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TRANSCRIPT
Nama : Isni Maretha
Nim : 03101403052
Kelas : B
Jurusan : Teknik Kimia Kampus Palembang
Mata Kuliah : Perpindahan Panas 1
Soal 12.1
Merancang penukar untuk sub-cool kondensat dari kondensor metanol dari 95oC hingga 40oC. Arus-tingkat metanol 100.000kg / h. Air payau akan ne digunakan sebagai pendingin, dengan kenaikan suhu 25oC hingga 40oC
Soal 12.2
Gunakan Bell’s Mehtod untuk menghitung koefisien transfer panas Shell Side dan pressure drop untuk design exchanger pada contoh 12.1 .
Diketahui :
Shell I.d (Ds)(ID) 894 mm
Bundle diameter (Db) 820 mm
Nember of tube (Nt) 918 mm
Tube O.d (Do) 20 mm
Pitch 1,25 ∆ (Pt) 25 mm
Tube Length (L) 4830 mm
Baffle Pitch (lb) 356 mm
Fluid flow rate on the shell side (Ws) 100.000 kg/s
Methanol density (ρ) 750 kg/m3
Viscosity (µ) 0,34 x 10-3 kg/m s
Heat Capacity (Cp) 2,84 kJ/kgoC
Thermal Conductivity (kf) 0,19 W/m oC
Hot fluid Cold fluid
T1= 95oC t1= 25oC
T2=40oC t2= 40oC
W=100.000 kg/h w= 68,9 kg/s
c= 2,84 kj/kg oC c= 4,2 kj/kg oC
µ=0,34 mNs/m2 µ=0,8 mNs/m2
k= 0,19 W/moC k= 0,59 W/moC
shell side Tube side
shell i.d (Ds) = 894mm Number and length (L) = 918mm;4830mm
Baffle space (lb) = 356mm OD (do) , Pitch 1,25 (pt) = 20mm, 25mm
Passes= 1 Passes = 2
Assumsi :
1 shell pass 2 or more tube3 passes
Penyelesaian:
W =100.000kgh
W =100.000 kg1 h
x1h
3600 s=27,8
kgs
1. Heat Balance
Cp Methanol = 2.84 Kj/kg℃
Q=W x Cp x ∆ T
Q=27,8Kgs
x 2,84kjkg℃ ( 95℃−40℃ )=4339,2 kj /s
2. True Temperature difference ∆t
Hot Fluid Cold Fluid Different
95 High Temperature 40 55
40 Low Temperatur 25 15
55 Different 15 40
(T1 – T2) (t2 – t1) (∆t2 - ∆t1)
LMTD=∆ T lm=(T 1−t 2 )−(T 2−t 1 )
lnT 1−t 2
T 2−t 1
=( 95℃−40℃ )−(40℃−25℃)
¿(95℃−40℃)(40℃−25℃)
=31℃
R=T 1−T 2
t 2−t 1
=95℃−40℃40℃−25℃
=3,67
S=t2−t1
T 1−t 1
=40℃−25℃95℃−25℃
=0,21
Dari grafik, didapat nilai Ft = 0,85.
∆ T m=F t x ∆ T lm=0,85x 31℃=26℃
3. Caloric Temperature Tc and tc
∆ tC
∆ tH
=∆ t1
∆ t 2
=(T 2−t1 )(T 1−t2 )
=( 40−25 )℃(95−40 )℃
=15℃55℃
=0,27℃
U 1=1 (1+2 (25 ) )=51;U 2=1 (1+2 ( 40 ) )=81
K c=U 2−U 1
U 1
=81−5151
=3051
=0,588
F c=¿0,424 (diperoleh dari fig.17 the caloric temperature factor Fc)
T c=T2+Fc (T 1−T 2 )=40+0,424 ( 95−40 )=63,32℃
t c=t1+Fc (t 2−t 1 )=25+0.424 (40−25 )=31,36℃
Cold Fluid (Tube side)
4. Flow Area AS
OD = 20mm = 0,79 in ≈34∈¿
ID = 16mm =0,629 inFlow Area at’ = 0,302 in2 (pada table 10)
at=no of tube x flow area per tubeno of passes
at=918 x 0,302¿2
144 x2=0,96 ft2 x 0,0929=0,090 m2
5. Mass Velocity Gt
W = QC p (t2−t1 )
= 4339,236 kj /s(4,2 kj /kg℃ ) (40℃−25℃ )
=68,9 kg /s
Gt=Wat
Gt=68,9 kg/ s
0,90 m2=765,56 kg /m2 s
6. Reynold Number (Re)
ℜ=ID xGt
μ
ℜ=16.10−3 m x765,56 kg /m2 s
0,8.10−3 Ns
m2
=15311,2
7. Identifikasi Nilai J H
Dari Grafik 24 di dapat J H=3,7 x10−3
8. Menghitung Nilai Prandtl Number
(Cp . μ /k )1/3=[ ( 4,2.103 j /kg℃ ) x ( 0,8.10−3 Ns/m2 )0,59W /m℃ ]
1/3
=1,8
9. Menghitung nilai hi /∅ t
hi
∅ t=J H ( k
D )(c .μk )
13
hi
∅ t=3,7.10−3( 0,59
Wm℃
0.016 m ) (1,8 )=0,25
10.Menghitung nilai hio /∅ t
hio
∅ t=
hi
∅ tx
IDdo
hio
∅ t=0,25 x
0,016 m0,02 m
=0,2
11. Tube Wall Temperature tw
twdari 10b = 32oC
µw = 0,66 mNs/m2
∅ t= (μ /μw )0,14=(0,8 mNs /m2/0,66 mNs /m2)0,14=1,02
12. Corrected Coefficient
hio=( hio
∅ t )∅ t
hio= (0,2 ) x (1,02 )=0,204
13. Coefficient Clean Over All, Uc
U c=(h io xho )(hio+ho )
=(2995 x 2573 )(2995+2573 )
=1384,004 kj /m2 s℃
14. Coefficient Overall Design Ud
a” = 0,3271 ft2 = 0,03 m2 (dari table 10, heat exchanger and condenser tube data)
A=No of tube x length x a
A=918. 10−3 x 4830. 10−3 x 0,03 m2=0,13m2
U d=Q
A x∆ t= 4339,2 kj /s
( 0,13 m2 x26℃ )=1283,73 kj /m2 s℃
koefisien transfer panas pada tube side : ht=h i x ℜ
ht=(0,25 x 1,02 ) x (1435,4 )=366,027 W /m2 s
15. Dirt Factor (Rd)
Rd=U C−U d
U CU d
=1384,004
kj
m2 s℃−1283,73 kj /m2 s℃
1384,004kj
m2 s℃ x 1283,73 kj /m2 s℃
=5,64 x10−5
Hot Fluid (shell side)
4b. Flow Area (A s¿
A s=ID xC ' B
144 PT
A s=894 mm x720 x 178 mm144 (25 mm )
=31826,4 mm2=0,0318264 m2
5b. Mess Velocity (G S )
GS=WAS
GS=100.000 kg
1hx
1 h3600 s
x1
0,0318264 m2=872,8
kgs
m2
6b. Reynold Number (Re)
De=1,120
x ( 252−0,917 x 202 )=14,201 mm=14,201.10−3m
μ=0,34 x10−3 kgm
s
ℜ=D e GS
μ=
14,201.10−3 m x 872,8kgs
m2
0,34 x10−3 kg /m s=36.454
7b. Identifikasi J H
J H=3,3 x10−3 (Dari Grafik)
8b. Menghitung Nilai Prandtl Number (Cp. μ/k )1/3
(Cp . μ /k )1/3=[ (2,84. 103 j /kg℃ ) x (0,34. 10−3 Ns/m2 )0,19 W /m℃ ]
1/3
=1,7
9b. menghitung nilai ho/∅ s
ho/∅ s=J H (k / De ) (Cp. μ/k )1/3
ho/∅ s=3,3. 10−3( 0,19W /m℃14,201m )(1,7 )=0,075
10b. Tube Wall Temperature tw
tw=tc+[ (ho /∅ s )/ (( hio /∅ t )+ (ho /∅ s )) x ( Tc−t c ) ]tw=31,36℃+ [ (0,07 ) / (0,2+0,07 ) x (63,32℃−31,36℃ ) ]=39,65℃
11b. Identifikasi nilai μw
Dari grafik 14, pada tw=32℃ untuk methanol 90% (x=12,3 ; y=11,8 ) ,
μw=0,37 mNs /m2
∅ s=( μ /μw )0,14=( 0,34mNs /m2
0,37mNs /m2 )0,14
=0,98
12b. corrected Coefficient
ho=( ho
∅ s )∅ s
ho=0,07 x 0,98=0,069
koefisien transfer panas pada shell-side,
hs=ho x ℜ=0,069 x 36454=2515,326 W /m2 s
13b. Pressure Drop (ΔP)
tube-side
untuk Re = 15311,2 (Jf = 4,4 x 10-3)
∆ Pt=2 x ( 8 x 4,3. 10−3 ) x [( 4,83. 10−3
16 )+25] x( 0,752
2 )=7,2 kpa=1,1 ps i
Shell-side
Untuk Re = 36.454 (Jf = 4x10-2)
∆ Pt=(8 x 4,3. 10−3 ) x ( 89414,4 ) [( 4,83.10−3
16 )] x ( (750 ) (1,162 )2 )=272 kpa=39 ps i
Pressure drop terlalu besar dapat dikurangi dengan meningkatkan baffle pitch.
∆ P s=272 kpa
4=68 kpa=10 psi