Lampiran Nm & Ne Sorbitol

LAMPIRAN A

PERHITUNGAN NERACA MASSA

Produksi ini menggunakan bahan baku tepung jagung dan hidrogen. Pendekatan yang

digunakan dalam perhitungan:

Kapasitas perancangan per tahun = 65.000 ton/tahun

Waktu operasi dalam 1 tahun = 330 hari

Laju alir produksi =

65.000 tontah un

x1000 kg

1 tonx

1 tah un11bulan

x1 bulan30 h ari

x1 h ari24 jam

= 8207,071 kg/jam

= 45,05121922 kmol/jam

1. Tangki Perebusan (M-01)

Fungsi : Melarutkan tepung jagung dengan air menjadi slurry

Neraca massa total : F1 + F2 = F3

Tabel A.1 Komposisi tepung jagung jenis srikandi putih

Komponen komposisiPati 0,7833Lemak 0,0238Protein 0,0789abu 0,0108Serat 0,0129

H2O 0,0924

Neraca massa komponen pati

F1pati = F3

pati

F1pati = F3

pati = 8760,481 kg/jam

F’ = 0,7833 x F1pati

F’ = 11184,069 kg/jam

F1lemak = F3

lemak

F1protein = F3

protein

F1serat = F3

serat

F1abu = F3

abu

F3H2O = 230,53 kmol/jam x 18 kg/kmol

= 4149,702 kg/jam

F1H2O + F2

H2O = F3H2O

F2H2O = F3

H2O - F1H2O

= 4149,702 – 1033,408

= 3116,294 kg/jam

Untuk H2O

F1H2O = 0,0924 x F’

= 1033,408 kg/jam

Untuk lemak

F1lemak = 0,0238 x F’

= 266,181 kg/jam

Untuk protein

F1protein = 0,0789 x F’

= 882,423 kg/jam

Untuk serat

F1serat = 0,0129 x F’

= 144,274 kg/jam

Untuk abu

F1abu = 0,0108 x F’

= 120,788 kg/jam

KomposisiInput (kg/Jam) Output (kg/Jam)

Aliran 1 Aliran 2 Aliran 3Pati 8760,481 8760,481Lemak 266,181 266,181Protein 882,423 882,423abu 120,788 120,788Serat 144,274 144,274H2O 1033,408 3116,294 4149,702

TOTAL 11207,556 3116,294 14323,849

14323,849

2. Reaktor Hidrolisa (R-01)

Fungsi : Menghidrolisa pati (C12H22O11) menjadi glukosa (C6H12O6) dengan katalis

HCl.

Neraca massa total : F3 + F4 = F5

Neraca Massa komponen :

F5glukosa = F7

glukosa

= 8306,779 kg/jam

N5glukosa =

8306,779 kg / jam180,16 kg /kmol

= 46,108 kmol/jam

Konversi pati = 90%

Xpati = N 3 pati−N 6 pati

N 3 pati

0,9 = 23,053 kmol / jam

N 3 pati

N3pati = 25,615 kmol/jam

F3pati = 25,615 kmol/jam x 342 kg/kmol

= 8760,481 kg/jam

Air yang bereaksi 10%

Xair = N 3 air – N 5 air

N 3 air

0,1 = 23,053 kmol / jam

N 3 pati

N3air = 230,538 kmol/jam

F3air = 230,538 kmol/jam x 18 kg/kmol

= 4149,702 kg/jam

C12H22O11 + H2O 2 C6H12O6

m : 25,615 230,53

r : 23,054 23,053 46,108

s : 2,562 207,485 46,108 kmol/jam

F3lemak = F5

lemak = 266,181 kg/jam

F3protein = F5

protein = 882,423 kg/jam

F3serat = F5

serat = 144,274 kg/jam

F3abu = F5

abu = 120,788 kg/jam

F5maltosa = F7

maltosa = 332,271 kg/jam F5dekstrin = F7

dekstrin = 4,984 kg/jam

F5pati = 2,562 kmol/jam x 342 kg/kmol = 876,048 kg/jam

F4 = 0,1 x F’

= 0,1 x 11184,069 kg/jam

= 1118,406 kg/jam

F4HCl = 0,37 x F4

= 0,37 x 1118,406 kg/jam

= 413,811 kg/jam

F4HCl = F5

HCl = 413,811 kg/jam

F4H2O = F4 - F4

HCl

= 1118,406 - 413,811

= 704,596 kg/jam

F3 = F3pati + F3

lemak + F3protein + F3

abu + F3serat + F3

air

= 8760,481 + 266,181 + 882,423 + 120,788 + 144,274 + 4149,702

= 14323,849 kg/jam

F5H2O = (F3+ F4) – (F5

pati + F5lemak + F5

protein + F5abu + F5

serat + F5glukosa + F5

maltosa +

F5dekstrin + F5

HCl )

= (14323,849 + 1118,406) – (876,048 + 266,181 + 882,423 + 120,788 +

144,274 + 8306,779 + 332,271 + 4,984 + 413,811)

= 4094,695 kg/jam

KomposisiInput (kg/Jam)

Output (kg/Jam)

Aliran 3 Aliran 4 Aliran 5Pati 8760,481 876,0481Lemak 266,181 266,181Protein 882,423 882,423Abu 120,788 120,788Serat 144,274 144,274Glukosa 8306,781Maltosa 332,271

Air 4149,702 704,596 4094,695Dekstrin 4,984HCl 413,811 413,811

TOTAL14323,849 1118,407

15442,26 15442,26

3. Filter Press (F-01)

Fungsi : Memisahkan sisa pati, protein, lemak, dan serat yang bercampur dalam

larutan glukosa.

Neraca Massa total : F6 = F8+ F7

Neraca massa komponen aliran 6:

F6pati = F5

pati = 876,04813 kg/jam

F6lemak = F5


F6protein = F5

proten = 418,346 kg/jam

F6abu = F5

abu = 69,047 kg/jam

F6serat = F5


F6glukosa = F5

glukosa = 8306,781 kg/jam

F6maltosa = F5

maltosa= 332,271 kg/jam

F6air = F5

air = 4094,695 kg/jam

F6dekstrin = F5

dekstrin = 4,984 kg/jam

F6HCl = F5

HCl = 413,811 kg/jam

Asumsi : pati, lemak, protein, abu dan serat 100% terbuang sedangkan glukosa,

maltose, dekstrin, HCl dan air yang ikut terbuang sebesar 0,1%.


F7pati = F6

pati = 876,04813 kg/jam

F7lemak = F6


F7protein = F6

proten = 418,346 kg/jam

F7abu = F6

abu = 69,047 kg/jam

F7serat = F6


F7glukosa = 0,1% x F6

glukosa

= 0,001 x 8306,781 kg/jam

= 8,307 kg/jam

F7air= 0,1% x F6

air

= 0,001 x 4094,695 kg/jam

= 4,095 kg/jam

F7maltosa= 0,1% x F6

maltosa

= 0,001 x 332,271 kg/jam

= 0,332 kg/jam

F7dekstrin= 0,1% x F6

dekstrin

= 0,001 x 4,984 kg/jam

= 0,005 kg/jam

F7HCl= 0,1% x F6

HCl

= 0,001 x 413,811 kg/jam

= 0,414 kg/jam



glukosa

= 0,999 x 8306,781 kg/jam

= 8298,474 kg/jam

F8air = 99,9% x F6

air

= 0,999 x 4094,695 kg/jam

= 4090,601 kg/jam

F8maltosa= 99,9% x F6

maltosa

= 0,999 x 332,271 kg/jam

= 331,939 kg/jam

F8dekstrin= 99,9% x F6

dekstrin

= 0,999 x 4,984 kg/jam

= 4,979 kg/jam

F8HCl = 99,9% x F6

HCl

= 0,999 x 413,811 kg/jam

= 413,397 kg/jam


Aliran 6 Aliran 7 Aliran 8Pati 876,048 876,04813Lemak 223,389 223,389Protein 418,346 418,346Abu 69,047 69,047Serat 89,355 89,355Glukosa 8306,781 8,307 8298,474Maltosa 332,271 0,332 331,939Air 4094,695 4,095 4090,601Dekstrin 4,984 0,005 4,979HCl 413,811 0,414 413,397

TOTAL 14828,7271689,338 13139,389

14828,727

4. Tangki Netralizer (TK-04)

Fungsi : Menetralkan asam didalam larutan glukosa.

Neraca Massa total : F8+ F9 = F10


F8glukosa = 8298,474 kg/jam

F8air = 4090,601 kg/jam

F8maltosa= 331,939 kg/jam

F8dekstrin= 4,979 kg/jam

F8HCl = 413,397 kg/jam

F8 = F8glukosa + F8

air + F8maltosa + F8

dekstrin + F8HCl

= 8298,474 + 4090,601 + 331,939 + 4,979 + 413,397

= 113139,389 kg/jam


N8HCl =

413,397 kg / jam36,5 kg /kmol

= 11,326 kmol/jam

Asumsi : HCl dan NaOH habis bereaksi sehingga yang bereaksi 11,326

kmol/jam.

HCl + NaOH NaCl + H2O

m : 11,326 11,326

r : 11,326 11,326 11,326 11,326

s : - - 11,326 11,326

F9NaOH = 11,326 kmol/jam x 40 kg/kmol

= 453,038 kg/jam

Konsentrasi NaOH yang digunakan adalah 95%

F9NaOH = 95% x F9

F9 = 453,038 kg / jam

0,95 = 476,882 kg/jam

F9air = F9

– F9NaOH

= 476,882 kg/jam - 453,038 kg/jam

= 23,844 kg/jam


Glukosa, maltose dan dekstrin tidak ikut bereaksi, sehingga :

F10glukosa = F8


F10maltosa= F8


F10dekstrin= F8

dekstrin= 4,979 kg/jam

F10NaCl = 11,326 kmol/jam x 58,5 kg/kmol

= 662,567 kg/jam

F10H2O = (F8 + F9) – (F10

glukosa + F10maltosa + F10

dekstrin + F10NaCl)

= (113139,389 + 476,882) – (8298,474 + 331,939 + 4,979 + 662,567 )

= 4318,312 kg/jam


Aliran 8 Aliran 9 Aliran 10Glukosa 8298,474 8298,474Maltosa 331,939 331,939Air 4090,601 23,844 4318,312Dekstrin 4,979 4,979HCl 413,397NaOH 453,038NaCl 662,567

TOTAL13139,389 476,882

13616,27113616,271

5. Dekanter (DK-01)

Fungsi : Memisahkan NaCl yang bercampur didalam larutan glukosa.

Neraca Massa total : F10 = F12+ F11





F10NaCl = 662,567 kg/jam

F10H2O = 4318,312 kg/jam


Asumsi : NaCl terpisahkan seluruhnya dari produk dan kemungkinan glukosa,

maltose, air dan dekstrin yang ikut terbuang adalah sebesar 0,1%.

F11NaCl = F10

NaCl = 662,567 kg/jam


glukosa

= 0,001 x 8298,474 kg/jam

= 8,298 kg/jam

F11maltosa = 0,1% x F10

maltosa

= 0,001 x 331,939 kg/jam

= 0,332 kg/jam

F11air = 0,1% x F10

air

= 0,001 x 4318,312 kg/jam

= 4,318 kg/jam

F11dekstrin = 0,1% x F10

dekstrin

= 0,001 x 4,979 kg/jam

= 0,005 kg/jam


F12glukosa= F10

glukosa- F11glukosa

= 8298,474 - 8,298

= 8290,175 kg/jam

F12maltosa= F10

maltosa - F11 maltosa

= 331,939 - 0,332

= 331,607 kg/jam

F12air= F10

air - F11air

= 4318,312 - 4,318

= 4313,993 kg/jam

F12dekstrin= F10

dekstrin - F11 dekstrin

= 4,979 - 0,005

= 4,974 kg/jam


Aliran 10 Aliran 11 Aliran 12Glukosa 8298,474 8,298 8290,175Maltosa 331,939 0,332 331,607Air 4318,312 4,318 4313,993Dekstrin 4,979 0,005 4,974NaCl 662,567 662,567

TOTAL 13616,271675,521 12940,750

13616,271

6. Decoloring (DC-01)

Fungsi : Menghilangkan zat warna yang ada didalam larutan glukosa dengan

menggunakan karbon aktif.

Neraca Massa total : F12+ F13 = F14


F12glukosa= 8290,175 kg/jam


F12air= 4313,993 kg/jam



Asumsi : 2,2% karbon aktif yang masuk dari umpan bahan baku.

F13karbon aktif= 2,2% x F’

= 0,022 x11184,069 kg/jam

= 246,050 kg/jam


F14glukosa= F12

glukosa= 8290,175 kg/jam

F14maltosa= F12


F14air= F12

air= 4313,993 kg/jam

F14dekstrin= F12


F14dekstrin= F12


KomposisiInput(kg/Jam)

Output(kg/Jam)

Aliran 12 Aliran 13 Aliran 14Glukosa 8290,175 8290,175Maltosa 331,607 331,607Air 4313,993 4313,993Dekstrin 4,974 4,974Karbon Aktif 246,050 246,050

TOTAL12940,750 246,050

13186,79913186,799

7. Filter Press (F-02)

Fungsi : Memisahkan karbon aktif yang bercampur didalam larutan glukosa.

Neraca Massa total : F14 = F15 + F16


F14glukosa= 8290,175 kg/jam


F14air= 4313,993 kg/jam

F14dekstrin = 4,974 kg/jam

F14karbon aktif= 246,050 kg/jam


Karbon aktif terpisahkan 100% sedangkan glukosa, maltose, air dan dekstrin

yang ikut terbuang sebesar 0,1%.

F16KA= F14

KA= 246,050 kg/jam


glukosa

= 0,001 x 8290,175 kg/jam

= 8,290 kg/jam

F16maltosa = 0,1% x F14

maltosa

= 0,001 x 331,607 kg/jam

= 0,332 kg/jam

F16air = 0,1% x F14

air

= 0,001 x 4313,993 kg/jam

= 4,314 kg/jam

F16dekstrin = 0,1% x F14

dekstrin

= 0,001 x 4,974 kg/jam

= 0,005 kg/jam


F15glukosa= F14

glukosa- F16glukosa

= 8290,175 - 8,290

= 8281,885 kg/jam

F15maltosa= F14

maltosa- F16maltosa

= 331,607 - 0,332

= 331,275 kg/jam

F15air= F14

air- F16air

= 4313,993 - 4,314

= 4309,679 kg/jam

F15dekstrin= F14

dekstrin - F16 dekstrin

= 4,974 - 0,005

= 4,969 kg/jam


Aliran 14 Aliran 16 Aliran 15Glukosa 8290,175 8,290 8281,885Maltosa 331,607 0,332 331,275Air 4313,993 4,314 4309,679Dekstrin 4,974 0,005 4,969Karbon Aktif 246,050 246,050

TOTAL 13186,799258,990 12927,809

13186,799

8. Evaporator (EV-01)

Fungsi : Memekatkan larutan glukosa hingga 78%.

Neraca Massa total : F15 = F17 + F18




F15air= 4309,679 kg/jam



Untuk memekatkan sirup glukosa menjadi 78% berarti harus menghilangkan

kandungan airnya sebesar 22%.

F15air= F18

air + F17air

F15air= (0,78 x F15

air) + F17air

0,22 F15air = F17

air

F17air = 0,22 x 4309,679 kg/jam

= 948,129 kg/jam

F17glukosa = F15


F17maltosa= F15


F17dekstrin= F15



F18air= F15

air - F17air

F18air= 4309,679 kg/jam - 948,129 kg/jam

= 3361,550 kg/jam


Aliran 15 Aliran 18 Aliran 17Glukosa 8281,885 8281,885Maltosa 331,275 331,275Air 4309,679 3361,550 948,129Dekstrin 4,969 4,969

TOTAL 12927,8093361,550 9566,259

12927,809

9. Reaktor Hidrogenasi (R-02)

19

20

21

Reaksi : C6H12O6 + H2 C6H14O6Berdasarkan US Patent No 4.322.569 perbandingan

mol glukosa dan hidrogen yaitu 1:1200.

Tabel L1 mutu dan uji glukosa

Komposis

i

persentasi

Glukosa 50%

Maltosa 2%

Air 47.97%

Dekstrin 0.03%

(Sumber: Departemen Perindustrian)

Basis 1kg sirup glukosa terdapat 50% glukosa

mol C6H12O6 = 0,5 kg/jam = 0,00277 kmol/jam

180,156 kg/kmol

mol H2 = 1200 x 0,00277

= 3,330446946 kmol/jam

Konversi glukosa menjadi sorbitol 98% dan yield 99%( US Patent No 4.322.569)

mol C6H14O6 = 98% x 0,00277

= 0,002719865 kmol/jam

C6H12O6 + H2 C6H14O6

m : 0,002775372 3,330446946

r : 0,002719865 0,002719865 0,002719865

s : 5,55074E-05 3,327727081 0,002719865

untuk kapasitas 65000 ton/ tahun maka

scale up = 45,05121922 = 16563,7703 kmol/jam

0,002719865

mol C6H12O6 = 16563,7703 x 0.002775372

= 45,97063185 kmol/jam

mol H2 = 45,97063185 x 1200

= 55164,75822 kmol/jam

mol C6H14O6 = 45,05121922 kmol/jam

reaksi yang terjadi:

C6H12O6 + H2 C6H14O6

m : 45,97063185 55164,75822

r : 45,05121922 45,05121922 45,05121922

s : 0,919412637 55119,707 45,05121922

reaksi samping :

C12H22O11 + H2 C12H24O11

m 0,967803907 55119,707

r 0,958125868 0,958125868 0,958125868

s 0,009678039 55118,74888 0,958125868

mol Maltosa = 331,2754061 = 0,967803907

342,296

mol Maltilol = 99% x 0,967803907

= 0,958125868

glukosa yang dibutuhkan = 45,97063185 x 180,156

= 8281,885152 kg/jam

Sirup glukosa yang dibutuhkan = 8281,885152 = 16563,7703 kg/jam

0,5

Hidrogen yang dibutuhkan = 55164,75822 x 2

= 111212,1526 kg/jam

Aliran F19

Glukosa = 50% x 16563,7703

= 8281,885152 kg/jam

Maltosa = 2% x 16563,7703

= 331,2754061 kg/jam

Air = 47,97% x 16563,7703

= 7945,640615 kg/jam

Dextrin = 0,03% x 16563,7703

= 4,969131091 kg/jam

Aliran F20

Hidrogen recycle = 75% x F26

= 75% x 111108.2858

= 83331.21435 kg/jam

Hidrogen make up = 111212,1526 - 83331,21435

= 27880.93823 kg/jam

Hidrogen = 83331.21435 + 83331.21435 kgk/jam

= 111212,1526 kg/jam

Aliran F21

Glukosa = 0,919412637 x 180,156

= 165,637703 kg/jam

Maltosa = 0,009678039 x 342,296

= 3,312754061 kg/jam

Hidrogen = 55118,74888 x 2

= 111119,3977 kg/jam

Sorbitol = 45,05121922 x 182,172

= 8207,070707 kg/jam

Maltilol = 0,958125868 x 344,312

= 329,8942338 kg/jam

Air = 47,97% x 16563,7703

= 7945,640615 kg/jam

Dextrin = 0,03% x 16563,7703

= 4,969131091 kg/jam

Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)F19 F20 F21

Glukosa 8281.885152 165.637703Maltosa 331.2754061 3.312754061Air 7945.640615 7945.640615Dekstrin 4.969131091 4.969131091Hidrogen 111212.1526 111119.3977Maltitol 329.8942338Sorbitol 8207.070707Subtotal 16563.7703 111212.1526 127775.9229Total 127775.9229 127775.9229

10. Separator (SE-01)

25

23

24

efisiensi separator = 99-100% (sumber : US. Patent No.4.322.569) dan efisiensi separator Flash drum =99,99 % (Suherman,2009)Aliran F23

Glukosa = 0,919412637 x 180,156

= 165,637703 kg/jam

Maltosa = 0,009678039 x 342,296

= 3,312754061 kg/jam

Hidrogen = 55118,74888 x 2

= 111119,3977 kg/jam

Sorbitol = 45,05121922 x 182,172

= 8207,070707 kg/jam

Maltilol = 0,958125868 x 344,312

= 329,8942338 kg/jam

Air = 7945,640615 kg/jam

Dextrin = 4,969131091 kg/jam

Aliran F25

Hidrogen = 99,99% x F25 Hidrogen= 111108.2858 kg/jam

Glukosa = 0,01% x F25 Glukosa= 0.01656377 kg/jam

Maltosa = 0,01% x F25 Maltosa= 0.000331275 kg/jam

Air = 0,01% x F25 Air= 0.794564061 kg/jam

Dekstrin = 0,01% x F25 Dekstrin= 0.000496913 kg/jam

Maltitol = 0,01% x F25 Maltitol= 0.032989423 kg/jam

sorbitol = 0,01% x F25 Sorbitol= 0.820707071 kg/jam

Aliran F24

Hidrogen = 111119,3977 - 111108.2858= 11.11193977 kg/jam

Glukosa = 165,637703 - 0.01656377= 165.6211393 kg/jam

Maltosa = 3.312754061 - 0.000331275= 3.312422785 kg/jam

Air = 7945,640615 - 0.794564061= 7944.846051 kg/jam

Dekstrin = 4,969131091 - 0.000496913= 4.968634178 kg/jam

Maltitol = 329.8942338 - 0.032989423= 329.8612444 kg/jam

sorbitol = 8207,070707 - 0.820707071

= 8206.25 kg/jam

Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)

F23 F25 F24Glukosa 165.637703 0.01656377 165.6211393Maltosa 3.312754061 0.00033127

53.312422785

Air 7945.640615 0.794564061

7944.846051

Dekstrin 4.969131091 0.000496913

4.968634178

Hidrogen 111119.3977 111108.2858

11.11193977

Maltitol 329.8942338 0.032989423

329.8612444

Sorbitol 8207.070707 0.820707071

8206.25

Subtotal 127775.9229 111109.9515

16665.97143

Total 127775.9229 127775.9229

11. Pressure Swing Adsorption (PSA-01)

31 25

30

Berdasarkan US Patent No 4.322.569 hidrogen yang dihasilkan dari unit PSA 75% Aliran 25Hidrogen = 99,99% x F25 Hidrogen

= 111108.2858 kg/jam Glukosa = 0,01% x F25 Glukosa

= 0.01656377 kg/jamMaltosa = 0,01% x F25 Maltosa

= 0.000331275 kg/jam

Air = 0,01% x F25 Air= 0.794564061 kg/jam

Dekstrin = 0,01% x F25 Dekstrin= 0.000496913 kg/jam

Maltitol = 0,01% x F25 Maltitol= 0.032989423 kg/jam

sorbitol = 0,01% x F25 Sorbitol= 0.820707071 kg/jam

Aliran 31Hidrogen = 75% x 111108.2858

= 83331.21435 kg/jam

Aliran 30

Glukosa = 0,01% x F25 Glukosa

= 0.01656377 kg/jam

Maltosa = 0,01% x F25 Maltosa

= 0.000331275 kg/jam

Air = 0,01% x F25 Air

= 0.794564061 kg/jam

Dekstrin = 0,01% x F25 Dekstrin

= 0.000496913 kg/jam

Maltitol = 0,01% x F25 Maltitol

= 0.032989423 kg/jam

sorbitol = 0,01% x F25 Sorbitol

= 0.820707071 kg/jam

Komponen Masuk (Kg/Jam) Keluar (Kg/Jam)F25 F31 30

Glukosa 0.01656377 0.01656377Maltosa 0.000331275 0.00033127

5Air 0.794564061 0.79456406

1Dekstrin 0.000496913 0.00049691

3Hidrogen 111108.2858 83331.2143

527777.0714

5Maltitol 0.032989423 0.03298942

3

Sorbitol 0.820707071 0.820707071

Subtotal 111109.9515 83331.21435

27778.7371

Total 111109.9515 111109.9515


28

26

27

Berdasarkan US Patent No 4.322.569 sorbitol yang terbentuk dari unit Evaporator sebesar 70%

Aliran 26

Glukosa = 165,637703 - 0.01656377

= 165.6211393 kg/jam

Maltosa = 3.312754061 - 0.000331275

= 3.312422785 kg/jam

Air = 7945,640615 - 0.794564061

= 7944.846051 kg/jam

Dekstrin = 4,969131091 - 0.000496913

= 4.968634178 kg/jam

Maltitol = 329.8942338 - 0.032989423

= 329.8612444 kg/jam

Sorbitol = 8207,070707 - 0.820707071

= 8206.25 kg/jam

F26 total = 165.6211393 + 3.312422785 + 7944.846051 + 4.968634178 +

329.8612444 + 8206.25

Massa umpan = 16654.85949 kg/jam

massa bahan kering = F27 total – F27 air

= 16654.85949 - 7944.846051

= 8710.013441 kg/jam

Fraksi air (Xf) = 8710.013441 = 0.522971295

16654.85949

Fraksi liquid (Xl) = 0,7

neraca massa komponen total

F . Xf = (L . Xl) + (V . Xv)

F . Xf = 16654.85949 x 0.522971295

= 8710.013441

V . Xv = 0

L = (F . Xf) / Xl = 8710.013441 = 12442.87634

0,7

jumlah air yang diuapkan

V = F – L = 16654.85949 - 12442.87634

= 4211.983148 kg/jam

Aliran F28

Air = 4211.983148 kg/jam

Aliran 27

Glukosa = 165,637703 - 0.01656377

= 165.6211393 kg/jam

Maltosa = 3.312754061 - 0.000331275

= 3.312422785 kg/jam

Air = 7944.846051 - 4211.983148

= 3732.862903 kg/jam

Dekstrin = 4,969131091 - 0.000496913

= 4.968634178 kg/jam

Maltitol = 329.8942338 - 0.032989423

= 329.8612444 kg/jam

Sorbitol = 8207,070707 - 0.820707071

= 8206.25 kg/jam

Komponen Masuk (kg/Jam) Keluar (Kg/Jam)F26 F28 F27

Glukosa 165.6211393 165.6211393Maltosa 3.312422785 3.312422785Air 7944.846051 4211.98314

83732.862903

Dekstrin 4.968634178 4.968634178Maltitol 329.8612444 329.8612444Sorbitol 8206.25 8206.25Subtotal 16654.85949 4211.98314

812442.87634

Total 16654.85949 16654.85949

LAMPIRAN B

PERHITUNGAN NERACA ENERGI

Basis perhitungan = 1 hari operasi

Satuan operasi = kJ/jam

Temperatur basis = 250C

Dari Reklaitis 1983 & Carl L Yaws .1999 diperoleh harga kapasitas panas

untuk masing – masing komponen yang terlibat dalam pembuatan sorbitol adalah

sebagai

berikut:

Nilai Cp pada temperatur 298.15K

Air Cp(l) J/mol K Cp(v) J/mol K Hidrogen (gas) J/mol Ka 18.2964 7.9857 25.399b 0.472118 0.00046332 2.02E-02c -0.00133878 1.4028E-06 -3.85E-05d 0.000001314 -6.5784E-10 3.19E-08e -8.76E-12(Sumber : Reklaitis 1983 & Carl L Yaws .1999)

Perhitungan Cp pada temperatur 298.15K metode Kopp’s Rule

elemen Cp(J/g atom C)Solid Liquid

C 7.5 12H 9.6 18O 17 25

komponen formula heat capacity (J/mol K)glukosa C6H12O6 2.89E+02

maltosa C12H22011 1021.311514

dekstrin (C6H10O5)10 22378.46604

maltitol C12H24O11 1072.705517

sorbitol C6H12O6 292.115453pati C6H10O5 223.5914333hidrogen H2 0.141706755

Untuk Abu

Cp(kJ/kmol.K)A 0,18B 0,000078

Cp = A + BT

Untuk HCl

Cpl(kJ/kmol.K)A 17,7227B 0,904261C -0,00564496D 1,13383E-05

Perhitungan Panas (dQ/dt)

Panas dapat dihitung dengan menggunakan rumus;

dQ/dt = ∑FCpdT (Reklaitis,1983)

Dimana:

dQ/dt = jumlah panas persatuan waktu, J/jam

F = jumlah massa masing – masing komponen persatuan waktu, kg/jam

Cp = kapasitas panas masing – masing komponen, kkal/kgoC

dT = Temperatur masing – masing komponen dikurangi temperatur referensi, oC

Perhitungan kapasitas panas :

Untuk padatan :

Pati (C12H22O11) = (12 x 7,5) + (22 x 9,6) + (11 x 17)

= 488,2 J/g-atom 0C x 342 g/mol

= 166964,4 J/mol 0C = 611,255 kJ/kmol 0K

Lemak (C15H31COOH) = (16 x 7,5) + ( 32 x 9,6) + (2 x 17)

= 461,2 J/g-atom 0C x 256 g/mol

= 118067,2 J/mol 0C = 432,24 kJ/kmol 0K

Serat ((C6H10O5)1000) = (6000 x 7,5) + ( 10000 x 9,6) + (5000 x 17)

= 226000 J/g-atom 0C x 162000 g/mol

= 20114000 J/mol 0C = 73637,19 kJ/kmol 0K

Protein (CH3(CHNH2)COOH)= (3 x 7,5) + ( 7 x 9,6) + (2 x 17) + (1 x 26)

= 149,7 J/g-atom 0C x 89 g/mol

= 24251400 J/mol 0C = 88784,18 kJ/kmol 0K

Untuk liquid :

Pati (C12H22O11) = (12 x 12) + (22 x 18) + (11 x 25)

= 815 J/g-atom 0C x 342 g/mol

= 278730 J/mol 0C = 1020,428 kJ/kmol 0K

Lemak (C15H31COOH) = (16 x 12) + ( 32 x 18) + (2 x 25)


= 209408 J/mol 0C = 766,64 kJ/kmol 0K

Serat ((C6H10O5)1000) = (6000 x 12) + ( 10000 x 18) + (5000 x 25)

= 377000 J/g-atom 0C x 162000 g/mol

= 6,107 x 1010 J/mol 0C

= 2,235 x 108 kJ/kmol 0K

Protein (CH3(CHNH2)COOH) = (3 x 12) + ( 7 x 18) + (2 x 25) + (1 x 33)


= 21805 J/mol 0C = 79,827 kJ/kmol 0K

1. Tangki Perebusan (M-01)

Temperatur basis = 250C = 298,15 K

Temperatur umpan = 300C = 303,15 K

Temperature produk = 500C = 323,15 K

Panas yang masuk (Qi) aliran 1 :

Qpati = F1 pati ∫298,15

303,15

Cp dT

= ¿) x 611,255 kJ/mol 0K x (303,15 - 298,15) K

= 78287,881 kJ/jam

Qair = F1air ∫298,15

303,15

Cp dT

= ¿) x 374,645 kJ/mol

= 21489,89 kJ/jam

Qprotein = F1 protein ∫298,15

303,15

Cp dT

= ¿) x 88784,18 kJ/kmol 0K x (303,15 - 298,15) K

= 4401416,34 kJ/jam

Qabu = F1abu ∫298,15

303,15

Cp dT

= ¿) x 1,017 kJ/kmol

= 1,614 kJ/jam

Qlemak = F1lemak ∫298,15

303,15

CpdT

= ¿) x 432,24 kJ/kmol 0K x (303,15 - 298,15) K

= 2247,164 kJ/jam

Qserat = F1 serat ∫298,15

303,15

Cp dT

= ¿) x 73637,19 kJ/kmol 0K x (303,15 - 298,15) K

= 327,900 kJ/jam

Qair umpan= F2air umpan ∫298,15

303,15

Cp dT

= ¿) x 374,64 kJ/kmol

= 64803,86 kJ/jam

Total panas yang masuk

dQ/dT =78287,881 + 21489,89 + 4401416,34 + 1,614 + 2247,164 + 327,900 +

64803,86 = 4568574,66 kJ/jam

Panas yang keluar (Qo) aliran 3 :


323,15

Cp dT

= ¿) x 611,255 kJ/kmol 0K x (323,15 - 298,15) K

= 391439,408 kJ/jam


323,15

Cp dT

= ¿) x 1878,589 kJ/kmol

= 432703,559 kJ/jam


323,15

Cp dT

= ¿) x 443920,922 kJ/kmol 0K x (323,15 - 298,15) K

= 22007081,7 kJ/jam


323,15

Cp dT

= ¿) x 0,067 kJ/kmol

= 0,106 kJ/jam


323,15

Cp dT

= ¿) x 2161,21 kJ/kmol 0K x (323,15 - 298,15) K

= 11235,82 kJ/jam


303,15

Cp dT

= ¿) x 368185,97 kJ/kmol 0K x (323,15 - 298,15) K

= 1639,50 kJ/jam

Total panas yang keluar

dQ/dT =391439,408 + 432703,559 + 22007081,7 + 0,106 + 11235,82 +

1639,50

= 22844100,1 kJ/jam

Agar temperatur menjadi 100oC maka perlu ditambahkan steam.

Panas yang dilepas = 22844100,1 - 4568574,66

= 18275525,43 kJ/jam

Cp air = 4,18 kJ/kg K

Q = m x Cp x dT

m = 87442,705 kg/jam

2. Reaktor (R-01)

Temperature slurry = 500C = 323,15 K

Komponen Qin(kJ/jam) Qout(kJ/jam)

Pati 78287,88156 391439,4078air 21489,89626 432703,5591protein 4401416,338 22007081,69abu 1,614947134 0,106535825lemak 2247,164655 11235,82328serat 327,9002753 1639,501376

64803,8618SUBTOTAL 4568574,657 22844100,09Steam 18275525,43TOTAL 22844100 22844100,09

Temperature HCl = 300C = 303,15 K

Temperature produk = 1350C = 408,15 K


Qpati = 391439,408 kJ/jam

Qair = 432703,559 kJ/jam

Qprotein = 22007081,7 kJ/jam

Qabu = 0,106 kJ/jam

Qlemak = 11235,82 kJ/jam

Qserat = 1639,50 kJ/jam


Qair = F4 air ∫298,15

323,15

Cp dT

= ¿) x 374,645 kJ/kmol

= 120009,386 kJ/jam

QHCl = F4 HCl ∫298,15

323,15

Cp dT

= ¿) x 437,38 kJ/kmol

= 4958,713 kJ/jam


dQ/dT = 391439,408 + 432703,559 + 22007081,7 + 0,106 + 11235,82 +

1639,50 + 120009,386 + 4958,713

= 22798393,3 kJ/jam

Panas yang keluar (Qi) aliran 5 :


408,15

CpdT

= ¿) x 611,255 kJ/kmol 0K x (408,15 - 298,15) K

= 287525,956 kJ/jam


408,15

Cp dT

= ¿) x 464,316 kJ/kmol

= 8563,443 kJ/jam


408,15

Cp dT

= ¿) x 443920,922 kJ/kmol 0K x (408,15 - 298,15) K

= 87063,156 kJ/jam


408,15

Cp dT

= ¿) x 22,83 kJ/kmol

= 36,244 kJ/jam


408,15

Cp dT

= ¿) x 2161,21 kJ/kmol 0K x (408,15 - 298,15) K

= 87684,247 kJ/jam


408,15

Cp dT

= ¿) x 368185,97 kJ/kmol 0K x (408,15 - 298,15) K

= 21903947,1 kJ/jam

Qglukosa = F5 glukosa ∫298,15

408,15

Cp dT

= ¿) x 288,882 kJ/kmol 0K x (408,15 - 298,15) K

= 1465204,79 kJ/jam

Qdekstrin = F5dekstrin ∫298,15

408,15

CpdT

= ¿) x 22378,466 kJ/kmol 0K x (408,15 - 298,15) K

= 6216621,44 kJ/jam

Qmaltosa = F5maltosa ∫298,15

408,15

Cp dT

= ¿) x 1021,311 kJ/kmol 0K x (408,15 - 298,15) K

= 1635,811 kJ/jam

QHCl = F5 HCl ∫298,15

408,15

Cp dT

= ¿) x 15273,036 kJ/kmol

= 173154,618 kJ/jam


dQ/dT = 287525,956 + 8563,443 + 87063,156 + 36,244 + 87684,247 +

21903947,1 + 1465204,79 + 6216621,44 + 1635,811 + 173154,618

= 30231436,8 kJ/jam

Tabel Hf 298

Komponen Hr298

C6H12O6 -1.065,121kJ/mol

C12H22O11 -1864,9762kJ/mol

H2O -13564,627kJ/mol

Maltosa -1962.17048kJ/mol

( Sumber: Carl L Yaws.1999 & metode verma dan doraiswamy)

Reaksi yang terjadi :

C12H22O11 + H2O 2 C6H12O6

ΔHf 298,15 =[n.ƩΔH°ƒ C6H12O6 - (n.ƩΔH°ƒ C12H22O11 + n.ƩΔH°ƒ H2O)]

ΔHf 298,15= -1065,121 kJ/mol

ΔH408,15 = (n. ƩCp.dT produk - n. ƩCp.dT reaktan)

ΔH408,15 = (n. Cp.dT C6H12O6 - (n. Cp.dT C12H22O11 + n. Cp. dT H2O))

= - 88,827 kJ/mol

ΔH reaksi = ΔHf 298,15 + ΔH408,15

= -1153,949 kj/mol

Q reaksi = ΔHr x mol bereaksi

Q reaksi = -1153,949 kJ/mol x 46,1088 mol/jam

Q reaksi = -53207,208 kJ/jam

Q in = Q out + Qreaksi + Q cooler (Felder, 2005)

Q cooler = Q in – (Q out + Qreaksi)

Q cooler = 22798393,3– (34103488,87+ (-53207,208))

Q cooler = -7525414,784 kJ/jam

Q cooler = m cooler x Cp air x (T cooler out – T cooler in)

Apabila Cp air = 4,181 kJ/kg.K

Tcooler in = 30 oC = 303,15 K

Tcooler out= 60 oC = 333,15 K

m cooler = Q cooler

Cp air x¿¿

m cooler = 60011,282 kg/jam


Pati 653468,0814 287525,9558air 120009,3867 154141,9752protein 22007081,69 87063,15641abu 0,106535825 36,24395165lemak 11235,82328 87684,24791Serat 1639,501376 21903947,06Glukosa 1465204,79Dekstrin 6216621,439Maltosa 1635,810827HCl 4958,713508 173154,6179SUBTOTAL 22798393,3 30377015,29Qpendingin -7.525.414,784Qreaksi -53.207,208TOTAL 22798393,3 22798393,3

3. Cooler (C-01)

T masuk = 135oC (408.15K)

T keluar = 60oC (333.15K)

T air pendingin in = 25oC (298.15K)

T air pendingin out = 50oC (323.15K)

Aliran 5


408,15

CpdT

= ¿) x 611,255 kJ/kmol 0K x (408,15 - 298,15) K

= 287525,956 kJ/jam


408,15

Cp dT

= ¿) x 8357,695 kJ/kmol

= 154141,975 kJ/jam


408,15

Cp dT

= ¿) x 443920,922 kJ/kmol 0K x (408,15 - 298,15) K

= 87063,156 kJ/jam


408,15

Cp dT

= ¿) x 22,83 kJ/kmol

= 36,244 kJ/jam


408,15

Cp dT

= ¿) x 2161,21 kJ/kmol 0K x (408,15 - 298,15) K

= 87684,247 kJ/jam


408,15

Cp dT

= ¿) x 368185,97 kJ/kmol 0K x (408,15 - 298,15) K

= 21903947,1 kJ/jam


408,15

Cp dT

= ¿) x 288,882 kJ/kmol 0K x (408,15 - 298,15) K

= 1465204,79 kJ/jam


408,15

CpdT

= ¿) x 22378,466 kJ/kmol 0K x (408,15 - 298,15) K

= 6216621,44 kJ/jam


408,15

Cp dT

= ¿) x 1021,311 kJ/kmol 0K x (408,15 - 298,15) K

= 1635,811 kJ/jam

QHCl = F5 HCl ∫298,15

408,15

Cp dT

= ¿) x 15273,036 kJ/kmol

= 173154,618 kJ/jam


dQ/dT = 287525,956 + 154141,975 + 87063,156 + 36,244 + 87684,247 +

21903947,1 + 1465204,79 + 6216621,44 + 1635,811 + 173154,618

= 30377015,3 kJ/jam

Aliran 6


333,15

Cp dT

= ¿) x 611,255 kJ/kmol 0K x (333,15 - 298,15) K

= 91485,534 kJ/jam

Qair = F6 air ∫298,15

333,15

Cp dT

= ¿) x 2633,43 kJ/kmol

= 4568,66 kJ/jam


333,15

Cp dT

= ¿) x 443920,922 kJ/kmol 0K x (333,15- 298,15) K

= 13133,1391 kJ/jam

Qabu = F6 abu ∫298,15

333,15

Cp dT

= ¿) x 5,428 kJ/kmol

= 4,926 kJ/jam


333,15

Cp dT

= ¿) x 2161,21 kJ/kmol 0K x (333,15 - 298,15) K

= 23414,34 kJ/jam


333,15

Cp dT

= ¿) x 368185,97 kJ/kmol 0K x (333,15 - 298,15) K

= 4316453,32 kJ/jam


333,15

Cp dT

= ¿) x 288,882 kJ/kmol 0K x (333,15 - 298,15) K

= 466201,524 kJ/jam

Qdekstrin = F6 dekstrin ∫298,15

333,15

Cp dT

= ¿) x 22378,466 kJ/kmol 0K x (333,15- 298,15) K

= 1978015,91 kJ/jam


333,15

CpdT

= ¿) x 1021,311 kJ/kmol 0K x (333,15 - 298,15) K

= 520,485 kJ/jam

QHCl = F6 HCl ∫298,15

333,15

CpdT

= ¿) x 3423,846 kJ/kmol

= 38817,087 kJ/jam


dQ/dT = 91485,534 + 4568,66 + 13133,1391 + 4,926 + 23414,34 +

4316453,32 + 466201,524 + 1978015,91 + 520,485 + 38817,087

= 6976614,94 kJ/jam

Panas yang diterima = 6976614,94 - 30377015,3

= -23400400,4 kJ/jam

Cp air = 4,18 kj/kg K

ΔT = -85 K

Air pendingin yang dibutuhkan

Q = m x Cp x dT

m = 65860,963 kg/jam


Pati 287525,9558 91485,5314air 154141,9752 48568,66784protein 87063,15641 13133,13912abu 36,24395165 4,926736389lemak 87684,24791 23414,34036

serat 21903947,06 4316453,322Glukosa 1465204,79 466201,5241Dekstrin 6216621,439 1978015,912Maltosa 1635,810827 520,4852633HCl 173154,6179 38817,08712SUBTOTAL 30377015,29 6976614,937Panas yang diserap -23400400,36TOTAL 6976614,937 6976614,937


Temperatur masuk = 80oC (353.15K)

Temperatur keluar = 110oC (383.15K)

Temperatur uap air = 100oC (383.15K)

Aliran 15

Qair = F15air ∫298,15

353.15

Cp dT

= ¿) x 4148,39 kJ/kmol

= 76280 kJ/jam


353.15

Cp dT

= ¿) x 288,882 kJ/kmol 0K x (353,15 - 298,15) K

= 730406,785 kJ/jam


353,15

Cp dT

= ¿) x 22378,466 kJ/kmol 0K x (353,15- 298,15) K

= 3772,10079 kJ/jam


353,15

Cp dT

= ¿) x 1021,311 kJ/kmol 0K x (353,15 - 298,15) K

= 707235,489 kJ/jam

Total panas masuk

dQ/dT = 76280 +730406,785 +3772,10079 +707235,489

= 1517694,44 kJ/jam

Aliran 18

Qair = F18air ∫298,15

383.15

Cp dT

= ¿) x 703,863 kJ/mol

= 131331,761 kJ/jam

Aliran 17

Qair = F17air ∫298,15

383.15

Cp dT

= ¿) x 6435,066 kJ/mol

= 338658,731 kJ/jam


383.15

Cp dT

= ¿) x 288,882 kJ/mol 0K x (383.15 - 298,15) K

= 1128810,49 kJ/jam


383.15

Cp dT

= ¿) x 22378,466 kJ/kmol 0K x (383.15- 298,15) K

= 5829,61 kJ/jam


383.15

Cp dT

= ¿) x 1021,311 kJ/kmol 0K x (383.15 - 298,15) K

= 84016,4882 kJ/jam

Total panas keluar

dQ/dT = 131331,761+338658,731 +1128810,49 +5829,61 + 84016,4882

= 1688647,08 kJ/jam

Panas yang dilepas = 1688647,08 - 1517694,44

= 170952,638 kJ/jam

Superheated steam pada 1atm 150oC, H(150) = 2776.3 kJ/kg

Saturated steam pada 1 atm, 100 oC, Hv(100) = 2691.3 kJ/kg

HL(100) = 461.3 kJ/kg (Reklaitis, 1983)

λ = [H(150) - Hv(100)] + [Hv(100) - HL(100)]

λ = 2315 kj/kg

m = Q yang dilepas / λ

m = 73,845 kg/jam


air(liquid) 76280,06521 338658,7315Glukosa 730406,785 1128810,486Dekstrin 3772,100794 5829,610317

Maltosa 707235,4888 84016,48822air(vap) 131331,7614SUBTOTAL 1517694,44 1688647,077Panas yang dilepas 170952,6376TOTAL 1688647,077 1688647,077

5. Cooler (C-02)

Temperatur masuk = 110oC (383,15K)

Temperatur keluar = 100oC (373,15K)

T air pendingin in = 25oC (298,15K)

T air pendingin out = 50oC(323,15K)

Aliran 17

Qair = 338658,731 kJ/jam

Qglukosa = 1128810,49 kJ/jam

Qdekstrin = 5829,61 kJ/jam

Qmaltosa = 84016,4882 kJ/jam

Total panas masuk

dQ/dT =338658,731 +1128810,49 +5829,61 + 84016,4882

= 1557315,32 kJ/jam

Aliran 19

Qair = F19air ∫298,15

373.15

Cp dT

= ¿) x 5670,725 kJ/kmol

= 298433,723 kJ/jam


373.15

Cp dT

= ¿) x 288,882 kJ/kmol 0K x (373.15 - 298,15) K

= 996009,252 kJ/jam


373.15

Cp dT

= ¿) x 22378,466 kJ/kmol 0K x (373.15- 298,15) K

= 5143,773 kJ/jam


373.15

Cp dT

= ¿) x 1021,311 kJ/kmol 0K x (373.15 - 298,15) K

= 74132,195 kJ/jam

Total panas keluar

dQ/dT = 298433,723 +996009,252 +5143,773 + 74132,195

= 1373718,94 kJ/jam

Panas yang diterima = 1373718,94 – 1557315,32

= -183596,37 kJ/jam


ΔT = -60 K


Q = m x Cp x dT

m = 732,0429 kg/jam


air 338658,7315 298433,7231Glukosa 1128810,486 996009,2522Dekstrin 5829,610317 5143,77381Maltosa 84016,48822 74132,19549SUBTOTAL 1557315,316 1373718,945Panas yang diserap -183596,3712TOTAL 1373718,945 1373718,945

6. Heater (HE-03)

Temperatur masuk = 30oC (303,15 K)

Temperatur keluar = 100oC (373,15 K)

Aliran 33

Hidrogen = F33x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 55164758,22 x 143,8791569

= 7937058902 J/jam

Aliran 20

Hidrogen = F20x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 55164758,22 x 2168.800684

= 1,19542E+11 J/jam

Agar temperatur menjadi 100oC maka perlu ditambahkan steam.

Panas yang dilepas = 1,19542E+11-7937058902

= 1,11605E+11J/jam

= 111604521,4 kj/jam


ΔT = 70 K

Steam yang dibutuhkan

Q = m x Cp x dT

m = 381423,518 kg/jam

komponen Masuk (J/jam) Keluar (J/jam)F33 F20

H2 7937058902 1.19542E+11steam 1.11605E+11total 1.19542E+11 1.19542E+11

7. Reaktor (R-02)

19 21

Temperatur masuk = 100oC (373,15 K)

Temperatur keluar = 145oC (418,15 K)

Glukosa = F19 x Cp x dT

= 45970,63185 x 288,8827677 x (373,15-298,15)

= 996009252,2 J/jam

Maltosa = F19 x Cp x dT

= 967,8039068 x 1021,311514 x (373,15-298,15)

= 74132195,49 J/jam

Dextrin = F19 x Cp x dT

= 275,8176671 x 22378,46604 x (373,15-298,15)

= 462928222,4 J/jam

Air = F19 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 4900,481445 x 5671.178789

= 27791506.43 J/jam

Hidrogen = F19 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 55164758,22 x 2168.800684

= 1.19641E+11

Total panas masuk

dQ/dT =(996009252,2+74132195,49+462928222,4+27791506,43+1.19641E+11)

= 1,21202E+11 J/jam

Aliran 21


= 919,4126371 x 288,8827677 x (418,15-298,15)

= 31872296,07 J/jam


= 9,678039068 x 1021,311514 x (418,15-298,15)

= 1186115,128 J/jam


= 275,8176671 x 22378,46604 x (418,15-298,15)

= 740685155,8 J/jam

Air = F21 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 4900,481445 x 9133,392257

= 44758019,29 J/jam


Toperasi

a+bT +c T2+d T3 dT

= 55118748,88 x 3478,586535

=1,91735E+11 J/jam

Sorbitol = F21 x Cp x dT

= 23836,14485 x 292,115453 x (418,15-298,15)

= 835548750,4 J/jam

Maltilol = F21 x Cp x dT

= 1810,894285 x 1072.705517 x (418,15-298,15)

= 233106754,9 J/jam

Total panas keluar

dQ/dT= (31872296,07+1186115,128+740685155,8+44758019,29+1,91735E+11

+835548750,4+233106754,9)

= 1,93622E+11 J/jam

Tabel Hf 298

reaktan ΔH°ƒ (kj/mol) produk ΔH°ƒ (kj/mol)C6H12O6 -1065.12088 C6H14O6 -1140H2 0C12H22O11

-1864.97616 C12H24O11 -1962.17048

H2 0( Sumber: Carl L Yaws.1999 & metode verma dan doraiswamy)

reaksi yang terjadi:

C6H12O6 + H2 C6H14O6

reaksi samping :

C12H22O11 + H2 C12H24O11

reaksi utama

ΔHf 298,15 =[n.ƩΔH°ƒ C6H14O6 - (n.ƩΔH°ƒ C6H12O6 + n.ƩΔH°ƒ H2)]

ΔHf 298,15= -74,87912 KJ/mol

reaksi samping

ΔHf 298,15=[n.ƩΔH°ƒ C12H24O11 - (n.ƩΔH°ƒ C12H22O11 + n.ƩΔH°ƒ H2)]

ΔHf 298,15= -97,19432 KJ/mol

ΔHf 298,15 total = (ΔH298,15 utama) + (ΔH298,15 samping)

ΔHf 298,15 total = -172,07344 KJ/mol

reaksi utama


ΔH = (n. Cp.dT C6H14O6 - (n. Cp.dT C6H12O6 + n. Cp. dT H2))

= -3,090664294 kj/mol

reaksi samping


ΔH = (n. Cp.dT C12H24O11 - (n. Cp.dT C12H22O11 + n.Cp. dT H2))

= 9,64586693 kj/mol

ΔH418,15 total = (ΔH418,15 utama) + (ΔH418,15 samping)

ΔH418,15 total = 6,555202636 kj/mol

ΔH reaksi = ΔHf 298,15 total + ΔH418,15 total

= -165,5182374 kj/mol

Q reaksi = ΔHr x mol bereaksi

Q reaksi = -165,5182374 kJ/mol x 23836,14485 mol/jam

Q reaksi = -3945316.682 kJ/jam

Q in = Q out + Qreaksi + Q cooler (Felder, 2005)

Q cooler = Q in – (Q out + Qreaksi)

Q cooler = 1.21202E+11 – (1.93622E+11 + (-3945316682))

Q cooler = -68474951533 J/jam

Q cooler = -68474951,53 kJ/mol

Q cooler = m cooler x Cp air x (T cooler out – T cooler in)

Apabila Cp air = 4,181 kJ/kg.K

Tcooler in = 30 oC = 303,15 K

Tcooler out= 60 oC = 333,15 K

m cooler = Q cooler

Cp air x¿¿

m cooler = 546052,2451kg/jam

komponen masuk (J/jam) keluar (J/jam)19 21

Glukosa 996009252.2 31872296.07Maltosa 74132195.49 1186115.128Dextrin 462928222.4 740685155.8Air 27791506.43 44758019.29H2 1.19641E+11 1.91735E+11Sorbitol 835548750.4Maltilol 233106754.9subtotal 1.21202E+11 1.93622E+11Q pendingin -68474951533Q reaksi -3945316682total 1.21202E+11

8. Cooler (C-03)

T masuk = 145oC (418.15K)

T keluar = 90oC (363.15K)

T air pendingin in = 25oC (298.15K)

T air pendingin out = 40oC (313.15K)

Aliran 22


= 919,4126371 x 288,8827677 x (418,15-298,15)

= 31872296,1 J/jam


= 9,678039068 x 1021,311514 x (418,15-298,15)

= 1186115,13 J/jam


= 275,8176671 x 22378,46604 x (418,15-298,15)

= 740685156 J/jam

Air = F22 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 4900,481445 x 9133,392257

= 44758019,3 J/jam


Toperasi

a+bT +c T2+d T3 dT

= 55118748,88 x 0,141706755

= 937283882 J/jam


= 23836,14485 x 292,115453 x (418,15-298,15)

= 835548750 J/jam


= 1810,894285 x 1072.705517 x (418,15-298,15)

= 233106755 J/jam

Total panas masuk

dQ/Dt =( 31872296,07+ 1186115,128+ 740685155,8+ 44758019,29

+ 937283882,1+ 835548750,4+233106755)

= 2824440974 J/jam

Aliran 23


= 919,4126371 x 288,8827677 x (363,15-298,15)

= 17264160,4 J/jam


= 9,678039068 x 1021,311514 x (363,15-298,15)

= 642479,028 J/jam


= 275,8176671 x 22378,46604 x (363,15-298,15)

= 401204459 J/jam

Air = F23 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 4900,481445 x 4908,964192

= 24056287,9 J/jam


Toperasi

a+bT +c T2+d T3 dT

= 55118748,88 x 0,141706755

= 7810699,02 J/jam


= 23836,14485 x 292,115453 x (363,15-298,15)

= 452588906 J/jam


= 1810,894285 x 1072.705517 x (363,15-298,15)

= 126266159 J/jam

Total panas keluar

dq/dT=(17264160,4+642479,028+401204459+24056287,9+7810699,02

+ 452588906+126266159 )

= 1029833151 J/jam

Panas yang diterima = 1029833151-2824440974

= -1794607823 J/jam

= -1794607.823 kj/jam


ΔT = -105 K


Q = m x Cp x dT

m = 4088,876333 kg/jam

masuk (J/jam) keluar (J/jam)komponen F22 F23

Glukosa 31872296.07 17264160.37Maltosa 1186115.128 642479.0275Dextrin 740685155.8 401204459.4Air 44758019.29 24056287.94H2 937283882.1 7810699.017Sorbitol 835548750.4 452588906.4Maltilol 233106754.9 126266158.9subtotal 6678715099air pendingin -5648881947Total 1029833151 1029833151


Steam 150 oC 28

26 27

Steam 100oC

Temperatur masuk = 90oC (363.15K)

Temperatur keluar = 110oC (383.15K)

Temperatur uap air = 110oC (383.15K)

Aliran 26


= 919,4126371 x 288,8827677 x (363,15-298,15)

= 17264160,4 J/jam


= 9,677071264 x 1021,311514 x (363,15-298,15)

= 642414,7796 J/jam


= 275,7900854 x 22378,46604 x (363,15-298,15)

= 401164339 J/jam

Air = F26 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 4899,991397 x 4908,964192

= 24053882,31 J/jam


= 23833,76124 x 292,115453 x (363,15-298,15)

= 452543647,6 J/jam


= 958,0300552 x 1072.705517 x (363,15-298,15)

= 66799468,17 J/jam

Total panas masuk

dq/dT= (17264160,4+642414,7796+401164339+24053882,31+452543647,6

+66799468,17)

= 962466185,7 J/jam

Aliran 28

Air = F28 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 2089277,355 x 703,8637317

= 1470566556 J/jam

Aliran 27


= 919,4126371 x 288,8827677 x (383,15-298,15)

= 22573952,1 J/jam


= 9,677071264 x 1021,311514 x (383,15-298,15)

= 840080,8657 J/jam


= 275,7900854 x 22378,46604 x (383,15-298,15)

= 524599520,2J/jam

Air = F27 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 2302.246764 x 6435,587376

= 14816310,21 J/jam


= 45046,71409 x 292,115453 x (383,15-298,15)

= 1118501510 J/jam


= 958,0300552 x 1072,705517 x (383,15-298,15)

= 87353150,69 J/jam

Total panas keluar

dq/dT = (1470566556+ 22573952,1+840080,8657+524599520+14816310,21

+1118501510+87353150,69 )

= 3239251080 J/jam

Panas yang dilepas = 3239251080-962466185,7

= 2276784894 J/jam

= 2276784,894 kj/jam

Superheated steam pada 1atm 150oC, H(150) = 2776.3 kJ/kg

Saturated steam pada 1 atm, 100 oC, Hv(110) = 2691.3 kJ/kg

HL(110) = 461.3 kJ/kg (Reklaitis, 1983)

λ = [H(150) - Hv(110)] + [Hv(110) - HL(110)]

λ = 2315 kj/kg

m = Q yang dilepas / λ

m = 983,4923949 kg/jam

Komponen Masuk(J/jam) Keluar (J/jam)F26 F28 F27

Glukosa 17262433.96 22573952.1Maltosa 642414.7796 840080.866Dextrin 401164339 524599520Air 24053882.31 1470566556 14816310.2Maltitol 66799468.17 87353150.7Sorbitol 452543647.6 1118501510subtotal 962466185.7steam 2276784894Total 3239251080 3239251080

10. Cooler (C-04)

Temperatur masuk = 110oC (383,15K)

Temperatur keluar = 30oC (303,15K)

T air pendingin in = 25oC (298,15K)

T air pendingin out = 40oC(313,15K)

Aliran 27


= 919,4126371 x 288,8827677 x (383,15-298,15)

= 22573952,1 J/jam


= 9,677071264 x 1021,311514 x (383,15-298,15)

= 840080,8657 J/jam


= 275,7900854 x 22378,46604 x (383,15-298,15)

= 524599520,2J/jam

Air = F27 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 2302.246764 x 6435,587376

= 14816310,21 J/jam


= 45046,71409 x 292,115453 x (383,15-298,15)

= 1118501510 J/jam


= 958,0300552 x 1072,705517 x (383,15-298,15)

= 87353150,69 J/jam

Total panas masuk

dq/dT = ( 22573952,1+840080,8657+524599520+14816310,21+1118501510

+87353150,69 )

= 1768684524 J/jam

Aliran 29


= 919,4126371 x 288,8827677 x (303,15-298,15)

= 1327879,535 J/jam


= 9,677071264 x 1021,311514 x (303,15-298,15)

= 49416,52151 J/jam


= 275,7900854 x 22378,46604 x (303,15-298,15)

= 30858795,31 J/jam

Air = F29 x ∫Tref

Toperasi

a+bT +c T2+d T3 dT

= 2302.246764 x 910,3465107

= 2095842,308 J/jam


= 45046,71409 x 292,115453 x (303,15-298,15)

= 65794206,48 J/jam


= 958,0300552 x 1072,705517 x (303,15-298,15)

= 5138420,629 J/jam

Total panas keluar

dq/dT = (1327879,535+49416,52151+30858795,31+2095842,308+65794206,48

+5138420,629)

= 105264560,8 J/jam

Panas yang diterima = 105264560,8 - 1768684524

= -1663419963 J/jam

= -1663419,963 kj/jam


ΔT = -80 K


Q = m x Cp x dT

m = 4974,341996 kg/jam

komponen Masuk(J/jam) Keluar(J/jam)F27 F29

Glukosa 22573952.1 1327879.54Maltosa 840080.866 49416.5215Dextrin 524599520 30858795.3Air 14816310.2 2095842.31Maltilol 87353150.7 5138420.63Sorbitol 1118501510 65794206.5subtotal 1768684524air pendingin -1663419963TOTAL 105264561 105264561

Lampiran Nm & Ne Sorbitol

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