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