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Teknik KimiaHeat Exchanger
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Heat Exchanger-01 (HE-01)
Tugas : memanaskan slurry organik ke Fermentor
Data kapasitas panas gas masing-masing komponen
Cp
kj/kmol.kKomponen ∫Cpdt
(kJ/kmol.K)0,007911111
0,005129139 C6H12O6 0,00791
0,006344697 C13H25O7N3S 0,005129139
C12H24O6 0,006344697
Komponen A B C D E
C2H4 32,0830 -0,0148 2,4774,E-04 -2,3766E-07 6,8274E-11
CH4 34,942 -0,039957 1,9184,E-04 -1,5303E-07 -3,9321E-11
C2H6 28,1460 0,0434 1,8946,E-04 -1,9082E-07 5,3349E-11
1. Neraca Massa Disekitar HE-01
cold Fluid : produk keluar separator
komposisi kg/jam lb/jam kmol/jam yi
C2H4 1959,38884 4320,452469,9781728
7 0,9994
CH4 0,56016148 1,23520,03501009
2 0,0005
C2H6 0,210060555 0,46320,00700201
8 0,0001
TOTAL 1960,159062 4322,150770,0201849
8 1,0000
2. Neraca Panas
H1 = H11 + QpendinginH11 =
Panas Masuk he
Qp = Beban panas yang harus dilepaskan steam
cold Fluida
Fase gas in T1 = 283,0000 K 10,0000 C
komposisi kmol/jam yi Cp, kj/kmol H1, kj/jam
C2H4 69,97817287 0,9994 -650,1413 -45495,6973
CH4 0,035010092 0,0005 -532,3953 -18,6392
C2H6 0,007002018 0,0001 -786,8818 -5,5098
TOTAL 70,02018498 1,0000 -1969,4183 -45519,8463
Fase gas out T2 = 543,0000 K 270,0000 C
komposisi kmol/jam yi Cp, kj/kmol H11, kj/jam
C2H4 69,97817287 0,9994 13.284,1100 929597,7443
CH4 0,035010092 0,0005 9.613,1104 336,5558841
C2H6 0,007002018 0,0001 16.519,0992 115,6670385
TOTAL 70,02018498 1,0000 39.416,3196 930049,9672
ta = 260,0000 K
Q = 975.569,8135 kj/jam 924661,654 btu/jam
hot fluid
steam
Suhu masuk = 626,0000 oF = 330,0000 oC
Suhu keluar = 572,0000 oF = 300,0000 oC
Tv = 599,0000 oF
λ = 564,2980 btu/lb Kern.Tbl.07
Beban Panas = 924661,6540 btu/jam
jumlah steam yang dibutuhkan = 1638,6052 lb/jam 743,2582 kg/jam
3. Δt
cold fluid (F) hot fluid (F) difference
F C F C
50,0000 10,0000 higher temp. 572,0000 300,0000 522,0000 T2
518,0000 270,0000 lower temp. 626,0000 330,0000 108,0000 T1
468,0000 difference 54,0000
1. Neraca Massa Disekitar HE-01
cold Fluid : produk keluar Rotary Filter
komposisi kg/jam lb/jam kmol/jam xi
C6H12O6 3589,587 7915,0393 19,94215 0,6199
C13H25O7N3S 2871,67 6332,0324 9,50884106 0,2956
C12H24O6 717,917 1583,0070 2,719382576 0,0845
TOTAL 7179,174 15830,0787 32,17037364 1,0000
2. Neraca Panas
H1 = H11 + QpendinginH11 =
Panas Masuk he
Qp = Beban panas yang harus dilepaskan steam
cold Fluida
Fase liquid in T1 = 300,0000 K 27,0000 C
komposisi kmol/jam xi Cp, kj/kmol H1, kj/jam
C6H12O6 19,94215 0,6199 0,0098 0,1956
C13H25O7N3S 9,50884106 0,2956 0,0030 0,0288
C12H24O6 2,719382576 0,0845 0,0011 0,0029
TOTAL 32,17037364 1,0000 0,0139 0,2273
Fase liquid out T2 = 308,0000 K 35,0000 C
komposisi kmol/jam xi Cp, kj/kmol H11, kj/jam
C6H12O6 19,94215 0,6199 0,0490 0,977969558
C13H25O7N3S 9,50884106 0,2956 0,0152 0,144159593
C12H24O6 2,719382576 0,0845 0,0054 0,014584629
TOTAL 32,17037364 1,0000 0,0696 1,13671378
ta = 8,0000 K
Q = 0,9094 kj/jam 0,861917316 btu/jam
hot fluid
steam
Suhu masuk = 212,0000 oF = 100,0000 oC
Suhu keluar = 86,0000 oF = 30,0000 oC
Tv = 149,0000 oF
λ = 950,0000 btu/lb Kern.Tbl.07
Beban Panas = 0,8619 btu/jam
jumlah steam yang dibutuhkan = 0,0009 lb/jam 0,0004 kg/jam
3. Δt
cold fluid (F) hot fluid (F) difference
F C F C
80,6000 27,0000 higher temp. 86,0000 30,0000 5,4000
95,0000 35,0000 lower temp. 212,0000 100,0000 117,0000
14,4000 difference 126,0000
LMTD = 36,2835 oF 2,3797 oC
ΔT LMTD= 36,2835 oF
* Temperatur : Ta = 87,8000 oF
ta = 149,0000 oF ΔT= 61,2000 oF
4. Menentukan Ud
Nilai Ud ditentukan pertama kali dari tabel 8 kern untuk aqueous solutions (less than 2 c.p 200-700)
Hot fluid = steam
Cold fluid = liquid
Ud = 250 Btu/lb.ft2.F
5. Luas transfer panas
Q = Ud.A.ΔTLMTD
= 0,0001 ft2 Double Pipe Exchanger (batasan A < 100 - 200 ft2) (Dipilih)
Shell and Tube Heat Exchanger (batasan A>200 ft2)
6. Lay Out HE
Dipilih alat penukar panas jenis double pipe exchanger dengan ukuran 2 x 1 1/4 IPS
Dari tabel 11, Kern diperoleh data dimensi pipa :Nominal pipe size
OD, inSchedule Number ID, in
Flow area per pipe, in2
Surface per lin ft, ft2/ft Weight per lin ft, lb steel
IPS, in Outside Inside
1,2500 1,6600 40,0000 1,3800 1,5000 0,4350 0,3620 2,28002,0000 2,3800 40,0000 2,0670 3,3500 0,6220 0,5420 3,6600
Menghitung harga tetapan perpindahan panas
Fluida Panas : Annulus, Steam
tU
QA
D
tU
QA
D
LaA
tN
Menentukan flow area :
D2 = 2,0670 in = 0,1723 ft
D1 = 1,6600 in = 0,1383 ft
aA =
aA = 0,0083 ft2
Equivalent diameter
DE = 0,0761 ft
Kecepatan Massa :
4
)( 122
2DD
a
1
2
12
2)(
D
DDD E
A
HA a
WG
GA = 0,1097 lb/(jam.ft2)
Bilangan Reynold :
μ = 91,9251 lb/ft.jam
REA = 0,0001
jh = 1,0 fig 24 Kern
c = 0,4100 btu/lb.F ....fig 2 Kern
k = 0,0137 btu/j.ft2.(F/ft) .....table 5 Kern
= 13,6469
AE
EA
GDR
31
.
k
c
ho = 2,4553 btu/j.ft2.F
(untuk air)
Fluida dingin : Pipa, Campuran multi komponen
Menentukan flow area :
D = 1,3800 in = 3,5052 cm
= 0,1150 ft
ap = 0,0104 ft2
kecepatan massa :
14,03
1
0
..
.
k
c
D
kjHho
4
2Da p
Gp = 1524817,0368 lb/(jam.ft2)
Bilangan Reynold :
μ = 92,7065 lb/ft.j
REP = 1891,4949
Dari fig. 24, Kern diperoleh :
JH = 7,000
c = 107,6715 btu/lb.F
k = 112,8147 btu/j.ft2.(F/ft)
a
WG C
PP
EP
GDR
= 4,4560
= 30599,5076 btu/j.ft2.F
wall temperature
tw = 87,8049 F 304,0027279 K
μw= 92,6617 lb/ft.hr
= 1,0001
Koreksi koeffisien
hio = 30601,5825 btu/j.ft2.F
= 25439,8698OD
IDhh ii 0
31
.
k
c
31
..
k
c
D
kjHhi
14,0
= 2,4551 btu/j.ft2.F
Asumsi Rd = 0,0050
= 0,4123
UD = 2,4253
= 0,0050 ft2.F.j/btu
00
0.0
hh
hhU
i
iC
DC
DCD UU
UUR
.
RdUU CD
11
Menghitung pressure Drop (ΔP) :
Fluida panas (steam) :
DE = 0,0339 ft
Bilangan Reynold :
REA = 0,0000405
Faktor Friksi :
)(' 12 DDD E
AE
AE
GDR
.''
42,0
264,00035,0
ERf
s = 1,0000 ........ table 6 kern
f = 18,4772 ρ = 62,5
ΔFA = 0,00000000019 ft
V = 0,000000488 ft/detik
E'D.2.g.2
L.2AG.f.4
AF
.3600AG
V
g
VFt .2
.32
ΔFt =0,00000000000001
1 ft
Pressure drop total :
ΔPA = 0,0000000001 psi
Fluida dingin (Campuran multi komponen) s = 1,0000..tabel 6 kern
faktor friksi : ρ = 62,5000 lb/ft3
s = 0,8050..tabel 6 kern
ρ = 50,3125
f = 0,0146
144
.tAA
FFP
42,0
264,00035,0
ERf
ΔFp = 29,42370767 ft
pressure drop total :
ΔPp = 12,77 psi
SUMMARY
0,0000 houtside 2,4553
Uc 2,4551
E
pp Dg
LGfF
'...2
...4 2
144. P
P
FP
Ud 2,4253
Rd calc 0,0050
Rd req. 0,0030
0,000000000084 DP calc. 12,771
10,0000 DP allow. 10,0000
1 micropoise= 0,0001 cp
10000 = 0,0001
μ liquid = A+BT+CT^2 (micropoise)
komponen yi micropoise cp
H2O 1 3600 0,36 (Figure 14, Kern)
μ liquid = A+BT+CT^2 (micropoise)
komponen A B C xi micropoise lb/ft.jam
H2O -10,2158 1792,5-
0,0000162 1 379999,1 91,9251
1. Bagian tube (cold fluid)
μ liquid = micro poise
komponen xi micropoisesampah organik (karbo,protein, lemak) 1,0000
38322,9000
K liquid = A+BT+CT^2 (W/m K) 304,15
komponen xi W/m.K
C6H12O6 0,6199 47,1350
C13H25O7N3S 0,2956 13,9345
C12H24O6 0,0845 4,1136
1,0000
Cp = (Kj/kg K)
komponen Cp(kJ/kg.K) xi
C6H12O6 268,4846 0,6199 1,424433,109
6
C13H25O7N3S 139,2655403 0,2956 1,549
C12H24O6 43,0486 0,0845 1,675
450,7988
Pada Tw 87,8049 F 304,0027 K
μ gas = A+BT+CT^2 (micropoise)
μ liquid = micro poise
komponen xi micropoisesampah organik (karbo,protein, lemak) 1,0000
38304,3437
viskositas cair
komponen A B C D Vis (cp)
H2O -10,2158 1792,5 0,01773-
0,00001263 0,4327
338,1500