Perhitungan Kinerja Heat Exchanger 116-C Secara Desain

Diketahui:

SHELL (COOLING WATER)

t1 = 33 oC

= 33 x 1,8

= 59,4 oF

t2 = 47 oC

= 47 x 1,8

= 84,6 oF

t2 – t1 = (84,6 – 59,4) oF........................................................................(Kern, Hal.85)

= 25,2 oF

w = 178.000 kg/hr ..........(Tabel Pengukuran Flow Cooling Water dengan Flow meter

KURITA)

= 178.000 kg/hr x 2,2046 Ib/kg

= 392418,8 Ib/hr

TUBE (SYN GAS)

T1 = 164,67 oC

= 164,67 oC x 1,8

= 296,406 oF

T2 = 63,53 oC

= 63,53 oC x 1,8

= 114,354 oF

T1 – T2 = (296,406 – 114,354) oF.............................................................(Kern, Hal.85)

= 182,052 oF

W = 35525,65 kg/hr ............(Tabel Pusri II Synthesis Loop Uprate Case Pada Rate

Operasi 100%)

W = 35525,65 kg/hr x 95,248

100 x 2,2046 Ib/kg

99

= 74598,0888 Ib/hr .......................................... (Tabel Rate Operasi Pada 95,248%)

1. PERHITUNGAN NERACA PANAS

Q = W x C x (T1-T2) = w x c x (t1-t2).......................(Persamaan 5.7 Kern, Hal. 88)

a. Pada Shell (Cooling Water)

Diketahui :

w = 178.000 kg/hr .............................(Tabel Pengukuran Flow Cooling Water)

= 178.000 kg/hr x 2,2046 Ib/kg

= 392418,8 Ib/hr

tavg = t 1+t 2

2

= (59,4+84,6)

2

= 72 oF

C = 4186 J/kg oC.............................(Tabel Pengukuran Flow Cooling Water)

= 4186 J/kg oC x 1 Kcal/ kgC4186 J /kgK

= 1 Kcal/ kg oC = 1 BTU/Ib oF

Maka,

Q = w x c (t2-t1) ..................................................(Persamaan 5.7 Kern, Hal. 88)

= 392418,8 Ib/hr x 1 BTU/IboF x (84,6-59,4) oF

= 9888933,6 BTU/hr

b. Pada Tube (Syn Gas)

Diketahui :

W = 35525,65 kg/hr .....................(Tabel Pusri II Synthesis Loop Uprate Case)

= 35525,65 kg/hr x 95,248

100 x 2,2046 Ib/kg

= 74598,0888 Ib/hr

Rate operasi = 95,248.........................................................(Tabel Rate Operasi)

100

Tavg = T 1+T 2

2

= (164,67+63,53)

2

= 205,38 oF

C = 3332,97 J/kg oK .................(Tabel Pusri II Synthesis Loop Uprate Case)

= 3332,97 J/kg oK x 1 Kcal/ kgC4186 J /kgK

= 0,7960 Kcal/kg oC = 0,7960 BTU/Ib oF

Maka,

Q = W x C (T1-T2) ................................................. (Persamaan 5.7 Kern, Hal. 88)

= 74598,0888 Ib/hr x 0,7960 BTU/Ib oF x (296,406 – 114,354) oF

= 10810262,08 BTU/hr

2. LMTD dan ∆t

Hot Fluida

(oF)Cold Fluida

(oF)

Difference

(oF)

T1 296,406 Higher

Temp

t1 59,4 237,006 ∆t2

T2 114,354 Lower

Temp

t2 84,6 29,754 ∆t1

182,052 Difference 25,2 207,252 (∆t2-∆t1)

(T1-T2) (t2-t1)

101

LMTD =

(T 1−t 1 )−(T 2−t 2)

ln( (T 1−t 1 )(T 2−t 2) ) ................................. (Persamaan 5.14 Kern,

Hal. 89)

=

(296,406−59,4 )−(114,354−84,6 ) F

ln( 296,406−59,4114,354−84,6 )F

= 99,8746 oF

R = T 1−T 2t 2−t 1

....................................................... (Persamaan 5.14 Kern, Hal.

149)

= (296,406−114,354)F

(84,6−59,4 ) F

= 7,2242 oF

S = t 2−t 1T 1−t 1

........................................................ (Persamaan 5.14 Kern, Hal.

149)

= 84,6−59,4 F

296,406−59,4 F

= 0,1063 oF

FT = 0,97 ........................................................................(Fig. 18 Kern Hal 828)

Sehingga,

∆t = LMTD x FT .....................................(Persamaan 7.42 Kern, Hal. 149)

= 99,8746 oF x 0,97

= 96,8783 oF

3. TEMPERATURE AVERAGE (Tavg dan tavg)

tavg = (t 1+t 2)

2

102

= (59,4+84,6 ) F

2

= 72 oF

Tavg = (T 1+T 2)

2

= (296,406+114,354 ) F

2

= 205,38 oF

4. LUAS ALIRAN (as dan at)

a. Pada Shell (Cooling Water)

Diketahui :

ID = 736 mm .................................................(Tabel Spesifikasi HE 116-C)

= 73,6 cm x 0,3937 in/cm = 28,9763 in

OD = 19 mm .................................................(Tabel Spesifikasi HE 116-C)

= 1,9 cm x 0,3937 in/cm = 0,74803 in

Tube pitch (PT) = 25mm ...............................(Tabel Spesifikasi HE 116-C)

= 2,5 cm x 0,3937in/cm = 0,9842 in

Jumlah baffles = 4 ...(Fig. Kellog Syn Gas Compressor Interstage Cooler)

Jarak baffles (B) = 3810 mm ........(Fig. Kellog Syn Gas Compressor

Interstage Cooler)

= 3810 mm/4

= 952,5 mm

= 95,25 cm x 0,3937in/cm = 37,4999 in

Tube clearance (C”) = PT –OD ................................................(Kern, 1950)

= 0,9842in – 0,74803 in

= 0,2362 in

Maka,

103

as = ID xC x B} over {144 x PT ¿ ...............................................

(Persamaan 7.1 Kern, Hal.138)

= 28,9763∈x0,2362∈x 37,4999∈ ¿144 x 0,9842∈¿¿

¿

= 1,8109 ft2

b. Pada Tube

Diketahui :

Jumlah Tube (Nt) = 254/shell.........................(Tabel Spesifikasi HE 116-C)

Jumlah Pass (n) = 2 .......................................(Tabel Spesifikasi HE 116-C)

Diameter Luar (OD) =19 mm .......................(Tabel Spesifikasi HE 116-C)

=1,9 cm x 0,3937 in/cm = 0,74803 in

BWG = 14 .....................................................(Tabel Spesifikasi HE 116-C)

Maka,

a’t= 0,268 in2........................................................(Tabel 10 Kern, Hal.843)

at = Nt x a ' t144 xn

..................................................(Persamaan 7.48 Kern,

Hal.150)

= 254/ shell x 0,268∈2

144 x 2

= 0,2363 ft3

5. KECEPATAN MASSA (Gs dan Gt)

a. Pada Shell

Diketahui : ws = 392418,8 Ib/hr

as = 1,8109 ft2

maka, Gs = wsas

.............................................(Persamaan 7.2 Kern, Hal

138)

104

= 392418,8lb /hr

1,8109 ft 2

= 216698,2164 Ib/hr ft2

b. Pada Tube

Diketahui : wt = 35525,65 Ib/hr

at = 0,2363 ft2

maka, Gt = w tat

..............................................(Persamaan 7.2 Kern, Hal

138)

¿ 35525,65 Ib /hr

0,2363 ft ²

= 150341,3034 Ib/hr ft2

6. BILANGAN REYNOLD (Re)

a. Pada Shell

Dik : tavg = 72oF

μ = 0,95 x 2,42 = 2,299 Ib/ft hr ...............(Fig.14 Kern, Hal. 823)

Ds = 0,95 in ft / 12 in = 0,0791 ft ...............(Fig.28 Kern, Hal. 845)

Res= Ds x Gs / μ ...............................(Persamaan 7.3 Kern, Hal 152)

= 0,0791 ft x 216698,2164 Ib/hr ft2 / 2,299 Ib/ft hr

= 7455,7759

b. Pada Tube

Dik : Tavg = 205,38oF

μ = 1,176 x 10-2 = 0,01176 lb/fthr ............(Fig. 15 Kern, Hal. 825)

= 0,01176 x 2,42 = 0,0284 lb/fthr

Dt = 0,584 in ............................................(Tabel.10 Kern, Hal. 843)

= 0,584 in x 1 ft/12in = 0,0486 ft

105

Ret = Dt x Gt / μ .................................(Persamaan 7.3 Kern, Hal 152)

= 0,0486 ft x 150341,3034 Ib/hr ft2 / 0,0284 Ib/ft hr

= 257274,2023

7. FAKTOR PERPINDAHAN PANAS

a. Pada Shell

Res = 7455,7759

JH = 50 ......................................................................(Fig.28 Kern, hal.838)

b. Pada Tube

Ret = 257274,2023

L = 4264mm ...............................................(Tabel Spesifikasi HE 116-C)

= 4,264 m / 0,0254 m/in = 167,8740 in x 0,0833 ft/in = 13,98 ft

D = 0,584 ft/12 = 0,0486 ft ...............................(Tabel.10 Kern, Hal.843)

L/D = 13,98 ft/0,0486 ft = 287,2602

jH = 320 ...............................................................(Fig.24 Kern, Hal.834)

8. KOEFISIEN PERPINDAHAN PANAS

a. Pada Shell

tavg = 72 oF

c = 1 BTU/lb oF ...................................................(Fig.4 Kern, Hal. 806)

k = 0,3491 BTU/hrft oF ....................................(Table.4 Kern, Hal. 800)

Maka :

(c x μ

k)1/3 ....................................................(Persamaan 6.15b Kern,

Hal.150)

= ( 1 BTU

lbFx 2,299 lb / fthr

0,3491 BTU / fthrF )1/3 = 1,8744

106

b. Pada Tube

Tavg = 205,38 oF

c = 1,3077 BTU/lb oF

k = 9,479 x 10-2 = 0,09479 BTU/hrft oF....(Tabel Spesifikasi HE 116-C)

Maka :

(c x μ

k)1/3 ....................................................(Persamaan 6.15b Kern,

Hal.150)

= ( 1,3077 BTU

lbFx0,0284 lb / fthr

0,09479 BTU / fthrF )1/3 = 0,7317

9. ho dan hio

a. Pada Shell

jH = 50 ......................................................................(Fig.28 Kern, hal.838)

k/Ds .........................................................(Persamaan 6.15b Kern, Hal.150)

= 0,3491 BTU/hrft oF/0,0791 ft

= 4,4134 BTU/hr oF

Maka :

ho = jH x kD

x ( c x μk )

1 /3

ɸs .........................(Persamaan 6.15 Kern,

Hal.122)

hoɸs

= jH x kD

x ( c x μk )

1 /3

hoɸs

= 50 x 4,4134 BTU/hr oF x 1,8744 = 413,6238 BTU/hr oF

b. Pada Tube

jH = 320 ...................................................................(Fig.24 Kern, Hal.834)

k/Dt .........................................................(Persamaan 6.15b Kern, Hal.150)

107

= 0,09479 BTU/hrft oF/0,0486 ft

=1,9504 BTU/hr oF

Maka :

hi = jH x kD

x ( c x μk )

1 /3

ɸt ..........................(Persamaan 6.15 Kern,

Hal.122)

hiɸt

= jH x kD

x ( c x μk )

1 /3

hiɸt

= 320 x 1,9504 BTU/hr oF x 0,7317

hi= 456,6744 BTU/hr oF

hio = hi x IDOD

.............................................(Persamaan 6.5 Kern, Hal.

105)

= 456,6744 BTU/hr oF x 28 , 9763∈ ¿0,74803∈¿¿

¿

= 17690,11415 BTU/hr oF

10. CLEAN OVERALL COEFICIENT, Uc

Uc = hio x hohio+ho

= 17690,11415

BTUhrF

x413,6238BTUhrF

17690,11415BTUhrF

+413,6238B TUhrF

= 404,1735 ....

(Persamaan 6.38

Kern, Hal. 121)

11. DESIGN OVERALL COEFFICIENT, UD

Diketahui: OD = ¾ in ..........................................(Tabel Spesifikasi HE 116-C)

BWG = 14 .........................................(Tabel Spesifikasi HE 116-C)

a” = 0,1963 ft2 / in ft ................................(Tabel.10 Kern, Hal.843)

L = 14 ft .............................................(Tabel Spesifikasi HE 116-C)

108

Nt = 254/shell.....................................(Tabel Spesifikasi HE 116-C)

Maka:

A = a” x L x Nt ..................................................................(Appendix Tabel.10)

= 0,1963 ft2/in ft x 14 ft x 254/shell

= 698,0428 ft2

Sehingga,

UD = Q

A x ∆ t ......................................................(Persamaan 6.11 Kern, Hal.

107)

= 10810262,08 BTU /hr

698,0428 ft ² x 96,8783 F

= 159,8555 BTU/hrft2 oF

12. FAKTOR PENGOTOR, Rd

Rd = Uc−UDUc x UD

.....................................................(Persamaan 6.13 Kern, Hal.

108)

= 404,1735 – 159,8555404,1735 x 159,8555

= 0,0037 BTU/hrft2 oF

13. PRESSURE DROP

a. Pada shell

Res = 7455,7759

f = 0,0012 ft2/in2 .................................................(Fig.29 Kern, Hal. 839)

s = 1,0 …...........................................................(Tabel.6 Kern, Hal. 808)

∆ Pt = f x Gs² x L x n

5,22 x 1010 x Ds x S x ɸs .......................(Persamaan 7.45 Kern, Hal.

148)

109

= 0,0012 ft ²/¿ ² x

(216698,2164 )2 Ibhr

f t 2 x 13,98 ft x2

5,22 x1010

x0,0791 ft x 1 x 1

= 0,3815 psi/14,7 = 0,0259 kg/cm2

b. Pada tube

Ret = 257274,2023

f = 0,0013 ft2/in2 ...................................................(Fig.29 Kern, Hal.839)

Dt = ID/12 ...............................................................................(Kern, 1950)

= 28,9763/12

= 2,4146 ft

No of crosses

N + 1 = 12L/B .....................................(Persamaan 7.43 Kern, Hal 147)

= 12 x 13,98 ft/ 37,4999 in

= 4,4736

s = ρgas

ρwater ............................................................................(Kern,

1950)

= 15,48 kg/mᶾ

1 kg/mᶾ

= 15,48 kg /mᶾ

s =15,48 kg/mᶾ62,5 kg/mᶾ

= 0,2476

∆PT = f x>² x Dt x (N+1)

5,22 x 1010 x Dt x S x ɸs ..............................(Persamaan 7.44 Kern,

Hal.147)

= 0,0013 ft ² /¿ ² x

(150341,3034 )2 Ibhr

f t 2 x 2,4146 ft x 4,4736

5,22 x1010

x0,0486 ft x0,2476 x 1

= 0,5052 psi

110

= 0,0343 kg/cm2

14. EFISIENSI KERJA HE

η = Panas yangdiserapPanas yang dilepas

x 100%

= 9888933,6

10810262,08 x 100%

= 91,47%

111