Mass and Energy Balance Evaporator
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Mass and Energy Balance of Biosystem
BE2202
BIOENGINEERING DEPARTMENT
SCHOOL OF LIFE SCIENCES AND TECHNOLOGY
BANDUNG INSTITUTE OF TECHNOLOGY
2016
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Group:
• Anca Awal Sembada 11214003
• Kevin Sumendap 11214033
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Multistage Evaporation
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Problems
In the four-stage evaporation system shown in Figure 2.24, a 50%
by weight sugar solution is concentrated to 65% by evaporating an
equal amount of water in each of the four stages. With a total input of
50.000 lb/h, a product stream of 35.000 lb/h is produced. Determine
the compositions of the intermediate streams.
With N2=N4, N4=N6, and N6=N8
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Sebelum melakukan penyelesaian neraca massa, dilakukan perhitungan terhadap nilai derajat kebebasan (DK), yang meliputi:
• DK pada Unit I
• DK pada Unit II
• DK pada Unit III
• DK pada Unit IV
• DK Process
• DK Keseluruhan
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Analisis Derajat Kebebasan (DK)
Unit I Unit II Unit III Unit IV Process Keseluruhan
Variabel Aliran
5 5 5 5 14 8
Neraca Independen
2 2 2 2 8 2
Data Aliran 1 0 0 1 2 2
Data Komposisi
1 0 0 1 2 2
Hubungan Pembantu
0 0 0 0 3 3
Derajat Kebebasan (DK)
1 3 3 1 -1 -1
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Setelah dilakukan analisis derajat kebebasan (DK), disimpulkan
bahwa penyelesaian yang akan kita ambil adalah penyelesaian unit
process dengan DK = -1.
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Karena DK = -1, maka pada sistem ada kelebihan pertelaan
(kelebihan data), sehingga agar DK = 0, salah satu data harus
dihilangkan agar persamaan neraca massa dapat diselesaikan. Data
yang dihilangkan adalah data aliran pada Products, 35.000 lb/h.
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Persamaan Neraca Independen Unit I
• Sugar Balances:
25.000 = Xs,3.N3
• Water Balances:
25.000 = N2 + Xw,3.N3
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Persamaan Neraca Independen Unit II
• Sugar Balances:
Xs,3.N3 = Xs,5.N5
• Water Balances:
Xw,3.N3 = N4 + Xw,5.N5
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Persamaan Neraca Independen Unit III
• Sugar Balances:
Xs,5.N5 = Xs,7.N7
• Water Balances:
Xw,5.N5 = N6 + Xw,7.N7
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Persamaan Neraca Independen Unit IV
• Sugar Balances:
Xs,7.N7 = 0,65.N9
• Water Balances:
Xw,7.N7 = N8 + 0,35.N9
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Persamaan Neraca Independen Total
• Sugar Balance
25.000 = Xs,3.N3
Xs,3.N3 = Xs,5.N5
Xs,5.N5 = Xs,7.N7
Xs,7.N7 = 0,65.N9 +
25.000 = 0,65.N9
Didapatkan nilai N9 = 38.461,54 lb/h
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Persamaan Neraca Independen Total
• Water Balance
25.000 = N2 + Xw,3.N3
Xw,3.N3 = N4 + Xw,5.N5
Xw,5.N5 = N6 + Xw,7.N7
Xw,7.N7 = N8 + 0,35.N9 +
25.000 = N2 + N4 + N6 + N8 + 0,35.N9
25.000 = N2 + N4 + N6 + N8+ + 13.461,54
11.538,46 = N2 + N4 + N6 + N8
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Karena, N2=N4, N4=N6, and N6=N8
Maka,
11.538,46 = N2 + N4 + N6 + N8
11.538,46 = 4N2
Sehingga, N2 = 2.884,62 lb/h
N4 = 2.884,62 lb/h
N6 = 2.884,62 lb/h
N8 = 2.884,62 lb/h
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Menentukan Komposisi Intermediate Stream
Stream 3 antara UNIT I dan UNIT II
• Sugar Balances:
25.000 = Xs,3.N3
N3 = 25.000
Xs,3
• Water Balances:
25.000 = N2 + Xw,3.N3
25.000 = 2.884,62 + (Xw,3.25000
Xs,3)
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Menentukan Komposisi Intermediate Stream
Stream 3 antara UNIT I dan UNIT II
25.000 = 2.884,62 + (Xw,3.25.000
Xs,3)
22.115,38 = Xw,3Xs,3
. 25.000
Xw,3Xs,3
= 0,8846
Water pada stream 3 didapatkan sebesar 0,4694 atau 46,9%
Sugar pada stream 3 didapatkan sebesar 0,5306 atau 53,1%
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Menentukan Komposisi Intermediate Stream
Stream 5 antara UNIT II dan UNIT III
Nilai N3
Nilai N3 dapat dicari dengan menggunakan persamaan neraca total UNIT 1 yaitu N1 = N2 + N3
50.000 = 2.884,62 + N3
N3 = 47.115,38 lb/h
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Menentukan Komposisi Intermediate Stream
Stream 5 antara UNIT II dan UNIT III
• Sugar Balances:
Xs,3.N3 = Xs,5.N5
(0,531).(47.115,38) = Xs,5.N5
25.018,27 = Xs,5.N5
N5 = 25.018,27
Xs,5
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Menentukan Komposisi Intermediate Stream
Stream 5 antara UNIT II dan UNIT III
• Water Balances:
Xw,3.N3 = N4 + Xw,5.N5
(0,469).(47.115,38) = 2.884,62 + Xw,5.N5
22.097,11 = 2.884,62 + Xw,5.N5
19.212,49 = Xw,5.N5
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Menentukan Komposisi Intermediate Stream
Stream 5 antara UNIT II dan UNIT III
19.212,49 = Xw,5.N5 dan N5 = 25.018,27
Xs,5
19.212,49 = Xw,5.25.018,27
Xs,5Xw,5Xs,5
= 0,7679
Water pada stream 5 didapatkan sebesar 0,4344 atau 43,4%
Sugar pada stream 5 didapatkan sebesar 0,5656 atau 56,6%
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Menentukan Komposisi Intermediate Stream
Stream 7 antara UNIT III dan UNIT IV
Nilai N7
Nilai N7 dapat dicari dengan menggunakan persamaan neraca total UNIT IV yaitu N7 = N8 + N9
N7 = 2.884,62 + 38.461,64
N7 = 41.346,26 lb/h
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Menentukan Komposisi Intermediate Stream
Stream 7 antara UNIT III dan UNIT IV
• Sugar Balance
Xs,7.N7 = 0,65.N9
Xs,7.(41.346,26) = 0,65.(38.461,64)
Xs,7.(41.346,26) = 25.000,07
Xs,7 = 0,6046, sehingga Xw,7 = (1-0,6046) = 0,3954
Water pada stream 7 didapatkan sebesar 0,3954 atau 39,5%
Sugar pada stream 7 didapatkan sebesar 0,6046 atau 60,5%
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Kesimpulan
Komposisi pada masing-masing Intermediate Stream adalah sebagai berikut:
Feed : Sugar 50% Stream 7 : Sugar 60,5%
Water 50% Water 39,5%
Stream 3 : Sugar 53,1% Product : Sugar 65%
Water 46,9% Water 35%
Stream 5 : Sugar 56,6%
Water 43,4%