𝑣𝑎𝑝 𝐼𝐼𝐼



Air yang diuapkan = 𝐿𝑢𝑎𝑠 𝑘𝑒𝑠𝑒𝑙𝑢𝑟𝑢 𝑕𝑎𝑛 x Total air yang diuapkan 1100



= 5200 x 25.598,31 ku = 5.415,028 ku Nira keluar badan III = Nira keluar badan II – air diuapkan di badan III = 18.223,34 ku – 5.415,028 ku = 12.808,31 ku d. Badan IV 𝐿𝑢𝑎𝑠 𝑒𝑣𝑎𝑝 𝐼𝑉



Air yang diuapkan = 𝐿𝑢𝑎𝑠 𝑘𝑒𝑠𝑒𝑙𝑢𝑟𝑢 𝑕𝑎𝑛 x Total air yang diuapkan 1100



= 5200 x 25.598,31 ku = 5.415,028 ku



Nira keluar badan IV = Nira keluar badan III – air diuapkan di badan IV = 12.808,31 ku - 5.415,028 ku = 7.393,285 ku Nira kental



= 7.393,285 ku



D. STASIUN MASAKAN DAN PEMUTARAN



Nira Kental



A



Putaran A1



Gula A



Stroop A



Putaran SHS



Gula SHS



Klare SHS



Masakan A Gula C Nira Kental



C



Putaran C



Masakan C



D



Putaran D1



Gula D1



Stroop C



Putaran D2 Klare D



Masakan D Tetes Neraca massa:  Masakan A → Gula A + Stroop A Gula A



→ SHS + Klare SHS



 Masakan C → Gula C + Stroop C  Masakan D → Gula D1 + tetes Gula D1 → Gula D2 + Klare D  Overall: Nira kental → SHS + tetes



Gula D2



Tabel . Data analisis stasiun masakan (13 Agustus 2015) Uraian Nira Kental Masakan A Stroop A Gula A SHS Klare SHS Masakan C Stroop C Gula C Masakan D Tetes (Molasse) Gula D1 Gula D2 Klare D



% brix 61,67 93,7 83,6 99,2 99,9 74,1 96,9 86 97,3 99,4 93,49 96 96,6 84,6



Perhitungan: 1) Neraca massa overall: Nira kental → SHS + Tetes BNK HKnk



BT HKT



BSHS HKSHS



Nira kental = 7.393,285 ku BNK



= bnk x Nira kental =



61,67 100



x 7.393,285 ku



= 4.559,439 ku BSHS



𝐻𝐾 .𝑛𝑘 −𝐻𝐾 .𝑇



= 𝐻𝐾 .𝑆𝐻𝑆 − 𝐻𝐾 .𝑇 x BNK



% pol 46,21 74,9 49,7 98,3 99,7 66,5 67,5 42 91,4 58,9 31,23 86 89 45,9



HK 74,9 80 59,4 99,1 99,8 89,7 69,6 48,9 93,9 59,2 33,4 89,5 92,2 54,2



74,9 −33,4



= 99,8− 33,4 x 4.559,439 ku = 2.849,649 ku BTetes



= BNK – BSHS = 4.559,439 ku - 2.849,649 ku = 1.709,79 ku



2) Masakan D a. Masakan D → Gula D1 + Tetes BmasakanD HKmasakanD



B gl D1 HK gl D1 BmasakanD



BT HKT 𝐻𝐾 .𝐷1−𝐻𝐾 .𝑇



= 𝐻𝐾 .𝐷1− 𝐻𝐾 .𝑚𝑎𝑠𝑎𝑘𝑎𝑛



𝐷



x BT



74 ,9−33,4



= 74,9− 59,2 x 1.709,79 ku = 3.165,65 ku = Bmasakan D – Btetes



B gl D1



= 3.165,65 – 1.709,79 ku = 1.455,861 ku b. Gula D1 → Gula D2 + Klare D Bgl.D1 HKgl.D1



Bgl.D2 HKgl.D2



BklareD HKklareD



Bgl.D2



𝐻𝐾 .𝐷1−𝐻𝐾 .𝑘𝑙𝑎𝑟𝑒𝐷



= 𝐻𝐾 .𝐷2− 𝐻𝐾 .𝑘𝑙𝑎𝑟𝑒𝐷 x Bgl.D1 89,5−54,2



= 92,2− 54,2 x 1.455,861 ku = 1.352,418 ku BklareD



= Bgl.D1 – Bgl.D2 = 1.455,861– 1.352,418 = 103,4427 ku



3) Masakan A a. Gula A → SHS + Klare SHS Bgula.A HKgula.A



Bklare.SHS HKklare.SHS Bgula.A



BSHS HKSHS 𝐻𝐾 .𝑘𝑙𝑎𝑟𝑒 .𝑆𝐻𝑆 −𝐻𝐾 .𝑆𝐻𝑆



= 𝐻𝐾 .𝑘𝑙𝑎𝑟𝑒 .𝑆𝐻𝑆 − 𝐻𝐾 .𝑔𝑢𝑙𝑎 .𝐴 x BSHS 89,7−99,8



= 89,7− 99,1 x 2.849,649 ku = 3.061,857 ku Bklare.SHS



= Bgula.A – BSHS = 3.061,857 – 2.849,649 ku = 212,208 ku



b. Masakan A → Gula A + Stroop A Bmasakan.A HKmasakan.A



Bstroop.A HKstroop.A



Bgula.A HKgula.A



Bmasakan.A



𝐻𝐾 .𝑠𝑡𝑟𝑜𝑜𝑝 .𝐴−𝐻𝐾 .𝑔𝑢𝑙𝑎𝐴



= 𝐻𝐾 ..𝑠𝑡𝑟𝑜𝑜𝑝 .𝐴− 𝐻𝐾 .𝑚𝑎𝑠𝑎𝑘𝑎𝑛 =



59,4−99,1 59,4− 80



.𝐴



x 3.061,857 ku



= 5900,764 ku BStroop.A



= Bmasakan.A – Bgula.A = 5900,764 – 3.061,857 ku = 2.838,907 ku



4) Masakan C Masakan C → Gula C + Stroop C Bmasakan.C HKmasakan.C



Bgl.C HKgl.C



BStroop.C HKStroop.C



Data analisa masakan C: VMasakan C



= 1.050 HL



BJMasakan C



= 1,3



Masakan C



= V x BJ = 1.050 x 1,3 = 1.365 ku



Bmasakan C



= =



𝑏.𝑚𝑎𝑠𝑎𝑘𝑎𝑛𝐶 100 96,9 100



x Masakan C



x 1.365 ku



= 1.322,685 ku



x Bgula.A



Bgl.C



=



𝐻𝐾 .𝑚𝑎𝑠𝑎𝑘𝑎𝑛 .𝐶−𝐻𝐾 .𝑆𝑡𝑟𝑜𝑜𝑝 .𝐶 𝐻𝐾 .𝑔𝑙 .𝐶− 𝐻𝐾 .𝑆𝑡𝑟𝑜𝑜𝑝 .𝐶 69,6− 48,92



x Bmasakan.C



= 93,9− 48,92 x 1.322,685 ku = 608,4351 ku BStroop.C = Bmasakan.C – Bgl.C = 1.322,685 – 608,4351 = 714,2499 ku