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Problems 64 - buku Hukum Perlindungan Anak dan Kekerasan Dalam Rumah Tangga (Khusus Hukum Pidana) (Universitas HKBP Nommensen)
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6.4 : 10.000 lb/hr of 57oAPI gasoline is cooled from 150 to 130oF by heating 42oAPI kerosene from 70 to 100oF. Pressure drops of 10 psi are allowable with a minimum dirt factor of 0.004. a. How many 2 ½- by 1 ¼-in IPS hairpins 20 ft long are required? b. How shall they be arranged? c. What is the final fouling factor? Pembahasan : 1. Neraca Panas (Heat Balance) Qpanas = Qdingin Gasoline Tav = Menentukan nilai c dengan Figure 4 c = 0,533 Btu/lb°F Q = W.c.(t2 – t1) = 10000.0,533.(150-130) = 106600 Btu/hr Kerosene tav = Menentukan nilai c dengan Figure 4 c = 0,48 Btu/lb°F w = Q /c.(t2 – t1) = 106600/0,48.(100-70) = 7402,78 Btu/hr
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0,533 0,480
2. Δt Hot Fluid 150
130
Higher Temperature Lower Temperature
Cold Fluid 100
Difference
70
60
50
-10
LMTD =
= 54,91
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3. Temperatur Kalorik Dari informasi buku Kern (hal. 111) untuk liquid hidrokarbon digunakan nilai Tc, tc dengan menentukan Kc dan Fc (Fig.17), namun Tav dan tav digunakan bila : 1. Kedua liquid (cold terminal) encer (< 1,0 cp) 2. Temperature range 50 – 100 C 3. Perbedaan temperatur < 50 C
4. Flow Area Hot Fluid (Gasoline) =
2 ½ in ID Tabel 11
=
1 ¼ in OD Tabel 11
Cold Fluid (Kerosene) D=
1 ¼ in ID Tabel 11
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5. Kecepatan Alir Massa Hot Fluid (Gasoline) =
Cold Fluid (Kerosene) = 6. Reynolds Hot Fluid (Gasoline) saat 140 F, = 0,426 cp
Fig.14
= 0,426 x 2,42 = 1,03 lb/ft.hr
L/D = 2 L/D = 360
Cold Fluid (Kerosene) saat 85 F, = 1,620 cp
Fig.14
= 1,620 x 2,42 = 3,92 lb/ft.hr
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1,620
0,426
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7. Dari fig.24 didapat, Hot Fluid (Gasoline) = 230 Cold Fluid (Kerosene) = 69
8. Hot Fluid (Gasoline) Saat 140 F, c = 0,533 k = 0,087 (fig.1) = 1,85 Cold Fluid (Kerosene) Saat 85 F. c = 0,48 k = 0,0817 (fig.1) = 2,85
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9. Hot Fluid (Gasoline)
1,85 219,17 Cold Fluid (Kerosene) 2,85 139,71
10. = 54,67 x 1,380/1,66 = 168,05 = 116,13
= 85 + Hot Fluid (Gasoline) = 1,16 (
)
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Cold Fluid (Kerosene)
= 3,12 (
)
= 211,04
1,290
0,480
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11. Clean Overall Coeficient =
= 107,53 Btu/hr(
12. Koefisien Design
13. Surface
a. Maka dibutuhkan 2 hairpin atau 80 lin ft b. Hairpin dihubungkan secara seri dengan panjang 20 ft c. A aktual = 80 x 0,435 = 34,8 =
= 55,78
=
Dapat dilihat Rd actual > 7,5% Rd min. Sehingga berdasarkan ketentuan batas perbedaan Rd actual dengan Rd min yang memenuhi jika lebih besar (5 – 10%), maka desain ini dapat diterima. Oleh karena itu, perhitungan pressure drop dapat dilanjutkan.
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Pressure Drop Hot Fuid
72 Dari fig 6 kern diperoleh
Cold Fluid
Dari fig. 6 kern diperoleh Sehingga,
Hot Fluid
Cold Fluid
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0,800
0,715
1. Hot fluid
Pressure Drop,
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Di sini, kami membagi aliran di tiap stream menjadi 2. Namun, bila melihat dari kecilnya pressure drop actual, maka sepertinya masing-masing stream cukup dibuat 1 aliran, sehingga dalam perhitungan sebelumnya Dirt Factor = 0,004.
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6-6 Diinginkan untuk memanaskan 7000 lb/hr anilin dari 100 ke 150 lb/hr panas yang memiliki suhu awal 185
menggunakan toluene 10000
Bila fouling factor 0,005, sedangkan pressure drop
yang diperkenankan adalah 10 psi dan pipa harpin 15 ft, IPS
x
in tersedia, tentukan jumlah
pipa harpin yang dibutuhkan dan susunannya serta Rd Final Heat Balance Cold fluid (anilin), Hot fluid (toluene),
(viskos dianggap konstan dari rentang T1 – T2)
Toluene Aniline
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⁄
Diperoleh nilai Maka,
⁄
dan
(Kern Hal 111) ⁄
LMTD Hot Fluid 185 147,02
Higher Temp Lower Temp
Cold Fluid
Diff
150
35
∆t2
100
47,02
∆t1
-12,02
∆t2 -∆t1
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Pengecekan viskositas pada
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Fluida panas (toluene) pada μ<
Fluida dingin (aniline) pada μ>
Caloric Temperature
o
API
Fluida di pipa bagian dalam (aniline) Dengan asumsi s konstan pada 1,02 diperoleh
Mencari nilai Kc
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7o API aniline (cold fluid) Dari data diatas diperoleh
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Dari data yang diplotkan pada sumbu x = 1,31 menyentuh garis Kc = 1,41 menghasilkan
(Chapter 5 Kern) 7o API aniline (cold fluid)
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Flow Area
Dengan menggunakan rumus
Untuk pipa luar (Anulus)
Untuk pipa dalam
(
)
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(Kern hal 111)
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Karena flow area annulus lebih besar maka komponen yang lajunya lebih besar diletakkan di annulus (toluene) dan sebaliknya laju yang lebih kecil di letakkan pada pipa dalam (aniline).
Menghitung nilai koefisien perpindahan panas konveksi (h)
c
(
)( )
⁄
(
)
= kapasitas panas (fig 4) = viskositas (fig 14)
k
= koefisien perpindahan panas konduksi (table 4) = keofisien yang diperoleh dengan mengkorelasikan terhadap Re (fig 24)
Penentuan viskositas
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Fluida panas (toluene) pada μ = 0,34 cp
⁄
μ=
Fluida dingin (aniline) pada μ = 2 cp
μ=
Penentuan mass velocity Fluida dingin (aniline) di pipa bagian dalam ⁄
⁄
Fluida panas (toluene) pada pipa anulus
⁄ Reynolds number Fluida pada pipa anulus (toluene)
⁄ Dengan
Fluida di pipa bagian dalam (toluene)
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⁄
C
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Nilai fluida pada pipa bagian dalam (aniline)
fluida pada pipa anulus (toluene)
Mencari nilai k
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( )
(
( )
(
⁄
)
⁄
Menghitung nilai h
⁄
⁄
dianggap konstan
)
cold at inner pipe (aniline) ( )
⁄
hot fluid at annulus (toluene) ( )
Pemeriksaan
⁄
pada permukaan diameter luar (OD)
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Penentuan nilai ⁄
⁄ ⁄
dan ⁄
Penentuan viskositas pada tw
fluida dingin (aniline) pada
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= 1,55cp ⁄
= (
)
(
)
,036
fluida panas (toluene) pada = 0,4 cp = (
Sehingga,
)
(
)
⁄
⁄
⁄
⁄
Menghitung panjang pipa yang dibutuhkan
Koefisien bersih keseluruhan
⁄
Koefisien desain keseluruhan , dengan asumsi Rd = 0,005
⁄
Panjang yang dibutuhkan
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Setidaknya dibutuhkan 6 buah pipa hairpin berukuran 15 ft disusun seri Luas permukaan secara actual yaitu
, sehingga koefisien actual
desain
C
⁄ C
Pengecekan Pressure Drop Pressure drop pada pipa annulus (toluene sebagai fluida panas)
Friction factor dari Mc Adams dan Seltzer dengan koreksi 10%
Dari table 6 kern diperoleh Sehingga,
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Pressure Drop,
Pressure drop pada inner pipe (aniline sebagai fluida dingin) Friction factor dari Mc Adams dan Seltzer dengan koreksi 10%
Dari table 6 kern diperoleh Sehingga,
Pressure Drop,
(by-pass aliran yang memiliki laju volumetric yang tinggi)
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6-8 Suatu cairan didinginkan dari 350 hingga 300 °F oleh cairan lain yang dipanaskan dari 290 hingga 315 °F. Bagaimana perbedaan suhu sebenarnya menyimpang dari LMTD jika (a) Fluida panas disusun secara seri dan fluida dingin mengalir dalam dua jalur paralel aliran berlawanan, (b) Fluida panas disusun secara seri dan fluida dingin mengalir dalam tiga jalur aliranberlawanan aliran paralel, (c) Kisaran fluida dingin dalam (a) dan (b) diubah menjadi 275 hingga 300 °F. Penyelesaian: Perbedaan suhu sebenarnya diberikan melalui persamaan ini
Untuk satu seri aliran panas dan n aliran dingin paralel, dapat ditunjukkan melalui Persamaan (6.35a)
Dimana
[(
)( )
⁄
]
a. Fluida panas disusun seri dan fluida dingin mengalir dalam dua jalur paralel aliran berlawanan. Heat Balance: Hot fluid = Cold fluid =
(n = 2, 2 aliran paralel)
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∆t
0 P
n
h
n ny
b. Fluida panas disusun secara seri dan fluida dingin mengalir dalam tiga jalur aliran-berlawanan aliran paralel (n = 3, 3 aliran paralel)
[(
)( )
[(
)(
⁄
)
⁄
]
[
Perbedaan suhu sebenarnya adalah:
c. Kisaran fluida dingin dalam (a) dan (b) diubah menjadi 275 hingga 300 °F. Heat Balance: Hot fluid = Cold fluid = Kasus a : Untuk satu aliran panas seri dan 2 aliran paralel
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Karena ∆t
0 P
n
h
n ny
Kasus b : Untuk aliran panas seri dan aliran fluida dingin dalam 3 aliran paralel
[( [(
)( ) )(
⁄
)
⁄
[
Perbedaan suhu sebenarnya adalah:
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] ] ]
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6.10) 6330 lb/hr of toluene is cooled from 160 to l00°F by heating amyl acetate from 90 to 100°F using 15-ft hairpins. The exchangers are 2-by 1/4-in. IPS. Allow-ing 10 psi pressure drops and providing a minimum dirt factor of 0.004 (a) how many hairpins are required, (b) how shall they be arranged, and (c) what is the final dirt factor? Penyelesaian : (1) Neraca Panas Toluena =
= 130
C = 0,44 Btu/(lb)(
Toluena (Fig.2)
Q = 6330 x 0,44 (160-100) = 167.112 Btu/hr
0,44
130
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Amil Asetat =
= 95 Amil Asetat
C = 0,48 Btu/(lb)( (Fig.2) W = = 5802,5 Ib/h Q =wxcx
0,48
x (0,48) (10) = 27852 Btu/hr
95
(2) LMTD Hot Fluid
Cold Fluid
Diff
160
Higher Temp
100
60
100
Lower Temp
90
10 50
LMTD =
=
=
= 27,9
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(3) Caloric Temperature = 130
Viskositas < 1 cp
= 95
=1
Hot Fluid: anulus (Toluena)
Cold Fluid: inner pipe, Acetate
(4) Flow area
(4) D = 1,38/12 = 0,115 ft
= 2067/12 = 0,1725 ft
=
= 1,66/ 12 = 0,138 ft
=
=
/4 /4 = 0,0104
)/
= 3,14 (
)/4
= 0,00826 (5) Mass Velocity =
/
= 5802,5/0,0104 = 557932, 692 lb/hr (
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=
)/
=(
(6) pada 95 , )/0,138
= 0,71
(Fig.14)
= 0,71 x 2,42
= 0,0762
= 1,71 lb/(ft)(hr)
(5) Mass Velocity = w/
Reynolds =
= 6330/0,00826 = 767.000 lb/(hr)(
= )
= 37521,78
(6) Reynold Number At 130 ,
= 0,41 Cp
=
(Fig. 14)
= 0,41 x 2,42 = 0,99 Btu/(ft) (hr = = 0,0762 (767.000)/0,99 = 59.000
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(7)
= 167 (Fig. 24)
(7)
(8) At = 130 , c = 0,44 Btu/(lb)( K = 0,085 Btu/(hr)(
(Fig. 2)
= 112 (Fig. 24)
(8) at = 100
)( /ft) (Table.4)
, c = 0,48 Btu/(lb)(
k = 0,084 Btu/(hr)(
=
(9)
)( /ft) (Table. 4)
=
= 1,725
= 2,137
=
(9)
=
= 167 (0,085/0,0762) (1,725) (1)
= 112 (0,084/0,115) (2,173) (1)
= 232 Btu/(hr)(
= 177,77 Btu/(hr)(
)( )
(10) Correct =
)( )
to surface at OD x ID/OD
= 177,77 x 1,38/1,66 = 147,78
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(Fig. 2)
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(11)
Clean overall coefficient , Uc: Uc =
(12)
=
= 90,27 Btu/(hr)(
Design overall coefficient, =
+
:
=
= 0,004
)( )
+ 0,004
= 66,3 Btu/(hr)(
)( )
Summary 232
h outside
101,17
70 54 Required Surface Q=
A
A= =
= 90,34
Dari tabel 11 untuk 1 ¼ in. IPS standart pipe outside surface = 0,435 Required Length =
, sehingga:
= 207,6 lin ft
Maka, Setidaknya dibutuhkan 7 buah pipa hairpins berukuran 15 ft disusun seri . luas surface area secara aktual menjadi 210 X 0,435 = 91,35 Design (
. Sehingga, Koefisien
): = 65, 56 Btu/(hr) (
=
)( )
Faktor pengotor: =
=
= 0,004
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Pressure Drop Toluena, Hot Fluid 1‟)
(Pers 6.4)
= 0,00714 s = 0,87,
= 62,5 x 0,87 =54,3
(Table. 6)
2‟)
3‟) (
)
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Amil Asetat (1)
(Table. 6) (2)
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Problem 6.12 100.000 lb/jam nitrobenzena harus didinginkan dari 325 hingga 275 °F dengan benzena yang dipanaskan dari 100 hingga 300 °F. Hairpins sepanjang dua puluh kaki berukuran 4 kali 3 inci. Pipa ganda IPS akan digunakan , dan penurunan tekanan 10 psi diperbolehkan . Diperlukan faktor kotoran minimal 0,004 . Berapa banyak hairpins yang dibutuhkan,Bagaimana mengaturnya, dan berapa factor kotoran akhir Dik : •
Nitrobenzene (Hot Fluid) W = 100.000 lb/jam T1 = 325oF T2 = 275oF
•
Benzene (Cold Fluid) t1 = 100oF t2 = 300oF
Dit: Berapa banyak hairpins yang dibutuhkan? Bagaimana mengaturnya? dan berapa factor kotoran akhir? Penyelesaian: 1. Hot Balanace (Hot Fluid) Perhitungan Tav, tav, C, Q, dan W Nitrobenzene (Hot Fluid) Tav= Tav= Tav= 300oF C = 0,445 Btu/lb (oF) from fig. 2 page 804 Q = W. C. (T1-T2) Q = 100.000 lb/hr. 0,445 Btu/lb (oF).(325 oF- 275 oF) Q = 100.000 lb/hr. 0,445 Btu/lb (oF).(50 oF) Q = 2.225.000 Btu/hr
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Benzene (Cold Fluid)
Benzene (Cold Fluid) tav= tav= tav= 200oF C = 0,48 Btu/lb (oF)
from fig. 2 page 804
w= w= w = 23177,08 lb/hr
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2. LMTD Hot Fluida
Cold FLuid
Diff
325 o F
Higher Temp
300 o F
25
Δt2
275 o F
Lower Temp
100 o F
175
Δt1
150
Δt2Δt1
„
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LMTD = = =
(
(
)
)
= = 76,923 oF = 77 oF
3. Caloric temperature
4. Flow area Hot fluid : annulus, Nitrobenzene D2 = ID Annulus D2 = D2 = 0,3355 ft D1 = OD inner pipe D1 =
(tabel 11. page 844)
(tabel 11. page 844)
D1 = 0,292 ft
= =
(
(
)
)
= = 0,0216 ft2
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Diameter Ekuivalen (De) De =
(
De =
(
)
(Eq. 6.3 Page 111) )
De = De = De = 5. Mass velocity (Ga) ANNULUS HOT FLUID Ga = Ga = Ga =
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6. Viskositas Pada T = 300oF, (X=10,6 ; Y= 16,2) (Fig. 14 Page 822) = 0,29 Cp (Fig. 14 page 823) = 0,29 x 2,24 = 0,7018 lb/ft.hr Bilangan Reynold, Rea = Rea = Rea = 621418
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7. Transfer Heat (Jh) = 1038
(Gambar 24 Page 824)
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8. Pada T= 300oF
C=0,445 Btu/lb 0F (Tabel 4 Page 800)
ka = k1 + (T-T1) ka = 0,095 + (300-86) ka = 0,095 + (214) ka = 0,095 + (- 0,012) ka = 0,083 Btu/hr.ft2.(oF/Ft)
= (
= = 1,254
)
(
)
(
)
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9. Koefisien film (ho) = jH( ) ( )
= 1038 (
)(1,254)
= 1038 x 0,8811 x 1,254 = 1147 Btu/hr. ft2 oF Cold fluid : inner pipe benzene 10. Flow area D = 3,068 D=
(tabel 11. page 844)
D = 0,256 ft
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Daerah aliran ap: = =
(
(
)
=0,051 ft2
)
11. Mass velocity (Gp) ANNULUS Cold FLUID Gp= Gp= Gp= 12. Viskositas Pada T = 200oF, ( X = 12,5 ; Y = 10,9) = 0,26 cp x 2,42 = 0,26 x 2,42 = 0,629 lb/ft.hr Bilangan Reynold,
Fig 14. Page 822 (Gambar 14 page 823)
Rep = Rep = Rep = 184.960
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13. jH = 425
(Gambar 24, Page 824)
14. Konduktivitas, k benzene
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Pada T= 200oF
C=0,48 Btu/lb 0F (Tabel 4 Page 800)
kp = k1+ (T-T1) kp = 0,092 + (200-86) kp = 0,092 + (-0,0106) kp = 0,081 Btu/hr.ft2.(oF/Ft)
= (
=
)
(
)
(
)
= 1,242
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15. Koefisien film (ho) = jH( ) ( ) = 425 (
)(1,242)
= 425 x 0,3164 x 1,242 = 167 Btu/hr. ft2 oF =
x
= 167 x = 167 x 0,877 = 146,5 16. Menentukan tw = tc + (
tw = 200 + (
tw = 200 + (
o
dan )(Tc-tc)
)(300-200)
)(100)
tw = 287,29 F Pada tw = 287,29 oF
a =( a =(
)
a =0,99
)
= 1147 = 1147 ho = 1135,53 Btu/hr. ft2 oF
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Pada tw = 287,29 oF
=( =(
)
= 1,07
)
= 167 = 167 hi = 178,69 Btu/hr. ft2 oF = 146,5 = 146,5 hio = 156,755 Btu/hr. ft2 oF
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17. Clean Overall coefficient ( Uc = (
)
Uc = (
Uc = (
)
)
Uc = 137,74 Btu/(hr.ft2 oF) 18. Design Overall coefficient ( Ud = (
)
Ud = (
)
Ud = (
)
Ud = 88,80 Btu/(hr.ft2 oF) 1135,53
156,755
h outside 137,74
Ud
88,80
19. Required Surface Q = UD A Δt A= (
A= (
A=
)
)
20. Required Length Required length = (
Required length = ( Required length =
)
)
tabel 11 page 844
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Dapat terpenuhi jika menghubungkan 9 hairpins berukuran 20 ft. Sehingga : 360 x 0,917 = 330,12 ft2 The actual design coefficient is UD= (
)
UD= 87,53 Btu/hr.ft2.oF RD= (
RD= (
RD= (
) )
)
RD= 0,0042 (hr.ft2.oF)/Btu
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7-7 43.200 lb/j 35oAPI distilat didinginkan dari 250 sampai 120
dengan menggunakan air
pendingin yang mengalami peningkatan suhu dari 85 hingga 120 . Untuk maksud tersebut digunakan 19 ⁄ in ID shell dengan jumlah tube 204 buah, OD tube = ⁄ “ 16 BWG. L = 16 ft dengan pitch bujur sangkar 1 in. Tube passes = 4 , baffle space = 5 in. Susunan seperti apa yang menghasilkan pressuredrop mendekati 10 psi, tentukan Rd, dan temperature optimum keluaran air
Dari penjabaran soal di atas, diperoleh data yang diketahui : Shell ID
Tube = 19,25 in
Jumlah dan panjang
= 204 dan 16 ft
Baffle space = 5 in
OD, BWG, Pitch
Passes
Passes
= ⁄ in, 16 BWG, 1 in
=1
=4
Hot fluid (35o API distilate), Cold fluid (water),
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⁄
Diperoleh nilai Maka,
dan
⁄ Kern Hal 111 ⁄
⁄
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LMTD
Hot Fluid 250 120
Higher Temp Lower Temp
Cold Fluid
Diff
120
130
∆t2
85
35
∆t1
95
∆t2 -∆t1
Mencari FT
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Diperoleh FT = 0,84
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Maka
Caloric temperature
Mencari nilai Kc
35o API distilate (hot fluid) Dari data diatas diperoleh
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Menentukan nilai
Dari data yang diplotkan pada sumbu x = 0,27 menyentuh garis Kc = 0,31 menghasilkan
Chapter 5 Kern 35o API distilate (hot fluid)
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Water (cold fluid fluid)
Flow Area
Shell (Cold Fluid-air)
Dimana:
ID
: Diameter dalam
C‟
: Tube space
B
: Baffle space
PT
: Tube Pitch
n‟
: Shell passes
Tube (Hot Fluid-35o API distilate)
Dimana: : Jumlah tube n
: Passes : Luas penampang pertube
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Karena flow area tube hamper sama dengan shell maka (42o API kerosene) diasumsikan terletak pada dan (34o API mid-continent crude) pada tube Menghitung nilai koefisien perpindahan panas konveksi (h)
c
(
)( )
⁄
(
)
= kapasitas panas (fig 2) = viskositas (fig 14)
k
= koefisien perpindahan panas konduksi (table 4)
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= keofisien yang diperoleh dengan mengkorelasikan terhadap Re (fig 24)
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Cold fluid (water) pada = 0,7 cp ⁄
=
Hot fluid (35o API distilate) pada = 1,8 cp =
Penentuan mass velocity
⁄
Cold fluid (air) at shell ⁄
⁄
Hot fluid (35o API distilate) at tube ⁄
⁄
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Reynolds number Fluid at shell (air)
Dengan
Dimana: de
: Diameter dalam
PT
: Tube Pitch
d0
: Diameter luar tube ⁄ , sehingga
Dengan ⁄
Fluid at tube (35o API distilate) ⁄ Dengan ⁄
⁄
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Nilai Fluid at shell (air)
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Fluid at tube (35o API distilate)
Mencari nilai k
⁄
⁄
( )
(
)
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Hot fluid at tube (35o API distilate)
( )
⁄
(
⁄
)
Menghitung nilai h (part 1) Hot fluid at tube (35o API distilate) ( ) ⁄
⁄
cold fluid at shell (air) ( )
⁄
Pemeriksaan
pada permukaan diameter luar (OD)
Penentuan nilai
⁄
dan
Hot fluid at annulus pipe ⁄
⁄
⁄
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Cold fluid (air) pada = 0,4 cp = (
)
(
)
⁄
Hot fluid (35o API distilate) pada = 1,7cp = (
Sehingga,
)
(
)
⁄
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⁄
⁄
Menghitung Rd
⁄
Clean overall coefficient
⁄
Design overall coefficient
Rd
⁄
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|
C
C
|
Pengecekan Pressure Drop Pressure drop pada tube (35o API distilate sebagai hot fluid)
Dengan
„
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Diperoleh
dan ⁄
Dengan
⁄
diperoleh
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Pressure drop pada shell (air sebagai cold fluid)
⁄
⁄
Dengan
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Diperoleh
dan
Outlet Optimum Temperature
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7-9 100.000 lb/jam larutan potassium iodide 20 persen harus dipanaskan dari 80 sampai 200 oF menggunakan steam pada 15 psig. Tersedia untuk servis adalah 10 in. ID 1-2 exchanger tanpa baffle memiliki 50 t
ng ¾ in. OD,16 BWG,16‟0” p nj ng i t
nt k
lint
np
pitch
segitiga (Triangular pitch) 15/16-in. Tekanan turun (pressure drops) dan faktor kotoran (dirt factor)?
Penyelesaian Diketahui : Wpotasium iodia
= 100.000 lb/hr
T1
= 250oF
T2
= 250oF
t1
= 80oF
t2
= 200oF
Tekanan steam
= 15 psig
ID
= 10 in
Tanpa baffle memiliki 50 tabung ¾ in OD
= 16
BWG
= 16‟0”
Pitch
= 15/16-in.Triangular pitch
Ditanya : Pressure drops dan dirt factor? Jawab : Shell side
Tube side
ID = 10 in
J ml h
Baffle space = half circles
OD, BWG, pitch = ¾ in, 16 BWG
Passes = 1
15/16-in.Triangular pitch
n p nj ng
Passes =2 (1) Heat Balance Weighted specific heat = 0,8 Cpwater + 0,2
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50, 16‟0”
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CpKI Weighted specific heat = 0,8 (1) + 0,2 (0,3) Weighted specific heat = 0,86 Btu/(lb)(oF) Potasium iodida, Q = W.C (t2 – t1) Q = 100.000 x 0,86 (200 – 80) Q = 10.320.000 Btu/hr Steam, Q ms hfg = mc Cpc (t2 – t1) Pabs = 14,7 + 15 = 29,7 psi = 30 psi Diperoleh dari tabel 7 buku Kern hfg = 945,3 Sehingga, 945,3 x ms = 100.000 x 0,86 (200 – 80) 945,3 x ms = 10.320.000 ms =
= 10917,169 lb/hr
(2)
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Hot Fluid
Cold Fluid
Diff
250
Temperatur Tinggi
200
50
250
Temperatur Rendah
80
170
0
Perbedaan
120
120
LMTD = LMTD = LMTD = LMTD = 98,36oF ……. P
R=
m n 5.14
i B k K n)
R= R=0
(3)Tc dan tc Tc =
=
tc = =
==
= 250oF = 140oF
Hot Fluid : tube side, steam
(4)Luas aliran, 0,302 in2 dari tabel 10 at =
……. P
m n 7.48 h l 152
i
k K n)
at = at = at = 0,053 ft2
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(5)Kecepatan massa Gt =
……. P
m n 7.2)
Gt = Gt = 205.984,340 lb/hr . ft2 (6)Pada Ta = 250oF 0,013 x 2,42 0,03146 lb/ft.hr (Gambar 15 terlampir) D=
=
= 0,05167 ft
(Tabel 10 terlampir)
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0,013
Maka : Ret =
…….. P
m n 3.6)
Ret = Ret = 338.309,31
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(7)Maka : jH
620……. G m
24 t l mpi )
(8)Pada Ta = 250oF C = 1,02 btu/lbm.oF K = 0,395 btu/h.ft.oF
(9)Maka nilai hio : = = 0,4 hi = jH .
……. P
.
hi = 620.
m n 6.15 )
. 0,4
hi = 1895,88 btu/h.ft2.oF hio = hi.
……
hio = 1895,88 .
m
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hio = 1567,26 btu/h.ft2.oF
Cold Fluid : shell side, Pottasium iodide 4‟) L
li n
as = (area of shell) – (area of tubes) ( ) ]–*
as = [
as = 0,392 ft2
+
5‟) K c p t n m Gs = Gs = Gs = 255.102,04 lb/hr.ft2 6‟) P
Ta = 140oF
0,98 x 2,42 2,37 l /ft.h ……. T l mpi ) Ds = Ds =
……. P
m n 6.3)
= 0,159 ft2
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Res = Res = Res = 17.114,44
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7‟) P
Ta = 140oF
jH = 64 8‟) Dik t h i
hw
nt k col fl i
= 140oF
ta =
Pada ta = 140oF k = 0,9 . 0,381 = 0,343 Btu/(hr)(ft2)(oF/ft) Jadi telah diketahui bahwa : k = 0,343 Btu/(hr)(ft2)(oF/ft) = 2,37 lb/(ft)(hr) c = 0,86 Btu/(lb)(oF) Sehingga : ( )
=(
)
= 1,98
9‟) ho = jH = jH = (64)
……. P
( )
m n 6.15 )
( )
(1,98)
= 273,36 Btu/(hr)(ft2)(oF)
10‟) M nc i nil i tw (Tc – tc) ……. P
tw = t c +
m n 5.31 )
(250 – 140)
tw = 140 + tw = 273,24oF 11‟) P
tw = 273,24oF
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=2 = 0,18 x 2,42 = 0,436 lb/(ft)(hr) =(
=(
)
= 1,32
12‟) Co ho =
)
ct co ffici nt ……. P
m n 6.36)
ho = 273,36 x 1,32 ho = 360,39 Btu/(hr)(ft2)(oF)
13‟) Cl n Ov
ll Co ffici nt Uc)
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Uc =
……. P
m n 6.38)
Uc = Uc = Uc = 293,01 Btu/(hr)(ft2)(oF) 14‟) D ign Overall Coefficient (UD) Berdasarkan tabel 10, pad OD = ¾ in, BWG = 16 Dip ol h xt n l
f c /ft, ”
0,1963 ft2/ft
Maka : A
j ml h x p nj ng x ”
A
50 x 16‟0” x 0,1963
UD =
……. P
157 ft2
m n 5.3)
UD = UD = 668,2847 Btu/(hr)(ft2)(oF) 15‟) Di t F cto Rd) Rd =
……. P
m n 6.13)
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Rd = Rd = Rd = -0,00191 Ringkasan 1567,26
houtside
360,39
Uc
293,0121
Ud
668,2847
Rd calculated
-0,00191
Pressure Drop (1) Spesific vol of steam from table 7 (Halaman 816 dari buku Kern)
Maka diperoleh pada saat Pabs 30 psi, v = 13,746 ft 3/lb
= 62,5 lb/ft3 Cari nilai specific gravity (s) s= s= s= s = 0,00116
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Res = 17.114,44 maka Diperoleh nilai f sesuai gambar 26 (halaman 836 dari buku Kern)
f = 0,00016 ft2/in2 = x
……. P
m n 7.45)
= x = 34,71 psi
1‟) =
……. P
m an 6.4)
=
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= 0,159 ft ……. P
=
m n 7.3)
= 0,159 x 255.102,04/3,14 = 12.917,58 Kemudian lihat gambar 26 (halaman 836) untuk mencari nilai f f = 0,00023 ft2/in2
2‟)
=
……. P
m n 7.45)
Buku Perry Tabel 2-1 (halaman 2-21), Potassium iodida Spesific gravity = 3,13 = = 0,00543 psi
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7-11 A 1-2 exchanger recovers heat from 10,000 lb/hr of boiler blowdown at 135 psig by heating raw water from 70 to 96 °F. Raw water· flows in the tubes. Available for the service is a 10.02 in. ID 1-2 exchanger having 52 tubes% in. OD, 16 BWG, 8'0" long. Baffies are spaced· 2 in. apart, and the bundle is arranged for two tube passes. What are the pressure drops and fouling factors? (Exchanger A 1-2 memulihkan panas dari 10.000 lb/jam blowdown boiler pada 135 psig dengan memanaskan air mentah dari 70 hingga 96 °F. Air mentah· mengalir di dalam tabung. Tersedia untuk layanan ini adalah 10,02 in. Penukar ID 1-2 memiliki 52 tabung% in. OD, 16 BWG, 8'0" panjang. Baffle diberi jarak· 2 in. terpisah, dan bundel diatur untuk dua lintasan tabung. Apa penurunan tekanan dan faktor pengotoran?) Dik: ;
(asumsi) Dit = pressure drops and fouling factors?
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Penyelesaian: 1‟) Hot B l nc hal 170, Kern 1965
01111
( ) (
)
( )
2‟) LMTD Fluida Panas
Fluida Dingin
Diff
135
Temparatur tertinggi
96
39
96
Temperatur terendah
70
26
13 LMTD =
⁄
Pers 5.14 Hal 89, Kern 1965
hal 170, Kern 1965
hal 170, Kern 1965
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Mencari Faktor Temperatur ( Dari grafik gambar 18. didapat nilai
sebesar 0,825 Pers 7.42, Hal 149, Kern 1965
3‟) C lo ic T mp
t Pers 5.28 , hal 96, Kern 1965
Pers 5.29 , hal 96, Kern 1965
Hot Fluida, Shell : Boiler/Steam 4‟) Flow Pers 7.1, Hal 138, Kern 1965
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(
( )
5‟) M
)
V l Pers 7.2, Hal 138, Kern 1965
6‟) m nc i vi ko it Pada
) menggunakan Fig. 14, Kern 1965 : didapat viskositas
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Mencari
menggunakan Fig. 28 Kern 1965 , didapat viskositas
pers 3.6, Hal 112, Kern 1965
7‟) H t t n
f cto jH)
Dari Fig. 28 Kern 1965 didapat jH = 59
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8‟) D i t l 4, K n 1965 ik t h i nil i k i 0,356 p h 86 Dilakukan interpolasi untuk mendapatkan nilai k air pada , didapat nilai k = 0,370 btu/(hr)( )( / ) ( )
⁄
(
9‟)
(
)
( )
⁄
⁄
pers 6.15b , Hal 112, Kern 1965
⁄
pers 6.15b , Hal 112, Kern 1965
)
9‟)
(
10‟) t
( ) )
w ll-temperature ⁄
pers 5.31, Hal 98, Kern 1965
11‟) P
12‟)
(
)
( )
pers 6.36, Kern 1965
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Cold Fluida,:Tube, Raw water 4) Flow area berdasarkan tabel 10. hal, 843, Kern 1965 ⁄ 5) Mass Vel
Pers 7.48, Hal 152, Kern 1965 ⁄ Pers 7.2, Hal 138, Kern 1965
6) mencari viskositas ( ) menggunakan Fig. 14, Kern 1965 : Pada
didapat viskositas
mencari D menggunakan tabel 10, Kern 1965 , didapat D = 0,62
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pers 3.6, Hal 112, Kern 1965
7) L/D = 8/0,052 = 153,84 jH = 23
menggunakan Fig. 24, Kern 1965
8) Pada c = 1 Btu/lb. =
k = interpolasi = =
x = k = 0,355 Btu / (j.Ft2.
/Ft)
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( ) 9)
10)
⁄
) ⁄ = 2,05
=(
=
(
.
( )
⁄
)
= 2661 Btu / (j.Ft2.
= 321,89 .
11) Pada Tw = 107,1
12)
(
= 321,89
pers 6.15b , Hal 112, Kern 1965
)
,
)
= 0,69 x 2,42 = 1,67 lb/ft.j (
)
= 321,89
= 321,89 x 1,18 = 369,21 Btu/(j.Ft2.
)
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Pressure Drop ( Penurunan Tekanan Fluida Panas 1‟) Res = F = 0,000268 Ft2 / in2 2‟) No. of c o
menggunakan Fig. 29, Kern 1965
,N+1
12L/B = 12 X 8/12 = 8
Ds = 10,02 / 12 = 0,83 ft
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3‟)
= = = 0,18 psi
Diterima
= 10 psi
hitung <
Maka perhitungan (Design) Dapat Diterima Pressure Drop ( Penurunan Tekanan Fluida dingin (1) Ret = F = 0,000148 Ft2 / in2 (2)
menggunakan Fig. 26, Kern 1965
=
= = 0,064 psi
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥
𝑥
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3‟)
,
V2/2g‟
0,0075 m ngg n k n Fig. 27, K n 1965
= (4n/s) (V2/2g‟) = (4x2)/2 x 0,0075 = 0.06 psi Diterima
= 10 psi
hitung <
Maka perhitungan (Design) Dapat Diterima
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13. Koefisien keseluruhan Bersih Uc
hio .ho hio ho
369,21(773,9) Uc 369,21 773,9 Uc 249,96 Btu/(J.ft 2 .o F)
(6.38)
14. Desain Koefisien Keseluruhan (U ) a '' 0,1963 ft
(Tabel 10)
A 52 x8 x0,1963 81,66 ft Q UD AT 390000 UD 180,6ft Btu/(hr)(ft 2 )( o F) (81,66)(26,45) 2
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15. Faktor Kotoran (Rd) Rd
Uc U D U cU D
(6.13)
249,96 180,6 249,96(180,6) Rd 0,0015 Rd ketentuan 0,0015 Rd
Rd hitung (0,0015) Rd ketentuan (0,0015)
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