Tugas Azas Teknik Kimia
Kelompok : 7
Oleh : Rio Agung Prakoso/ 110405015
Mutiara Mendopa/ 110405027
Halimah Tusak Diyah/ 140405037
Ekanza Rizqandy Kamny/ 140405056
Eka Mustaqim Barus/ 120405018
Departemen Teknik Kimia
Fakultas Teknik
Universitas Sumatera Utara
Tahun Ajaran 2015/ 2016
2.28 In the plant flowsheet considered in problem 2.27, recovery of soluble as product is
quite low. In an effort to improve recovery, suppose the waste solution from the press is
recycled back to the percolators. However, to reduce the possible release of bitter-tasting
material during pressing, the dewatering rate is decreased so that a slurry of 40% insolubles
is generated. As shown in the flowsheet figure P2.28 to handle the higher water content
slurry, the drier operation is also adjusted to preduce a 62,5% insolubles coffe grounds.
Calculate the recovery ratio resulting from this modification. Assume that the feed coffee
composition is 0% water and 32,7% insolubles and allow the feed water-to-coffee ratio to
be variable. The soluble-to-water ratios in both streams leaving the press are the same.
H2O (1)
Kopi (2)
F1
F2
2
W1 = 0,327 (solubles)
Perkolator
(8)
Separator
aqua
F7
7
W5
(3)
W 35 =0,28
F3
W 1 =0,2
Press
(4)
F4
Aqua
(6)
Drier
F6
6
5
aqua
W54
W1 =
(5)
H2O
F5
exract
F8
W85 =0,
Tabel Derajat Kebebasan
Perkolator
press
Drier
Proses
Overall
separator
Variabel alur bebas
10
17
Pers. Neraca bebas
-Laju alir
-komposisi
Hub.pembantu
Spesifikasi alur
Press
Basis
F3=1000
N.M Total = F3= F4+ F7
1000 = F4 + F7
N.M Komponen
-Solubles
w 5 F3 =
w5
w 5 F3=
w 5 F4 +
0,2 x 1000 = 0,4 x F4 + 0
200
F4 = 0,4
w 5 F7
w45 F4 +
0,28 x 1000 =
-Insolubles :
F4 +
w 75 F7
7
w 5 F7
= 500
Maka: F7= 1000-F4
= 1000-500
= 500
W 75
W 75
( 1W 75 ) = ( 0,6W 45 )
3
4
F3 W 5 = F4 W 5
7
F7 W 5
1000.0,28 = 500 W5
- 500.
280-500 W75
W54 =
500
5
W5
28-50 W 75
= 50
W7
5
1W 7
28-50 W 75
= 50
W57
1- W 57
28-50 W 5
= 2-50
50
x 2-50 W 75
7
7 2
2 W 5 - 50( W 5 = 28-70
80 W5 = 28
W 75 = 0,35
4
W5
0,35
4
=
10,35
0,6- W 5
4
0,21-0,35 W 5 =
W5 . 0,35
W75 - 50( W75 2
W 5 = 0,21
Perkolator
N.M total = F1+F2+F7=F3 +F8
= F1+F2 + 500 = 1000 + F8
2
7
Solubles : (1- W 1 )F2 + W 5 F7
3
8
= W 5 F3 + W 5 F8
(1-0,327)F2 + (0,35 x 500) = (0,28 x 1000) + (0,35 x F8)
0,673 F2 + 175
= 280 + 0,375 F8
0,673 F2
= 105 + 0,35 F8
Insolubles: W 1 F2 = W 1 F3
0,327 F2
= 0,2 x 1000
F2
= 611,6208
Maka F8 = 0,673 F2
= 105 x 0,35 F8
0,673 x 611,6208 = 105 + 0,35 F8
411,6208 105
F8
= 0,35 F8
= 876,0594
F1+F2+F7 = F3 +F8
F1
= F3 +F8- F2+F7
= 1000 + 876,0594 + 611,6208-500
= 764,4386
Drier
N.M Total
= F4
500
= F5 + F6
= F5 + F6
= F4 W 1
Insolubles
= F5 W 1 + F6 W 1
F4 = F5 + F6
500 x 0,4 = F5.0 + F6 x 0,625
Aqua
200
= 0,625 F6
F5 = 500-320
F6
= 320
F5 = 180
= (1-
W54 -
W14 ) F4
(1-0,21-0,4)500
195
= F5 + (1-
= 180 + (1-
W65
8
W61 ) F6
W65
180+120+195
= 320
W5 x F
W65 -
W65 -0,625) 320
= 180 + 120 -320
W5
% recovery =
F5 = F4 - F6
= 0,328125
( 1- W12) F2
x 100
0,35 x 876,0594
( 10,327 ) 611,6208
x 100
= 0,74491 x 100
= 74,4491
3.23 In the sulfuric acid industry, oleum is aterm used for 100% acid containing free,
unreacted SO3 dissolved in the acid. A 20% oleum, for example, is 20 lb SO 3 in 80 lb 100%
acid, per 100 lb mix. A contact sulfuric acid plant produces 10 tons 20% oleum and 40 tons
98% acid per day (see flowsheet, figure P3.23). The oleum tower is fed with a portion of
the acid from the 98% tower. The 98% tower is a fed with 97% acid, obtained by diluting
part of the 98% output. The gas fed to the oleum tower analyzed 10.8% SO 3 calculate the
amount of acid to be fed to each tower per day.
F4
W 4H 2 so 4 =
0,98
F5= 40.000
(4)
W H 2 so 4 =
(5)
0,97
Menara
(1)
(3)
Oleum
Gas
N1
Menara
Asam
3
SO 3
(8)
Diluter
(7)
(6)
F6
(2)
F7
6
H2SO4
F2 =
10000
W gas
F H 2O
N SO 3=0,108
2
SO 3
(9)
W H 2 so 4 =
W H 2 so 4 =
0,98
Mixing
=
0,98
(10
) 10
F = 40.000
2
W H 2 so 4 =
0,58 asam
Reduksi yang terjadi pada menara oleum dan menara
SO3 + H2O
r1
H2SO4
SO3 + H2O
r2
H2SO4
Hubungan pembatas (2-1) (3-1) = 2
Asumsi: gas buang mengandung SO3
Tabel Derajat Kebebasan
oleum
m.asam
Variabel alur bebas
8+1
7+1
Pers. Neraca bebas
diluter
mixing
proses
over all
18+2
8+1
12
2(+1)
1(+1)
1(+1)
Spesifikasi alur
-Laju alir
-komposisi
Hub.pembantu
F8H 2O ,
N 9gas , r
Strategi penyelesaian
1. Neraca massa pada keseluruhan N1,
2. Oleum DK= 2-1 (N1) = 0
N
Acid DK = 2-1 )=1
8
Diluter DK = 1-1 ( F H 2O = 0
3
N SO 3 ,
3. Oleum F4,
N gas , r, asumsi td gas buang mengandung SO3
4. DK = 0-0 = 0
5. Diluter = F7,F5
3
6. Asam DK = 1-3 ( N SO 3 , N gas , F3) + 2 = 0
= 2-2 (F4,F7) = 0
7. Mixing
Neraca massa over all
H2SO4 = 0,98 x 400 + 0,8 x 10.000
SO3
= 98 r
47.200
= 98 r
= 481,633
= 0,2 x 10.000 = 0,108
N 1 x 80 - 80r
2000
= 8,64 N1 80r
N1
= 4651,043
Neraca massa pada oleum
H2SO4
8000 = 0,98 F4 + 98 r1
H2O
= 0,02 F4 18 r1
r1
SO3 : 80
= 8,613
F4 =7346,9388
N 3SO 3 + 2000 = 80 x 4691,0431x 0,108 80 r1
80
N 3SO 3
= 38350,6124 80 x 8,1633
N 3SO 3
Gas
N 3gas
= 473,469
= 0,892 x 4691,0431
= 4184,411
Neraca massa pada Deluter
H2SO4
= 0,97 F5 = 0,987
H2O
= 0,03 F5 = 0,02 F7 + 9462,388
F5
= 928000,024 F7 = 918530,636
F5
= F7 +
N 8H 2 O
928000,024 = 918530,636 + 949,388
928000,024 = 928000,024
Neraca massa Asam
H2SO4
= 0,98 F6 = 0,97 x 928000,024 + 98 x r2
H2O
= 0,02 F6 = 0,03 + 928000,024 18 x r2
F6
= 965877,55
r2= 473,469
F6 = F7 + F10 + F4
965877,55 = 918530,636 + 4000 + 7346,9388 = 965877,57
Proses
Over all
H2SO4
0,98 x 4000 + 0,8 x 1000 = 98 r
SO3
= 0,2 x 10000
= 0,108 N1 x 80-80r
H2O
= 0,02 x 40000
Gas
N ggas
F6H 2O
- 18 x r
= 0,892 x N1
