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