Luas permukaan luar a ″ = 0.2618 ft
2
ft Tabel 10. Kern, 1965
Jumlah tube, 344
, ft
ft 0.2618
ft 12
ft 1,0798
a L
A N
2 2
t
= ×
= ×
= buah
Nilai terdekat adalah 16 buah dengan ID shell = 8 in Tabel 9. Kern, 1965 b. Koreksi U
D
Koefisien menyeluruh kotor t
A Q
U
D
Δ ⋅
= dimana,
Nt L
a A
× ×
=
A = 0.2618 × 12 × 2 = 6,2832 ft
2
186 ,
17 190
6,2832 20517,304 =
⋅ =
D
U Btu h ft
2
F Penentuan R
D
design:
1. Flow Area a
a. shell side
Pt 144
B C
ID a
s
× ×
× =
Kern, 1965 Keterangan:
C’ = 1.25 – 1 = 0.25 in B = 4 in
0.0555 1
144 4
25 .
8 =
× ×
× =
s
a ft
2
b. tube side
n 144
a Nt
a
t t
× ×
=
a’t = 0.421 Tabel 10, Kern, 1965
0.0234 2
144 0.421
16 =
× ×
=
t
a ft
2
2. Mass Velocity G
a. shell side
s
a W
Gs =
Kern, 1965
205371,549 0.0555
11398,121 = =
Gs lbh ft
2
b. tube side
t
a W
Gt =
Kern, 1965
45 ,
828591 0.0234
19389,04 = =
Gt lbh ft
2
3. Koefisien Perpindahan Panas
a. shell side
asumsi awal h
o
= 300 Btuhr ft
2
F
b. tube side
untuk steam, h
io
= 1500 Btu ft
2
F Temperatur
dinding T
w
T
w
=
c c
c
t T
ho hio
ho t
− +
+
T
w
= 234.38
202.85 392
300 1500
300 202.85
= −
+ +
o
F Δt
w
= T
w
– t
c
= 234.38 – 202.85 = 31,53
o
F dari fig. 15.11, Kern, 1965, nilai h
o
300, maka h
o
= 300 Btuhr ft
2
F
4. Koefisien perpindahan panas menyeluruh bersih
Uc
o io
o io
h h
h h
Uc +
× =
250 300
1500 300
1500 Uc
= +
× =
Btu h ft
2
F
5. Faktor Pengotor
R
D
D C
D C
D
U U
U U
R ⋅
− =
0.0542 186
, 17
250 186
, 17
250 =
⋅ −
=
D
R
R
D
hitung ≥ R
D
ketentuan, maka spesifikasi dapat diterima.
6. Pengecekan nilai flux
20000 A
Q
19001,022 1,0798
20517,304 =
nilai flux 20000, maka perhitungan memenuhi. Perhitungan Pressure Drop :
a. Shell side
ΔP
s
diabaikan
b. Tube side
μ ×
= Gt
D Re
t
D = ID tube = 0.732 in Tabel 10. Kern, 1965
3 ,
1232782 0.041
45 ,
828591 0.73212
Re =
× =
t
untuk R
e
= 3
, 1232782 , f = 0.00008 ft
2
in
2
Fig.26, Kern, 1965
t 10
2 t
t
s ID
10 22
. 5
N L
G f
P φ
⋅ ⋅
⋅ ⋅
⋅ ⋅
⋅ =
Δ Kern, 1965
483 ,
1 0.857
0.73212 10
22 .
5 2
12 45
, 828591
0.00008
10 2
= ⋅
⋅ ⋅
⋅ ⋅
⋅ ⋅
= Δ
t
P psi
2 r
g 2
V s
n 4
P ⋅
= Δ
untuk G
t
= 45
, 828591
,
2
g 2
V
= 0.001 Fig.27, Kern, 1965
0.009 0.001
857 .
2 4
= ⋅
⋅ =
Δ
r
P psi
r t
T
P P
P Δ
+ Δ
= Δ
psi P
T
492 ,
0,009 483
, =
+ =
Δ
ΔP
T
yang diperbolehkan adalah ≤ 10 psi, maka ΔP
s
dapat diterima.
LC.19 Pompa Reboiler P-108
Jenis : centrifugal pump
Kondisi operasi Temperatur
= 92,61
C Densitas larutan
ρ = 1,102 kgL = 68,7954 lbmft
3
Viskositas larutan μ
= 0,171 cp
= 0,00011 lb
m
ft ⋅s
Laju alir massa F =
5180,964 kgjam = 2,8783 lb
m
s Laju alir volumetrik, Q = ρ
F =
3
lbmft 68,7954
lbms 2,8783
= 0,0418 ft
3
s
0,13 0,45
f opt
i,
ρ 3,9Q
D =
0,13 3
0,45 3
lbmft 68,7954
s ft
3,90,0418 =
= 1,6198 in = 0,1349 ft Ukuran spesifikasi pipa
: Brownell, 1959
Ukuran pipa nominal =
0,5 in Schedule
pipa =
40 Diameter dalam ID
= 0,622 in
= 0,0518 ft Diameter luar OD
= 0,84 in
= 0,0699 ft Luas penampang dalam A
t
= 0,304 in
= 0,00211 ft
2
Kecepatan linier,
= =
t
A Q
v
2 3
ft 0,00211
s ft
0,0418
= 19,8104 fts Bilangan Reynold,
232 ,
641785 lbmft
0,00011 ft
0,0518 fts
19,8104 lbmft
68,7954 μ
D v
ρ N
3 Re
= ⋅
= =
Karena N
Re
2100, maka aliran turbulen. Untuk pipa sainlessl steel diperoleh
0,25 Nre
0,079 f
=
= 0,0028 esposito, 1994 Instalasi pipa:
Pipa lurus 15 ft ; ft
0,0518 .s
lbm.ftlbf 32,174
ft 15
2 fts
19,8104 20,0028
F
2
= = 365,6377 ft.lbflbm
1 elbow 90 ,
2 .s
lbm.ftlbf 32,174
2 fts
19,8104 10,75
F
2
= = 0,2307 ft.lbflbm
1 gate valve, 2
.s lbm.ftlbf
32,174 2
fts 19,8104
10,19 F
2
= = 0,0584 ft.lbflbm
1 kontraksi, 2
.s lbm.ftlbf
32,174 21
fts 19,8104
0,55 F
2
= = 0,1692 ft.lbflbm
1 ekspansi, 2
.s lbm.ftlbf
32,174 21
fts 19,8104
11 F
2
= = 0,3076 ft.lbflbm
Total Friksi : Σ F = 366,4036 ft.lbflbm
Kerja pompa : F
Σ Pv
Δ c
g α
2 2
v Δ
c g
g ΔZ
W +
+ ⎟
⎟ ⎠
⎞ ⎜
⎜ ⎝
⎛ +
= peters, 2004
Tinggi pemompaan, Δz = 5 ft
Static head ,
m f
c
lb lb
ft 5
g g
Δz ⋅
= Velocity head
, g
2 Δv
c 2
= ; Pressure head, ρ
P Δ
= 0 Maka : W = 371,4036 ft.lbflbm
Daya pompa : 026
, 1068
ft 68,7954lbm
s ft
0,0418 ft.lbflbm
371,4036 ρ
Q W
P
3 3
= =
= efisiensi pompa 80
: Hp
2743 ,
2 8
, 550
1068,026 P
= =
x Digunakan pompa dengan daya standar 2,5 Hp.
Daya motor : 2,9412Hp
85 ,
2,5Hp =
= : digunakan motor 3 Hp
LC.20 Bak Penampung cake Filter Press I B-101
Bentuk : persegi panjang
Kondisi Operasi :
Tekanan : 1 atm
Suhu : 25
C Laju alir massa
: 90,197 kgjam ρ bahan
: 1333.33 kgL Faktor Kelonggaran : 20
Perhitungan : a. Volume Bak
Volume fltrat, V
l
=
3
33 ,
1333 1
90,197 m
kg jam
jam kg
× = 0,0676 m
3
Volume cake
1 hari proses = 24 × 0,0676 = 1,6224 m
3
Volume bak, V
b
= 1 + 0,2 × 1,6224 m
3
= 1,9469 m
3
b. Ukuran Bak Penampung Direncanakan, p : l : t = 1 : 1 : 23
Vb = p × l × t = 23 × x
3
x =
9469 ,
1 2
3
3
×
x = 1,4039 m maka,
panjang = 1,4039 m lebar = 1,4039 m
tinggi = 0,9359 m
LC.21 Bak Penampung cake Filter Press II B-102
Bentuk : persegi panjang
Kondisi Operasi :
Tekanan : 1 atm
Suhu : 25
C Laju alir massa
: 247,336 kgjam ρ bahan
: 1204,819 kgL Faktor Kelonggaran : 20
Perhitungan : a. Volume Bak
Volume fltrat, V
l
=
3
819 ,
1204 1
247,336 m
kg jam
jam kg
× = 0,2053 m
3
Volume cake
1 hari proses = 24 × 0,2053 = 4,9272 m
3
Volume bak, V
b
= 1 + 0,2 × 4,9272 m
3
= 5,9126 m
3
c. Ukuran Bak Penampung Direncanakan, p : l : t = 1 : 1 : 23
Vb = p × l × t = 23 × x
3
x =
9126 ,
5 2
3
3
×
x = 4,2637 m maka,
panjang = 4,2637 m lebar = 4,2637 m
tinggi = 2,8424 m
LC.22 Heater H-101
Jenis : shell and tube exchanger
Deskripsi HE :
Tabel Deskripsi
Heater
DESCRIPTION Unit
SHELL SIDE TUBE SIDE
Cold Fluid Hot Fluid
1
Fluid Type Camp. Etanol
Steam In Out In
Out Temperature T
°C 30.00 80.00 120.00
100.00 2
o
F 86.00 176.00
248.00 212.00
3
Total Flow W
kgh 1467,774 14007,100
lbh 3229,103 30815,620
4
Total Heat kJh
344046,819
Transfer Q Btuh
326092,183 5
Pass
1 2
Length L
ft - 16
6 in
- 192
7 OD Tubes
in - 0.75
8
BWG -
10 9
Pitch Square
in - 1
Mencari Δt
1 2
1 2
t t
ln t
t LMTD
Δ Δ
Δ −
Δ =
Kern, 1988
F LMTD
o
84 86
248 176
212 ln
86 248
176 212
= ⎟⎟⎠
⎞ ⎜⎜⎝
⎛ −
− −
− −
=
Koreksi LMTD CMTD CMTD
Δt = LMTD × Ft
1 2
2 1
t t
T T
R −
− =
= 0,4
86 176
248 212
= −
−
1 1
1 2
t T
t t
S −
− =
= 0,55
86 248
86 176
= −
−
Dikarenakan R = 0, maka F
t
= 1 CMTD
Δt = 84 × 1 = 84 F
Caloric Temperature T
c
dan t
c
354 2
212 248
2 T
T T
2 1
c
= +
= +
=
F
219 2
86 176
2 t
t t
2 1
c
= +
= +
=
F
Menghitung jumlah tubes yang digunakan
Dari Tabel 8. Kern, 1965, untuk heater fluida dingin medium organic- fluida panas steam, diperoleh U
D
= 50 – 100, faktor pengotor R
d
= 0.003. Diambil U
D
= 77 Btujam ⋅ft
2
⋅°F
a. Luas permukaan untuk perpindahan panas,
2 D
ft 19,338
219 77
326092,183 Δt
U Q
A =
× =
× =
Luas permukaan luar a ″ = 0.1963 ft
2
ft Tabel 10. Kern, 1965
Jumlah tube, 16
, 6
ft ft
0.1963 ft
16 ft
19,338 a
L A
N
2 2
t
= ×
= ×
= buah
Nilai terdekat adalah 52 buah dengan ID shell = 10 in Tabel 9. Kern, 1965 b. Koreksi U
D
Dirt Overall Heat Transfer Coefficient t
A Q
U
D
Δ ⋅
= A = 0.1963 × 16 × 52 = 163,32 ft
2
769 ,
23 84
163,32 326092,183 =
⋅ =
D
U Btu h ft
2
F Penentuan R
D
design:
1 Flow Area a
a. shell side
